Transcript Slide 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, [email protected]
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials
Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Ross Fu <[email protected]>;
Fan Liu <[email protected]>
Ch120a-Goddard-L18
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Ch120a1
Last Time
Ch120a-Goddard-L18
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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)1
Sc++
(3d)1
Ti
V
Cr
Mn
(4s)2(3d)2
(4s)2(3d)3
(4s)1(3d)5
(4s)2(3d)5
Ti ++
V ++
Cr ++
Mn ++
(3d)2
(3d)3
(3d)4
(3d)5
Fe
Co
Ni
(4s)2(3d)6
(4s)2(3d)7
(4s)2(3d)8
Fe ++
Co ++
Ni ++
(3d)6
(3d)7
(3d)8
Cu
(4s)1(3d)10
Cu++
(3d)10
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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
yz
xz
xy
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Exchange stabilizations
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Skip energy stuff
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9
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>
SKIP for now
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
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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
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Details of deriving energy: one electron terms
Using H(1,2) = h(1) + h(2) +1/r12 + 1/R
SKIP for now
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 =Ch120a-Goddard-L03
haa + hbb + Jab + 1/R
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12
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
SKIP for now
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.
Summarizing, the energy of the Aψaψb wavefunction for H2 is
E = 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 positive
SKIP for now
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
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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 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
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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.
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Energy for 2 electrons with opposite spin
Now consider the exchange term for spin orbitals with opposite
spin
SKIP for now
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.
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
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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
SKIP for now
But when the spinorbitals have the opposite spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=
The total energy is
Eab = 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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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
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Consider further the case for spinorbtials with opposite spin
The wavefunction
SKIP for now
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]
Leads directly to
3E
ab = 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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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
SKIP for now
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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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
SKIP for now
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)>
Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>
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energy of the singlet wavefunction - details
1E
= <ab|H|(ab+ba)>/<ab|(ab+ba)>
SKIP for now
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)
Ch120a-Goddard-L03
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23
Ferrous FeII
x2-y2 destabilized by
heme N lone pairs
z2 destabilized by
5th ligand imidazole
or 6th ligand CO
y
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x
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Summary 4 coord and 5 coord states
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Out of plane motion of Fe – 4 coordinate
Ch120a-Goddard-L18
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Add axial base
N-N Nonbonded
interactions push
Fe out of plane
is antibonding
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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
Ch120a-Goddard-L18
This makes all three available as
possible ground states depending
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onIII,the
6threserved
ligand
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all rights
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
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Bond O2 to Mb
Ch120a-Goddard-L18
Simple VB
structures 
get S=1 or triplet
state
In fact MbO2 is
singlet
Why?
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change in exchange terms when Bond O2 to Mb
O2ps
O2pp
Assume perfect 10 K
7 Kdd
dd
VB spin pairing
5*4/2 up spin 4*3/2
Then get 4 cases
+
Thus average Kdd is down spin 2*1/2
(10+7+7+6)/4 =7.5
Ch120a-Goddard-L18
7 Kdd
6 Kdd
4*3/2
+
2*1/2
3*2/2
+
3*2/2
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31
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 q
What happened?
Binding to q would have DH = -33 + 44 = + 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
Ch120a-Goddard-L18
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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
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
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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{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)}
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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
0.07Ch120a-Goddard-L18
Mnd+0.71Mnsp+1.20H
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39
Bond angle at a transition metal
For two p orbitals expect 90°, HH nonbond repulsion increases it
What angle do two
d orbitals want
H-Ti-H plane
76°
Ch120a-Goddard-L18
Metallacycle plane
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40
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).
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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 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
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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
(3d)2
(4s)(3d)5
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+
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
0.07Ch120a-Goddard-L18
Mnd+0.71Mnsp+1.20H
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Example (Cl)2VH3
+ resonance
configuration
Ch120a-Goddard-L18
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45
Example ClMometallacycle
butadiene
Ch120a-Goddard-L18
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Example [Mn≡CH]2+
Ch120a-Goddard-L18
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47
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
Ch120a-Goddard-L18
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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
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49
Compare chemistry of column 10
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50
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:Ch120a-Goddard-L18
(5d)8(6s)2 3F excited
state
atWilliam
77.9A. Goddard
kcal/mol
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2011
III, all rights reserved
51
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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52
Ground state configurations for column 10
Ni
Ch120a-Goddard-L18
Pd
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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
Ch120a-Goddard-L18
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54
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?
© copyright 2011 William A. Goddard III, all rights reserved
Ch120a-Goddard-L18
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Step 1: examine GVB orbitals for (PH3)2Pt(CH3)
Ch120a-Goddard-L18
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Analysis of GVB wavefunction
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Alternative models for Pt centers
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Ch120a-Goddard-L18
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energetics
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Not agree with
experiment 62
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Possible explanation: kinetics
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63
Consider reductive elimination of HH,
CH and CC from Pd
Conclusion:
HH no barrier
CH modest barrier
CC large barrier
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64
Consider oxidative addition of
HH, CH, and CC to Pt
Ch120a-Goddard-L18
Conclusion:
HH no barrier
CH modest barrier
CCreserved
large barrier 65
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Summary of barriers
This explains why CC coupling not occur for Pt while CH
and HHcoupling is fast
But why?
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66
How estimate the size of barriers (without calculations)
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67
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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68
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
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resonance© stabilization
ofA. TS
III,high
barier
Ch120a-Goddard-L18
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Goddard
all rights
reserved
Examine CH coupling at transition state
Ch120a-Goddard-L18
H can overlap
both CH3 and Pd
sd hybrid
simultaneously
but CH3 cannot
thus get ~ ½
resonance
stabilization of TS
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© copyright 2011 William A. Goddard III, all rights reserved
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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