Transcript Slide 1

Lecture 14 February 5, 2010
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
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: Wei-Guang Liu <[email protected]>
Ted Yu <[email protected]>
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© copyright 2010 William A. Goddard III, all rights reserved
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Course schedule
Friday Feb. 5, 2pm L14 TODAY(caught up)
Midterm given out on Friday. Feb. 5, due on Wed. Feb. 10
It will be five hour take home with 30 min. break, open notes
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Last time
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Separated atom limit
MO notation
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Separated atoms notation
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Separated
atoms limit
Note that in
each case we
get one bonding
combination (no
new nodal
plane) and one
antibonding
combination
(new nodal
plane,
red lines)
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At large R 2ps  better bonding than 2pp
In earlier lectures we considered the strength of
one-electron bonds where we found that
Since the overlap of ps orbitals is obviously higher than pp
We expect that
bonding
antibonding
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Summarizing united atom limit
Note for 3d, the splitting is
3ds < 3dp < 3dd
Same argument as for 2p
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Correlation diagram for Carbon row homonuclear diatomics
C2 N O
2
2
United
atom
limit
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F2
O2+ separated
+
N
2
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Homonuclear Diatomics Molecules – the valence bond view
Consider bonding two Ne atoms together
Clearly there will be repulsive interactions as the doubly
occupied orbitals on the left and right overlap, leading to
repulsive interactions and no bonding. In fact as we will
consider later, there is a weak attractive interaction
scaling as -C/R6, that leads to a bond of 0.05 kcal/mol,
but we ignore such weak interactions here
The symmetry of this state is 1Sg+
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Halogen dimers
Next consider bonding of two F atoms. Each F has 3
possible configurations (It is a 2P state) leading to 9
possible configurations for F2. Of these only one leads
to strong chemical binding
This also leads to a 1Sg+ state.
Spectroscopic properties are
listed below .
Note that the bond
energy decreases for
Cl2 to Br2 to I2, but
increases from F2 to
Cl2. we will get back to
this later.
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Di-oxygen or O2 molecule
Next consider bonding of two O atoms. Each O has 3 possible
configurations (It is a 3P state) leading to 9 possible
configurations for O2. Of these one leads to directly to a double
bond
This suggests that the
ground state of O2 is a
singlet state.
At first this seemed plausible, but by the late 1920’s Mulliken
established experimentally that the ground state of O2 is
actually a triplet state, which he had predicted on the basis of
molecular orbitial (MO) theory.
This was a fatal blow to VB theory, bringing MO theory to the
fore, so we will consider next how Mulliken was able to figure
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out in the 1920’s without
theWilliam
aid A.ofGoddard
computers.
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III, all rights reserved
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O2 MO configuration
2
For O2 the ordering of the MOs
4
Is unambiguous
2
(1pg)2
Next consider states of (1pg
)2
2
2
2
2
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States based on (p)2
Have 4 spatial combinations
Which we combine as
where x and y denote
px and py
φ1, φ2 denote the
angle about the axis
and F is independent of φ1, φ2
Rotating about the axis by an angle g, these states transform as
DSS+
D+
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States arising from (p)2
Adding spin we get
MO theory
explains the triplet
ground state and
low lying singlets
O2
Energy
(eV)
1.636
(p)2
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Ground state
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0.982
0.0
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Using the correleation diagram
In order to use the correlation
2
diagram to predict the states of
diatomic molecules, we need to
2
have some idea of what effective
4
R to use (actually it is the
effective overlap with large R
small S and small R large S).
Mulliken’s original analysis [Rev.
Mod. Phys. 4, 48 (1932)] was
roughly as follows.
1. N2 was known to be
nondegenerate and very strongly
bound with no low-lying excited
states
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Choices for N2
2
4
4
2
2
2
2
2
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N2 MO configurations
This is compatible with several
orderings of the MOs
Largest R
2
2
4
2
4
4
2
2
2
2
Smallest R
2
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N2+
But the 13 electron molecules BeF, BO, CO+, CN, N2+
Have a ground state with 2S symmetry and a low lying 2S
sate.
