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1
Module 2.2: The Modern Atomic Theory and Quantum Mechanics
Veronica Calindas, R.C.Adamson University
This lecture is based on textbooks authored by Chang, Brown & Holme, and Redmore
Modern Atomic Theory (from Dalton’s Hypotheses)1. All matter is composed of atoms. The
atom is the smallest body that retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in nuclear reactions.
Modern Atomic Theory (from Dalton’s Hypotheses)3. All atoms of an element have the same
number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number. A sample of the element is treated as though its atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more elements in specific ratios.
The Atom and its Subatomic Particles
Properties of Subatomic Particles
Properties of the Three Key Subatomic Particles Ch
arge
Mass Rel
ative
1+
0
1-
Absolute (C)*
+1.60218 x 10-19
0
-1.60218 x 10-19
Relative (amu)†
1.00727
1.00866
0.00054858
Absolute (g)
1.67262 x 10-24
1.67493 x 10-24
9.10939 x 10-28
Location
in the Atom
Nucleus
Outside
Nucleus
Nucleus
Name (Symbol)
Electron (e-)
Neutron (n0)
Proton (p+)
Table 2.2
* The coulomb (C) is the SI unit of charge. † The atomic mass unit (amu) equals
1.66054 x 10-24 g.
Atomic Number, Mass Number and Isotopes
Atomic Number (Z) = number of protons present in the nucleus of each atom of an element.
Mass Number (A) = total number of neutrons and protons present in the nucleus of an atom of an element.
Atomic Number, Mass Number and Isotopes
Atomic Number, Mass Number and Isotopes In some cases, atoms that have the same
atomic number but different mass numbers exist. These are called isotopes.
Sample Problem:
(1)Give the number of protons, neutrons and electrons in each of these species:
a) 82Pb b) 29Cu
c) 80Hg d) 80Hg
ANSWERS: a) 82, 125, 82 b)29, 34, 29 c) 80, 119, 80, d) 80, 120, 80
207 63
199
200
Molecules and IonsMolecules – an aggregate of at least two
atoms in a definite arrangement held together by chemical forces (chemical bonds)-may contain atoms of the same element or atoms of two or more elements.
• Ions – an atom or a group of atoms that has either a net positive or net negative charge.- only e- are either lost or gained during chemical changes
MoleculesMonoatomic molecules – those that exist
as single atoms
e.g. Group 8A Noble Gases
Diatomic molecules – those that contains only two atoms.
e.g. N2, O2, and most of Group 7A Halogens
Polyatomic molecules – those that contain more than two atoms
IonsCation (X+) – an ion with positive net
charge due to loss of electron
Anion (X-) – an ion with negative net charge due to gaining of electron
Na11 protons11 electrons
Na+ 11 protons10 electrons
Cl17 protons17 electrons Cl- 17 protons
18 electrons
IonsMonoatomic ions– an ion that contains
only one atom
e.g. Na+, Mg2+, Fe3+, S2-, N3-
Polyatomic ions – an ion containing more than one atom
e.g. OH-, CN-, NH4+
Sample Problem:(1) Give the number of protons, neutrons
and electrons in each of these species:
a) K+ b) Mg2+
c) Fe3+ d) Br-
e) Mn2+ f) C-4
ANSWERS: a) 11, 12, 10 b) 12, 12, 10 c) 26, 29, 23 d) 35, 45, 36
e) 25, 30, 23 f) 6, 6, 10
The Duality of ElectronErwin Schrödinger wrote a complicated
mathematical equation that incorporates both particle behavior and wave behavior of an electron with respect to its probable location in the space of the system.
Schrödinger’s equation for hydrogen atom gave birth to a new era in physics and chemistry, this new field is called quantum mechanics
Eψ = -h2
8π2μ
{∂2ψ∂x2 +∂2ψ
∂y2+∂2ψ∂z2
{
+V(x,y,z)ψ
The Schrödinger Equation Wave function (Y) describes:
1 . energy of e- with a given Y
2 . probability of finding e- in a volume of space
Schrödinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.
