1
CHAPTER 2
ATOMIC STRUCTURE
21 Bohrrsquos Atomic Model
22 Quantum Mechanical Model
23 Electronic Configuration
2
21 Bohrrsquos Atomic
Model
3
Bohrrsquos Atomic Model
At the end of this topic students should be able to-
4
In 1913 a young Dutch physicist
Niels Bohr proposed a theory of
atom that shook the scientific world
The atomic model he described
had electrons circling a central
nucleus that contains positively
charged protons
Bohr also proposed that these orbits can only
occur at specifically ldquopermittedrdquo levels only
according to the energy levels of the electron
and explain successfully the lines in the
hydrogen spectrum
BOHRrsquoS ATOMIC MODELS
5
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
H
Nucleus
(proton) H11
BOHRrsquoS ATOMIC
POSTULATES
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
2
21 Bohrrsquos Atomic
Model
3
Bohrrsquos Atomic Model
At the end of this topic students should be able to-
4
In 1913 a young Dutch physicist
Niels Bohr proposed a theory of
atom that shook the scientific world
The atomic model he described
had electrons circling a central
nucleus that contains positively
charged protons
Bohr also proposed that these orbits can only
occur at specifically ldquopermittedrdquo levels only
according to the energy levels of the electron
and explain successfully the lines in the
hydrogen spectrum
BOHRrsquoS ATOMIC MODELS
5
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
H
Nucleus
(proton) H11
BOHRrsquoS ATOMIC
POSTULATES
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
3
Bohrrsquos Atomic Model
At the end of this topic students should be able to-
4
In 1913 a young Dutch physicist
Niels Bohr proposed a theory of
atom that shook the scientific world
The atomic model he described
had electrons circling a central
nucleus that contains positively
charged protons
Bohr also proposed that these orbits can only
occur at specifically ldquopermittedrdquo levels only
according to the energy levels of the electron
and explain successfully the lines in the
hydrogen spectrum
BOHRrsquoS ATOMIC MODELS
5
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
H
Nucleus
(proton) H11
BOHRrsquoS ATOMIC
POSTULATES
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
4
In 1913 a young Dutch physicist
Niels Bohr proposed a theory of
atom that shook the scientific world
The atomic model he described
had electrons circling a central
nucleus that contains positively
charged protons
Bohr also proposed that these orbits can only
occur at specifically ldquopermittedrdquo levels only
according to the energy levels of the electron
and explain successfully the lines in the
hydrogen spectrum
BOHRrsquoS ATOMIC MODELS
5
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
H
Nucleus
(proton) H11
BOHRrsquoS ATOMIC
POSTULATES
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
5
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
H
Nucleus
(proton) H11
BOHRrsquoS ATOMIC
POSTULATES
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
6
2 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC
POSTULATES
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
7
3 At ordinary conditions the electron is at the ground state
(lowest level) If energy is supplied electron absorbed
the energy and is promoted from a lower energy level to
a higher ones (Electron is excited)
4 Electron at its excited states is unstable It will fall back
to lower energy level and released a specific amount of
energy in the form of light The energy of the photon
equals the energy difference between levels
BOHRrsquoS ATOMIC POSTULATES
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
8
1 Electron moves in circular orbits about the nucleus In
moving in the orbit the electron does not radiate any
energy and does not absorb any energy
1 The energy of an electron in a hydrogen atom is
quantised that is the electron has only a fixed set of
allowed orbits called stationary states
[ orbit = stationary state = energy level = shell ]
BOHRrsquoS ATOMIC MODELS
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
9
Ground state
the state in which the electrons have their lowest energy
Excited state
the state in which the electrons have shifted from a lower
energy level to a higher energy level
Energy level
energy associated with a specific orbit or state
Energy levels in an atom
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
10
The energy of an electron in its level is given by
RH (Rydberg constant) or A = 218times10-18J
n (principal quantum number) = 1 2 3 hellipinfin (integer)
Note
n identifies the orbit of electron
Energy is zero if electron is located infinitely far from nucleus
Energy associated with forces of attraction are taken to be
negative (thus negative sign)
THE ENERGY LEVEL
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
11
At the end of this topic students should be able to-
Level No Learning Outcomes
C2 21 d Describe the formation of line spectrum of hydrogen atom
C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and
Pfund series
C3 21 f Calculate the energy change of an electron during transition
ΔE = RH (1n12 - 1n2
2)
where RH = 218 x 10-18 J
C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition
ΔE = hν where ν= cλ
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
12
Emission Spectra
Emission Spectra
Continuous
Spectra
Line
