Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy

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1 Lecture 13, October 31, 2016 Transition metals, Pd and Pt Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday 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 Assistant: Shane Flynn Recitation: Mon 3pm

2 Bonding at a transition metaal Bonding to a transition metals can be quite covalent. Examples: (Cl 2 )Ti(H 2 ), (Cl 2 )Ti(C 3 H 6 ), Cl 2 Ti=CH 2 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 TiCl 2 group has ~ same electronegativity as H or CH 3 The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s A{[(Tidσ)(H1s)+ (H1s)(Tidσ)](αβ βα)} 2

2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH

3 GVB orbitals for bonds to Ti Ti dσ character, 1 elect H 1s character, 1 elect Covalent 2 electron TiH bond in Cl 2 TiH 2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH bond in CH 4 Csp 3 character 1 elect H 1s character, 1 elect 3

4 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 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 d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 4

5 Bond angle at a transition metal For two p orbitals expect 90, HH nonbond repulsion increases it What angle do two d orbitals want H-Ti-H plane 76 Metallacycle plane 5

Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or dσ character to a ligand along the z axis Can we make a 2 nd bond, also of pure dσ character

6 Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or dσ character to a ligand along the z axis Can we make a 2 nd bond, also of pure dσ character (rotationally symmetric about the ζ axis) to a ligand along some other axis, call it ζ. For pure p systems, this leads to θ = 90 For pure d systems, this leads to θ = 54.7 (or ), this is ½ the tetrahedral angle of (also the magic spinning angle for solid state NMR). 6

7 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 3 F state (7) and a 3 P state (3), with 3 F lower. This is because the electron repulsion between say a d xy and d x2-y2 is higher than between sasy d z2 and d xy. Best is dσ with dδ because the electrons are farthest apart This favors θ = 90, but the bond to the dδ orbital is not as good Thus expect something between 53.7 and 90 Seems that ~76 is often best 7

8 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 (3d) 2 (4s)(3d) 5 Now make bond to less electronegative ligands, H or CH 3 Use 4s if available, otherwise use d orbitals 8

9 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 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 d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 9

10 Example (Cl) 2 VH 3 + resonance configuration 10

11 Example ClMometallacycle butadiene 11

12 Example [Mn CH] 2+ 12

13 Summary: start with Mn + s 1 d 5 dy2 σ bond to H1s dx2-x2 non bonding dyz π bond to CH dxz π bond to CH dxy non bonding 4sp hybrid σ bond to CH 13

14 Summary: start with Mn + s 1 d 5 dy2 σ bond to H1s dx2-x2 non bonding dyz π bond to CH dxz π bond to CH dxy non bonding 4sp hybrid σ bond to CH 14

15 Compare chemistry of column 10 15

16 Ground state of group 10 column Pt: (5d) 9 (6s) 1 3 D ground state Pt: (5d) 10 (6s) 0 1 S excited state at 11.0 kcal/mol Pt: (5d) 8 (6s) 2 3 F excited state at 14.7 kcal/mol Ni: (5d) 8 (6s) 2 3 F ground state Ni: (5d) 9 (6s) 1 3 D excited state at 0.7 kcal/mol Ni: (5d) 10 (6s) 0 1 S excited state at 40.0 kcal/mol Pd: (5d) 10 (6s) 0 1 S ground state Pd: (5d) 9 (6s) 1 3 D excited state at 21.9 kcal/mol Pd: (5d) 8 (6s) 2 3 F excited state at 77.9 kcal/mol 16

17 Salient differences between Ni, Pd, Pt 2 nd row (Pd): 4d much more stable than 5s Pd d 10 ground state 3 rd row (Pt): 5d and 6s comparable stability Pt d 9 s 1 ground state 17

Salient differences between Ni, Pd, Pt Ni Pd Pt 2 nd row (Pd): 4d much more stable than 5s Pd d 10 ground state 3 rd row (Pt): 5d and 6s comparable stability Pt d 9 s 1 ground state 4s more stable

18 Salient differences between Ni, Pd, Pt Ni Pd Pt 2 nd row (Pd): 4d much more stable than 5s Pd d 10 ground state 3 rd row (Pt): 5d and 6s comparable stability Pt d 9 s 1 ground state 4s more stable than 3d 5s much less stable than 4d 6s, 5d similar stability 3d much smaller than 4s (No 3d Pauli orthogonality) Huge e-e repulsion for d 10 Differential shielding favors n=4 over n=5, stabilize 4d over 5s d 10 4d similar size to 5s (orthog to 3d,4s Relativistic effects of 1s huge decreased KE contraction stabilize and contract all ns 18 destabilize and expand nd

19 Next section Theoretical Studies of Oxidative Addition and Reductive Elimination: J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 6928 (1984) wag 190 Reductive Coupling of H-H, H-C, and C-C Bonds from Pd Complexes J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 8321 (1984) wag 191 Theoretical Studies of Oxidative Addition and Reductive Elimination. II. Reductive Coupling of H-H, H-C, and C-C Bonds from Pd and Pt Complexes J. J. Low and W. A. Goddard III Organometallics 5, 609 (1986) wag

20 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 20

21 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? 21

22 Step 1: examine GVB orbitals for (PH 3 ) 2 Pt(CH 3 ) 22

23 Analysis of GVB wavefunction 23

24 Alternative models for Pt centers 24

25 25

26 26

27 energetics Not agree with experiment 27

28 Possible explanation: kinetics 28

29 Consider reductive elimination of HH, CH and CC from Pd Conclusion: HH no barrier CH modest barrier CC large barrier 29

30 Consider oxidative addition of HH, CH, and CC to Pt Conclusion: HH no barrier CH modest barrier CC large barrier 30

31 Summary of barriers This explains why CC coupling not occur for Pt while CH and HHcoupling is fast But why? 31

32 How estimate the size of barriers (without calculations) 32

33 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 33

34 Examine CC coupling at transition state Can orient the CH 3 to obtain good overlap with Pd sd hybrid OR can orient the CH 3 to obtain get good overlap with the other CH 3 But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get resonance stabilization of TS high barier 34

35 Examine CH coupling at transition state H can overlap both CH 3 and Pd sd hybrid simultaneously but CH 3 cannot thus get ~ ½ resonance stabilization of TS 35

36 Now we understand Pt chemistry But what about Pd? Why are Pt and Pd so dramatically different 36

37 Pt goes from s 1 d 9 to d 10 upon reductive elimination thus product stability is DECREASED by 12 kcal/mol Using numbers from QM 37

38 Ground state configurations for column 10 Ni Pd Pt 38

39 Pd goes from s 1 d 9 to d 10 upon reductive elimination thus product stability is INCREASED by 20 kcal/mol Using numbers from QM Pd and Pt would be ~ same 39

40 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 40

41 Summary energetics Conclusion the atomic character of the metal can control the chemistry 41

Lecture 16 February 20 Transition metals, Pd and Pt

Lecture 16 February 20 Transition metals, Pd and Pt Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course