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1
The Electronic The Electronic Structure of AtomsStructure of Atoms
4.14.1 The Electromagnetic SpectrumThe Electromagnetic Spectrum
4.24.2 Deduction of Electronic Structure Deduction of Electronic Structure from Ionization Enthalpiesfrom Ionization Enthalpies
4.34.3 The Wave-mechanical Model of the The Wave-mechanical Model of the AtomAtom
4.4 Atomic Orbitals4.4 Atomic Orbitals
44
2
The electronic structure of atomsThe electronic structure of atoms Chapter 4 The electronic structure of atoms (SB p.80)
Two sources of evidence : -
(a) Study of atomic emission spectra of the elements
(b) Study of ionization enthalpies of the elements
3
Atomic Emission Spectra原子放射光譜
Spectra plural of Spectrum
Arises from light emitted from individual atoms
4
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
Speed of light (ms1) = Frequency Wavelength
c = 3108 ms1
5
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
Visible light
6
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
Ultraviolet light
Increasing energy
7
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
X-rays Increasing energy
8
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
Gamma rays Increasing energy
9
Frequency (Hz / s1)
Wavelength (m)
Decreasing energy
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
Infra-red light
10
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
red
violet
Microwave & radio waves
Decreasing energy
11
Types of Emission SpectraTypes of Emission Spectra4.1 The electromagnetic spectrum (SB p.82)
1. Continuous spectra
E.g. Spectra from tungsten filament and sunlight
2. Line Spectra
E.g. Spectra from excited samples in discharge tubes
12
Continuous spectrum of white Continuous spectrum of white lightlight
Fig.4-5(a)
4.1 The electromagnetic spectrum (SB p.82)
13
Line spectrum of hydrogenLine spectrum of hydrogen
Fig.4-5(b)
4.1 The electromagnetic spectrum (SB p.83)
14
How Do Atoms Emit Light ?
H2(g)Electric discharge
Or HeatingH(g)
Hydrogen atom in ground state
means
its electron has the lowest energy
15
Energy
Atom in ground state
H(g)Electric discharge
Or HeatingH*(g)
Excited Ground
Atoms in excited state
16
H*(g) H(g)Ground Excited
Energy
Atom in ground state
Atoms in excited state
Not Stable
h
Atom returns to ground state
h
17
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = hPlanck’s equation
18
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = hPlanck : Nobel laureate Physics,
1918
19
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = hE = energy of the emitted light = E2 –
E1
E2
E1
20
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = h = Frequency of the emitted light
E2
E1
21
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = hh (Planck’s constant) = 6.63 1034 Js
E2
E1
22
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = h
h = 6.63 1034 Js
E2
E1
Energy cannot be absorbed or emitted by an atom in any arbitrary amount.
23
Energy
Atom in ground state
Atoms in excited state
Atom returns to ground state
h
E = h
h = 6.63 1034 Js
E2
E1
Energy can only be absorbed or emitted by an atom in multiples of 6.631034 J.
24
4.1 The electromagnetic spectrum (SB p.84)
Characteristic Features of the Characteristic Features of the Hydrogen Emission Line Hydrogen Emission Line Spectrum Spectrum 1. The visible region – The Balmer Series
green
red
violetultraviole
t
Convergence limit
25
Q.1
νhE ν
λc
λ
hcE
= 3.031019 J
m 10656.3
ms 103.00Js 106.639
1834
26
J103.03
E E19
2n3n
energy of one photon emitted
27
Q.2
(a) The spectral lines come closer at higher frequen
cy and eventually merge into a c
ontinuum(連續體 )
(b) n = n = 2
28
(c) The electron has been removed
from the atom. I.e. the atom
has been ionized.
Q.2e1
H(g) H+(g) + e
29
2nn corresponds to the last spectral line of the Balmer series
30
The Complete Hydrogen Emission The Complete Hydrogen Emission SpectrumSpectrum
UV
Visible IR
4.1 The electromagnetic spectrum (SB p.83)
31
Q.3
(a) The spectral lines in each series get closer at higher frequency.
(b) Since the energy levels converge at higher level, the spectral lines also converge at higher frequency.
