Lecture 13 February 5, 2014
Homonuclear diatomics
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, wag@wag.caltech.edu
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials
Science, and Applied Physics,
California Institute of Technology
Teaching Assistants:Sijia Dong <sdong@caltech.edu>
Samantha Johnson <sjohnson@wag.caltech.edu>
Ch120a-Goddard-L11
© copyright 2011 William A. Goddard III, all rights reserved
Ch120a1
Homonuclear Diatomics
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
2
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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4
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 orbital (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
thisCh120a-Goddard-L12
out in the 1920’s without
theWilliam
aid A.ofGoddard
computers.
© copyright 2011
III, all rights reserved
5
The homonuclear diatomic correlation diagram
Mulliken knew the ordering of the atomic orbitals and
considered how combinations of the atomic orbitals
would change as the nuclei were pushed together to
eventually form a united atom.
First consider the separate atoms limit where there is a
large but finite distance R separating the atoms.
The next slide shows the combinations formed from 1s,
2s, and 2p orbitals.
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6
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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7
Splitting of levels
General nodal arguments allow us to predict that
But which is lower of
and which is lower of
Here the nodal plane
arguments do not help
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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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9
Separated atom limit
MO notation
Ch120a-Goddard-L12
Separated atoms notation
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10
Homonuclear Diatomics Molecules – Ne2
compare the VB and MO views
Consider bonding two Ne atoms together
3 anti
bonds
2
MO
4
4
VB
3 bonds
2
2
1s22s22p6=10 e
1s22s22p6=10 e
20 e
MO view 3 antibonds cancel 3 bonds,
thus no net bonding
2
Generally bond ~ 1/(1+S)
Antibond ~ 1/(1-S)
Where S= overlap. Thus net is
antibonding left and right overlap, leading
to repulsive interactions and no bonding.
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2
2
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11
Homonuclear Diatomics Molecules – F2
compare the VB and MO views
Consider bonding two Ne atoms together
2 anti
bonds
0
MO
4
4
VB
3 bonds
2
2
1s22s22p5=9 e
1s22s22p5=9 e
20 e
MO view 2 antibonds cancel 2 bonds, but
still net single bond
2
2
2
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12
Homonuclear Diatomics Molecules – O2
compare the VB and MO views
Consider bonding two Ne atoms together
1 anti
bonds
0
MO
2
4
VB
3 bonds
2
2
1s22s22p5=9 e
1s22s22p5=9 e
20 e
MO view 1 antibonds cancel 1 bonds, but
still net double bond
But now we have 2 electrons to put into
the two pgx and pgy orbitals, can get either
singlet or triplet states
Lets examine this in more detail
2
2
2
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13
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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14
States arising from (p)2
Adding spin we get
MO theory
explains the triplet
ground state and
low lying singlets
Ch120a-Goddard-L12
xy-yx better than xy+yx
Because no chance for both
electrons at same spot (like
Hund’s rule on spin of atoms)
Triplet lower by Kxy, singlet
higher by Kxy
xx-yy better than xx+yy
Because decrease chance for
both electrons at same spot
(like Hund’s L rule atoms)
Triplet lower by Kxy, singlet
higher by Kxy
Ground state
© copyright 2011 William A. Goddard III, all rights reserved
15
States arising from (p)2
Adding spin we get
MO theory
explains the triplet
ground state and
low lying singlets
MO
exact
1.636
O2
(p)2
0.982
Energy (eV)
Ch120a-Goddard-L12
Ground state
© copyright 2011 William A. Goddard III, all rights reserved
0.0
16
States arising from (p)2
MO
exact
1.636
(p)2
Ground state
Spin and parity
forbidden
0.982
0.0
Spin and
angular
momentum
forbidden
Mulliken knew that these splittings would be ~ 1 eV in
infrared region, so he looks near dusk to get the longest
path of the sunlight through the atmosphere and saw the
transition
Immediately MO big winner and VB big loser
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17
First excited configuration of O2 (pu)3(pg)3
(1pg)2
(1pu)3 (1pg)3
0
2
Ground
4
excited
2
MO
1S +
u
1D
(1pu)3
(1pg)3
2
u
3S u
2
3S +
u
Only dipole allowed
transition from 3Sg-
1S u
3D
u
2
Strong transitions (dipole allowed) DS=0 (spin)
2
-  SSg
S
or
P
but
S
u
u
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18
The states of O2 molecule
Moss and Goddard JCP 63, 3623 (1975)
(pu)3(pg)3
(pu)4(pg)2
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19
Homonuclear Diatomics Molecules – N2
compare the VB and MO views
Consider bonding two Ne atoms together
0 anti
bonds
0
MO
0
4
VB
3 bonds
2
2
1s22s22p3=7 e
1s22s22p3=7 e
MO view 0 antibonds
20 e
2
get triple bond
Get singlet state, 1Sg+
2
2
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20
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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21
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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23
Ch120a-Goddard-L12
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24
Homonuclear Diatomics Molecules – N2+
compare the VB and MO views
0 anti
bonds
VB
0
MO
0
3
2.5 bonds
2
2
1s22s22p3=7 e
1s22s22p3=6 e
20 e
This suggests that N2+
3
1
1
2P
u ground
2S + 1st
g
2S + 2nd
u
2
state
exc. state
exc. state 2
But really ground state is 2Sg+, 1st exc. State is 2Pu
2
copyright
2011 William
A. Goddard III, all rights reserved
Something wrong©with
order
of orbitals
Ch120a-Goddard-L12
25
United atom limit
Next consider the limit in which the two nuclei are fused
together to form a united atom
For N2 this would lead to a Si atom.
