CHEM2915

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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

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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

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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

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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

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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


x

y

x

z

y

z

x

y

y

z

x

= orientation of orbital

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Magnetic Quantum Number

• Orbital quantum number, l = n-1, n-2, n-3, 0

• Magnetic quantum number, ml = l, l 1, l – 2, …, -l


= 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

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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

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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

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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

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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

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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 …

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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

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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

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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

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• 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

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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

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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)

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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

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CHEM2915