Thermochemistry (4 lectures)
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Transcript Thermochemistry (4 lectures)
CHEM2915
Introduction to the Electronic Structure
of Atoms and Free Ions
A/Prof Adam Bridgeman
Room: 222
Email: [email protected]
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, 0)
(0,1)
1
(1, 0) (0,1)
ML
0
(0, 0)
-
-
-
-
-
-
-
(0, 0) (1, -1) (-1,1)
-1
-2
(-1, 0) (0, -1)
-
- -
- -
(1, 0)
(0,1)
- -
(0, 0) (1, -1) (-1,1)
(1, -1) (-1,1)
-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, 0)
(0,1)
1
(1, 0) (0,1)
ML
0
(0, 0)
-
-
-
-
-
-
-
(0, 0) (1, -1) (-1,1)
-1
-2
(-1, 0) (0, -1)
-
- -
- -
(1, 0)
(0,1)
- -
(0, 0) (1, -1) (-1,1)
(1, -1) (-1,1)
-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
ML
0
(1,1)
-
(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