Lecture 15, November 6, 2015: Transition metals, Pd and Ptwag.caltech.edu/home/ch120/Lectures/Ch125a_FA2015/Ch125-120-L15-TM... · Ch120a-Goddard-L18 © copyright 2011 William A

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L18 Ch125a-Ch120a-

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Ch 125a: Elements of Quantum Chemistry with Applications to Chemical Bonding and Properties of

Molecules and Solids

William A. Goddard, III, [email protected] 316 Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Teaching Assistants: Ch125a Sijia Dong <[email protected]> Ch120a Kurtis Carsch < [email protected] >

Lecture 15, November 6, 2015: Transition metals, Pd and Pt

Room 115 BI Hours: 11-11:50am Monday, Wednesday, Friday

Ch 120a:Nature of the Chemical bond

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L18 2


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{[(Tidσ)(H1s)+ (H1s)(Tidσ)](αβ−βα)}


GVB orbitals for bonds to Ti

Covalent 2 electron TiH bond in Cl2TiH2

Covalent 2 electron CH bond in CH4

Ti dσ character, 1 elect H 1s character, 1 elect

Csp3 character 1 elect H 1s character, 1 elect

Think of as bond from Tidz2 to H1s


But TM-H bond can also be s-like

Cl2TiH+

ClMnH

Ti (4s)2(3d)2

The 2 Cl pull off 2 e from Ti, leaving a d1 configuration

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

Ti-H bond character 1.07 Tid+0.22Tisp+0.71H

Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H


Bond angle at a transition metal

For two p orbitals expect 90°, HH nonbond repulsion increases it

H-Ti-H plane

76°

Metallacycle plane

What angle do two d orbitals want


Best bond angle for 2 pure Metal bonds using d orbitals

Assume that the first bond has pure dz2 or dσ character to a ligand along the z axis

Can we make a 2nd 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 125.3°), this is ½ the tetrahedral angle of 109.7 (also the magic spinning angle for solid state NMR).


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 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


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


But TM-H bond can also be s-like

Cl2TiH+

ClMnH

Ti (4s)2(3d)2

The 2 Cl pull off 2 e from Ti, leaving a d1 configuration

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

Ti-H bond character 1.07 Tid+0.22Tisp+0.71H

Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H


Example (Cl)2VH3

+ resonance configuration


Example ClMo-metallacycle butadiene


Example [Mn≡CH]2+


Summary: start with Mn+ s1d5

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


Summary: start with Mn+ s1d5

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


Compare chemistry of column 10


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

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

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


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


Salient differences between Ni, Pd, Pt

Ni Pd Pt

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 d10

4d similar size to 5s (orthog to 3d,4s

Differential shielding favors n=4 over n=5,

stabilize 4d over 5s d10

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

Relativistic effects of 1s huge decreased KE contraction stabilize and contract all ns destabilize and expand nd

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L18

Next section

20

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 206


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


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?


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


Analysis of GVB wavefunction


Alternative models for Pt centers




energetics

Not agree with experiment


Possible explanation: kinetics


Consider reductive elimination of HH, CH and CC from Pd

Conclusion: HH no barrier

CH modest barrier CC large barrier


Consider oxidative addition of HH, CH, and CC to Pt

Conclusion: HH no barrier

CH modest barrier CC large barrier


Summary of barriers

But why?

This explains why CC coupling not occur for Pt while CH and HHcoupling is fast


How estimate the size of barriers (without calculations)


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


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


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


Now we understand Pt chemistry

But what about Pd?

Why are Pt and Pd so dramatically different


Pt goes from s1d9 to d10 upon reductive elimination thus product stability is DECREASED by 12 kcal/mol

Using numbers from QM


Ground state configurations for column 10

Ni Pd Pt


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


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


Summary energetics

Conclusion the atomic character of the metal can

control the chemistry

Documents

Lecture 15, November 6, 2015: Transition metals, Pd and Ptwag.caltech.edu/home/ch120/Lectures/Ch125a_FA2015/Ch125-120-L15-TM... · Ch120a-Goddard-L18 © copyright 2011 William A