1. Slide 1/21 CHEM2915 A/Prof Adam Bridgeman Room: 222 Email: A.Bridgeman@chem.usyd.edu.au www.chem.usyd.edu.au/~bridge_a/chem2915 Introduction to the Electronic Structure of Atoms and Free Ions

2. Slide 2/21 Where Are We Going…? Week 10: Orbitals and Terms  Russell-Saunders coupling of orbital and spin angular momenta  Free-ion terms for p 2 Week 11: Terms and ionization energies  Free-ion terms for d 2  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

3. 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 m l 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 mv 2 H H-like r Ze 

4. 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 m l 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, m l = l, l  1, l – 2, …, -l l = 0: sl = 1: p l = 2: d = number of nodal planes

5. Slide 5/21 Orbital Quantum Number Orbital quantum number, l = n-1, n-2, n-3, 0 Magnetic quantum number, m l = l, l  1, l – 2, …, -l  e.g. l = 2 gives m l = 2, 1, 0, -1, -2: so 2l + 1 = five d-orbitals = number of nodal planes = orientation of orbital

6. Slide 6/21 Magnetic Quantum Number Orbital quantum number, l = n-1, n-2, n-3, 0 Magnetic quantum number, m l = 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 m l = 2 m l = 1 m l = 0 m l =  2 m l =  1

7. Slide 7/21 Spin Quantum Number All electrons have spin quantum number, s = +½ Magnetic spin quantum number, m s = s, s  1 = + ½ or –½ : 2s+1 = 2 values

8. 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 1s 2, C 1s 2 2s 2 2p 2 r ij e2e2  H H-like = riri Ze    ½ mv i 2 + i i for every electron i Neglects interaction between electrons  e - / e - repulsion is of the same order of magnitude as H H-like  i≠j

9. Slide 9/21 Many Electron Atoms – p 1 Configuration A configuration like p 1 represents 6 electron arrangements with the same energy  there are three p-orbitals to choose from as l =1  electron may have up mlml 01 mlml msms microstate 1 +½+½ 1 ½½ 0 +½+½ 0 ½½ 11 +½+½ 11 ½½ ( 1 ) + or down spin

10. Slide 10/21 Many Electron Atoms – p 2 Configuration A configuration like p 2 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 mlml 0 1 mlml 0 1

11. Slide 11/21 Many Electron Atoms – L For a p 2 configuration, both electrons have l = 1 but may have m l = 1, 0, -1 L is the total orbital angular momentum m l2 = 1 m l =  1 m l1 = 1  L max = l 1 + l 2 = 2  L min = l 1 - l 2 = 0  L = l 1 + l 2, l 1 + l 2 – 1, … l 1 – l 2 = 2, 1 and 0  For each L, M L = L, L-1, … -L L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I …

12. Slide 12/21 Many Electron Atoms – S Electrons have s = ½ but may have m s = + ½ or - ½ S is the total spin angular momentum  S max = s 1 + s 2 = 1  S min = s 1 - s 2 = 0  S = s 1 + s 2, s 1 + s 2 – 1, … s 1 – s 2 = 1 and 0  For each S, M S = S, S-1, … -S

13. Slide 13/21 Many Electron Atoms – p 2 L = 2, 1, 0  for each L: M L = L, L  1, L – 2, …, -L  for each L, there are 2L+1 functions S = 1, 0  for each S: M S = 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: L 2S+1 L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … “ singlets ” : 1 D, 1 P, 1 S “ triplets ” : 3 D, 3 P, 3 S

14. Slide 14/21 Microstates – p 2 For example, 3 D has L = 2 and S = 1 so:  M L = 2, 1, 0, -1, -2 and M S = 1, 0, -1  five M L values and three M S values: 5 × 3 = 15 wavefunctions with the same energy ( m l1, m l2 ) +/- M L = m l1 + m l2 M S = m s1 + m s2

15. MSMS p2p2 10 2 1 MLML 0 -2

16. 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 p 2  Microstates such as and are not allowed For example, for p 2  6 ways of placing 1 st electron, 5 ways of placing 2 nd electron (Pauli)  Divide by two because of indistinguishabillty:

17. MSMS p2p2 10 2 1 MLML 0 -2 Pauli forbidden Indistinguishable

18. Slide 18/21 Working Out Allowed Terms 1.Pick highest available M L : there is a term with L equal to this M L Highest M L = 2  L = 2  D term 2.For this M L : pick highest M S : this term has S equal to this M 2 Highest M S = 0  S = 0  2S+1 = 1: 1 D term 3.Term must be complete: For L = 2, M L = 2, 1, 0, -1, -2 For S = 0, M s = 0 4.Repeat 1-3 until all microstates are used up a.Highest M L = 1  L = 1  P term b.Highest M S = 1  S = 1  2S+1 = 3: 3 P term c.Strike out 9 microstates (M S = 1, 0, -1 for each M L = 1, 0, -1) d.Left with M L = 0  L = 0  S term e.This has M S = 0  S = 0  2S+1 = 1: 1 S term } for each value of M s, strike out microstates with these M L values

19. MSMS p2p2 10 2 1 MLML 0 -2 1D1D 3P3P 1S1S

20. Slide 20/21 Check The configuration p 2 gives rise to 15 microstates These give belong to three terms:  1 D is composed of 5 states (M S = 0 for each of M L = 2, 1, 0, -1, -2  3 P is composed of 9 states (M S = 1, 0, -1 for each of M L = 1, 0, -1  1 S is composed of 1 state (M S = 0, M L = 0)  5 + 9 + 1 = 15 The three terms differ in energy:  Lowest energy term is 3 P as it has highest S (unpaired electrons)

21. Slide 21/21 Summary Configurations For many electron atoms, the H H-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 d 2