Spectra
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
13
Continuous Spectrum
A spectrum consists all wavelength components
(containing an unbroken sequence of frequencies) of the
visible portion of the electromagnetic spectrum are
present
It is produced by incandescent solids liquids and
compressed gases
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
14
Regions of the Electromagnetic Spectrum
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
15
When white light from
incandescent lamp is passed
through a slit then a prism it
separates into a spectrum
The white light spread out into
a rainbow of colours produces
a continuous spectrum
The spectrum is continuous in
that all wavelengths are
presents and each colour
merges into the next without a
break
FORMATION OF CONTINUOUS SPECTRUM
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
16
Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous amp discrete lines produced by
excited atoms and ions as the electrons fall back to a lower energy
level The radiation emitted is only at a specific wavelength or
frequency It means each line corresponds to a specific wavelength
or frequency
Line spectrum are composed of only a few wavelengths giving a
series of discrete line separated by blank areas
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
17
prismfilm
The emitted light (photons) is then separated into its components by
a prism Each component is focused at a definite position according
to its wavelength and forms as an image on the photographic plate
The images are called spectral lines
FORMATION OF ATOMIC LINE SPECTRUM
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
18
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
En
erg
y
When an electrical discharge is passed through a sample of
hydrogen gas at low pressure hydrogen molecules decompose to
form hydrogen atoms
Radiant energy (a
quantum of energy)
absorbed by the atom (or
electron) causes the
electron to move from a
lower-energy state to a
higher-energy state
Hydrogen atom is said to
be at excited state (very
unstable)
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
19
FORMATION OF ATOMIC LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6n = infin
En
erg
y
When the electrons fall
back to lower energy
levels radiant energies
(photons) are emitted in
the form of light
(electromagnetic radiation
of a particular frequency or
wavelength)
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
20
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum λE
Energy
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
21
FORMATION OF ATOMIC LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5n = infin
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
λ E
Energy
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
22
Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n = infin
Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
23
Exercise Complete the following table
Series n1 n2
Spectrum
region
Lyman 1 234hellip Ultraviolet
Balmer 2 345hellip Visible light
Paschen 3 456hellip Infrared
Brackett 4 567hellip Infrared
Pfund 5 678hellip Infrared
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
24
The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series
Exercise
Line
spectrumE λ
Specify the increasing order of the radiant energy
frequency and wavelength of the emitted photon
Which of the line that corresponds to
i) the shortest wavelength
ii) the lowest frequency
A B C D E
ν
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
25
Describe the transitions of electrons that lead to
the lines W and Y respectively
Solution
Line
spectrum
W Y
Exercise
Balmer series
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
26
Homework
Calculate En for n = 1 2 3 and 4 Make a one-
dimensional graph showing energy at different
values of n increasing vertically On this graph
indicate by vertical arrows transitions that lead to
lines in
a) Lyman series
b) Paschen series
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
27
In Lyman series the frequency of the convergence of
spectral lines can be used to find the ionisation energy of
hydrogen atom
IE = hνinfin
The frequency of the first line of the Lyman series gt the
frequency of the first line of the Balmer series
Significance of Atomic Spectra
Lyman Series
Line
spectrum λ E
Balmer Series
νinfin
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
28
Line
spectrum
ABCDE
Exercise
Paschen series
Solution
Which of the line in the Paschen series corresponds to the
longest wavelength of photon
Describe the transition that gives rise to the line
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
29
Radiant energy emitted when the electron moves from
higher-energy state to lower-energy state is given by the
difference in energy between energy levels
Energy calculation
ΔE = Ef - Ei
where
Thus
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
30
The amount of energy released by the electron is called a photon of energy
A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength
whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency
Energy calculation
ΔE = hν
Wherec (speed of light) = 300times108 ms-1
Thus
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
31
n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level A specific amount of energy
