1. William A. Goddard, III, wag@wag.caltech.edu
Lecture 20 February 22, 2010
Pd,Pt ox-add, red-elim; Bonds to MH+
Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy
William A. Goddard, III,
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
Ted Yu

2. Remaining Course schedule
Monday Feb. 22, 2pm L20, normal, TODAY
Wednesday Feb. 24, 2pm L21, normal
Friday Feb. 26, 2pm L22, move to 115BI
Friday Feb. 26, 3pm L23, move to 115BI (make up for Mar. 1)
Monday, Mar. 1 no class (wag at review in Maryland)
Wednesday, Mar. 3 no class (wag at review in Maryland)
Friday Mar. 5, 2pm, L24 (make up for March 3)
Friday Mar 5 3pm L25 (catching up to March 5)
Monday, Mar. 8 2pm L26, caught up, normal
Wednesday, Mar. 10 no class (wag at conference in Chicago)
Thursday Mar. 11, 2pm, L27 Last lecture (make up for Mar. 10)
Final available Thursday March 11
Final due back Friday March 19

3. Last time

4. How predict character of Transition metal bonds?
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
(4s)(3d)5
(3d)2
Now make bond to less electronegative ligands, H or CH3
Use 4s if available, otherwise use d orbitals

5. 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

6. Example (Cl)2VH3
+ resonance configuration

7. Compare chemistry of column 10

8. Ground state of group 10 column
Pt: (5d)9(6s)1 3D ground state
Pt: (5d)10(6s)0 1S excited state at 11.0 kcal/mol
Pt: (5d)8(6s)2 3F excited state at 14.7 kcal/mol
Ni: (5d)8(6s)2 3F ground state
Ni: (5d)9(6s)1 3D excited state at 0.7 kcal/mol
Ni: (5d)10(6s)0 1S excited state at 40.0 kcal/mol
Pd: (5d)10(6s)0 1S ground state
Pd: (5d)9(6s)1 3D excited state at 21.9 kcal/mol
Pd: (5d)8(6s)2 3F excited state at 77.9 kcal/mol

9. 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

10. Why are Pd and Pt so different
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

11. Why is CC coupling so much harder than CH coupling?
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?

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

13. Analysis of GVB wavefunction

14. Alternative models for Pt centers

17. Not agree with experiment
energetics
Not agree with experiment

18. New material

19. Possible explanation: kinetics

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

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

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

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

24. 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

25. 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

26. 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

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

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

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

30. 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

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

32. 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

33. Ground states of neutral atomsSc
(4s)2(3d) Ti
(4s)2(3d)2 V
(4s)2(3d)3 Cr
(4s)1(3d)5 Mn
(4s)2(3d)5 Fe
(4s)2(3d)6 Co
(4s)2(3d)7 Ni
(4s)2(3d)8 Cu
(4s)1(3d)10
Sc+
(4s)1(3d)1
Ti+
(4s)1(3d)2 V+
(4s)0(3d)3
Cr+
(4s)0(3d)5
Mn+
(4s)1(3d)5
Fe+
(4s)1(3d)6
Co+
(4s)0(3d)7
Ni+
(4s)0(3d)8
Cu+
(4s)0(3d)10
Sc++
(3d)1
Ti ++
(3d)2
V ++
(3d)3
Cr ++
(3d)4
Mn ++
(3d)5
Fe ++
(3d)6
Co ++
(3d)7
Ni ++
(3d)8
Cu++
(3d)10

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

35. 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 Tc Re
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

36. 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

37. 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

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

43. 1st row

45. Schilling

46. Steigerwald