Benzene:
1
2
6
3
5
4
We will solve Schodinger equation for this molecule by considering only p-orbitals of
six carbons under the Huckel approximation. Huckel approximation, though quite
crude, provides very useful results.
Since -bonding in planar molecules has different symmetry from the -bonding,
corresponding molecular orbitals separate in the Hamiltonian -do not have cross
off-diagonal element and thus can be solved separately. Besides, -bonding is much
stronger than -bonding, - and  ∗ molecular orbitals lie within the  −  ∗ gap. Thus
HOMO and LUMO orbitals are either of ,  ∗ -type or nonbonding n-type. All that allows
considering only p z -orbitals in conjugated molecules for treating their spectroscopic
features, while s and p x ,p y orbitals will be responsible primarily for -bonding, i.e.
molecule’s shape.
Huckel approximation can be summarized with following statements:
1)  p i |p j   ij - all overlap integrals are zero
2)  p i |H|p i   - diagonal elements equal atomic energies of carbon p z electrons
3)  p i |H|p j  neighbors - the interaction energy is nonzero only for neighboring
carbons.
Both,  and , are negative and can be parameterized for a typical conjugated carbon
and carbon-carbon bond. Such a parameterization can be also introduced for hetero
atoms, S, O, N
If we write the secular equation for benzene
det
−E

0
0
0


−E

0
0
0
0

−E

0
0
0
0

−E

0
0
0
0

−E

 0;

0
0
0

−E
4
2
2
4
−4  9  − E − 6  − E   − E 6  0.
6
The solutions can be found E    , E   − 2, E   − , E    2. For an arbitrary
cyclic molecule with identical conjugated bonds, the solutions can be written as
E    2 cos 2
n k and graphically represented as enegries horizontally leveled at the
corners of the appropriate polygon symmetrically placed on one of its corners:
1
"Graphical solution" to cyclic aromatics
E =  + 2 cos(2k/n)
One can see that there are degenerate solutions but symmetries of the appropriate
orbitals can be found by finding appropriate eigen functions.
Let’s do that analysis for benzene using group symmetry:
Benzene belongs to the ooint group:D 6h When six p orbitals are taken as the basis
set for a representation of this group we obtain Γ  , The character matrix is also shown
on the top. C ′2 ’s and  v ’s were chosen to go through carbon atoms and C ′′ 2 ’s and  d
’s go in between.
D 6h
E 2C 6 2C 3 C 2 3C ′2 3C ′′ 2
i
A 1g
1
1
1
1
1
1
1
1
1
1
1
1
A 2g
1
1
1
1
−1
−1
1
1
1
1
−1
−1
B 1g
1
−1
1
−1
1
−1
1
−1
1
−1
1
−1
B 2g
1
−1
1
−1
−1
1
1
−1
1
−1
−1
1
E 1g
2
1
−1
−2
0
0
2
1
−1 −2
0
0
E 2g
2
−1
−1
2
0
0
2
−1
−1
2
0
0
A 1u
1
1
1
1
1
1
−1 −1
−1 −1
−1
−1
A 2u
1
1
1
1
−1
−1
−1 −1
−1 −1
1
1
B 1u
1
−1
1
−1
1
−1
−1
1
−1
1
−1
1
B 2u
1
−1
1
−1
−1
1
−1
1
−1
1
1
−1
E 1u
2
1
−1
−2
0
0
−2 −1
1
2
0
0
E 2u
2
−1
−1
2
0
0
−2
1
1
−2
0
0
Γ
6
0
0
0
−2
0
0
0
0
−6
0
2
Γ 3N 36
0
0
0
−4
0
0
0
0
12
0
4
2S 3 2S 6  h 3 d 3 v
2
That representation can be reduced into irreducible for this point group:
Γ   B 2g  E 1g  A 2u  E 2u
A bit faster approach is to recognize that D 6h D 6  C i and thus use a smaller group,
D 6 , instead:
D 6 E 2C 6 2C 3 C 2 3C ′2 3C ′′ 2
A1
1
1
1
1
1
1
A2
1
1
1
1
−1
−1
B1
1
−1
1
−1
1
−1
B2
1
−1
1
−1
−1
1
E1
2
1
−1
−2
0
0
E2
2
−1
−1
2
0
0
Γ
6
0
0
0
−2
0
That representation can be reduced into irreducible for this point group:
Γ   B 2  E 1  A 2  E 2 which can be easily assigned g and u subscript based on the
required negative character with respect to operation  h in D 6h . Thus Γ  
B 2g  E 1g  A 2u  E 2u
We have not found the eigen functions nor energies yet, just their symmetries. Now
the task is easier than before.
We have two options:
a)Trained eye can draw the nodal planes and assign the molecular orbitals by analysis
of their characters with respect to appropriate symmetry operations as shown below.I
prematuraly put energies next to the MO, but note that encreaing number of the nodal
planes corresponds to increasing energy of MO. Also, a 2u with energy   2 is the
lowest because both,  and , are negative. The approach is handy but might be
confusing when trying to find those c 1 and c 2 coefficients to finish the construction.
3
c1
+
-c1
-c1
-c1
--c 2
c1
+
c1
+
c1
c1
-c2
+
c1
-c1
-
c1
+
-
c1
-c1
-
c2
c1
-c2
-c1
+
c1
+
-
--c
b2g