is absorbed
ΔE = hν = E1-E3 (+ve)
Electron falls from higher to lower energy level
A photon of energy is released
ΔE = hν = E3-E1 (-ve)
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
32
Energy level diagram for the hydrogen atom
Pote
ntial energ
y
n = 1
n = 2
n = 3
n = 4
n = infin
Energy
released
Energy
absorbed
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
33
Exercises
1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)
1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom
3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom
(answer 155 x 10-19J)
4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
34
At the end of this topic students should be able to-
Level No Learning Outcomes
C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series
1 λ = RH (1n12 - 1n2
2)
where RH = 1097 x 107 m-1 and n1ltn2
C3 21 i Calculate the ionisation energy of hydrogen atom from
Lyman series
C2 21 j State the limitation of Bohrrsquos atomic model
C2 21 k State the dual nature of electron using the Brogliersquos
postulate and Hesseinbergrsquos uncertainty
principle
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
35
Wavelength emitted by the transition of electron
between two energy levels is calculated using
Rydberg equation
Rydberg Equation
RH = 1097 times 107 m-1
λ = wavelength
Since λ should have a positive value thus n1 lt n2
where
1λ = RH (1ni2 ndash 1nf
2)
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
36
Calculate the wavelength in nanometers of the spectrum
of hydrogen corresponding to ni = 2 and nf = 4 in the
Rydberg equation
Example
Solution
Rydberg equation
1λ = RH (1ni2 ndash 1nf
2)
ni = 2 nf = 4
RH = 1097 x 10m7
1λ = RH (122 ndash 142)
= RH(14-116)
λ = 486m x 102 m
= 486nm
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
37
Use the Rydberg equation to calculate the wavelength of the
spectral line of hydrogen atom that would result when an
electron drops from the fourth orbit to the second orbit then
identified the series the line would be found
Example
Solution
1λ = RH (1n12 ndash 1n2
2)
n1 = 2 n2 = 4
1λ = 1097 x 107 (122 ndash 142)
λ = 486 x 10-7 m
= 486 nm
e dropped to the second orbit (n=2)
gtgtgt Balmer series
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
38
EXAMPLE 3
Calculate the wavelengths of the fourth line in the
Balmer series of hydrogen
n1 = 2 n2 = 6
RH = 1097 x 107m-1
λ = 410 x 10-7 m
RH 22 62
1 11=
λ
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
39
Different values of RH and its usage
1 RH = 1097 times 107 m-1
RH n21
n22
1 11=
λ
RH = 218 x 10-18 J
n1 lt n2
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
40
EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom
ΔE = 458 x 10-19 J
ΔE = (663 times 10-34Js)X(300times108 ms-1)
RH n21
n22
1 11=
λ
521097 x 107
22
1 11
=λ
1
λ= 02303 X 107 m-1
X (02303 X 107 m-1)
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
41
Calculate what is
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum
in Lyman series
Wave number = 1wavelength
EXERCISE
For Lyman series n1 = 1
amp n2 = infin
Ans
i 9116 x10-8m
ii 329 x1015 s-1
iii 10970 X 107 m-1
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
42
Definition Ionization energy is the minimum energy
required to remove one mole of electron from one mole
of gaseous atomion
M (g) rarr M+ (g) + e ΔH = +ve
The hydrogen atom is said to be ionised when electron
is removed from its ground state (n = 1) to n = infin
At n = infin the potential energy of electron is zero here
the nucleus attractive force has no effect on the electron
(electron is free from nucleus)
Ionization Energy
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
43
n1 = 1 n2 = infin
∆E = RH (1n12 ndash 1n2
2)
= 218 X 10 -18 (112 ndash 1 infin 2)
= 218 X 10 -18 (1 ndash 0)
= 218 X 10 -18 J
Ionisation energy
= 218 X 10 -18x 602 X 1023J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Example
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
44
λ1
1 st lineConvergent limit
Finding ionisation energy experimentally
λinfin
Ionisation energy is determined by detecting
the wavelength of the convergence point
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
45
1097 1066 1052 1027 974 822
wave number (x106 m-1)
The Lyman series of the spectrum of hydrogen is shown
above Calculate the ionisation energy of hydrogen from
the spectrum
Example
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
46
ΔE = hcλ
=h x c λ = h x c x wave no
= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1
= 21806x 10-20 J
= 218 x 10-18J
Ionisation energy
= 218 X 10 -18x 602 X 1023 J mol-1
=1312 x 106 J mol-1
= 1312 kJ mol-1
Solution
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
47
Compute the ionisation energy of hydrogen atom in kJ molminus1
Exercise
Solution
J
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
48
The weakness of Bohrrsquos Theory
1 His theory could not be extended to predict the energy
levels and spectra of atoms and ions with more than
one electron It only can explain the hydrogen spectrum
or ions contain one electron eg He+ Li2+
1 Electrons are restricted to orbit the nucleus at certain
fixed distances
1 It cannot explain for the dual nature of electron
1 It cannot explain for the extra lines formed in the
hydrogen spectrum
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
49
Davisson amp Germer observed the diffraction of