32
Rydberg Equation
22H b
1a1
Rλ
1
Relates wavelength of the emitted light of hydrogen atom with the electron transition
33
22H b
1a1
R1
λ
1= wave number of the emitted light= number of waves in a unit length
e.g. 1m 100 1
100 waves in 1 meter
νλ
1
34
c
λ
1
22H b
1a1
R1
λ
1
35
22H b
1a1
Rλ
1
Electron transition : b a,
a, b are integers and b > a
a represents the lower energy level to which the electron is dropping back
b represents the higher energy levels from which the electron is dropping back
36
22H b
1a1
R1
Balmer series,
a = 2
b = 3, 4, 5,…
37
22H b
1a1
R1
Lyman series,
a = 1
b = 2, 3, 4,…
2 3 4 5 6
38
22H b
1a1
R1
Paschen series,
a = 3
b = 4, 5, 6,…
4 5 6 7
39
22H b
1a1
R1
RH is the Rydberg constant
40
Q.4
0.000
0.012
0.016
0.020
0.028
0.040
0.063
0.111
2.742.602.572.522.442.302.061.52(106
m1)
1
2b1
41
2b1
λ
1
106 m1
16m102.74
41
R1
21
R H22H
y-intercept
RH = 1.096107 m1
42
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
4.1 The electromagnetic spectrum (SB p.84)
Discrete spectral lines
energy possessed by electrons within hydrogen atoms cannot
be in any arbitrary quantities but only in specified amounts
called quanta.
43
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
4.1 The electromagnetic spectrum (SB p.84)
Only certain energy levels are allowed for the electron in a hydrogen atom.
The energy of the electron in a hydrogen atom is quantized.
44
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
4.1 The electromagnetic spectrum (SB p.84)
45
Niels Bohr (1885-1962)
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
Bohr’s Atomic Model of Hydrogen
Nobel Prize Laureate in Physics, 1922
46
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
Nobel Prize Laureate in Physics, 1922
for his services in the investigation of the structure of atoms and of the radiation emanating from them
47
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
Bohr’s model of H atom
48
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
1. The electron can only move around the nucleus of a hydrogen atom in certain circular orbits with fixed radii.
Each allowed orbit is assigned an integer, n, known as the principal quantum number.
49
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
2. Different orbits have different energy levels.
An orbit with higher energy is further away from the nucleus.
50
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
3. Spectral lines arise from electron transitions from higher orbits to lower orbits.
E2
E1
n = 2
n = 1
51
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
For the electron transition E2 E1,
E = E2 – E1 = h
E2
E1
n = 2
n = 1
the energy emitted is related to the frequencyof light emitted by the Plank’s equation :
52
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
4. In a sample containing numerous excited H* atoms,
different H* atoms may undergo different kinds of electron transitions to give a complete emission spectrum.
53
4 3
5 2
4 2
3 2
4 1
3 1
1876
434.1
486.1
656.7
97.32
102.6
121.7
1876
434.3
486.5
656.6
97.28
102.6
121.62 1
Wavelength Determined by Experiment (nm)
Wavelength Predicted by Bohr (nm)
Electronic Transition
54
Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum
5. The theory failed when applied to elements other than hydrogen (multi-electron systems)
55
Illustrating Bohr’s Theory
First line of Lyman series, n = 2 n = 1
E = E2 – E1
E2
E1
n = 2
n = 1
By Planck’s equation,
hc
hE
56
E = E2 – E1
E2
E1
n = 2
n = 1
By Rydberg’s equation,
hc
hE
2
22
1
111
nnR
1
hcE
2
22
1
11
nnhcR
57
2
22
1
11
nnhcR
hcE
E = E2 – E1
E2
E1
n = 2
n = 122
21 n
hcR
n
hcRE
2
12
2 n
hcR
n
hcR12 EE
58
All electric potential energies have negative signs except E
E1 < E2 < E3 < E4 < E5 < …… < E = 0
All are negative
59
Balmer series,
Transitions from higher levels to n = 2
22
1
2
1
nhcRE n = 3, 4, 5, …
Visible region
60
Lyman series,
Transitions from higher levels to n = 1
22
1
1
1
nhcRE n = 2, 3, 4,…
More energy released
Ultraviolet region
61
Paschen series,
Transitions from higher levels to n = 3
22
1
3
1
nhcRE n = 4, 5, 6,…
Less energy released
Infra-red region
62
E = E2 – E1 = h
4.1 The electromagnetic spectrum (SB p.87)
UVvisible
IR
63
Q.5 n = 100 n = 99
122
7
100
1
99
110096.1
1
m
m2105.4
Energy of the light emitted is extremely small.
