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Slide 1/15 Where Are We Going…? Week 6: Orbitals and Terms  Russell-Saunders coupling of orbital and spin angular momenta  Free-ion terms for p 2 Week.

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Presentation on theme: "Slide 1/15 Where Are We Going…? Week 6: Orbitals and Terms  Russell-Saunders coupling of orbital and spin angular momenta  Free-ion terms for p 2 Week."— Presentation transcript:

1 Slide 1/15 Where Are We Going…? Week 6: Orbitals and Terms  Russell-Saunders coupling of orbital and spin angular momenta  Free-ion terms for p 2 Week 7: Terms and ionization energies  Free-ion terms for d 2  Ionization energies for 2p and 3d elements Week 8: Terms and levels  Spin-orbit coupling  Total angular momentum Week 9: Levels and ionization energies  j-j coupling  Ionization energies for 6p elements

2 Slide 2/15 d 2 and f 2 For p 2  6 ways of placing 1 st electron, 5 ways of placing 2 nd electron (Pauli)  Divide by two because of indistinguishabillty:  Number of combinations for 2 identical objects with 6 choices: n C 2 For d 2  10 ways of placing 1 st electron, 9 ways of placing 2 nd electron (Pauli)  Divide by two because of indistinguishabillty: For f 2  14 ways of placing 1 st electron, 13 ways of placing 2 nd electron (Pauli)  Divide by two because of indistinguishabillty:

3 MSMS d2d2 10 4 3 2 1 0 MLML -2 -3 -4

4 Slide 4/15 d 2 and f 2 Terms The configuration d 2 gives rise to 45 microstates These give belong to five terms:  1 G (L = 4: 2L+1 = 9, S = 0: 2S+1 = 1  9 × 1 = 9 states  3 F (L = 3: 2L+1 = 7, S = 1: 2S+1 = 3  7 × 3 = 21 states  1 D (L = 2: 2L+1 = 5, S = 0: 2S+1 = 1  5 × 1 = 5 states  3 P (L = 1: 2L + 1 = 3, S = 1: 2S+1 = 3  3 × 3 = 9 states  1 S (L = 0: 2L +1 = 1, S = 0: 2S+1 = 1  1 × 1 = 1 states  9 + 21 + 5 + 9 + 1 = 45 The configuration f 2 gives rise to 91 microstates These give belong to seven terms:  1 I, 3 H, 1 G, 3 F, 1 D, 3 P, 1 S L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I …

5 Slide 5/15 Hund’s First and Second Rules 1 st Rule: the term with the highest value of S lies lowest  lower repulsion between electrons with parallel spins 2 nd Rule: if the first rule is ambiguous, the term with the maximum L is lowest  lower repulsion between electrons with like-rotation p2:1D3P1Sd2:3F1D3P1Sf2:3H1G3F1D3P1Sp2:1D3P1Sd2:3F1D3P1Sf2:3H1G3F1D3P1S

6 Slide 6/15 Ground Terms Quickly Draw out 2l+1 boxes and label with m l values  Place electrons in boxes to maximize S (1 st Rule)  Occupy from left to right to maximize L (2 nd Rule)  Add m s to get S and m l to get L 0 1 p1:p1: p2:p2: L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … p3:p3: p4:p4: L = 1, S = ½  2S+1 = 2  2P 2P L = 1, S = ½ + ½ = 1  2S+1 = 3  3P 3P L = 0, S = 3 × ½ = 3/2  2S+1 = 4  4S 4S L = 1, S = ½ + ½ = 1  2S+1 = 3  3P 3P

7 Slide 7/15 p n Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 10LSground term p0p0 00 1S1S p1p1 11/2 2P2P p2p2 11 3P3P p3p3 03/2 4S4S p4p4 11 3P3P p5p5 11/2 2P2P p6p6 00 1S1S

8 d n Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 210-2LSground term d0d0 00 1S1S d1d1 21/2 2D2D d2d2 31 3F3F d3d3 33/2 4F4F d4d4 22 5D5D d5d5 05/2 6S6S d6d6 22 5D5D d7d7 33/2 4F4F d8d8 31 3F3F d9d9 21/2 2D2D d 10 00 1S1S

9 Slide 9/15 Ionization Energies: (i) Hund’s 1 st Rule Overall increase across each period  Increase in e - / nuclear attraction acts to increase IE and  Increase in e - / e - repulsion acts to lower IE  Increase in nuclear charge is generally more important  Decrease between p 3 and p 4 (“half filled shell stability”) p-block ionization energies: M  M +

10 Slide 10/15 p n  p n-1 Effect of ionization 10LSΔSΔS p1p1 11/2 p2p2 11 p3p3 03/2 p4p4 11 p5p5 11/2 p6p6 00 Overall increase across each period  Increase in e - / nuclear attraction acts to increase IE and  Increase in e - / e - repulsion acts to lower IE  Increase in nuclear charge is in general more important  BUT ionization of p 4 increases S  large reduction in repulsion and so ionization is easier -1/2 +1/2

11 Slide 11/15 Ionization Energies: (ii) Hund’s 2 nd Rule Overall increase with lower IE for d 6 than for d 5 (1 st Rule)  d 1 -d 5 and d 6 -d 10 are not linear (“1/4 and 3/4 shell effects”) 3d-block ionization energies: M 2+  M 3+

12 d n Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 210-2LSΔSΔSΔLΔL d0d0 00 d1d1 21/2-1/2 d2d2 31 d3d3 33/2-1/2 d4d4 22 d5d5 05/2-1/2 d6d6 22+1/2 d7d7 33/2+1/2 d8d8 31 d9d9 21/2+1/2 d 10 00+1/2 0 +1 +2 -2 0 +1 +2 harder easier

13 Slide 13/15 Ionization Energies: f n “ half filled shell ” “ 3/4 shell ” “ 1/4 shell ” Lanthanide ionization energies: M 2+  M 3+

14 Slide 14/15 Summary Ground terms Occupy orbital m l values to maximize S and L Ionization energy and Hund’s 1 st Rule “Half filled shell stability” is really reflection of higher e - /e - repulsion when electrons have to pair Ionization energy and Hund’s 2 nd Rule “1/4” and “3/4 shell stability” is really a reflection of higher e - /e - repulsion when electrons do not orbit in the same direction Task! Work out ground terms and explain IE for f n elements


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