In between these two 2S states is a 2P state with spin
orbital splitting that implies a p3 configuration
This implies that
Is the ground configuration for N2 and that the low lying
states of N2+ are
This agrees with
the observed
spectra
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Correlation diagram for Carbon row homonuclear diatomics
C2 N O
2
2
United
atom
limit
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F2
O2+ separated
+
N
18
2
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1s and 2s cases
B
A
B
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A
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Bond
Anti
BO
1
2
2.5
3
2.5
2
1
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More about O2
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First excited configuration
(1pg)2
Ground configuration
(1pu)3 (1pg)3 excited configuration
1S +
u
1D
(1pu)3 (1pg)3
u
3S u
Only dipole allowed
transition from 3Sg-
3S +
u
1S u
3D
u
Strong transitions (dipole allowed) DS=0 (spin)
-  SSg
S
or
P
but
S
u
u
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The states of O2 molecule
Moss and Goddard JCP 63, 3623 (1975)
(pu)3(pg)3
(pu)4(pg)2
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Role of O2 in atmosphere
Moss and Goddard JCP 63, 3623 (1975)
Strong
Get 3P +
1D O atom
Weak
Get 3P +
3P O atom
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Implications
UV light > 6 eV (l < 1240/6 = 207 nm) can dissociate O2 by
excitation of 3Su+ which dissociates to two O atom in 3P state
UV light > ~7.2 eV can dissociate O2 by excitation of 3Suwhich dissociates to one O atom in 3P state and one in 1D
(maximum is at ~8.6 eV, Schumann-Runge bands)
Net result is dissociation of O2 into O atoms
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Regions of the
atmosphere
mesosphere
O + hn  O+ + eHeats from light
stratopause
O + O2  O3
100
altitude (km)
O2 + hn  O + O
O3 + hn  O + O2
Heats from light
tropopause
50
stratosphere
30
20
10
troposphere
Heated from earth
200
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300
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ionosphere
night
Heaviside-Kennelly layer
Reflects radio waves to allow
long distance communications
D layer day
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nightglow
At night the O atoms created
during the day can recombine to
form O2
The fastest rates are into the
Herzberg states, 3Su+ 1Su-
3D
u
Get emission at ~2.4 eV, 500 nm
Called the nightglow (~ 90 km)
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Problem with MO description: dissociation
3S g
state: [(pgx)(pgy)+ (pgy) (pgx)]
As R∞ (pgx)  (xL – xR) and (pgy)  (yL – yR)
Get equal amounts of {xL yL and xR yR} and {xLyR and xR yL}
Ionic: [(O-)(O+)+ (O+)(O-)]
covalent: (O)(O)
But actually it should dissociate to neutral atoms
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Back to valence bond (and GVB)
Four ways to combine two 3P states of O to form a s bond
bad
Closed shell
Open shell
Each doubly occupied orbital
overlaps a singly occupied
orbital, not so repulsive
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Looks good because make p
bond as in ethene, BUT have
overlapping doubly occupied
orbitals  antibonding
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Analysis of open shell configurations
Each can be used to form a singlet state or a triplet state, e.g.
Singlet: A{(xL)2(yR)2[(yL)(xR) + (xR)(yL)](ab-ba)}
Triplet: A{(xL)2(yR)2[(yL)(xR) - (xR)(yL)](ab+ba)} and aa, bb
Since (yL) and (xR) are orthogonal, high spin is best (no
chance of two electrons at same point) as usual
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VB description of O2
+
+
+
Must have resonance of two
VB configurations
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Bond H to O2
Bring H toward px on Left O
Overlap doubly
occupied (pxL)2
thus repulsive
Overlap singly
occupied (pxL)2
thus bonding
Get HOO bond angle ~ 90º
S=1/2 (doublet)
Antisymmetric with respect to plane:
A” irreducible representation (Cs
group)
2A”
state
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Bond weakened by ~ 51 kcal/mol
due
to A.loss
in III,Oall2 rights
resonance
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William
Goddard
reserved
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Bond 2nd H to HO2 to form hydrogen peroxide
Bring H toward py on right O
Expect new HOO bond angle ~ 90º
Expect HOOH dihedral ~90º
Indeed H-S-S-H:
HSS = 91.3º and HSSH= 90.6º
But H-H overlap leads to steric effects for HOOH,
net result:
HOO opens up to ~94.8º
HOOH angle  111.5º
trans structure, 180º only 1.2 kcal/mol higher
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Rotational barriers
7.6 kcal/mol Cis
barrier
HOOH
1.19 kcal/mol
Trans barrier
HSSH:
5.02 kcal/mol trans barrier
7.54 kcal/mol cis barrier
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Compare bond energies (kcal/mol)
O2 3Sg-
119.0
50.8
HO-O
68.2
17.1
HO-OH
51.1
67.9
H-O2
HOO-H
51.5
85.2
Interpretation:
OO s bond = 51.1 kcal/mol
OO p bond = 119.0-51.1=67.9 kcal/mol (resonance)
Bonding H to O2 loses 50.8 kcal/mol of resonance
Bonding H to HO2 loses the other 17.1 kcal/mol of resonance
Intrinsic H-O bond is 85.2 + 17.1 =102.3
compare CH3O-H: HO bond is 105.1
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Bond O2 to O to form
ozone
Require two OO s bonds get
States with 4, 5, and 6 pp
electrons
Ground state is 4p case
Get S=0,1
but 0 better
Goddard
et al Acc. Chem. Res.
