Transcript Slide 1

Lecture 15 February 15, 2013
Transition metals
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-L17
© copyright 2011 William A. Goddard III, all rights reserved
Ch120a1
Transition metals
(4s,3d) Sc---Cu
(5s,4d) Y-- Ag
(6s,5d) (La or Lu), Ce-Au
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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Hemoglobin
Blood has 5 billion erythrocytes/ml
Each erythrocyte contains 280 million
hemoglobin (Hb) molecules
Each Hb has MW=64500 Dalton
(diameter ~ 60A)
Four subunits (a1, a2, b1, b2) each with
one heme subunit
Hb
Each subunit resembles myoglobin (Mb)
which has one heme
Ch120a-Goddard-L17
Mb
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The action is at the heme or Fe-Porphyrin molecule
Essentially all action occurs at the heme, which is basically an
Fe-Porphyrin molecule
The rest of the Mb
serves mainly to
provide a
hydrophobic
envirornment at the
Fe and to protect
the heme
Ch120a-Goddard-L17
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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.
Ch120a-Goddard-L17
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Energies of the 5 Fe2+ d orbitals
x2-y2
z2=2z2-x2-y2
yz
xz
xy
Ch120a-Goddard-L17
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Exchange stabilizations
Ch120a-Goddard-L17
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Ferrous FeII
x2-y2 destabilized by
heme N lone pairs
z2 destabilized by
5th ligand imidazole
or 6th ligand CO
y
Ch120a-Goddard-L17
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x
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Four
coordinate FeHeme – High
spin case,
S=2 or q
The 5th axial ligand will destabilize q2 since dz2 is doubly
occupied
A pi acceptor would stabilize q1 wrt q2
Bonding O2 to 5 coordinate will stabilize q3 wrt q1
Future
discuss only q1
and denote
as
q
Ch120a-Goddard-L17
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2011 William
A. Goddard
III, all rights reserved
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Four coordinate Fe-Heme – Intermediate spin, S=1 or t
Ch120a-Goddard-L17
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Four coordinate Fe-Heme – Low spin case, S=0 or s
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Out of plane motion of Fe – 4 coordinate
Ch120a-Goddard-L17
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13
Add axial base
N-N Nonbonded
interactions push
Fe out of plane
is antibonding
Ch120a-Goddard-L17
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14
Summary 4 coord and 5 coord states
Ch120a-Goddard-L17
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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-L17
This makes all three available as
possible ground states depending
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onIII,the
6threserved
ligand
© copyright 2011 William A. Goddard
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
Ch120a-Goddard-L17
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Bond O2 to Mb
Ch120a-Goddard-L17
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-L17
7 Kdd
6 Kdd
4*3/2
+
2*1/2
3*2/2
+
3*2/2
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Bonding O2 to Mb
Exchange loss
on bonding O2
Ch120a-Goddard-L17
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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
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Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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compare bonding of CO and O2 to Mb
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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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)}
Ch120a-Goddard-L17
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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-L17
Mnd+0.71Mnsp+1.20H
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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-L17
Metallacycle plane
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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).
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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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-L17
Mnd+0.71Mnsp+1.20H
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Example (Cl)2VH3
+ resonance
configuration
Ch120a-Goddard-L17
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Example ClMometallacycle
butadiene
Ch120a-Goddard-L17
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Example [Mn≡CH]2+
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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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
Ch120a-Goddard-L17
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