e2u

1
c1
+
-c1
c1
e1g

a2u

-
+
c1
-c1
b) More straight forward (also a bit more tedious) approach is by using the projection
operator:
P  ∑ R   R ∗ O R
Again, we’ll resort to a smaller group, D 6 , instead of D 6h . Then projection operators
would look shorter :0). For example:
P A 2  1O E  1O C 6  1O C −1
 1O C −1
 1O C 3  1O C 2 − 1O C ′2 1 − 1O C ′2 2 − 1O C ′2 3 − 1O C ′′2 1 − 1O
6
3
Using which on p 1 results in
P A 2 p 1  p 1  p 2  p 6  p 3  p 5  p 4  p 1  p 5  p 3  p 2  p 4  p 6  2p 1  p 2  p 3  p 4  p 5 
After normalizing we obtain: 4 a 2u   1 p 1  p 2  p 3  p 4  p 5  p 6 . The same way it
6
can be dobe with others.
Some attention is required for doubly degenerate species. The projection operator for
E2 :
P E 2  2O E − 1O C 6 − 1O C −1
− 1O C −1
− 1O C 3  2O C 2 . when applied to say p 1 gives:
6
3
E2
P p 1  2p 1 − p 2 − p 6 − p 5 − p 3  2p 4 . That gives only one symmetry adapted orbital,
 5  1 2p 1 − p 2 − p 6 − p 5 − p 3  2p 4 
2 3
Another has to be found by generating a similar one after applying P E 2 say to p 2 :
P E 2 p 2  2p 2 − p 3 − p 1 − p 6 − p 4  2p 5
and making it orthogonal to  5 :
 6  2p 2 − p 3 − p 1 − p 6 − p 4  2p 5  −  5   5 |2p 2 − p 3 − p 1 − p 6 − p 4  2p 5 
 2p 2 − p 3 − p 1 − p 6 − p 4  2p 5  −
1
2 3
2
2p 1 − p 2 − p 6 − p 5 − p 3  2p 4 −6 
 32 p 2 − p 3  p 5 − p 6 
− 1O C −1
− 1O C 3 − 2O C 2 and
Analogously:P E 1  2O E  1O C 6  1O C −1
6
3
4
P E 1 p 1  2p 1  p 2  p 6 − p 5 − p 3 − 2p 4 ;  2 
P E 1 p 2  2p 2  p 3  p 1 − p 6 − p 4 − 2p 5 and
 3  2p 2  p 3  p 1 − p 6 − p 4 − 2p 5 −

