Lecture 19 February 16, 2011 Transition metals:Pd and Pt

1. Lecture 19 February 16, 2011 Transition metals:Pd and Pt Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>

2. Last time

3. Compare chemistry of column 10

4. Ground state of group 10 column Pt: (5d)9(6s)13D ground state Pt: (5d)10(6s)01S excited state at 11.0 kcal/mol Pt: (5d)8(6s)23F excited state at 14.7 kcal/mol Ni: (5d)8(6s)23F ground state Ni: (5d)9(6s)13D excited state at 0.7 kcal/mol Ni: (5d)10(6s)01S excited state at 40.0 kcal/mol Pd: (5d)10(6s)01S ground state Pd: (5d)9(6s)13D excited state at 21.9 kcal/mol Pd: (5d)8(6s)23F excited state at 77.9 kcal/mol

5. Salient differences between Ni, Pd, Pt Ni Pd Pt 2nd row (Pd): 4d much more stable than 5s  Pd d10 ground state 3rd row (Pt): 5d and 6s comparable stability  Pt d9s1 ground state 5s much less stable than 4d 6s, 5d similar stability 4s more stable than 3d Differential shielding favors n=4 over n=5, stabilize 4d over 5s d10 3d much smaller than 4s (No 3d Pauli orthogonality) Huge e-e repulsion for d10 Relativistic effects of 1s huge decreased KE  contraction  stabilize and contract all ns  destabilize and expand nd 4d similar size to 5s (orthog to 3d,4s

6. Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why is CC coupling so much harder than CH coupling?

7. Step 1: examine GVB orbitals for (PH3)2Pt(CH3)

8. Analysis of GVB wavefunction

9. Alternative models for Pt centers

11. energetics Not agree with experiment

12. Possible explanation: kinetics

13. Consider reductive elimination of HH, CH and CC from Pd Conclusion: HH no barrier CH modest barrier CC large barrier

14. Consider oxidative addition of HH, CH, and CC to Pt Conclusion: HH no barrier CH modest barrier CC large barrier

15. Summary of barriers This explains why CC coupling not occur for Pt while CH and HHcoupling is fast But why?

16. How estimate the size of barriers (without calculations)

17. Examine HH coupling at transition state Can simultaneously get good overlap of H with Pd sd hybrid and with the other H Thus get resonance stabilization of TS  low barrier

18. Examine CC coupling at transition state Can orient the CH3 to obtain good overlap with Pd sd hybrid OR can orient the CH3 to obtain get good overlap with the other CH3 But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get resonance stabilization of TS  high barier

19. Examine CH coupling at transition state H can overlap both CH3 and Pd sd hybrid simultaneously but CH3 cannot thus get ~ ½ resonance stabilization of TS

20. Now we understand Pt chemistry But what about Pd? Why are Pt and Pd so dramatically different

21. new

22. Pt goes from s1d9 to d10 upon reductive eliminationthus product stability is DECREASED by 12 kcal/mol Using numbers from QM

23. Pd goes from s1d9 to d10 upon reductive eliminationthus product stability is INCREASED by 20 kcal/mol Using numbers from QM Pd and Pt would be ~ same

24. Thus reductive elimination from Pd is stabilized by an extra 32 kcal/mol than for Pt due to the ATOMIC nature of the states The dramatic stabilization of the product by 35 kcal/mol reduces the barrier from ~ 41 (Pt) to ~ 10 (Pd) This converts a forbidden reaction to allowed

25. Summary energetics Conclusion the atomic character of the metal can control the chemistry

26. Examine bonding to all three rows of transition metals Use MH+ as model because a positive metal is more representative of organometallic and inorganic complexes M0 usually has two electrons in ns orbitals or else one M+ generally has one electron in ns orbitals or else zero M2+ never has electrons in ns orbitals

27. Ground states of neutral atoms

28. Bond energies MH+ Re Mo Au Cr Cu Ag

29. Exchange energies: Mn+: s1d5 For high spin (S=3) A[(d1a)(d2a)(d3a)(d4a)(d5a)(sa)] Get 6*5/2=15 exchange terms 5Ksd + 10 Kdd Responsible for Hund’s rule Ksd Kdd Mn+ 4.8 19.8 Tc+ 8.3 15.3 Re+ 11.9 14.1 kcal/mol Form bond to H, must lose half the exchange stabilization for the orbital bonded to the H A{(d1a)(d2a)(d3a)(d4a)(sdba)[(sdb)H+H(sdb)](ab-ba)} sdb is a half the time and b half the time

30. Ground state of M+ metals Mostly s1dn-1 Exceptions: 1st row: V, Cr-Cu 2nd row: Nb-Mo, Ru-Ag 3rd row: La, Pt, Au

31. Size of atomic orbitals, M+ Valence s similar for all three rows, 5s biggest Big decrease from La(an 57) to Hf(an 72 Valence d very small for 3d

32. Charge transfer in MH+ bonds electropositive 1st row all electropositive 2nd row: Ru,Rh,Pd electronegative 3rd row: Pt, Au, Hg electronegative electronegative

37. 1st row

39. Schilling

40. Steigerwald

43. 2nd row

50. 3rd row