william a. goddard, iii, wag@wag.caltech.edu 316 beckman institute, x3093
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© 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, wag@wag.caltech.edu316 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 <fur@caltech.edu>; Fan Liu <fliu@wag.caltech.edu>
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Last Time
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Transition metals
Aufbau
(4s,3d) Sc---Cu
(5s,4d) Y-- Ag
(6s,5d) (La or Lu), Ce-Au
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Transition metals
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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.
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Energies of the 5 Fe2+ d orbitals
x2-y2
z2=2z2-x2-y2
xy
xz
yz
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Exchange stabilizations
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Skip energy stuff
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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
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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
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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
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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
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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
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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
© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 16
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.
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© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 17
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
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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
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© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 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
© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 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 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.
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© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 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
\
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© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 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>]=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)>
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© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L03 © copyright 2011 William A. Goddard III, all rights reserved 23
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)
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© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 24
Ferrous FeII
x
y
z2 destabilized by 5th ligand imidazole
or 6th ligand CO
x2-y2 destabilized by heme N lone pairs
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Summary 4 coord and 5 coord states
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Out of plane motion of Fe – 4 coordinate
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Add axial base
N-N Nonbonded interactions push Fe out of plane
is antibonding
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 30
Bond O2 to Mb
Simple VB structures get S=1 or triplet state
In fact MbO2 is singletWhy?
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 31
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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 32
Bonding O2 to Mb
Exchange loss on
bonding O2
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
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compare bonding of CO and O2 to Mb
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New material
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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 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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)}
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 39
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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 40
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 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).
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 44
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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 45
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 s bond to H1sdx2-x2 non bondingdyz p bond to CHdxz p bond to CHdxy non bonding4sp hybrid s bond to CH
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 49
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
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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 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
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Ground state configurations for column 10
Ni Pd Pt
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 55
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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 56
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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© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 61
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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 barrierCC large barrier
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Consider oxidative addition of HH, CH, and CC to Pt
Conclusion: HH no barrier
CH modest barrierCC 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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L18 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
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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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stop
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