lecture 18 february 14, 2011 transition metals:pd and pt

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

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

Lecture 18 February 14, 2011

Transition metals:Pd and Pt

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

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

California Institute of Technology

Teaching Assistants: Wei-Guang Liu <[email protected]>Caitlin Scott <[email protected]>


Last time


Transition metal atoms


Transition metals

Aufbau

(4s,3d) Sc---Cu

(5s,4d) Y-- Ag

(6s,5d) (La or Lu), Ce-Au


Transition metals


Ground states of neutral atoms

Sc (4s)2(3d)1

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++ (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


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 sp2 orbital pointing at it.


Energies of the 5 Fe2+ d orbitals

x2-y2

z2=2z2-x2-y2

xy

xz

yz


Exchange stabilizations


Ferrous FeII

x

y

z2 destabilized by 5th ligand imidazole

or 6th ligand CO

x2-y2 destabilized by heme N lone pairs


Summary 4 coord and 5 coord states


Out of plane motion of Fe – 4 coordinate


Add axial base

N-N Nonbonded interactions push Fe out of plane

is antibonding


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

Free atom to 4 coord to 5 coord


Bonding of O2 with O to form ozone

O2 has available a p orbital for a bond to a p orbital of the O atom

And the 3 electron system for a bond to a p orbital of the O atom


Bond O2 to Mb

Simple VB structures get S=1 or triplet state

In fact MbO2 is singletWhy?


change in exchange terms when Bond O2 to Mb

O2p

O2p

10 Kdd

5*4/2

7 Kdd

4*3/2 +

2*1/2

6 Kdd

3*2/2 +

3*2/2

7 Kdd

4*3/2 +

2*1/2

Assume perfect VB spin pairing

Then get 4 cases

up spin

down spinThus average Kdd is (10+7+7+6)/4

=7.5


Bonding O2 to Mb

Exchange loss on

bonding O2


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 H = -33 + 44 = + 11 kcal/molInstead 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


Bond CO to Mb

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


compare bonding of CO and O2 to Mb


GVB orbitals for bonds to Ti

Covalent 2 electron TiH bond in Cl2TiH2

Covalent 2 electron CH bond in CH4

Ti d 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


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{[(Tid)(H1s)+ (H1s)(Tid)]()}


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 configurationH bonds to 4s because of exchange stabilization of d5

Ti-H bond character1.07 Tid+0.22Tisp+0.71H

Mn-H bond character0.07 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 d character to a ligand along the z axis

Can we make a 2nd bond, also of pure d character (rotationally symmetric about the axis) to a ligand along some other axis, call it

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).


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 d with d because the electrons are farthest apart

This favors = 90°, but the bond to the d 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 configurationH bonds to 4s because of exchange stabilization of d5

Ti-H bond character1.07 Tid+0.22Tisp+0.71H

Mn-H bond character0.07 Mnd+0.71Mnsp+1.20H


Example (Cl)2VH3

+ resonance configuration


Example ClMo-metallacycle butadiene


Example [Mn≡CH]2+


Summary: start with Mn+ s1d5

dy2 bond to H1sdx2-x2 non bondingdyz bond to CHdxz bond to CHdxy non bonding4sp hybrid bond to CH


new


Compare chemistry of column 10


Ground state of group 10 column

Pt: (5d)9(6s)1 3D ground statePt: (5d)10(6s)0 1S excited state at 11.0 kcal/molPt: (5d)8(6s)2 3F excited state at 14.7 kcal/mol

Pd: (5d)10(6s)0 1S ground statePd: (5d)9(6s)1 3D excited state at 21.9 kcal/molPd: (5d)8(6s)2 3F excited state at 77.9 kcal/mol

Ni: (5d)8(6s)2 3F ground stateNi: (5d)9(6s)1 3D excited state at 0.7 kcal/molNi: (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


Ground state configurations for column 10

Ni Pd Pt


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 barrierCC large barrier


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

Conclusion: HH no barrier

CH modest barrierCC 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


stop

lecture 18 february 14, 2011 transition metals:pd and pt

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