64
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
Microwave
65
Q.5 n = 100 n = 99
122
7
100
1
99
110096.1
1
m
m2105.4
Microwave region
Energy of the light emitted is extremely small.
66
Q.6
67
121323 EEE
Energy of the first line n = 3
n = 2
n = 1
68
121323 EEE
Energy of the first line
121323 hhh
121323 Hz13106.2463.292
= 45.71013 Hz
69
121424 EEE
Energy of the second line n = 4
n = 2
n = 1
70
121424 EEE
121424 hhh
121424 Hz13106.2463.308
= 61.71013 Hz
Energy of the second line
71
121525 EEE
n = 5
n = 2
n = 1
Energy of the third line
72
121525 EEE
121525 hhh
121525 Hz13106.2467.315
= 69.11013 Hz
Energy of the third line
73
Convergence Limits and Ionization EnthalpiesIonization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n = )
X(g) X+(g) + e
Units : kJ mol1
74
Convergence Limits and Ionization Enthalpies
Convergence Limits
The frequency at which the spectral lines of a series merge.
75
Q.7
Ionization enthalpy
1E 11 hE
Convergence limit of Lyman series
H(g) H+(g) + e
n = 1
n =
76
Na
Li
He
n = 3 n =
n = 2 n =
n = 1 n =
n = 1 n =
n = 3
n = 2
n = 1
n = 1H
Electron transition for ionization of atom
Energy Level of e to be removed in ground state of
atomAtom
Lyman series ?
Balmer series ?
Paschen series ?
77
Q.9
Ionization enthalpy of helium
111 hEE
= (6.6261034 Js)(5.291015 s1)(6.021023 mol1)
= 2110 KJ mol1
> Ionization enthaply of hydrogen
78
E
E2
E1
E2’
E1’
HeH
Ionization enthalpy : He > H
Relative positions of energy levels depend on the nuclear charge of the atom.
79
The uniqueness of atomic emission The uniqueness of atomic emission spectraspectra
No two elements have identical atomic spectra
4.1 The electromagnetic spectrum (SB p.89)
80
atomic spectra can be used to identify unknown elements.
4.1 The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission The uniqueness of atomic emission spectraspectra
81
Flame Test
NaCl(s) Na(g) + Cl(g)heat
atomization
Na(g) + Cl(g) Na*(g) + Cl*(g)heat
Na*(g) + Cl*(g) Na(g) + Cl(g) hNa + hCl
82
Flame Test
The most intense yellow light is observed.
83
Q.10
hc
hE ν
m10450)ms 10Js)(3.0010(6.626
9
1834
= 4.421019 J
84
Bright lines against a dark background
Dark lines against a bright background
Emission vs AbsorptionEmission spectrum of hydrogen
(nm)
Absorption spectrum of hydrogen
(nm)
visible
visible
85
Absorption spectra are used to determine the distances and chemical compositions of the invisible clouds.
86
The Doppler effect
Lower pitch heard
Higher pitch heard
The frequency of sound waves from a moving object (a) increases when the object moves towards the observer.(b) decreases when the object moves away from the observer.
87
Redshift and the Doppler effect
Frequency of light waves emanating from a moving object decreases when the light source moves away from the observer. wavelength increases spectral lines shift to the red side with longer wavelength redshift
88
Atomic Absorption Spectra
redshiftMoving at higher speed
Moving at lower speed
89
Redshift
Left : - Absorption spectrum from sunlight.
Right : - Absorption spectrum from a supercluster of distant galaxies
90
Atomic Absorption Spectra
Only
atomic emission spectrum of hydrogen
is required in A-level syllabus !!!
91
4.4.22 Deduction of Deduction of
Electronic Electronic Structure from Structure from
Ionization Ionization EnthalpiesEnthalpies
92
Evidence of the Existence of Evidence of the Existence of ShellsShells
For multi-electron systems,
Two questions need to be answered
93
Evidence of the Existence of Evidence of the Existence of ShellsShells
1.How many electrons are allowed to occupy each electron shells ?
2.How are the electrons arranged in each electron shell ?
