Chemistry 431
Lecture 11
The Born-Oppenheimer
Born Oppenheimer Approximation
The electronic structure of molecules
NC State University
Valence-bond ((VB)) orbitals for diatomics
We consider the orbital on atoms A and B. Following the
text we shall represent electron 1 on atom A as A(1).
A(1) iis an atomic
t i wave ffunction.
ti
Lik
Likewise,
i
electron
l t
2 iis
located on atom B and the wave function is B(2).
A(1)B(2) represents the electrons localized on each atom
and is not a bonding picture. If the electrons trade places
we have A(2)B(1)
A(2)B(1), which is also not a bonding situation
situation.
However, the combination A(1)B(2) + B(2)A(1) is a bonding
i t
interaction.
ti
B
By th
the same llogic
i A(1)B(2) - B(2)A(1) iis an
anti-bonding interaction.
We must consider the anti-symmetry issue. In this description
Only the anti-bonding interaction is anti-symmetric.
The role of spin in the VB picture
If we include electron spin then we have two possible spin
wave functions for the electrons:
α = spin
i up
β = spin down
Therefore, the spin wave function for the two electrons
can be either symmetric
α(1)α(2)
β(1)β(2)
α(1)β(2) + α(2)β(1)
or anti-symmetric
α(1)β(2)
(1)β(2) - α(2)β(1)
(2)β(1)
Note that there are three symmetric combinations, and only
one anti-symmetric combination. If we combine the spin and
spatial parts of the wave functions we can satisfy the antisymmetry requirement.
Total VB wave function for diatomics
The combined wave functions explain the fact that spin
pairing results in a singlet for the bonding combination and
a triplet
t i l t for
f the
th anti-bonding
ti b di combination.
bi ti
Anti-bonding combination (triplet)
[A(1)B(2) - B(2)A(1)][α(1)α(2)]
[A(1)B(2) - B(2)A(1)][β(1)β(2)] (2)
[A(1)B(2) - B(2)A(1)][α(1)β(2) + α(2)β(1)]
Bonding combination (singlet)
[A(1)B(2) + B(2)A(1)][
B(2)A(1)][α(1)β(2)
(1)β(2) - α(2)β(1)]
(2)β(1)]
Extension of VB to polyatomics
If we consider more than two nuclei the same principles can
be applied. However, this approach does not explain the
observed
b
d molecule
l
l geometries
t i off polyatomic
l t i molecules.
l
l
For example, H2O does not have a 90o bond angle as
predicted by simple VB theory.
This led to the idea of a hybrid orbital
orbital.
sp 3
h 1 = s + px + py + pz
h 2 = s – px – py + pz
h 3 = s – px + py – pz
h 4 = s + px – py – pz
sp 2
h 1 = s + 1 py
2
3p – 1 p
h2 = s +
y
2 y
2
3p – 1 p
h3 = s –
y
2 y
2
Note that these are all orthogonal.
sp
h 1 = s + px
h 2 = s – px
A note on orthogonality
The hydrogen atom wave functions are orthogonal to one
another. This means that they have no overlap. We can
examine
i some specific
ifi examples.
l
Th
The overlap
l iintegral
t
l off
px and pz can be written as follows:
p xp zdτ = 0
all space
We look back to the solution of the hydrogen atom to find
The mathematical forms of these p orbitals (and the volume
element dτ).
3
pz =
px =
cosθ
4π
3 sinθcosφ
4π
dτ = sinθdθdφ
A note on orthogonality
We can write all of this out:
2π
π
3 cosθ
4π
p z* p xdτ =
all space
0
0
= 3
4π
π
2π
cosφdφ
d
cosθsin
i θdθ
d
2
0
0
2π
= 3
4π
3 sinθcosφsinθdθdφ
4π
0
z2dz = 0
cosφdφ
0
0
let z = sin θ , dz = – cos θ
Of course the orbitals are normalized so
p z* p zdτ
d =1
all space
A note on normalization of sp hybrids
We apply this logic to the two sp hybrid orbitals.
Are they normalized? Are they orthogonal?
h 2* h 2 dτ =
=
s *s dτ
d –
(s – p z) *(s – p z)dτ
s *p z dτ
d –
p z*s dτ
d +
p z*p zdτ
d
=1+1=2
OK. The text book gave us non-normlized sp orbitals.