6, 368 (1973)
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sigma GVB orbitals ozone
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Pi GVB orbitals ozone
Some delocalization of central Opp pair
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Increased overlap
between L and R Opp
due to central pair
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Bond O2 to O to form ozone
lose O-O p resonance, 51 kcal/mol
New O-O s bond, 51 kcal/mol
Gain O-Op resonance,<17 kcal/mol,assume 2/3
New singlet coupling of pL and pR orbitals
Total splitting ~ 1 eV = 23 kcal/mol, assume ½
stabilizes singlet and ½ destabilizes triplet
Expect bond for singlet of 11 + 12 = 23 kcal/mol, exper = 25
Expect triplet state to be bound by 11-12 = -1 kcal/mol,
probably
between +2
and -2
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2010 William A. Goddard III, all rights reserved
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Alternative view of bonding in ozone
Start here with 1-3 diradical
Transfer electron from central doubly
occupied pp pair to the R singly occupied
pp.
Now can form a p bond the L singly
occupied pp.
Hard to estimate
strength
of bond
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New material
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Ring ozone
Form 3 OO sigma bonds, but pp pairs overlap
Analog: cis HOOH bond is 51.1-7.6=43.5
kcal/mol. Get total bond of 3*43.5=130.5 which
is 11.5 more stable than O2.
Correct for strain due to 60º bond angles = 26
kcal/mol from cyclopropane. Expect ring O3 to
be unstable with respect to O2 + O by ~14
kcal/mol,
But if formed it might be rather stable with
respect various chemical reactions.
Ab Initio Theoretical Results on the Stability of Cyclic Ozone
L. B. Harding and W. A. Goddard III
J. Chem. Phys. 67, 2377 (1977) CN 5599
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Photochemical smog
High temperature combustion: N2 + O2  2NO
Thus Auto exhaust  NO
2 NO + O2  2 NO2
NO2 + hn  NO + O
O + O2 + M  O 3 + M
O3 + NO  NO2 + O2
Get equilibrium
Add in hydrocarbons
NO2 + O2 + HC + hn  Me(C=O)-OO-NO2
peroxyacetylnitrate
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More on N2
The elements N, P, As, Sb, and Bi all have an (ns)2(np)3
configuration, leading to a triple bond
Adding in the (ns) pairs, we show
the wavefunction as
This is the VB description of N2, P2, etc. The
optimum orbitals of N2 are shown on the next slide.
The MO description of N2 is
Which we can
draw as
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GVB orbitals of N2
Re=1.10A
R=1.50A
R=2.10A
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Hartree Fock
Orbitals N2
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The configuration for C2
1
1
2
4
4
4
1
2
3
2
2
2
2
2
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The configuration for C2
Si2 has this configuration
1
1
2
4
4
4
1
2
3
2
2
2
From 1930-1962 the 3Pu was
thought to be the ground
state
2
2
1S + is ground state
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Now
50
Ground state of C2
MO configuration
Have two strong p bonds,
but sigma system looks just like Be2 which leads to a bond of ~ 1
kcal/mol
The lobe pair on each Be is activated to form the sigma bond.
