CHEM2915
Introduction to the Electronic Structure
of Atoms and Free Ions
A/Prof Adam Bridgeman
Room: 222
Email: A.Bridgeman@chem.usyd.edu.au
www.chem.usyd.edu.au/~bridge_a/chem2915
Slide 1/21
Where Are We Going…?
• Week 10: Orbitals and Terms
 Russell-Saunders coupling of orbital and spin angular momenta
 Free-ion terms for p2
• Week 11: Terms and ionization energies
 Free-ion terms for d2
 Ionization energies for 2p and 3d elements
• Week 12: Terms and levels
 Spin-orbit coupling
 Total angular momentum
• Week 13: Levels and ionization energies
 j-j coupling
 Ionization energies for 6p elements
Slide 2/21
Revision – Atomic Orbitals
•
•
For any 1 e- atom or ion, the Schrödinger equation can be solved
The solutions are atomic orbitals and are characterized by n, l and ml
quantum numbers
Hy = Ey
•
E is energy of the orbital y
•
H is the ‘Hamiltonian’ - describing the forces operating:
 Kinetic energy due to motion
 Potential energy due attraction to nucleus
 Total Hydrogen-like Hamiltonian
½ mv2
Ze
- r
HH-like
Slide 3/21
Atomic Orbitals - Quantum Numbers
•
•
For any 1-e- atom or ion, the Schrödinger equation can be solved
The solutions are atomic orbitals and are characterized by n, l and ml
quantum numbers
 Principal quantum number, n = 1, 2, 3, 4, 5, 6, …
 Orbital quantum number, l = n-1, n-2, n-3, 0 = number of nodal planes
 Magnetic quantum number, ml = l, l -1, l – 2, …, -l
l = 0: s
l = 1: p
l = 2: d
Slide 4/21
Orbital Quantum Number
•
Orbital quantum number, l = n-1, n-2, n-3, 0 = number of nodal planes
•
Magnetic quantum number, ml = l, l -1, l – 2, …, -l = orientation of orbital
 e.g. l = 2 gives ml = 2, 1, 0, -1, -2: so 2l + 1 = five d-orbitals
y
z
x
z
x
y
y
z
x
y
x
Slide 5/21
Magnetic Quantum Number
•
Orbital quantum number, l = n-1, n-2, n-3, 0 = number of nodal planes
related to magnitude of orbital angular momentum
•
Magnetic quantum number, ml = l, l -1, l – 2, …, -l = orientation of orbital
related to direction of orbital angular momentum
ml = 2
ml = 1
ml = 0
l=2
ml = -1
ml = -2
Slide 6/21
Spin Quantum Number
•
All electrons have spin quantum number, s = +½
•
Magnetic spin quantum number, ms = s, s -1 = +½ or –½: 2s+1 = 2 values
Slide 7/21
Many Electron Atoms
•
•
For any 2-e- atom or ion, the Schrödinger equation cannot be solved
The H-like approach is taken for every electron i
HH-like = S ½
mvi2
i
•
+S
i
Ze
- r
i
Treatment leads to configurations
 for example: He 1s2, C 1s2 2s2 2p2
•
Neglects interaction between electrons
 e- / e- repulsion is of the same
order of magnitude as HH-like
Si≠j
e2
rij
Slide 8/21
Many Electron Atoms – p1 Configuration
•
A configuration like p1 represents 6 electron arrangements with the same
energy
 there are three p-orbitals to choose from as l =1
 electron may have up or down spin
1
0
ml
-1
ml
ms
1
+½
1
-½
0
+½
0
-½
-1
+½
-1
-½
microstate
+
(1)
Slide 9/21
Many Electron Atoms – p2 Configuration
•
A configuration like p2 represents even more electron arrangements
•
Because of e- / e- repulsion, they do not all have the same energy:
 electrons with parallel spins repel one another less than electrons with
opposite spins
 electrons orbiting in the same direction repel one another less than
electrons with orbiting in opposite directions
lower in energy than
lower in energy than
1
0
ml
-1
1
0
-1
ml
Slide 10/21
Many Electron Atoms – L
•
For a p2 configuration, both electrons have l = 1 but may have ml = 1, 0, -1
•
L is the total orbital angular momentum
Lmax = l1 + l2 = 2
ml1 = 1
ml1 = 1
Lmin = l1 - l2 = 0
L = l1 + l2, l1 + l2 – 1, … l1 – l2
= 2, 1 and 0
For each L, ML = L, L-1, … -L
L: 0, 1, 2, 3, 4, 5, 6 …
ml2 = 1
ml = -1
code: S, P, D, F, G, H, I …
Slide 11/21
Many Electron Atoms – S
•
Electrons have s = ½ but may have ms = + ½ or - ½
•
S is the total spin angular momentum
Smax = s1 + s2 = 1
Smin = s1 - s2 = 0
S = s1 + s2, s1 + s2 – 1, … s1 – s2
= 1 and 0
For each S, MS = S, S-1, … -S
Slide 12/21
Many Electron Atoms – p2
•
L = 2, 1, 0
 for each L: ML = L, L -1, L – 2, …, -L
L: 0, 1, 2, 3, 4, 5, 6 …
code: S, P, D, F, G, H, I …
 for each L, there are 2L+1 functions
•
S = 1, 0
 for each S: MS = S, S-1, S-2, …. –S
 for each S, there are 2S+1 functions
Wavefunctions for many electron atoms are characterized by L and S
and are called terms with symbol:
2S+1
L
“singlets”: 1D, 1P, 1S
“triplets”: 3D, 3P, 3S
Slide 13/21
Microstates – p2
•
For example, 3D has L = 2 and S = 1 so:
 ML = 2, 1, 0, -1, -2 and MS = 1, 0, -1
 five ML values and three MS values: 5 × 3 = 15 wavefunctions with the
same energy
+/- +/( ml1, ml2 )
ML = ml1 + ml2
MS = ms1 + ms2
Slide 14/21
MS
p2
1
2
0
 