3
2
p2 
3
2
B 2g  1 
E 1g
3
2
p3 −
2 
3 
A 2u  4 
E 2u  5 
6 
1
6
p5 −
3
2
p6 
3
2
2p 1  p 2 − p 3 − 2p 4 − p 5  p 6 6 
p 2  p 3 − p 5 − p 6 
2p 1  p 2 − p 3 − 2p 4 − p 5  p 6 
1
p 2  p 3 − p 5 − p 6 
2
1
p 1  p 2  p 3  p 4
6
1
2
2
2p 1  p 2 − p 3 − 2p 4 − p 5  p 6 
p 1 − p 2  p 3 − p 4  p 5 − p 6 
1
2 3
1
2 3
1
2 3
1
2 3
 p5  p6
2p 1 − p 2 − p 3  2p 4 − p 5 − p 6 
p 2 − p 3  p 5 − p 6 
So we found the eigen functions which graphically are represented below with shaded
areas corresponding to a negative sign and the size of each circle resembling the
value of coefficient in the eigen function.
b2g
e2u
e1g
a2u
The energies of these MO can be found by evaluating appropriate   i |H| i :
E(b 2g ) 
  1 |H| 1  1 p 1 − p 2  p 3 − p 4  p 5 − p 6 |H| 1 p 1 − p 2  p 3 − p 4  p 5 − p 6  
6
6
5
 p 1 |H|p 1 − p 2  p 3 -p 4 p 5

1
6
  p 3 |H| p 1
− p 6   −  p 2 |H|p 1 − p 2  p 3 -p 4 p 5 -p 6   
− p 2  p 3 − p 4  p 5 -p 6   −  p 4 |H| p 1 -p 2
 p 5 |H| p 1 -p 2 p 3
 p3 − p4  p5 − p6   
− p 4  p 5 − p 6   −  p 6 |H|p 1 -p 2 p 3 -p 4
 p5 − p6 
matrix element for highlighted orbitals are zero

 p 1 |H|p 1 − p 2 − p 6   −  p 2 |H|p 1 − p 2  p 3     p 3 |H|−p 2  p 3 − p 4   −
1
6

−  p 4 |H|p 3 − p 4  p 5     p 5 |H|−p 4  p 5 − p 6   −  p 6 |H|p 1  p 5 − p 6  
 16 6 − 2   − 2
Energies of other orbitals are calculated the same way:
Eb 2g 
1 
Ee 1g 
2 
3 
Ea 2u    4 
Ee 2u 
1
6
p 1 − p 2  p 3 − p 4  p 5 − p 6 
1
2 3
2p 1  p 2 − p 3 − 2p 4 − p 5  p 6 
1
p 2  p 3 − p 5 − p 6 
2
1
p 1  p 2  p 3  p 4
6
5 
1
2 3
6 
1
2
 p5  p6
2p 1 − p 2 − p 3  2p 4 − p 5 − p 6 
p 2 − p 3  p 5 − p 6 
b2g
-2b2g 0.408(p1 p2p3p4p5p6)
e2u
-e2u(1) 0.289(2p1 p2p3p4p5p6); e2u(2) 0.5( p2p3p5p6)
e1g
e1g(1) 0.289(2p1 p2p3p4p5p6); e1g(2) 0.5( p2p3p5p6)
a1u
2a1u 0.408(p1 p2p3p4p5p6)