The net result is no net contribution to bond from sigma
electrons. It is as if we started with HCCH and cut off the Hs
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C2, Si2,
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Low-lying states of C2
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Include B2, Be2, Li2, Li2+
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Re-examine the energy for H2+
For H2+ the VB wavefunctions were
Φg = (хL + хR) and
Φu = (хL - хR) (ignoring normalization)
where H = h + 1/R. This leads to the energy for the bonding state
eg = <L+R|H|L+R>/ <L+R|L+R> = 2 <L|H|L+R>/ 2<L|L+R>
= (hLL + hLR)/(1+S) + 1/R
And for the antibonding state
eu = (hLL - hLR)/(1-S) + 1/R
We find it convenient to rewrite as
eg = (hLL + 1/R) + t/(1+S)
eu = (hLL + 1/R) - t/(1-S)
where t = (hLR - ShLL) includes the terms that dominate the
bonding and antibonding character of these 2 states
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The VB interference or resonance energy for H2+
The VB wavefunctions for H2+
Φg = (хL + хR) and Φu = (хL - хR) lead to
eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx
eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux
where t = (hLR - ShLL) is the VB interference
or resonance energy and
ecl = (hLL + 1/R) is the classical energy
As shown here the t dominates the bonding
and antibonding of these states
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Analysis of classical and interference energies
The classical energy, ecl = (hLL + 1/R), is the total energy of the
system if the wavefunction is forced to remain an atomic orbital
as R is decreased.
The exchange part of the energy is the change in the energy
due to QM interference of хL and хR, that is the exchange of
electrons between orbitals on the L and R nuclei
The figure shows that ecl is weakly antibonding with little
change down to 3 bohr whereas the exchange terms start
splitting the g and u states starting at ~ 7 bohr.
Here the bonding of the g state arises solely from the
exchange term, egx = t/(1+S) where t is strongly negative,
while the exchange term makes the u state hugely repulsive,
eux = -t/(1-S)
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Analysis of classical and interference energies
egx = t/(1+S) while eux = -t/(1-S)
Consider first very long R, where S~0
Then egx = t while eux = -t
so that the bonding and antibonding effects are similar.
Now consider a distance R=2.5 bohr =
1.32 A near equilbrium
Here S= 0.4583
t= -0.0542 hartree leading to
egx = -0.0372 hartree while
eux = + 0.10470 hartree
ecl = 0.00472 hartree
Where the 1-S term in the denominator
makes the u state 3 times as
antibonding
as the g ©state
is2010
bonding.
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William A. Goddard III, all rights reserved
61
Analytic results - details
Explicit calculations (see appendix A of chapter 2) leads to
S = [1+R+ R2/3] exp(-R)
ecl = - ½ + (1 + 1/R) exp(-2R)
t = -[2R/3 – 1/R] exp(-R) – S(1+1/R) exp(-2R)
t ~ -[2R/3 – 1/R] exp(-R) neglecting terms of order exp(-3R)
Thus for long R, t ~ -2S/R
That is, the quantity in t dominating the bond in H2+ is
proportional to the overlap between the atomic orbitals.
At long R this leads to a bond energy of the form
t~ -(2/3) R exp(-R)
That is the bond strength decreases exponentially with R.
t has a minimum at ~ R=2 bohr, which is the optimum R.
But S continues to increase until S=1 at R=0.
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Contragradience
The above discussions show that the interference or exchange
part of KE dominates the bonding, tKE=KELR –S KELL
This decrease in the KE due to overlapping orbitals is
dominated by
tx = ½ [< (хL). ((хR)> - S [< (хL)2>
Dot product is
хL
large and negative
in the shaded
region between
atoms, where the L
and R orbitals have
opposite slope
(congragradience)
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хR
63
The VB exchange energies for H2
For H2, the classical energy is slightly attractive, but again the
difference between bonding (g) and anti bonding (u) is
essentially all due to the exchange term.
