Lecture 18 February 14, 2011 Transition metals:Pd and Pt

1. Lecture 18 February 14, 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. Transition metal atoms

4. Transition metals Aufbau (4s,3d) Sc---Cu (5s,4d) Y-- Ag (6s,5d) (La or Lu), Ce-Au

5. Transition metals

6. Ground states of neutral atoms

7. The heme group The net charge of the Fe-heme is zero. The VB structure shown is one of several, all of which lead to two neutral N and two negative N. Thus we consider that the Fe is Fe2+ with a d6 configuration Each N has a doubly occupied sp2s orbital pointing at it.

8. Energies of the 5 Fe2+ d orbitals x2-y2 z2=2z2-x2-y2 yz xz xy

9. Exchange stabilizations

10. Ferrous FeII y x x2-y2 destabilized by heme N lone pairs z2 destabilized by 5th ligand imidazole or 6th ligand CO

11. Summary 4 coord and 5 coord states

12. Out of plane motion of Fe – 4 coordinate

13. Add axial base N-N Nonbonded interactions push Fe out of plane is antibonding

14. Free atom to 4 coord to 5 coord Net effect due to five N ligands is to squish the q, t, and s states by a factor of 3 This makes all three available as possible ground states depending on the 6th ligand

15. Bonding of O2 with O to form ozone O2 has available a ps orbital for a s bond to a ps orbital of the O atom And the 3 electron p system for a p bond to a pp orbital of the O atom

16. Bond O2 to Mb Simple VB structures  get S=1 or triplet state In fact MbO2 is singlet Why?

17. change in exchange terms when Bond O2 to Mb O2ps O2pp 10 Kdd 7 Kdd 5*4/2 4*3/2 + 2*1/2 Assume perfect VB spin pairing Then get 4 cases 7 Kdd 6 Kdd up spin 4*3/2 + 2*1/2 3*2/2 + 3*2/2 Thus average Kdd is (10+7+7+6)/4 =7.5 down spin

18. Bonding O2 to Mb Exchange loss on bonding O2

19. Modified exchange energy for q state But expected t binding to be 2*22 = 44 kcal/mol stronger than q What happened? Binding to q would have DH = -33 + 44 = + 11 kcal/mol Instead the q state retains the high spin pairing so that there is no exchange loss, but now the coupling of Fe to O2 does not gain the full VB strength, leading to bond of only 8kcal/mol instead of 33

20. Bond CO to Mb H2O and N2 do not bond strongly enough to promote the Fe to an excited state, thus get S=2

21. compare bonding of CO and O2 to Mb

22. GVB orbitals for bonds to Ti H 1s character, 1 elect Ti ds character, 1 elect Covalent 2 electron TiH bond in Cl2TiH2 Think of as bond from Tidz2 to H1s Csp3 character 1 elect H 1s character, 1 elect Covalent 2 electron CH bond in CH4

23. Bonding at a transition metaal Bonding to a transition metals can be quite covalent. Examples: (Cl2)Ti(H2), (Cl2)Ti(C3H6), Cl2Ti=CH2 Here the two bonds to Cl remove ~ 1 to 2 electrons from the Ti, making is very unwilling to transfer more charge, certainly not to C or H (it would be the same for a Cp (cyclopentadienyl ligand) Thus TiCl2 group has ~ same electronegativity as H or CH3 The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s A{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)}

24. But TM-H bond can also be s-like Cl2TiH+ Ti (4s)2(3d)2 The 2 Cl pull off 2 e from Ti, leaving a d1 configuration Ti-H bond character 1.07 Tid+0.22Tisp+0.71H ClMnH Mn (4s)2(3d)5 The Cl pulls off 1 e from Mn, leaving a d5s1 configuration H bonds to 4s because of exchange stabilization of d5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H

25. Bond angle at a transition metal H-Ti-H plane 76° Metallacycle plane For two p orbitals expect 90°, HH nonbond repulsion increases it What angle do two d orbitals want

26. Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure dz2 or ds character to a ligand along the z axis Can we make a 2nd bond, also of pure ds character (rotationally symmetric about the z axis) to a ligand along some other axis, call it z. For pure p systems, this leads to  = 90° For pure d systems, this leads to  = 54.7° (or 125.3°), this is ½ the tetrahedral angle of 109.7 (also the magic spinning angle for solid state NMR).

27. Best bond angle for 2 pure Metal bonds using d orbitals Problem: two electrons in atomic d orbitals with same spin lead to 5*4/2 = 10 states, which partition into a 3F state (7) and a 3P state (3), with 3F lower. This is because the electron repulsion between say a dxy and dx2-y2 is higher than between sasy dz2 and dxy. Best is ds with dd because the electrons are farthest apart This favors  = 90°, but the bond to the dd orbital is not as good Thus expect something between 53.7 and 90° Seems that ~76° is often best

28. How predict character of Transition metal bonds? (4s)(3d)5 (3d)2 Start with ground state atomic configuration Ti (4s)2(3d)2 or Mn (4s)2(3d)5 Consider that bonds to electronegative ligands (eg Cl or Cp) take electrons from 4s easiest to ionize, also better overlap with Cl or Cp, also leads to less reduction in dd exchange Now make bond to less electronegative ligands, H or CH3 Use 4s if available, otherwise use d orbitals

29. But TM-H bond can also be s-like Cl2TiH+ Ti (4s)2(3d)2 The 2 Cl pull off 2 e from Ti, leaving a d1 configuration Ti-H bond character 1.07 Tid+0.22Tisp+0.71H ClMnH Mn (4s)2(3d)5 The Cl pulls off 1 e from Mn, leaving a d5s1 configuration H bonds to 4s because of exchange stabilization of d5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H

30. Example (Cl)2VH3 + resonance configuration

31. Example ClMo-metallacycle butadiene

32. Example [Mn≡CH]2+

33. Summary: start with Mn+ s1d5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH

34. Summary: start with Mn+ s1d5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH

35. new

36. Compare chemistry of column 10

37. 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

38. Salient differences between 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

39. Ground state configurations for column 10 Ni Pd Pt

40. Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why are Pd and Pt so different

41. 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?

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

43. Analysis of GVB wavefunction

44. Alternative models for Pt centers

47. energetics Not agree with experiment

48. Possible explanation: kinetics

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

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