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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 Ch120a-Goddard-

L01

1

Nature of the Chemical Bond with applications to catalysis, materials

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

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

Lecture 16 February 20Transition metals, Pd and Pt

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday

Teaching Assistants: Ross Fu <[email protected]>; Fan Liu <[email protected]>

mailto:[email protected]

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 2

Last Time


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)1Ti (4s)2(3d)2V (4s)2(3d)3Cr (4s)1(3d)5Mn (4s)2(3d)5Fe (4s)2(3d)6Co (4s)2(3d)7Ni (4s)2(3d)8Cu (4s)1(3d)10

Sc++ (3d)1Ti ++ (3d)2V ++ (3d)3 Cr ++ (3d)4Mn ++ (3d)5Fe ++ (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 s orbital pointing at it.


Energies of the 5 Fe2+ d orbitals

x2-y2

z2=2z2-x2-y2

xy

xz

yz


Exchange stabilizations


Skip energy stuff

© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 10

Consider the product wavefunctionΨ(1,2) = ψa(1) ψb(2) And the HamiltonianH(1,2) = h(1) + h(2) +1/r12 + 1/RIn the details slides next, we deriveE = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>E = haa + hbb + Jab + 1/R where haa =<a|h|a>, hbb =<b|h|b>

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12

Represent the total Coulomb interaction between the electron density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0

Energy for 2 electron product wavefunction

SKIP for now


Details in deriving energy: normalization

First, the normalization term is <Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)> Which from now on we will write as<Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalizedHere our convention is that a two-electron function such as <Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or <ψb(2) ψb(2)> are assumed to be over just one electron and we ignore the labels 1 or 2

SKIP for now


Using H(1,2) = h(1) + h(2) +1/r12 + 1/RWe partition the energy E = <Ψ| H|Ψ> asE = <Ψ|h(1)|Ψ> + <Ψ|h(2)|Ψ> + <Ψ|1/R|Ψ> + <Ψ|1/r12|Ψ>Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant<Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> =

= <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> =≡ haa

Where haa≡ <a|h|a> ≡ <ψa|h|ψa>Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> =

= <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> =≡ hbb

The remaining term we denote as Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy isE = haa + hbb + Jab + 1/R

Details of deriving energy: one electron terms

SKIP for now


The energy for an antisymmetrized product, A ψaψb

The total energy is that of the product plus the exchange term which is negative with 4 partsEex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa > - < ψaψb|1/r12|ψb ψa >The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+ <ψa|ψb><ψb|h(2)|ψa

>+ <ψa|ψb><ψb|ψa>/R But <ψb|ψa>=0Thus all are zeroThus the only nonzero term is the 4th term:-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy (or the 2-electron exchange) since it arises from the exchange term due to the antisymmetrizer.Summarizing, the energy of the Aψaψb wavefunction for H2 isE = haa + hbb + (Jab –Kab) + 1/R

SKIP for now


The energy of the antisymmetrized wavefunction

The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positiveEee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0

This follows since the integrand is positive for all positions of r1 and r2 then

We derived that the energy of the A ψa ψb wavefunction is

E = haa + hbb + (Jab –Kab) + 1/R

Where the Eee = (Jab –Kab) > 0

Since we have already established that Jab > 0 we can conclude thatJab > Kab > 0

SKIP for now


Separate the spinorbital into orbital and spin parts

Since the Hamiltonian does not contain spin the spinorbitals can be factored into spatial and spin terms. For 2 electrons there are two possibilities:Both electrons have the same spinψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)]So that the antisymmetrized wavefunction isAψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]==[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)]Also, similar results for both spins downAψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]==[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)]

Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb>We see that the spatial orbitals for same spin must be orthogonal


Energy for 2 electrons with same spinThe total energy becomesE = haa + hbb + (Jab –Kab) + 1/R where haa ≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb> where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>

We derived the exchange term for spin orbitals with same spin as followsKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)> ≡ Kab

where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>Involves only spatial coordinates.

SKIP for now


Now consider the exchange term for spin orbitals with opposite spinKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)> = 0Since <a(1)|b(1)> = 0.

Energy for 2 electrons with opposite spin

Thus the total energy isEab = haa + hbb + Jab + 1/R With no exchange term unless the spins are the same

Since <ψa|ψb>= 0 = < Φa| Φb><a|b> There is no orthogonality condition of the spatial orbitals for opposite spin electronsIn general < Φa| Φb> =S, where the overlap S ≠ 0

SKIP for now


Summarizing: Energy for 2 electronsWhen the spinorbitals have the same spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]The total energy isEaa = haa + hbb + (Jab –Kab) + 1/R

But when the spinorbitals have the opposite spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=The total energy isEab = haa + hbb + Jab + 1/R With no exchange term

Thus exchange energies arise only for the case in which both electrons have the same spin

SKIP for now


Consider further the case for spinorbtials with opposite spin

Neither of these terms has the correct permutation symmetry separately for space or spin. But they can be combined[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)]

Which describes the Ms=0 component of the triplet state

[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(2)][a(1)b(2)]

Which describes the Ms=0 component of the singlet state

Thus for the ab case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry


Consider further the case for spinorbtials with opposite spin

The wavefunction[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)] Leads directly to 3Eab = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)] [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)]These three states are collectively referred to as the triplet state and denoted as having spin S=1 The other combination leads to one state, referred to as the singlet state and denoted as having spin S=0[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]We will analyze the energy for this wavefunction next.

SKIP for now


Consider the energy of the singlet wavefunction

[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba)The next few slides show that 1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)Where the terms with S or Kab come for the exchange

\

SKIP for now


energy of the singlet wavefunction - details

[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba)1E = numerator/ denominator Where numerator =<(ab+ba)(ab-ba)|H|(ab+ba)(ab-ba)> =

=<(ab+ba)|H|(ab+ba)><(ab-ba)|(ab-ba)>denominator = <(ab+ba)(ab-ba)|(ab+ba)(ab-ba)>Since <(ab-ba)|(ab-ba)>= 2 <ab|(ab-ba)>=

2[<a|a><b|b>-<a|b><b|a>]=2We obtainnumerator =<(ab+ba)|H|(ab+ba)> = 2 <ab|H|(ab+ba)> denominator = <(ab+ba)|(ab+ba)>=2 <ab|(ab+ba)>

Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

SKIP for now


energy of the singlet wavefunction - details

1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

Consider first the denominator

<ab|(ab+ba)> = <a|a><b|b> + <a|b><b|a> = 1 + S2

Where S= <a|b>=<b|a> is the overlap

The numerator becomes

<ab|(ab+ba)> = <a|h|a><b|b> + <a|h|b><b|a> +

+ <a|a><b|h|b> + <a|b><b|h|a> +

+ <ab|1/r12|(ab+ba)> + (1 + S2)/R

Thus the total energy is 1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)

SKIP for now


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 ps orbital for a s bond to a ps orbital of the O atom

And the 3 electron p system for a p bond to a pp 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

O2ps

O2pp

10 Kdd

5*4/27 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 qWhat happened?Binding to q would have DH = -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


New material


GVB orbitals for bonds to Ti

Covalent 2 electron TiH bond in Cl2TiH2

Covalent 2 electron CH bond in CH4

Ti ds 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{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)}


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 ds character to a ligand along the z axis

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

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 ds with dd because the electrons are farthest apart

This favors = 90°, but the bond to the dd 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 s bond to H1sdx2-x2 non bondingdyz p bond to CHdxz p bond to CHdxy non bonding4sp hybrid s bond to CH


Summary: start with Mn+ s1d5

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


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


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 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 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

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