-Ex/(1 - S2)
+Ex/(1 + S2)
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1E
g
3E
u
= Ecl + Egx
= Ecl + Eux
Each energy is
referenced to
the value at
R=∞, which is
-1 for Ecl, Eu, Eg
0 for Exu and Exg
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64
Analysis of classical and exchange energies for H2
For H2 the VB energies for the bonding state (g, singlet) and
antibonding (u, triplet) states are
1E = Ecl + E x
g
g
3E = Ecl + E x
u
u
Where Ecl = <ab|H|ab>/<ab|ab> = haa + hbb + Jab + 1/R
Egx = Ex/(1 + S2)
Eux = - Ex/(1 - S2)
where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2
Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St contains the 1e part
T2 = {Kab –S2Jab} contains the 2e part
The one electron exchange
compared to the H2+ case
for H2 leads to
egx ~ +t/(1 + S)
1x
2
Eg ~ +2St /(1 + S )
x ~ -t/(1 - S)
e
u
Eu1xCh120a-Goddard-L14
~ -2St /(1 - S2) © copyright 2010 William A. Goddard III, all rights reserved
65
Analysis of the VB exchange energy, Ex
where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2
Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St
Where t = (hab – Shaa) contains the 1e part
T2 = {Kab –S2Jab} contains the 2e part
Clearly the Ex is
dominated by T1
and clearly T1 is
dominated by the
kinetic part, TKE.
Thus we can
understand
bonding by
analyzing just the
KE part if Ex
Ch120a-Goddard-L14
T2
T1
Ex
TKE
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66
Analysis of the exchange energies
The one electron exchange
for H2 leads to
Eg1x ~ +2St /(1 + S2)
Eu1x ~ -2St /(1 - S2)
which can be compared to
the H2+ case
egx ~ +t/(1 + S)
eux ~ -t/(1 - S)
For R=1.6bohr (near Re), S=0.7
Eg1x ~ 0.94t vs. egx ~ 0.67t
Eu1x ~ -2.75t vs. eux ~ -3.33t
For R=4 bohr, S=0.1
Eg1x ~ 0.20t vs. egx ~ 0.91t
Eu1x ~ -0.20t vs. eux ~ -1.11t
E(hartree)
Eu1x
Consider a very small R
with S=1. Then
Eg1x ~ 2t vs. egx ~ t/2
so that the 2e bond is twice
as strong as the 1e bond
1x
E
g
but at long R, the 1e bond is
R(bohr)
67
stronger
than the 2e bond
Ch120a-Goddard-L14
© copyright 2010 William A. Goddard III, all rights reserved
Van der Waals interactions
For an ideal gas the equation of state is given by
pV =nRT
where p = pressure; V = volume of the container
n = number of moles; R = gas constant = NAkB
NA = Avogadro constant; kB = Boltzmann constant
Van der Waals equation of state (1873)
[p + n2a/V2)[V - nb] = nRT
Where a is related to attractions between the particles,
(reducing the pressure)
And b is related to a reduced available volume (due to
finite size of particles)
Ch120a-Goddard-L14
© copyright 2010 William A. Goddard III, all rights reserved
68
London Dispersion
The universal attractive term postulated by van der Waals was
explained in terms of QM by Fritz London in 1930
The idea is that even for spherically symmetric atoms such as
He, Ne, Ar, Kr, Xe, Rn the QM description will have
instantaneous fluctuations in the electron positions that will
lead to fluctuating dipole moments that average out to zero.
The field due to a dipole falls off as 1/R3 , but since the
average dipole is zero the first nonzero contribution is from 2nd
order perturbation theory, which scales like
-C/R6 (with higher order terms like 1/R8 and 1/R10)
Consequently it is common to fit the interaction potentials to
functional froms with a long range 1/R6 attraction to account
for London dispersion (usually refered to as van der Waals
attraction) plus a short range repulsive term to acount for short
Range
Pauli Repulsion)
Ch120a-Goddard-L14
© copyright 2010 William A. Goddard III, all rights reserved
69
Noble gas dimers
s
Ar2
Re
De
Ch120a-Goddard-L14
LJ 12-6
E=A/R12 –B/R6
= De[r-12 – 2r-6]
= 4 De[t-12 – t-6]
r= R/Re
t= R/s
where s = Re(1/2)1/6
=0.89 Re
© copyright 2010 William A. Goddard III, all rights reserved
70
Remove an electron from He2
Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Two bonding and two antibonding  BO= 0
Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½
Get 2Su+ symmetry.
Bond energy and bond distance similar to H2+, also BO = ½
Ch120a-Goddard-L14
© copyright 2010 William A. Goddard III, all rights reserved
71
stop
Ch120a-Goddard-L14
© copyright 2010 William A. Goddard III, all rights reserved
72