Existence of Shells
Existence of Subshells
94
Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n = )
X(g) X+(g) + e
Ionization enthalpyIonization enthalpy
95
Evidence of the Existence of Evidence of the Existence of ShellsShells
Successive ionization enthalpies
Q. 11
96
Q.11(a)
Be(g) Be+(g) + e IE1
Be+(g) Be2+(g) + e IE2
Be2+(g) Be3+(g) + e IE3
Be3+(g) Be4+(g) + e IE4
97
Q.11(b)
IE1 < IE2 < IE3 < IE4
Positive ions with higher charges attract electrons more strongly.
Thus, more energy is needed to remove an electron from positive ions with higher charges.
98
Q.11(c)
21060149051758900
IE4
(kJ mol1)
IE3
(kJ mol1)
IE2
(kJ mol1)
IE1
(kJ mol1)
99
Q.11(c)
(i) The first two electrons are relatively easy to be removed.
they experience less attraction from the nucleus,
they are further away from the nucleus and occupy the n = 2 electron shell.
100
Q.11(c)
(ii) The last two electrons are very difficult to be removed.
they experience stronger attraction from the nucleus,
they are close to the nucleus and occupy the n = 1 electron shell.
101
Electron Diagram of Beryllium
Second Shell
First Shell
102
Energy Level Diagram of Beryllium
n =2
n= 1
103
E
IE1 = E - E2
E2
E1
Be
104
E
IE1
E2
E1
Be
E2’
E1’
Be+
IE2
IE1 < IE2
105
E
IE1
E2
E1
Be
E2’
E1’
Be+
E2’’
E1’’
Be2
+
IE2
IE1 < IE2 << IE3
IE3
106
E
IE1
E2
E1
Be
E2’
E1’
Be+
E2’’
E1’’
Be2
+
E2’’’
E1’’’
Be3
+
IE1 < IE2 << IE3 < IE4
IE3IE2 IE4
107
The Concept of Spin(自旋 )
Spin is the angular momentum intrinsic to a body.
E.g.Earth’s spin is the angular momentum associated with Earth’s rotation about its own axis.
自轉
108
On the other hand,
orbital angular momentum of the Earth is associated with its annual motion around the Sun (公轉 )
109
Subatomic particles like protons and electrons possess spin properties.
i.e. they have spin angular momentum.
But their spins cannot be associated with rotation since they display both particle-like and wave-like behaviours.
110
Paired electrons in an energy level should have opposite spins.
Electrons with opposite spins are represented by arrows in opposite directions.
Q.12
111
4 groups of electrons
Q.12
112
n = 4
n = 3 n = 2
n = 1 Which group of
electrons is in the first shell ?
Q.12
113
2, 8, 8, 2
2882
Q.12
114
n = 4
n = 2
n = 3
n = 1
Q.12
115
Evidence of the Existence of Evidence of the Existence of ShellsShells
4.2 Deduction of electronic structure from ionization enthalpies (p.91)
2, 8, 1
116
n = 2
n = 3
n = 1
Na
117
Variation of IE1 with Atomic NumberEvidence for Subshell
118
Only patterns across periods are discussedRefer to pp.27-29 for further discussion
119
1.A general in IE1 with atomic number across Periods 2 and 3.13.(a)
120
2 – 3 - 3 13.(a)
121
2.IE1 value : Group 2 > Group 3
3. IE1 value : Group 5 > Group 6
13.(a)
122
2.Peaks appear at Groups 2 & 5
3. Troughs appear at Groups 3 & 6
13.(a)
123
2,1
2,2
2,3
2,4
2,5
2,6
2,7
2,8
2,8,1
2,8,2
2,8,32,8,4
2,8,5
2,8,6
2,8,7 2,8,8
On moving across a period from left to right,1. the nuclear charge of the atoms (from +3 to +10 or +11 to +18)2. electrons are being removed from the same shell
es removed experience stronger attraction from the nucleus.
13.(b)
124
IE1 : B(2,3) < Be(2,2)
The electron removed from B occupies a subshell of higher energy within the n = 2 quantum shell
13.(b)
125
2s
2p
1s
n =
IE2s
IE2p
IE2s(Be) > IE2p(B)
IE1 : Be > B
13.(b)
126
IE1 : Al(2,8,3) < Mg(2,8,2)
The electron removed from Al occupies a subshell of higher energy within the n = 3 quantum shell
13.(b)
127
3s
3p
2p
n =
IE3s
IE3p
IE3s(Mg) > IE3p(Al)
IE1 : Mg > Al
13.(b)
128
3s
3p
2p
n =
Mg(12)
Al(13)
IE3s
IE3p
129
It is more difficult to remove an electron from a half-filled p subshell
13.(b)
IE1 : N(2,5) > O(2,6) ;
P(2,8,5) > S(2,8,6)
130
2s
2p
n =
IEN
O(8)N(7)
IEO
>First IE
131
2s
2p
n =
IEN
O(8)N(7)
IEO
>First IEThe three electrons in the half-filled 2p subshell occupy three different orbitals (2px , 2py , 2pz). repulsion between electrons is minimized.