However, if we multiply each orbital by a normalization
constant
t t , then
th they
th will
ill b
be normalized.
li d
1
2
(s – p z) *(s
(s – p z)dτ = 1 ∴h 2 = 1 (s – p z)
2
A note on orthogonality of sp hybrids
We continue to address the question: Are they orthogonal?
1
2
h 1* h 2 dτ
d = 1 ((s + p z) *(s
*( – p z)dτ
)d
2
= 1 s *s dτ – 1 s *p z dτ + 1 p z*s dτ – 1
2
2
2
2
= 1 – 1 =0
2 2
p z*p zdτ
They have zero overlap.
overlap They are orthogonal.
orthogonal This concept
applies not only to hybrid orbitals in VB theory but to the
linear combinations of orbitals in molecular orbital theory.
Heteronuclear diatomics
In the molecular orbital picture we will consider various
linear combinations of atomic orbitals. The simplest case
i a di
is
diatomic,
t i which
hi h h
has ttwo orbitals:
bit l
ψ = c AA + c BB
This wave function will be normalized if
c 2A + c B2 = 1
In fact, for a homonuclear diatomic these coefficients must
be the same by symmetry so there is not much work to do.
H
However,
if we consider
id HF
HF, th
then clearly
l l th
there will
ill b
be a nett
displacement of electrons towards fluorine. We can think
about sharing of electrons (covalency), but that sharing
is not necessarily equal (i.e. cA is not equal to cB).
Heteronuclear diatomics
To make the HF example more precise we can consider
the valence orbitals that interact.
H
F
ψ = c 1s H + c 2p zF
We already know that 1s orbital of hydrogen is 13.6 eV
below the ionization limit
limit. If we are given that the pz orbital
of fluorine is 18.6 eV below the ionization limit then we
can see that these two will not contribute equally to the
b di iinteraction.
bonding
t
ti
1s H
2pz F
Electronegativity
The electronegativity of an element is a measure of its ability
to attract electrons to itself. The Pauling scale of electronegativity
ti it iis b
based
d on b
bond
d di
dissociation
i ti energies.
i
If
We call the electronegativity of an atom A, χA then
|χ A – χ B| = 0.102
D(A – B) – 1 D(A – A) + D(B – B)
2
The greater the electronegativity,
electronegativity the greater the polar
character of the chemical bond. The consequence of this
charge asymmetry is that the molecule has a dipole moment.
Dipole moment
The dipole moment of a molecule is a measure of its tendency
to align in an externally applied electric field.
No field E = 0
Random orientation
Applied field E = V/d
Dipoles tend to align
Thermal fluctuations tend to cause the dipoles to be random.
Thus at high temperature the dipoles are harder align.
Dipole moment
A dipole on a molecule is caused by displacement of charge
within the molecule. The molecular charge is not affected,
b t th
but
the di
distribution
t ib ti iis nott symmetrical.
ti l
Negative
Positive
We can define the dipole moment as a charge displaced
through a distance.
μ = qd
where q is the charge and d is the distance. So for example,
if one electron is transferred over a distance of 1 Å the dipole
moment is (1 electron)(1 Å) = 1 eÅ. We can also write this
in units of Coulomb-meters. 1 eÅ = 1.62 x 10-29 Cm.
Dipole moments due to a charge cloud
We have seen that we cannot consider the electron to be a
discrete unit that can be placed somewhere in space, but
R th it iis a charge
Rather
h
cloud
l d ((so tto speak).
k) It iis iin an orbital.
bit l
We return to the idea of using wave functions to calculate
average properties. In the case of the dipole moment we
need a dipole operator. This is an operator that represents
the charge displacement.
μ = ez
We have assumed that the charge moves in the z direction.
U i thi
Using
this definition
d fi iti th
the quantum
t
mechanical
h i l di
dipole
l momentt
is:
μ=
Ψ*ezΨdτ
Ψ
ezΨdτ
Typical dipole moments
The commonly-used unit for dipole moments is the Debye
30 Cm
1D
Debye
b =3
3.33
33 x 10-30
C
4.8 Debye = 1 eÅ
For the series HF, HCl, HBr, HI we can see a trend in the
dipole moment that follows the electronegativity. However,
the bond length also plays a role since the dipole moment
depends on both charge and distance.