electrons when a beam of electrons was directed at a
nickel crystal Diffraction patterns produced by
scattering electrons from crystals are very similar to
those produced by scattering X-rays from crystals This
experiment demonstrated that electrons do indeed
possess wavelike properties
Thus can the lsquopositionrsquo of a wave be specified
Point to Ponder
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
50
de Brogliersquos Postulate
In 1924 Louis de Broglie proposed that not only light but all
matter has a dual nature and possesses both wave and
corpuscular properties De Broglie deduced that the particle
and wave properties are related by the expression
h = Planck constant (J s)
m = particle mass (kg)
μ = velocity (ms)
λ = wavelength of a matter wave
λ
=
h
m
μ
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
51
Heisenbergrsquos Uncertainty Principle
It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain
Stated mathematically
where Δx = uncertainty in measuring the position
Δp = uncertainty in measuring the momentum
= Δmv
h = Planck constant
h
4
π
Δx Δp ge
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
52
22 QUANTUM
MECHANICAL MODEL
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
53
At the end of this topic students should be able to-
Level No Learning Outcomes
C1 22 a Define the term orbital
C2 22 b Explain all four quantum numbers of an electron in an orbital
i) principal quantum number n
ii) angular momentum quantum number ℓ
iii) magnetic quantum number m
iv) electron spin quantum number s
C2 22 c Sketch the 3-D shapes of sp and d orbitals
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
54
Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron
Definition
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
55
Each of the electrons in an atom is described and
characterised by a set of four quantum numbers namely
a) principal quantum number n
b) angular momentum quantum number ℓ
c) magnetic quantum number m
d) electron spin quantum number s
Quantum Numbers
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
56
The value of n determines the energy of an orbital and thereby
the energy of the electron in that particular orbital
The principal quantum number may have only integral values n
=1 2 3 hellip infin
Principal Quantum Number n
n 1 2 3 4
shell K L M N
Orbital size
Energy increases
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
57
Angular Momentum Quantum Number ℓ
- Alternative name Subsidiary Azimuthal Orbital
Quantum Number
- The value of ℓ indicates the shape of the atomic orbital (AO) the
types of orbitals and the angular momentum of the electron
- The allowed values of ℓ are 0 1 2hellip (nminus1)
Letters are assigned to different numerical values of ℓ
Numerical value of ℓ Symbol
0
1
2
3
Orbital shape
s
p
d
f
spherical
dumbbell
cloverleaf
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
58
Angular Momentum Quantum Number ℓ
- ℓ is dependant on n (ie 0 le ℓ lt n)
If n = 1 ℓ can only be 0 (s-orbital)
If n = 2 ℓ can be 0 or 1 giving rise to two subshells
(s and p-orbitals) of slightly different energy
If n = 3 ℓ can be 0 1 or 2 (there are three subshells
(s p and d-orbitals)
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
59
Magnetic Quantum Number m
The direction or orientation of the magnetic field is determined
by the value of m
Possible values of m depend on the value of For a given
m can be minusℓ hellip 0 hellip + ℓ
(minus ℓ le m le + ℓ)
If ℓ = 0 m can only be 0 rArr one orbital in s-subshell
If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell
If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
60
The number of m values indicates the number of orbitals in
a subshell with a particular value
The values of n = 2 and = 1 indicate that we have a 2p-
subshell and in this subshell we have three 2p-orbitals
(because there are three values of m given by -1 0 and
+1)
Magnetic Quantum Number m
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
61
Electron Spin Quantum Number s
The value of s determines the direction of spinning motions of an
electron (either clockwise or counter clockwise) which is spinning
on its own axes as Earth does
The electron spin quantum number has a value of
+1
2-
1
2or
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
62
Atomic orbitals with the same energy (ie the same value
of n and ℓ) are said to be degenerated Therefore there
are (2 ℓ +1) degenerate orbitals for each value of ℓ
The maximum number of electrons in a particular energy
level n is given by the expression as follows
max no of eminus = 2n2
Points to Remember
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
63
Shell nℓ
(ℓltn)
Orbital
notation
m
(-ℓ le m le +ℓ)
No of
degenerated
orbitals
K
L
M
2
1
3
Exercise Complete the following table
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
64
Exercise
State whether or not each of the following symbols is an
acceptable designation for an atomic orbital Explain what
is wrong with the unacceptable symbols
b) 6g
a) 2d
c) 7s
d) 5i
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
65
Shape of Atomic Orbitals
a) s orbitals
Spherical shape with the nucleus at the centre
The probability of finding electrons at the distance r from the nucleus is the same from all direction
When ℓ = 0
As n increases s orbital
gets larger
