slide 1/21 chem2915 a/prof adam bridgeman room: 222 email: [email protected]...
TRANSCRIPT
Slide 1/21
CHEM2915
A/Prof Adam BridgemanRoom: 222
Email: [email protected]
www.chem.usyd.edu.au/~bridge_a/chem2915
Introduction to the Electronic Structure of Atoms and Free Ions
Slide 2/21
Where Are We Going…?
• Week 10: Orbitals and TermsRussell-Saunders coupling of orbital and spin angular momentaFree-ion terms for p2
• Week 11: Terms and ionization energiesFree-ion terms for d2 Ionization energies for 2p and 3d elements
• Week 12: Terms and levelsSpin-orbit couplingTotal angular momentum
• Week 13: Levels and ionization energiesj-j couplingIonization energies for 6p elements
Slide 3/21
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
H = E
• E is energy of the orbital
• 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
HH-like
rZe
Slide 4/21
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
Magnetic quantum number, ml = l, l 1, l – 2, …, -l
l = 0: s l = 1: p l = 2: d
= number of nodal planes
Slide 5/21
x
y
x
z
y
z
x
y
y
z
x
Orbital Quantum Number
• Orbital quantum number, l = n-1, n-2, n-3, 0
• Magnetic quantum number, ml = l, l 1, l – 2, …, -l
e.g. l = 2 gives ml = 2, 1, 0, -1, -2: so 2l + 1 = five d-orbitals
= number of nodal planes
x
y
x
z
y
z
x
y
y
z
x
= orientation of orbital
Slide 6/21
Magnetic Quantum Number
• Orbital quantum number, l = n-1, n-2, n-3, 0
• Magnetic quantum number, ml = l, l 1, l – 2, …, -l
= number of nodal planes
= orientation of orbital
related to magnitude of orbital angular momentum
related to direction of orbital angular momentum
l = 2
ml = 2
ml = 1
ml = 0
ml = 2
ml = 1
Slide 7/21
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 8/21
Many Electron Atoms
• For any 2-e- atom or ion, the Schrödinger equation cannot be solved• The H-like approach is taken
• Treatment leads to configurations
for example: He 1s2, C 1s2 2s2 2p2
rij
e2
HH-like = ri
Ze ½ mvi2 +
ii
for every electron i
• Neglects interaction between electrons
e- / e- repulsion is of the same order of magnitude as HH-like
i≠j
Slide 9/21
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
ml
01 -1
mlms microstate
1 +½
1 ½
0 +½
0 ½
1 +½
1 ½
( 1 )+
or down spin
Slide 10/21
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
ml
01 -1
ml
01 -1
Slide 11/21
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
ml2 = 1
ml = 1
ml1 = 1 ml1 = 1Lmax = l1 + l2 = 2
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 …
code: S, P, D, F, G, H, I …
Slide 12/21
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 13/21
Many Electron Atoms – p2
• L = 2, 1, 0
for each L: ML = L, L 1, L – 2, …, -L
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:
L2S+1
L: 0, 1, 2, 3, 4, 5, 6 …
code: S, P, D, F, G, H, I …
“singlets”: 1D, 1P, 1S
“triplets”: 3D, 3P, 3S
Slide 14/21
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
MS
p2 1 0 -1
2
1
ML 0
-1
-2
( , )1 1
( , )1 1
( , )1 1
( , )1 1
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )1 1
( , )1 1
( , )1 1 ( , )1 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0 ( , )0 1
( , )1 0 ( , )0 1
( , )1 1 ( , )1 1
( , )0 0
( , )1 1 ( , )1 1
( , )0 0
( , )1 1 ( , )1 1
( , )0 0
( , )1 1
( , )1 1
( , )0 0
Slide 16/21
• Electrons are indistinguishable
Microstates such as and are the same
BUT
Microstates such as and are different
Pauli Principle and Indistinguishabity
• The Pauli principle forbids two electrons having the same set of quantum numbers. Thus for p2
Microstates such as and are not allowed
( , )1 1
( , )1 1
( , )1 1
( , )1 1
( , )1 1
• For example, for p2
6 ways of placing 1st electron, 5 ways of placing 2nd electron (Pauli)
Divide by two because of indistinguishabillty:
( , )1 1
6 515
2
MS
p2 1 0 -1
2
1
ML 0
-1
-2
( , )1 1
( , )1 1
( , )1 1
( , )1 1
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0
( , )1 1
( , )1 1
( , )1 1 ( , )1 1
( , )1 0
( , )0 1
( , )1 0
( , )0 1
( , )1 0 ( , )0 1
( , )1 0 ( , )0 1
( , )1 1 ( , )1 1
( , )0 0
( , )1 1 ( , )1 1
( , )0 0
( , )1 1 ( , )1 1
( , )0 0
( , )1 1
( , )1 1
( , )0 0
Pauli forbidden
Indistinguishable
Slide 18/21
Working Out Allowed Terms
1. Pick highest available ML: there is a term with L equal to this ML
• Highest ML = 2 L = 2 D term
2. For this ML: pick highest MS: this term has S equal to this M2
• Highest MS = 0 S = 0 2S+1 = 1: 1D term
3. Term must be complete:
• For L = 2, ML = 2, 1, 0, -1, -2
• For S = 0, Ms = 0
4. 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
} for each value of Ms, strike out microstates with these ML values
MS
p2 1 0 -1
2
1
ML 0
-1
-2
( , )1 0
( , )1 0
( , )1 0
( , )1 0
( , )1 1
( , )1 1
( , )1 0
( , )1 0
( , )1 0
( , )1 0
( , )1 1
( , )1 1
( , )1 1 ( , )1 1
( , )0 0
1D
3P
1S
Slide 20/21
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 21/21
Summary
Configurations• For many electron atoms, the HH-like gives rise to
configurations• Each configuration represents more than one
arrangementsTerms• The arrangements or microstates are grouped into terms
according to L and S values• The terms differ in energy due to interelectron repulsionNext week• Hund’s rules and ionization energies
Task!• Work out allowed terms for d2