Application of the variation principle to a
h t
heteronuclear
l
di t i molecule
diatomic
l
l
The name of game is to find the coefficients cA and cB since
these will determine the nature of the chemical bond and
properties such as the dipole moment. We will use the
variation principle to do this. Starting with our trial wave
function.
ψ = c AA + c BB
The energy is
ψ *Hψdτ
E=
ψ *ψdτ
Minimize the energy (it will always be
greater
t than
th the
th true
t
energy))
We can use calculus to find the minimum of the function
function.
Here we will minimize with respect to the coefficients cA and
cB. We can call these the variational parameters. Since
There are two coefficients, there are two equations,
∂E = 0 , ∂E = 0
∂c A
∂c B
Let’s write out each of the integrals in terms of the coefficients,
ψ *Hψdτ
E=
ψ *ψdτ
Plug
g in the wave function and work out
ψ *Hψdτ =
c AA + c BB H c AA + c BB dτ
= c 2A AHAdτ + c Ac B AHBdτ+ c Ac B BHAdτ + c B2 BHBdτ
= c 2Aα A + 2c
2 Ac Bβ + c B2 α B
c AA + c BB c AA + c BB dτ = c 2A A 2dτ + 2c Ac B ABdτ + c B2 B 2dτ
= c 2A + 2c Ac BS + c B2
We have made the following definitions
αA =
AHAdτ , α B =
β=
AHBdτ =
BHBdτ
BHAdτ
Energy
Resonance
Take the derivatives and set up the
secular
l equation
ti
The energy
gy is
c 2Aα A + 2c Ac Bβ + c B2 α B
E=
c 2A + 2c Ac BS + c B2
Now take the derivative with respect to each coefficient
coefficient.
∂E =
∂c A
∂E =
∂c B
2 c Aα A – c AE + c Bβ – c BSE
2
A
c + 2c Ac BS + c
2
B
=0
2 c Bα B – c BE + c Aβ – c ASE
2
A
c + 2c Ac BS + c
2
B
c A α A – E + c B β – SE = 0
c B α B – E + c A β – SE = 0
=0
Take the derivatives and set up the
secular
l equation
ti
In matrix form the secular equation
q
is:
α A – E β – SE c A
=0
β – SE α B – E c B
The equations have a solution if the determinant vanishes:
α A – E β – SE
=0
β – SE α B – E
This is relativelyy easyy to solve exactlyy for a homonuclear
diatomic, where αA = αΒ.
For the heteronuclear case (and all other problems we will
work here) we can make the assumption that S = 0.
Solve the secular equation with
zero overlap
l
Expand
p
the determinant as shown and solve for the energy:
gy
αA – E β
= α A – E αB – E – β 2 = 0
β αB – E
E 2 – α A + α B E + α Aα B – β 2 = 0
E=
E=
α A + αB ±
α A + αB ±
α 2A + α 2B + 2α Aα B – 4α Aα B + 4β 2
2
α A – αB
2
2
+ 4β 2
This leads to a geometric solution where
2β
ζ = 1 arctan α – α
A
B
2
Wave functions and energies for
th heteronuclear
the
h t
l
di t i
diatomic
Expand the determinant as shown and solve for the energy:
E– = α B – β tan ζ ,
E+ = α A + β tan ζ ,
ψ = – A sinζ + B cosζ
ψ = A cosζ + B sinζ
As the energy different between αA and αB increases the
Magnitude of ζ decreases. In other words when αA – αB >> β
The orbitals essentially just the unmixed atomic orbitals.
Therefore, bonding will be strongest when αA ~ αB. The
approximate values of the energy are
β2
E– = α B +
α A – αB
β2
E+ = α A –
α A – αB
where we have used the fact that β < 0.
Application
pp
to p
polyatomics
y
The same principles apply to polyatomic molecules.
general form of the molecular orbitals is:
The g
ψj =
Σcχ
N
i=1
ij
i
A simple and useful kind of system to treat is are linear
and cyclic π systems.
Polyenes
Aromatic hydrocarbons
Huckel theory
Basic assumptions
- Overlap S = 0.