Shape of s orbital
with different n
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
66
b) p orbitals
When ℓ = 1
dumbbell shaped
three p-orbitals px py and pz
correspond m of -1 0 and +1
As n increases the p-orbitals get larger
All p-orbitals have a node at the nucleus
Shape of Atomic Orbitals
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
67
Shape of Atomic Orbitals
shape four d orbitals have four lobes (perpendicular)
one d orbital has two major lobes along z axis
and a donut-shaped girdles the centre
When ℓ = 2
m = -2 -1012
the orbitals are dyz dxz dxy dx2-y2 dz2
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
68
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
69
Shape of Atomic Orbitals
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
70
Electronic Configuration
At the end of this topic students should
be able to-
Level No Learning Outcomes
C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos
Exclusion Principle
C3 23 b Predict the electronic configuration of atoms and
monotaomic ions using spdf notation
C3 23 c Justify the anomalous electronic configurations of
chromium and copper
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
71
Representing Electronic Configuration
Method 1 Orbital diagram
O8
1s 2s 2p
Method 2 spdf notation
O8 1s 2s 2p2 2 4
box
platform
Concentric circle
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
72
Rules for Assigning Electrons to Orbitals
i) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy
The order of filling orbitals is
1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s
1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
1s 2s 2p
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
73
Relative Energy Level of Atomic Orbitals
en
erg
y
n=1
n=2
n=3
n=4
1s
2s 2p
3s
4s
3p
4p
3d
4d
en
erg
y
n=1
n=2
n=3
n=4
1s
2s2p
3s
4s
3p
4p
3d
4d5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
74
ii) Pauli Exclusion Principle
Rules for Assigning Electrons to Orbitals
No two electrons in an atom can have the same four
quantum numbers (n m s)
1s
a b c
e(a)
e(b)
e(c)
n ℓ m s
1 0
01
0
01
0
0
12
12
12
( )
)(
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
75
iii) Hundrsquos Rule
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy eg three p-orbitals and five d-
orbitals) contain an electron do the electrons begin to
occupy these orbitals in pairs The electrons in half-filled
orbitals have the same spins that is parallel spins
2p
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
76
Indicate which of the following orbital diagrams are
acceptable or unacceptable for an atom in ground state
Explain what mistakes have been made in each and draw
the correct orbital diagram
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
77
Draw lsquoelectrons-in-boxesrsquo diagram of the electronic
configuration of titanium Ti (Z = 22) Also write the ground-
state electronic configurations for Ti and Ti2+ ion
Exercise
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
IMPORTANT
In an empty atom the 4s orbital has a lower
energy compared to that of the 3d orbital That is
why electrons fill the 4s orbital first before filling
the 3d orbital
However once electrons isare added to the 3d
orbital the 4s electrons are repelled to a higher
energy level The 3d orbitals now have lower
energy than 4s
78
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
79
Points to remember
The electronic configuration of atom or monatomic ion at
ground state
rArr Distribution of electrons obeys Aufbau principle Pauli
exclusion principle and Hundrsquos rule
Each atomic orbital can only accommodate a maximum of 2
electrons
Atomic orbital is a 3-D region in space around the nucleus
where there is a high probability of finding an electron
Assigning electrons to subshells
s-orbital rArr a max of 2 electrons (ns2)
p-orbitals rArr a max of 6 electrons (np6)
d-orbitals rArr a max of 10 electrons (nd10)
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
80
The Anomalous Electronic Configurations of
Cr and Cu
Cr and Cu have electron configurations which are
inconsistent with the Aufbau principle The anomalous
are explained on the basis that a filled or half-filled orbital
is more stable
Element Expected Observedactual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
81
24Cr 18[Ar]
The actual orbital notation
24Cr 18[Ar]
Half filled orbital is more stable
(possesses an extra added stability)
3d 4s
3d 4s
Chromium predicted orbital notation
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
82
Copper predicted orbital notation
Cu [Ar]
The actual orbital notation
Cu [Ar]
4s3d
3d 4s
Full filled orbital is more stable
(possesses an extra added stability)
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
83
z = 21
z = 30
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
84
Write the ground-state electronic configuration and
explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
Exercise
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
Writing Electronic Configuration for Negative Ion
Add electron according to Aufbau Principle
Example
i Cl-
ii O2-
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
Writing Electronic Configuration for Positive Ions
Remove electron from the outermost orbital (largest value of n)
Example
i Mg2+
i K+
i Fe2+
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