Here we get just the normal atomic aufbau states
1s < 2s < 2p < 3s < 3p < 4s,3d < 4p etc
But now we consider an itty bity elongation of the Si
nucleus toward two N nuclei and how the atomic states
get perturbed
For the 1s orbital all that happens
is that the energy goes up (less
electron density on the nuclei)
and the symmetry becomes sg
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26
2s and 2p united atom orbitals
Similarly 2s just
goes to 2sg (and
a lower binding)
But the 2p case is more
interesting
For the 2ps state the splitting of
the nuclei lead to increased
density on the nuclei and hence
increased binding while for 2pp
there is no change in density
Thus 2psu < 2ppu
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27
Summarizing united atom limit
Note for 3d, the splitting is
3ds < 3dp < 3dd
Same argument as for 2p
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Summary more united atom levels
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29
Correlation diagram for Carbon row homonuclear diatomics
C2 N O
2
2
United
atom
limit
Ch120a-Goddard-L12
F2
O2+ separated
+
N
30
2
© copyright 2011 William A. Goddard III, all rights reserved atom limit
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
© copyright 2011 William A. Goddard III, all rights reserved
31
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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32
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
Ch120a-Goddard-L12
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33
Correlation diagram for Carbon row homonuclear diatomics
C2 N O
2
2
United
atom
limit
Ch120a-Goddard-L12
F2
O2+ separated
+
N
34
2
© copyright 2011 William A. Goddard III, all rights reserved atom limit
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
Ch120a-Goddard-L12
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35
The configuration for C2
1
1
2
4
4
4
1
2
3
2
2
2
2
2
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36
The configuration for C2
1
1
2
4
4
4
1
2
3
2
2
2
From 1930-1962 the 3Pu was
2
thought to be the ground
state. Now exper. Ground
2
1
+
state
is Sg is (0.09ev© lower)
Ch120a-Goddard-L12
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37
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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38
C2, Si2,
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Ch120a-Goddard-L12
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Low-lying states of C2
Ch120a-Goddard-L12
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Ch120a-Goddard-L12
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Ch120a-Goddard-L12
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43
The configuration for Si2
Si2 has this configuration
C2 has this configuration
1
1
1
2
4
4
4
3
2
2
2
2
Effectively Si2 has
lower overlap than C2
(toward Sep. atom limit)
Ch120a-Goddard-L12
2
2
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44
Do B2, Be2, Li2, Li2+
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45
MO and VB view of He dimer, He2
MO view
ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Net BO=0
VB view
ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2
Substitute sg = R + L and sg = R - L
Get ΨMO(He2) ≡ ΨMO(He2)
Ch120a-Goddard-L12
Pauli  orthog of R
to L  repulsive
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46
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)
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47
Noble gas dimers
No bonding at the VB or MO level
Only simultaneous electron
correlation (London attraction) or
van der Waals attraction, -C/R6
s
Ar2
Re
De
Ch120a-Goddard-L12
LJ 12-6 Force Field
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 2011 William A. Goddard III, all rights reserved
48
London Dispersion
The weak binding in He2 and other noble gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He 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)
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49
London Dispersion
The weak binding in He2 and other nobel gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He 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 forms with a long range 1/R6 attraction to account
for London dispersion (usually referred to as van der Waals
attraction) plus a short range repulsive term to account for
short
Range Pauli Repulsion)
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50
Remove an electron from He2 to get He2+
MO view
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
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51
Remove an electron from He2 to get He2+
MO view
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
VB view
Substitute sg = R + L and sg = L - R
Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)]
= (L)2(R) - (R)2(L)
-
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52
He2+
+
2S +
g
(sg)1(su)2
2S +
u
(sg
)2(s
u)
BO=0.5
MO good for discuss spectroscopy,
VB good for discuss chemistry
Check H2 and H2+ numbers
Ch120a-Goddard-L12
He2 Re=3.03A
De=0.02 kcal/mol
No bond
H2 Re=0.74xA
De=110.x kcal/mol
BO = 1.0
H2+ Re=1.06x A
De=60.x kcal/mol
BO = 0.5
© copyright 2011 William A. Goddard III, all rights reserved
53
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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54