 -
- 
- -
(1, 1)
(1, 1)
(1, 1)
(1, 1)
 -
- 
(1, 0)
(0, 1)
 
1
 
(1, 0) (0, 1)
0
- 

(0, 0)


-

-
- 
-

-





-2

( -1, 0) (0, -1)

- -
- -
(1, 0)
(0, 1)
- -
(0, 0) (1, -1) (-1, 1)
(1, -1) (-1, 1)

(0, 1)
 -
(0, 0) (1, -1) (-1, 1)

-1
 -
(1, 0)
 
ML
-1

(-1, -1)
-
-


-
( -1, 0) (0, -1)

-
-
-
-
(0, 0)
-
-
(1, -1) (-1, 1)

( -1, 0) (0, -1)
-
-
-

(-1, -1) (-1, -1)
-
-
( -1, 0) (0, -1)
-
-
(-1, -1)
Pauli Principle and Indistinguishabity
•
The Pauli principle forbids two electrons having the same set of quantum
numbers. Thus for p2
 
-  Microstates such as (1, 1) and (-1, -1) are not allowed
•
Electrons are indistinguishable
 
 
 Microstates such as (1, -1) and (-1, 1) are the same
BUT
- 
 -
 Microstates such as (1, -1) and (1, -1) are different
•
For example, for p2
 6 ways of placing 1st electron, 5 ways of placing 2nd electron (Pauli)
 Divide by two because of indistinguishabillty: 6  5  15
2
Slide 16/21
Pauli forbidden
MS
Indistinguishable
p2
1
2
0
 
 -
- 
- -
(1, 1)
(1, 1)
(1, 1)
(1, 1)
 -
- 
(1, 0)
(0, 1)
 
1
 
(1, 0) (0, 1)
0
- 

(0, 0)


-

-
- 
-

-





-2

( -1, 0) (0, -1)

- -
- -
(1, 0)
(0, 1)
- -
(0, 0) (1, -1) (-1, 1)
(1, -1) (-1, 1)

(0, 1)
 -
(0, 0) (1, -1) (-1, 1)

-1
 -
(1, 0)
 
ML
-1

(-1, -1)
-
-


-
( -1, 0) (0, -1)

-
-
-
-
(0, 0)
-
-
(1, -1) (-1, 1)

( -1, 0) (0, -1)
-
-
-

(-1, -1) (-1, -1)
-
-
( -1, 0) (0, -1)
-
-
(-1, -1)
Working Out Allowed Terms
1.
Pick highest available ML: there is a term with L equal to this ML
•
2.
For this ML: pick highest MS: this term has S equal to this M2
•
3.
4.
Highest ML = 2  L = 2  D term
Highest MS = 0  S = 0  2S+1 = 1: 1D term
Term must be complete:
•
For L = 2, ML = 2, 1, 0, -1, -2
•
For S = 0, Ms = 0
}
for each value of Ms, strike out
microstates with these ML values
Repeat 1-3 until all microstates are used up
a.
Highest ML = 1  L = 1  P term
b.
Highest MS = 1  S = 1  2S+1 = 3: 3P term
c.
Strike out 9 microstates (MS = 1, 0, -1 for each ML = 1, 0, -1)
d.
Left with ML = 0  L = 0  S term
e.
This has MS = 0  S = 0  2S+1 = 1: 1S term
Slide 18/21
1D
MS
3P
1S
p2
1
0
 -
2
(1, 1)
 -
 
1
ML
0
(1, 0)
(1, 0)
-2
- -
(1, 0)
- 
(1, 0)


(1, -1)
 -



( -1, 0)
-

-
(0, 0) (1, -1) (-1, 1)

-1
-1

( -1, 0)

-
-
-
(1, -1)
-
( -1, 0)
-
-
-
(-1, -1)
( -1, 0)
Check
•
The configuration p2 gives rise to 15 microstates
•
These give belong to three terms:
 1D is composed of 5 states (MS = 0 for each of ML = 2, 1, 0, -1, -2
 3P is composed of 9 states (MS = 1, 0, -1 for each of ML = 1, 0, -1
 1S is composed of 1 state (MS = 0, ML = 0)
 5 + 9 + 1 = 15
•
The three terms differ in energy:
 Lowest energy term is 3P as it has highest S (unpaired electrons)
Slide 20/21
Summary
Configurations
• For many electron atoms, the HH-like gives rise to
configurations
• Each configuration represents more than one
arrangements
Terms
• The arrangements or microstates are grouped into terms
according to L and S values
• The terms differ in energy due to interelectron repulsion
Next week
• Hund’s rules and ionization energies
Task!
• Work out allowed terms for d2
Slide 21/21