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.
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© 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, wagoddard3@gmail.com 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 <sdong@caltech.edu> Ch120a Kurtis Carsch < kcarsch@caltech.edu >
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
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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)}
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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
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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
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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
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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).
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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
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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
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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
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Example (Cl)2VH3
+ resonance configuration
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Example ClMo-metallacycle butadiene
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Example [Mn≡CH]2+
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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
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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
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Compare chemistry of column 10
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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
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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
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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
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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
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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
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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?
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Step 1: examine GVB orbitals for (PH3)2Pt(CH3)
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Analysis of GVB wavefunction
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Alternative models for Pt centers
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energetics
Not agree with experiment
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Possible explanation: kinetics
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Consider reductive elimination of HH, CH and CC from Pd
Conclusion: HH no barrier
CH modest barrier CC large barrier
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Consider oxidative addition of HH, CH, and CC to Pt
Conclusion: HH no barrier
CH modest barrier CC large barrier
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Summary of barriers
But why?
This explains why CC coupling not occur for Pt while CH and HHcoupling is fast
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How estimate the size of barriers (without calculations)
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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
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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
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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
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Now we understand Pt chemistry
But what about Pd?
Why are Pt and Pd so dramatically different
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Pt goes from s1d9 to d10 upon reductive elimination thus product stability is DECREASED by 12 kcal/mol
Using numbers from QM
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Ground state configurations for column 10
Ni Pd Pt
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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
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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
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Summary energetics
Conclusion the atomic character of the metal can
control the chemistry
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