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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55
1s and 2s cases
B
A
B
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A
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56
More on O2
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57
Exitation energies (eV) to O2 excited states
vertical
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58
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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© copyright 2011 William A. Goddard III, all rights reserved
59
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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60
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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61
ionosphere
night
Heaviside-Kennelly layer
Reflects radio waves to allow
long distance communications
D layer day
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62
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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63
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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64
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
Ch120a-Goddard-L12
Looks good because make p
bond as in ethene, BUT have
overlapping doubly occupied
orbitals  antibonding
© copyright 2011 William A. Goddard III, all rights reserved
65
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
Ch120a-Goddard-L12
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66
GVB wavefunction of triplet O2: sigma orbitals
(O2sL)2
O2pzL
bond
O2pzR
(O2sR)2
R=4 bohr
R=3 bohr
Re=2.28 bohr
Get orthogonal to O2s on other center
Causes some (2s-lpz) to stay orthogonal to bond pair
Ch120a-Goddard-L12Moss,
Bobrowicz,
Goddard
63, 4632
© copyright 2011
William JCP
A. Goddard
III, all (1975)
rights reserved
67
GVB wavefunction of triplet O2: pi orbitals
(OpxL
)2
O2pxR
Spin
paired
O2pyL
(OpyR)2
R=4 bohr
R=3 bohr
Re=2.28 bohr
Get orthogonal to O2pp on other center
Doubly occupied orbtial delocalizes (bonding)
Ch120a-Goddard-L12Moss,
Bobrowicz,
Goddard
63, 4632
© copyright 2011
William JCP
A. Goddard
III, all (1975)
rights reserved
68
GVB orbitals at Re
Problem: one VB configuration
not enough
+
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
69
VB description of O2
+
+
+
Must have resonance of two
VB configurations
Ch120a-Goddard-L12
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70
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
Ch120a-Goddard-L12
Looks good because make p
bond as in ethene, BUT have
overlapping doubly occupied
orbitals  antibonding
© copyright 2011 William A. Goddard III, all rights reserved
71
Bond energies
5.2 eV
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
72
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
Ch120a-Goddard-L12
Bond weakened by ~ 51 kcal/mol
due
toA.loss
inIII,Oall2 rights
resonance
© copyright 2011
William
Goddard
reserved
73
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
74
Stopped Feb. 5, 2014
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
75
Add material for O2 + C2H4 (sing and trip)
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
76
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
77
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
78
Add material for O2 + C2H4 (sing and trip)
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
79
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)
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
80
sigma GVB orbitals ozone
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
81
Pi GVB orbitals ozone
Some delocalization of central Opp pair
Ch120a-Goddard-L12
Increased overlap
between L and R Opp
due to central pair
© copyright 2011 William A. Goddard III, all rights reserved
82
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
Ch120a-Goddard-L12
© copyright
2011 William A. Goddard III, all rights reserved
83
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
84
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
85
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
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
86
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-L12
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87
Noble gas dimers
No bonding at the VB or MO level
Only simultaneous electron
correlation (London attraction) or
van der Waals attraction, -C/R6
s
Ar2
Re
De
Ch120a-Goddard-L12
LJ 12-6 Force Field
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
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88
London Dispersion
The weak binding in He2 and other noble gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He 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)
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
89
London Dispersion
The weak binding in He2 and other nobel gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He 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 forms with a long range 1/R6 attraction to account
for London dispersion (usually referred to as van der Waals
attraction) plus a short range repulsive term to account for
short
Range Pauli Repulsion)
Ch120a-Goddard-L12
© copyright 2011 William A. Goddard III, all rights reserved
90