132
3s
3p
n =
IEP
S(16)P(15)
IES
>First IEThe three electrons in the half-filled 3p subshell occupy three different orbitals (3px , 3py , 3pz).
repulsion between electrons is minimized.
133
The removal of an electron from O or S results in a half-filled p subshell with extra stability.(p.28, part 3)
Misleading !!!
134
O(2,6) O+(2,5)
2p
Is the 2p energy level of O+ lower or higher than that of O ?
Electrons in O+ experience a stronger attractive force from the nucleus.Not because of the half-filled 2p subshell
135
2p
As a whole, O+ is always less stable than O.It is because O(g) has one more electron than O+(g) and this extra electron has a negative potential energy.
O(2,6) O+(2,5)
136
Conclusions : -
(a) Each electron in an atom is described by a set of four quantum numbers.
(b) No two electrons in the same atom can have the same set of quantum numbers.
The quantum numbers can be obtained by solving the Schrodinger equation (p.16).
137
Quantum Numbers
1.Principal quantum number (n)
n = 1, 2, 3,…
related to the size and energy of the principal quantum shell.
E.g. n = 1 shell has the smallest size and electrons in it possess the lowest energy
138
Quantum Numbers
= 0, 1, 2,…,n-1
2.Subsidiary quantum number ( )
related to the shape of the subshells.
139
Quantum Numbers2. Subsidiary quantum number ( )
= 0, 1, 2,…,n-1
= 0 spherical s subshell= 1 dumb-bell p subshell
= 2 d subshell= 3 f subshell
complicated shapes
140
Each principal quantum shell can have one or more subshells depending on the value of n.
If n = 1,
possible range of :
name of subshell : 1s
0 to (1-1)
0No. of subshell : 1
141
Each principal quantum shell can have one or more subshells depending on the value of n.
0 to (2-1) 0, 1
If n = 2,
possible range of :
names of subshells : 2s, 2p No. of subshells : 2
142
Each principal quantum shell can have one or more subshells depending on the value of n.
names of subshells : 3s, 3p, 3d
0 to (3-1) 0, 1, 2
If n = 3,
possible range of :
No. of subshells : 3
143
Each principal quantum shell can have one or more subshells depending on the value of n.
names of subshells : 4s, 4p, 4d, 4f
0 to (4-1) 0, 1, 2, 3
If n = 4,
possible range of :
No. of subshells : 4
144
Quantum Numbers
3.Magnetic quantum number (m)
related to the spatial orientation of the orbitals in a magnetic field.
m = ,…0,…+
145
Each subshell consists of one or more orbitals depending on the value of
Possible range of m : 0 to +0 0
No. of orbital : 1
Name of orbital : s
If = 0 (s subshell),
146
Possible range of m : 1 to +1 1 ,0,+1
No. of orbitals : 3
Names of orbitals : px, py, pz
Each subshell consists of one or more orbitals depending on the value of
If = 1 (p subshell),
147
Possible range of m : 2 to +2 2, 1, 0, +1, +2
No. of orbitals : 5
Names of orbitals : 222 zyxyzxzxy d,d,d,d,d
Each subshell consists of one or more orbitals depending on the value of
If = 2 (d subshell),
148
Possible range of m : 3 to +3 3, 2, 1, 0, +1, +2, +3
No. of orbitals : 7
Names of orbitals : Not required in AL
Each subshell consists of one or more orbitals depending on the value of
If = 3 (f subshell),
149
Total no. of orbitals in a subshell = 2 + 1
Energy of subshells : -
s < p < d < f
150
4.Spin quantum number (ms)
2
1 or
2
1ms =
They describe the spin property of the electron, either clockwise, or anti-clockwise
151
4.Spin quantum number (ms)
Each orbital can accommodate a maximum of two electrons with opposite spins
152
Q.14(a)
32n = 4
18n = 3
8
1+3+5+7=16
1+3+5=9
1+3=4
4s, 4p, 4d, 4f
3s, 3p, 3d
2s, 2pn = 2
21(1s)1sn = 1
Total no. of
electrons
Total no. of orbitalsSubshells
Principal quantum
shell
153
14(b)
The total number of electrons in a principal quantum shell = 2n2
154
Q.15(a)
The two electrons of helium are in the 1s orbital of the 1s subshell of the first principal quantum shell.