The Four Quantum Numbers Derived from Schrödinger’s
equation for the hydrogen atom Quantum numbers are required
to describe the distribution of e- in hydrogen and other atoms, and the behavior of a specific e-.
17
The Four Quantum Numbers1. Principal Quantum Number (n)2. Azimuthal or Angular
Momentum Quantum Number (l )
3. Magnetic Quantum Number (ml )
4. Spin Quantum Number (ms)
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Principal Quantum Number (n)
Relates to the average distance of an e- from the nucleus in a particular orbital
Determines the energy of an orbital
Values are whole numbers only
19
The Schrödinger Equation
20
= Y fn(n, l, ml, ms)
principal quantum number n Where n = 1, 2, 3, 4, ….
n=1 n=2 n=3
distance of e- from the nucleus
Where 90% of thee- density is foundfor the 1s orbital
Principal Quantum Number (n)
Angular Momentum Quantum Number (l)
Determines the shape of an orbital
The value of l depends on the value of n (from 0 to n-1)◦ 0 s spherical◦ 1 p dumb-bell◦ 2 d clover leaf◦ 3 f
Subshell – one or more atomic orbitals having the same n and l values
22
The Schrödinger Equation
23
= Y fn(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
Thus, if n = 1, then l = 0 n = 2, then l = 0 or
1 n = 3, then l = 0, 1,
or 2Shape of the “volume” of space that the e- occupies
Orbital ‘Shapes’ Based on l Values
l = 0 (s orbitals)
l = 1 (p orbitals)
Orbital ‘Shapes’ Based on l Values
l = 2 (d orbitals)
Orbital ‘Shapes’ Based on l Values
l = 3 (f orbitals)
Magnetic Quantum Number (ml)
Describes the orientation of the orbital in space
Values depend on the l (values are denoted as –l, 0, +l)
For a subshell of quantum number l, there is a total of 2l + 1 atomic orbitals within that subshell. Atomic orbitals within the same subshell have essentially the same energy.
27
The Schrödinger Equation
28
= Y fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of lml = -l, …., 0, …. +l
orientation of the orbital in space
if l = 1 (p orbital), then ml = -1, 0, or 1if l = 2 (d orbital), then ml = -2, -1, 0, 1, or 2
Orbital ‘Orientation’ Based on ml Values
ml = -1 ml = 0 ml = 1
Orbital ‘Orientation’ Based on ml Values
ml = -2 ml = -1 ml = 0 ml = 1 ml = 2
Atomic Orbitals
31
Atomic Orbitals
32
Spin Quantum Number (ms)
33
Emission spectra showed that the lines can be split by applying an external magnetic field.
According to Electromagnetic Theory, a spinning charge generates a magnetic spin, which causes the e- to behave like a magnet.
There are 2 possible values for motion of e-: clockwise (+ ½ ) or counter clockwise (– ½ )
The Schrödinger Equation
34
= Y fn(n, l, ml, ms)
spin quantum number ms where the only possible values are either +½ or -½
ms = -½ms = +½
Sample Problem:List the values of n, l, and ml for
orbitals of the following subshells:
(1) 4d (2) 6p(3) 4s(4) 5f
ANSWERS: (1) n=4, l=2, ml=7 ,(2) n=6, l=1, ml=3, (3) n=4, l=0, ml=1, (4) n=5, l=3, ml=7
Sample Problem:What is the total number of
orbitals associated with the following principal number?
(1) n = 3(2) n = 5(3) n = 2(4) n = 4
ANSWERS: (1) 9; (2) 19; (3) 7; (4) 16
Electron ConfigurationDescribes how the electrons are
distributed among the various atomic orbitals.