- Carbon Coulomb integrals are α for π e-.
- Resonance integrals β for neighbor atoms.
o non-neighbor
o e g bo ato
atoms.
s
- β = 0 for
Form of secular determinant
-d
diagonal
ago a e
elements
e e s α - E.
- off-diagonal elements β for adjacent atoms.
- off-diagonal
g
elements β = 0 for others.
Ethene
For ethene there are two carbon atoms and two
π electrons. Huckel theory
y
2π electrons
ignores the σ electrons.
H
H
The secular determinant is
C
H
C
H
2
α–E
β
2
= α–E –β =0,E=α±β
β
α–E
Bonding
g energy
gy and electronic
spectra from a simple theory
The energy solutions give an MO diagram
E- = α - β
LUΜΟ
Lowest unoccupied
molecular orbital
E2p = α
E+ = α + β
ΗΟΜΟ
Highest occupied
molecular orbital
The highest
g
occupied
molecular orbital of ethene
π
The lowest unoccupied
molecular orbital of ethene
π∗
π∗
1 π node
d
Huckel theory for butadiene
Each carbon contributes α.
The resonance integral for each is β.
α–E
β
0
0
β
α–E β
0
0
β
α–E β
0
0
β α–E
We know how to expand a 4x4 determinant. However,
g 4th order polynomial.
y
One
We cannot solve the resulting
approach that makes this tractable is to use symmetry.
A second approach is to use the free electron model (FEM).
Application
pp
of the free electron
model to polyenes
Application of the free electron model to a polyene involves
the assumption that the "box" contains the n atoms of the
polyene. This is shown for butadiene below. The wavefunction coefficients are derived from the amplitude of the
sine function obtained from the solution of the particle in a
box. The particle in a box solutions are:
ψj =
j
2 sin jπx
a
a
where a is the length of the box and j is the quantum number
for a given state.
Application
pp
of the free electron
model to polyenes
If we imagine that an electron is placed in a box that contains
N atoms at positions na/N along the box then the free electron
model (FEM) states that the wavefunction coefficients for a
polyene will be given by:
Ψj =
c nj =
Σ
N
n=1
c njφn
2 sin jnπ
N
N+1
This is illustrated for butadiene below. The four orbitals
shown correspond to the four π orbitals of butadiene.
FEM MOs for butadiene
This is the lowest π orbital. It has no nodes.
FEM MOs for butadiene
This is the highest occupied π orbital. It has one node.
FEM MOs for butadiene
This is the lowest unoccupied π orbital. It has two nodes.
FEM MOs for butadiene
This is the highest π orbital. It has three nodes.
Wave functions for butadiene
Here we graphically represent the
varying coefficients and sign of the
Wave function the results for butadiene.
Ψ4 = 0.41φ1 – 0.67φ2 + 0.67φ3 – 0.41φ4
Ψ3 = 0.67φ1 – 0.41φ2 – 0.41φ3 + 0.67φ4
Ψ2 = 0.67φ1 + 0.41φ2 – 0.41φ3 – 0.67φ4
Ψ1 = 0.41φ1 + 0.67φ2 + 0.67φ3 + 0.41φ4
φ1
φ2
φ3
φ4
cn =
jjnπ
2 sin
i
N
N+1
An electronic wavefunction
corresponds
d tto each
h energy llevell
NODES
3
2
1
0
We can construct molecular
orbitals of benzene using the
six electrons in π orbitals
H
C
H
H
H
C
C
C
C
H
C
H
Benzene Structure
Electrons are
spin-paired
Electronic Energy Levels
The Perimeter Model
The benzene ring has D6h symmetry.
The aromatic ring has 6 electrons.
electrons
The π system approximates circular
electron path.
1
imφ
Φ=
e
2π
-5
5
-44
4
-33
3
-2
Δm=3
-11
2
Δm=1
1
m=0
The Perimeter Model
The porphine ring has D4h symmetry.
The aromatic ring has 18 electrons.
electrons
The p system approximates circular
electron path.
-5
N
N
5
Δm=9
Δm=1
-44
4
-33
3
-2
-11
2
N
N
1
imφ
Φ=
e
2π
1
m=0