1st electron : n = 1, l = 0, m = 0, ms = 21
2nd electron : n = 1, l = 0, m = 0, ms = 21
155
Q.15(b)
There are only 2 electrons in the 3rd quantum shell.
In the ground state, these two electrons should occupy the 3s subshell since electrons in it have the lowest energy.
156
Q.15(b)The two outermost electrons of magnesium are in the 3s orbital of the 3s subshell of the third principal quantum shell.
1st electron : n = 3, l = 0, m = 0, ms = 21
2nd electron : n = 3, l = 0, m = 0, ms = 21
157
4.4.33 The Wave-The Wave-
mechanical mechanical Model of the Model of the
AtomAtom
158
Models of the Atoms
1. Plum-pudding model by J.J. Thomson (1899)
2. Planetary(orbit)model by Niels Bohr (1913)
3. Orbital model by E. Schrodinger (1920s)
159
Electrons display both particle nature and wave nature.
Particle nature : mass, momemtum,…
Wave nature : frequency, wavelength, diffraction,…
160
Wave as particles : photons
hmc 2
hE By Planck(1)2mcE By Einstein(2)
h
chmc
(3)
Wave propertyParticle
property :momentum of a photon
161
Evidence :
photoelectric effect by Albert Einstein (1905)
Nobel Prize Laureate in Physics, 1921
162
Particles as waves L. De Broglie (1924)
Nobel Prize Laureate in Physics, 1929
h
c
νhmc
(3)
h
mv (4)
163
h
mc
h
mv
Any particle (not only photon) in motion (with a momentum, mv) is associated with a wavelength
(3)
(4)
164
Q.16
λ
hmv
mv
h
Electron
m10631
34
105.1100.5101.9
1063.6
λ
165
Q.16
λ
hmv
mv
hλ
Helium atom
m11327
34
102.7104.1106.6
1063.6
λ
166
Q.16
λ
hmv
mv
hλ
m107.3ms 10.5Kg 86Js106.63 37
1-
34
λ
100m world record holder
s 9.52s m 10.5
m 1001
167
Wave nature of electrons (1927)Wave nature of electrons (1927)
4.3 The Wave-mechanical model of the atom (p.95)
Evidence
of electron inter-atomic spacing in metallic crystals (1010m)
168
For very massive ‘particles’,
The wavelength associated with them (1037m) are much smaller than the dimensions of any physical system.
Wave properties cannot be observed.
169
LStanding waves only have certain allowable modes of vibration
Similarly, electrons as waves only have certain allowable energies.
= 2L
= L
= L32
Standing waves in a
cavity
170
The uncertainty principle
Heinsenberg
Nobel Prize Laureate in Physics, 1932
171
The uncertainty principle
It is impossible to simultaneously determine the exact position and the exact momentum of an electron.
4))((
hpx
uncertainty in position measurement
uncertainty in momentum measurement
172
high energy photon
Small and light electron
The momentum of electron would change greatly after collision
The uncertainty principle
173
low energy photonSmall and
light electron
The position of electron cannot be located accurately
The uncertainty principle
174
high energy photon
No problem in macroscopic world !
The uncertainty principle
175
Implications : -
The concept of well-defined orbits in Bohr’s model has to be abandoned.
We can only consider the probability of finding an electron of a ‘certain’ energy and momentum within a given space.
176
Schrodinger Equation
Nobel Prize Laureate in Physics, 1933
de Broglie : electrons as waves
Use wave functions () to describe electrons
Pronounced as psi
177
Schrodinger Equation
Nobel Prize Laureate in Physics, 1933
Heisenberg : Uncertainty principle
Probability (2) of finding electron at a certain position < 1.