A shorthand way of writing the quantum numbers for a specific atom
Electron Configuration: Orbital DiagramExample: Hydrogen atom: n=1, l=0, ml=0
1s1principal quantum
number n angular momentumquantum number l
number of electronsin the orbital or subshell
1s1
H
Then, the arrow-box configuration would yield:
Upward spin = +½Downward spin = -½
Rules for Writing an Electron Configuration1) From the modern Atomic Theory, in a
neutral atom, the number of protons equal to the number of electrons.
If Atomic number = number of protons and Number of protons = number of electrons,
then Atomic number = number of electron
(for atoms with no charge)
Only 2 electrons can occupy any subshell.
Rules for Writing an Electron Configuration2) No two electrons in the same atom can
have the same four quantum numbers (Pauli exclusion principle).
◦ If two electrons have the same n, l, and ml then they MUST have different values for ms – meaning, they must have opposite spins
Example:
1s2
He
1s21s2
(a) (b) (c)
Paramagnetism – when the two electrons have parallel spins
Paramagnetic substances are attracted to magnets.
The Pauli Exclusion Principle
1s2
He
1s2
Diamagnetism – when the two electrons in a single orbital have opposite spins
Diagmagnetic substances are slightly repelled by magnets.
He
1s2
REMEMBER: Any atom with an odd number of electrons must be paramagnetic. On the other hand, atoms that have an even number of electrons can either be paramagnetic or diamagnetic.
The Pauli Exclusion Principle
3) The most stable arrangement of electrons in subshells is the one with the greatest parallel spins (Hund’s Rule of Multiplicity).
Example: Nitrogen (Z=7) is 1s2 2s2 2p3, therefore the electron distribution as denoted by the arrow-box configuration would be
Rules for Writing an Electron Configuration
N
1s2 2s2 2px 2py 2pz
Example: Be (Z=4) is 1s2 2s2, therefore a diamagnetic
Rules for Writing an Electron Configuration
Be
1s2 2s2
Example: Si (Z=14) is 1s2 2s2 2p6 3s2 3p2 and thus, a paramagnetic
Si
1s2 2s2 2px 2py 2pz 3s2 3px 3py 3pz
4) An electron occupies the lowest energy orbital first before going to the next energy level (Aufbau principle).
Rules for Writing an Electron Configuration
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
H 1 electron
H 1s1
He 2 electrons
He 1s2
Li 3 electrons
Li 1s22s1
Be 4 electrons
Be 1s22s2
B 5 electrons
B 1s22s22p1
C 6 electrons
? ?
46
C 6 electrons
C 1s22s22p2
N 7 electrons
N 1s22s22p3
O 8 electrons
O 1s22s22p4
F 9 electrons
F 1s22s22p5
Ne 10 electrons
Ne 1s22s22p6
47
The Shielding Effect in Many-Electron Atom1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s <
4d < 5p < 6s
The 1s lies at lower energy level than 2s and 2p in a many-electron atom • This means that the 2s and 2p electrons are partly ‘shielded’ from the nucleus’s attractive force by the much-closer 1s electron.
•Because the stability of an electron is determined by the strength of its attraction to the nucleus, it follows that it will require less energy to remove an electron from 2p orbital than it would to remove the electron in 2s.
Problem:Write the electronic
configuration of the following:1. Ne (Z= 10)2. Sc (Z= 21)3. Ru (Z= 44)4. Pb (Z= 82)5. W (Z= 74)
Valence ElectronsElectrons which occupies the
outermost orbitals of an electron or the valence shell
Responsible for the reactions that an atom of an element can undergo
50
04/12/2023 General Chemistry for Engineers 51
04/12/2023 General Chemistry for Engineers 52
7.804/12/2023 53General Chemistry for Engineers
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2 [Ne] 1s22s22p6
What are the possible quantum numbers for the last (outermost) electron in Cl?
Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½54
04/12/2023 55General Chemistry for Engineers
Classification of Groups of Elements in the Periodic Table Accd’g to Type of Outermost Subshell filled