Quantum Chemistry
Lecture 5
The electronic structure of molecules
Absorption spectroscopy
Fluorescence spectroscopy
NC State University
3.5 Selective absorption and emission by
(source: P&O fig 4.2)
atmospheric gases
Absorption by gases in the atmosphere
5500 oK
Electronic
266 oK
Vibrational
Rotational
The electronic absorption of oxygen
LUMO
HOMO
p*
p
Potential energy surfaces oxygen dimer
In addition to transitions
Between states, absorption
Of light can also lead to
O – O bond breaking.
This is important for
Production of oxygen atoms
Needed to make ozone.
Schumann-Runge bands 175 – 200 nm
Proc. R. Soc. Lond. A, 226, 509-521, (1954)
The electronic absorption of ozone
LUMO
p*
HOMO
non
The importance of the ozone “continuum”
The fact that ozone can photodissociate leads to an intense
Absorption of UV radiation over a wide spectral range.
The ozone filter is very effective and protects the surface of
Earth from receiving UV radiation below about 270 nm.
Molecular Orbital Theory
In MO theory electrons are treated as
including the entire molecule.
– Each MO is built up from a linear
combination of atomic orbitals (LCAO).
MO = i
c ii where i are atomic orbitals
=1
I
– The coefficients are optimized by the selfconsistent field (SCF) method.
– The variational principle justifies
minimization of the energy by adjustment of
the coefficients ci.
The molecular orbitals in a diatomic
molecule are formed from linear
combinations of atomic orbitals
+
=
Bonding s
s
=
s
Anti-Bonding s*
Constructive overlap between two
atomic orbitals gives rise to a
bonding state
NO NODES
+
s
=
s
=> 1/s + s)
Destructive overlap between two
atomic orbitals gives rise to an
anti-bonding state
ONE NODE
s
=
s
=> 1/s - s)
For diatomic hydrogen we consider
the s and s* molecular orbitals in the
following energy diagram
s*
s
s
s
MO treatment of H2
anti-bonding
E1s
E1s
bonding
The two electrons must have opposite spins,a and b.
The wave function must be anti-symmetric with respect
to electron exchange.

MO = b(1) b(2)
Spatial part
1 a(1)b(2) – a(2)b(1)
2
Spin part (anti-symmetric)
The wavefunction is composed
of electronic and nuclear parts
 =  electronic nuclear
Total
Electronic
Nuclear
The wavefunction represents the probability
amplitude of electrons and nuclei.
The wave equation can be separated
into electronic and nuclear parts
Hamiltonian
Energy Operator
Eigenvalue
Energy value
H elec = Eelec
H nucl = Enucl
Wavefunctions
The electronic part
For diatomic hydrogen we consider
the s and s* molecular orbitals in the
following energy diagram
s*
s
s
s
Absorption of visible or ultraviolet
radiation leads to electronic transitions
Transition
moment
s*
s
s
The change in
nodal structure
also implies a
change in orbital
angular momentum.
s
The highest occupied molecular
orbital of ethylene
p
The lowest unoccupied molecular
orbital of ethylene
p*
p*
1 p node
Separation of electronic and nuclear
parts of the transition moment
The transition moment, -e<1|q|2> can be
separated into the electronic wavefunction 
that depends on q and the nuclear wavefunction  that does not.
e q =e  q  v  1v 2v 
These enter the rate expression as the square
e  q    1v 2v 
2
2
v
2
2
= M 12 FC
The electronic transition moment
The electronic transition moment is
M12 = -e<1|q|2>
Light will be absorbed when the electric vector
is aligned with the transition moment.
Light will not be absorbed when the electric
vector is perpendicular to the transition moment.
Transition moment of lowest
p-p* transition of ethylene
p
p*
The transition moment is perpendicular to the
change in nodal structure. Electromagnetic
polarized along this direction
will give the maximum transition probability.
Calculate the transition moment
for a p-p* transition using a
simple model
The p and p * states are:
1
1
p =
1 + 2 and  p * =
1 – 2
2
2
The transition moment is:
M p–p* = e pxp*d
The magnitude of the transition
moment for C2H4 can be
calculated from a simple model
M p–p* = e
2
1 + 2 x 1 – 2 d
=e
1 x1d –
2
= e x1 – x2
2
2 x2d
where x1 - x2 is the C=C bond length of 1.35 Å.
One charge displaced through 1 Å has a dipole
moment of 4.8 D. Mp-p* = 3.24 D for C2H4.
The nuclear part
The Franck-Condon factor is due
to the overlap of ground state v=0
with excited state v’=0’, 1’, etc.
Excited state
0-0’
Ground state
The Franck-Condon factor is due
to the overlap of ground state v=0
with excited state v’=0’, 1’, etc.
Excited state
0-1’
Ground state
The Franck-Condon factor is due
to the overlap of ground state v=0
with excited state v’=0’, 1’, etc.
Excited state
0-2’
Ground state
The Franck-Condon factor is due
to the overlap of ground state v=0
with excited state v’=0’, 1’, etc.
Excited state
0-3’
Ground state
The Franck-Condon factor is due
to the overlap of ground state v=0
with excited state v’=0’, 1’, etc.
Excited state
0-4’
Ground state
Based on the FC factors we can
construct a “stick” spectrum
0-1’
0-0’
0-2’
0-3’
0-4’
Calculated assuming E(0-0’) = 8000 cm-1 and
vibrational mode of 1000 cm -1. 1 eV = 8065.6 cm-1.
The Franck-Condon factor determines
the envelop of the absorption lineshape
D
D
D
S = D2/2
S is electron-phonon
coupling
D is nuclear
displacement
Analytical expression for the FC factor
The Franck-Condon factor is a vibrational overlap term.
It depends on nuclear displacement, which is treated using
the parameter S, the electron-vibration (or electron-phonon)
coupling. The big the displacement the bigger is S.
There is a progression of lines with relative intensities
given by:
The delta function gives a stick spectrum spaced by energy
equal to the vibrational frequency. This function is also
known as a Poisson distribution. For large S this function
approaches a Gaussian shape.
The absorption cross section, sA
The absorption cross section has units of area (cm2).
It gives a probability for absorption. We have discussed
the probability in terms of the transition dipole moment.
M12 and shape in terms of the Franck-Condon factor, FC.
The absorption cross section is proportional to the well
known extinction coefficient.
The extinction coefficient has units of M-1cm-1.
Beer-Lambert Law
–A
I = I 010
A =   Cd
A is the absorbance.
() is the extinction coefficient.
The unit of () is M-1cm-1.
C is the concentration (M).
d is the pathlength (cm).
The exponential attenuation I0
of the intensity is shown
in the Figure.
xx
I’
dx
I’+dI’
d
I
Fluorescence usually occurs
after vibrational relaxation
1. Absorption
2. Vibrational
relaxation
3.Fluorescence
The Franck-Condon principle:
Transitions are vertical in both
absorption and emission
1. Absorption
2. Vibrational
relaxation
3.Fluorescence
The Franck-Condon factor is
the same for absorbance and
fluorescence
1. Absorption
2. Vibrational
relaxation
3.Fluorescence
This leads to a “mirror image”
relationship between absorption
and fluorescence bands
Fluorescence
Absorption
Energy
0-3’
0-4’
0-2’ 0-1’ 0’-0 0’-1 0’-2
0-0’
Wavelength
0’-3
0’-4