178
0)(8
2
2
2
2
2
2
2
2
VEh
m
zyx
: wave function
m : mass of electron
h : Planck’s constant
E : Total energy of electron
V : Potential energy of electron
Schrodinger Equation
179
0)(8
2
2
2
2
2
2
2
2
VEh
m
zyx
Schrodinger Equation
The equation can only be solved for certain i and Ei
sinxdxsinxd
2
2
180
Schrodinger Equation
The wave function of an 1s electron is
0
23
21
0
1)1( a
Zr
ea
Zs
Z : nuclear charge (Z = 1 for Hydrogen)
a0 : Bohr radius = 0.529Å (1Å = 1010m)
r : distance of electron from the nucleus
Radius of 1s orbit of Bohr’s model
181
Schrodinger Equation
The allowed energies of H atom are given by
n = 1, 2, 3,…222
0
24
8 hn
ZmeE
n is the principal quantum number,
All other terms in the expression are constants
182
n = 1, 2, 3,…222
0
24
8 hn
ZmeE
2
1
nE
0Erefer to p.6 of notes
183
2
22
1
11
nnhcR
hcE
E = E2 – E1
E2
E1
n = 2
n = 122
21 n
hcR
n
hcRE
2
12
2 n
hcR
n
hcR12 EE
184
22is the is the probability of finding an electron at of finding an electron at a particular point in space. (electron density)a particular point in space. (electron density)
4.4 Atomic orbitals (p.98)
22(1(1s)s)
Probability never becomes zero
There is no limit to the size of an atom
185
contour diagram
relative probability of finding the electron
at the nucleus
as r 2
186
In practice, a boundary surface is chosen such that within which there is a high probability (e.g. 90%) of finding the electron.
187
The electron spends 90% of time within the boundary surface
188
A 3-dimensional time exposure diagram.
The density of the dots represents the probability of finding the electron at that position.
189
The 3-dimensional region within which there is a high probability of finding an electron in an atom is called an atomic orbital.
Each atomic orbital is represented by a specific wave function().
The wave function of a specific atomic orbital describes the behaviour of the electron in the orbital.
190
Total probability of finding the electron within the ‘shell’ of thickness dr
= 24r2dr
electron density within the ‘shell’
total volume of the ‘shell’
2 is the probability of finding the electron per unit volume
dr is infinitesimally small
191
22 4 r 2 as r , 4r2 as r a maximum at 0.529 Å
Orbital Model
192
0.529Å r (Å)
1
probability
Bohr’s Orbit Model
193
Total probability of finding the electron within the ‘shell’ of thickness dr
= 24r2dr
The sum of the probabilities of finding the electron within all ‘shells’ = 1
1dr4πψr
0r
2
2r
194
22 4 r
The total area bounded by the curve and the x-axis = 1
Check Point 4-3Check Point 4-3
1dr4πψr
0r
2
2r
195
ss Orbitals Orbitals4.4 Atomic orbitals (p.98)
196
nodal surface
There is no chance of finding the electron on the nodal surface.
197
198
Probabilities of finding the electron at A or B > 0
A
B
Probability of finding the electron at C = 0
C
How can the electron move between A & B ?
199
can be considered as the amplitude of the wave.
2 is always 0
200
2 = 0
201
pp Orbitals Orbitals
Two lobes along an axis
4.4 Atomic orbitals (p.100)
202
For each 2p orbital, there is a nodal plane on which the probability of finding the electron is zero.
4.4 Atomic orbitals (p.100)
yz plane xz plane xy plane
203
4.4 Atomic orbitals (p.101)
dd Orbitals Orbitals
Four lobes along two
axes
Two lobes & one belt
Four lobes between two
axes
204
Grand Orbital Table
http://www.orbitals.com/orb/orbtable.htm#table1
205
The END
206
Some insects, such as bees, can see light of shorter wavelengths than humans can. What kind of
radiation do you think a bee sees?
Back
4.1 The electromagnetic spectrum (SB p.82)
Ultraviolet radiationAnswer
207
What does the convergence limit in the Balmer series correspond to?
Back
4.1 The electromagnetic spectrum (SB p.87)
The convergence limit in the Balmer series corresponds to the electronic transition from
n = to n = 2.
Answer
208
4.1 The electromagnetic spectrum (SB p.88)
Given the frequency of the convergence limit of the Lyman series of hydrogen, find the ionization enthalpy of hydrogen.
