1. 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 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 Assistants: Ross Fu
Fan Liu
Ch120a-Goddard-L01

2. Last Time

3. Transition metals Aufbau (4s,3d) Sc---Cu (5s,4d) Y-- Ag
(6s,5d) (La or Lu), Ce-Au

4. Transition metals

5. Ground states of neutral atomsSc
(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

6. 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.

7. Energies of the 5 Fe2+ d orbitals
x2-y2
z2=2z2-x2-y2 yz xz xy

8. Exchange stabilizations

9. Skip energy stuff

10. Energy for 2 electron product wavefunction
Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian
H(1,2) = h(1) + h(2) +1/r12 + 1/R
In the details slides next, we derive
E = < 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
SKIP for now

11. Details in deriving energy: normalization
SKIP for now
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 normalized
Here 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

12. Details of deriving energy: one electron terms
Using H(1,2) = h(1) + h(2) +1/r12 + 1/R
We partition the energy E = <Ψ| H|Ψ> as
E = <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 is
E = haa + hbb + Jab + 1/R
SKIP for now

13. 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 parts
Eex=-< 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>=0
Thus all are zero
Thus 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.
SKIP for now
Summarizing, the energy of the Aψaψb wavefunction for H2 is
E = haa + hbb + (Jab –Kab) + 1/R

14. The energy of the antisymmetrized wavefunction
The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positive
Eee =∫ (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 that
Jab > Kab > 0
SKIP for now

15. 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 spin
ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)]
So that the antisymmetrized wavefunction is
Aψa(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 down
Aψa(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

16. Energy for 2 electrons with same spin
The total energy becomes
E = 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)>
SKIP for now
We derived the exchange term for spin orbitals with same spin as follows
Kab ≡ <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.

17. Energy for 2 electrons with opposite spin
Now consider the exchange term for spin orbitals with opposite spin
Kab ≡ <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)>
= 0
Since <a(1)|b(1)> = 0.
SKIP for now
Thus the total energy is
Eab = 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 electrons
In general < Φa| Φb> =S, where the overlap S ≠ 0

18. Summarizing: Energy for 2 electrons
When the spinorbitals have the same spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]
The total energy is
Eaa = 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)]=
Eab = haa + hbb + Jab + 1/R
With no exchange term
SKIP for now
Thus exchange energies arise only for the case in which both electrons have the same spin

19. 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

20. 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
SKIP for now
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.

21. 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

22. 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>]=2
We obtain
numerator =<(ab+ba)|H|(ab+ba)> = 2 <ab|H|(ab+ba)>
denominator = <(ab+ba)|(ab+ba)>=2 <ab|(ab+ba)>
SKIP for now
Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

23. energy of the singlet wavefunction - details
SKIP for now
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)

24. Ferrous FeII x2-y2 destabilized by heme N lone pairs
z2 destabilized by 5th ligand imidazole or 6th ligand CO x y

25. Summary 4 coord and 5 coord states

26. Out of plane motion of Fe – 4 coordinate

27. N-N Nonbonded interactions push Fe out of plane
Add axial base
N-N Nonbonded interactions push Fe out of plane
is antibonding

28. Free atom to 4 coord to 5 coord
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

29. 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

30. Simple VB structures  get S=1 or triplet state
Bond O2 to Mb
Simple VB structures  get S=1 or triplet state
In fact MbO2 is singlet
Why?

31. change in exchange terms when Bond O2 to Mb
O2ps
O2pp
10 Kdd
5*4/2
7 Kdd
4*3/2 + 2*1/2
Assume perfect VB spin pairing
Then get 4 cases
7 Kdd
6 Kdd
up spin
4*3/2 + 2*1/2
3*2/2 + 3*2/2
Thus average Kdd is ( )/4 =7.5
down spin

32. Exchange loss on bonding O2
Bonding O2 to Mb
Exchange loss on bonding O2

33. 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 DH = = + 11 kcal/mol
Instead 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

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

35. compare bonding of CO and O2 to Mb

36. New material

37. GVB orbitals for bonds to Ti
Ti ds character, 1 elect
H 1s character, 1 elect
Covalent 2 electron TiH bond in Cl2TiH2
Think of as bond from Tidz2 to H1s
Csp3 character 1 elect
H 1s character, 1 elect
Covalent 2 electron CH bond in CH4

38. 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)}

39. But TM-H bond can also be s-like
Cl2TiH+
Ti (4s)2(3d)2
The 2 Cl pull off 2 e from Ti, leaving a d1 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 d5s1 configuration
H bonds to 4s because of exchange stabilization of d5
Mn-H bond character
0.07 Mnd+0.71Mnsp+1.20H

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

41. 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 (also the magic spinning angle for solid state NMR).

42. 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

43. 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

44. But TM-H bond can also be s-like
Cl2TiH+
Ti (4s)2(3d)2
The 2 Cl pull off 2 e from Ti, leaving a d1 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 d5s1 configuration
H bonds to 4s because of exchange stabilization of d5
Mn-H bond character
0.07 Mnd+0.71Mnsp+1.20H

45. Example (Cl)2VH3
+ resonance configuration

46. Example ClMo-metallacycle butadiene

47. Example [Mn≡CH]2+

48. Summary:
start with Mn+ s1d5
dy2 s bond to H1s
dx2-x2 non bonding
dyz p bond to CH
dxz p bond to CH
dxy non bonding
4sp hybrid s bond to CH

49. Summary:
start with Mn+ s1d5
dy2 s bond to H1s
dx2-x2 non bonding
dyz p bond to CH
dxz p bond to CH
dxy non bonding
4sp hybrid s bond to CH

50. Compare chemistry of column 10

51. 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
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
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

52. 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

53. Ground state configurations for column 10Ni Pd Pt

54. 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

55. 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 are Pd and Pt so different

56. Why is CC coupling so much harder than CH coupling?
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?

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

58. Analysis of GVB wavefunction

59. Alternative models for Pt centers

62. Not agree with experiment
energetics
Not agree with experiment

63. Possible explanation: kinetics

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

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

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

67. How estimate the size of barriers (without calculations)

68. 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

69. 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

70. 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

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

72. stop