Frequency of the convergence limit = 3.29 1015 Hz
Planck constant = 6.626 10-34 J s
Avogadro constant = 6.02 1023 mol-1
Answer
209
4.1 The electromagnetic spectrum (SB p.88)
Back
For one hydrogen atom,
E = h
= 6.626 10-34 J s 3.29 1015 s-1
= 2.18 10-18 J
For one mole of hydrogen atoms,
E = 2.18 10-18 J 6.02 1023 mol-1
= 1312360 J mol-1
= 1312 kJ mol-1
The ionization enthalpy of hydrogen is 1312 kJ mol-1.
210
4.1 The electromagnetic spectrum (SB p.88)
The emission spectrum of atomic sodium is studied. The wavelength of the convergence limit corresponding to the ionization of a sodium atom is found. Based on this wavelength, find the ionization enthalpy of sodium.
Wavelength of the convergence limit = 242 nm
Planck constant = 6.626 10-34 J s
Avogadro constant = 6.02 1023 mol-1
Speed of light = 3 108 m s-1
1 nm = 10-9 m Answer
211
4.1 The electromagnetic spectrum (SB p.88)
Back
For one mole of sodium atoms,
E = hL
=
=
= 494486 J mol-1
= 494 kJ mol-1
The ionization enthalpy of sodium of 494 kJ mol-1.
Lhc
m 10242mol10 6.02 s m 10 3 s J 106.626
9-
123-1834
212
(a)Given the successive ionization enthalpies of boron, plot a graph of the logarithm of successive ionization enthalpies of boron against the number of electrons removed. Comment on the graph obtained.
Successive I.E. (in kJ mol-1): 800, 2400, 3700, 25000, 32800 Answer
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
213
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
(a)
The first three electrons of boron are easier to be removed because they are in the outermost shell of the atom. As the fourth and fifth electrons are in the inner shell, a larger amount of energy is required to remove them.
214
(b) Give a sketch of the logarithm of successive ionization enthalpies of potassium against no. of electrons removed. Explain your sketch. Answer
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
215
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
(b) There are altogether 19 electrons in a potassium atom. They are in four different energy levels. The first electron is removed from the shell of the highest energy level which is the farthest from the nucleus, I.e. the fourth (outermost) shell. It is the most easiest to be removed. The second to ninth electrons are removed from the third shell, and the next eight electrons are removed from the second shell. The last two electrons with highest ionization enthalpy are removed from the first (innermost) shell of the atom. They are the most difficult to be removed.
216
(c)There is always a drastic increase in ionization enthalpy whenever electrons are removed from a completely filled electron shell. Explain briefly. Answer
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
(c) A completely filled electron shell has extra stability. Once an electron is removed, the stable electronic configuration will be destroyed. Therefore, a larger amount of energy is required to remove an electron from such a stable electronic configuration.
Back
217
(a)What are the limitations of Bohr’s atomic model?
(b)Explain the term “dual nature of electrons”.
(c) For principal quantum number 4, how many sub-shells are present? What are their symbols?
Back4.3 The Wave-mechanical model of the atom (p.97)
(a) It cannot explain the more complicated spectral lines observed in emission spectra other than that of atomic hydrogen. There is no experimental evidence to prove that electrons are moving around the nucleus in fixed orbits.
(b) Electrons can behave either as particles or a wave.
(c) When n = 4, l = 0, 1, 2 and 3, there are 4 sub-shells. The symbols are 4s, 4p, 4d and 4f respectively.
Answer
218
(a)Distinguish between the terms orbit and orbital.
(b)Sketch the pictorial representations of an s orbital and a p orbital. What shapes are they?
4.4 Atomic orbitals (p.101)
(a) “Orbit” is the track or path where an electron is revolving around the nucleus. “Orbital” is a region of space in which the probability of finding an electron is very high (about 90 %).
(b) s orbital is spherical in shape whereas p orbital is dumb-bell in shape.
Answer
219
(c)How do the 1s and 2s orbitals differ from each other?
(d) How do the 2p orbitals differ from each other?
4.4 Atomic orbitals (p.101)
Answer
(c) Both 1s and 2s orbitals are spherical in shape, but the 2s orbital consists of a region of zero probability of finding the electron known as a nodal surface.
(d) There are three types of p orbitals. All are dumb-bell in shape. They are aligned in three different spatial orientations designated as x, y and z. Hence, the 2p orbitals are designated as 2px, 2py and 2pz.
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