We see that MO energies are symmetric with respect to value of . All planar
conjugated hydrocarbons with such property are called alternant. Alternant
hydrocarbons can be recognized by their ability to have all carbons labeled in two
colors alternatingly (thus the name), i.e. when each neighbor has a different color.
1
The ground state configuration is a 2u  2 e 1g  4  X A 1g . Note that this ground state
energy is lower than that of cyclohexatriene (benzene with three localized double
bonds). The effect of sharing -electrons, the so-called delocalization or resonance
energy, equals 2.
The lowest excited state configuration is a 2u  2 e 1g  3 e 2u  1 (with energy 2 above the
6
ground state) from which following states can be constructed:
e 1g   e 2u   B 1u  B 2u  E 1u . each of which can be either a singlet or a triplet, i.e.:
X:1A1g
b2g
b2g
b2g
e2u
e2u
e2u
e1g
e1g
e1g
a1u
a1u
a1u
1
B2u, 1B1u, 1E1u
3
B2u, 3B1u, 3E1u
At this level of comlexity we cannot choose which excited state is the lowest energy,
1
B 1u , 1 B 2u , or 1 E 1u (all we know is that they each have corresponding triplet of lower
energy 3 B 1u  1 B 1u , 3 B 2u  1 B 2u , and 3 E 1u  1 E 1u ). We have to analyse other higher
energy one-electron excited state configurations and find if there are any of the same
symmetry as the three under consideration. Configurational interaction between
configurations of the same symmetry lowers the appropriate lower energy state while
increasing the corresponding high energy state energy.
The next one electron excited state configurations (with energy 3)
are:a 2u  2 e 1g  3 b 2g  1 and a 2u  1 e 1g  4 e 2u  1 .
Neither of them has a proper symmetry. The former makes: e 1g  b 2g  e 2g , while the
latter: a 2u  e 2u  e 2g
3
E2g
3
E2g
4
B2u
We shall include even higher one electron excited state configuration with energy 4.
There is only one, a 2u  1 e 1g  4 b 2g  1 , which produces a 2u  b 2g  b 2u . Thus, out of the
three states, 1 B 1u , 1 B 2u , and 1 E 1u , only 1 B 2u has two configurations. Consequently, it
will be the lowest energy state.
7
Since in the D 6h group the dipole moment components behave like E 1u (x,y) and A 2u
1
1
(z), the only dipole allowed transition is C E 1u − X A 1g which indeed shows a high
extinction coefficient, while the lower energy transitions, 1 B 1u and 1 B 2u , are
electronically forbidden. They are observed anyway but due to so call vibronic
coupling.
Before that let’s consider electric dipole allowed transitions first. The matrix element
for the transition moment is:
M e′v′e"v"   ∗e ′ v ′ || e ′′ v ′′   ∗e ′ || e ′′   ∗v ′ | v ′′  M e′e"   ∗v ′ | v ′′1   ∗v ′ | v ′′2 . . .
1
2
The selection rules in this case are defined by the selection rules for electronic part
M e′e" and the Franck-Condon factor,   ∗v ′ | v ′′  2 , for vibrational part. The latter would
be nonzero if that integrand has the totally symmetric species, which means that for
totally symmetric vibrations all vibrations are allowed:
Δv  0, 1, 2, 3, . . .
For all other vibrations the totally symmetric integrand appears only for even
difference in vibrational number, i.e. when:
8
Δv  0, 2, 4, . . .
In cases like the lowest energy transition in benzene, which is electronically forbidden,
the only way the matrix element for the transition moment would be nonzero is to step
back from the Born-Oppenheimer approximation and consider mixing nuclear
(vibrational) and electronic coordinates, so called vibronic coupling.
The electronic Hamiltonian parametrically depends on on vibrational coordinates:
e
H e  H 0e  ∑ i ∂H
Q i . .  H 0e  H ′
∂Q i
The excited state wavefunction  0f becomes accordinly mixed with other zero-order
electronic states :
 e ′   0f  ∑ k c k  0k
where the degree of mixing, c k 
 0k |H ′ | 0f 
E 0f −E 0k
, depends on both, the vibronic coupling
element   0k |H ′ | 0f  and the energy separation between the states E 0f − E 0k . The
electronic transition moment becomes:
M e′e"   e ′ || e ′′   0f || 0e ′′   ∑ k c k   0k || 0e ′′ 
While the first term (0-0 transition) is zero, the other terms might be not. Then the
intensity of transition would become nonzero, or as it is often referred to as intensity
is borrowed from a neigboring transition. That is why B 1u transition, beeing closer
to the allowed E 1u , is more intense than B 2u . The energy factor E 0f − E 0k  is not the only
one responsible for nonzero c k , the other part ,   0k |H ′ | 0f , enforces additional
selection rules:
e
  0k |Q i | 0f 
  0k |H ′ | 0f  ∂H
∂Q i
Q i 0
or Δv  13, 5, . . .  for vibronicaly active mode, the mode for which the symmetry of
0
Γ  k  Γ Q i matches that of the ground state X. The rules for other vibrational states
would follow the rules of ordinary Franck-Condon factor described above. In case of
1
benzene, vibronically allowed transition should be a 2u and e 1u . For the A B 2u state it
translates to a reqire for either b 1g b 1g  B 2u  B 2u  or e 2g e 2g  B 2u  E 1u . There are
123-630 vibrational modes in benzene, which for the ground state would be realized
in 20 fundamental frequencies:
2a 1g  a 2g  a 2u  2b 1u  2b 2g  2b 2u  e 1g  3e 1u  4e 2g  2e 2u . From these modes only e 2g
satisfy the selection rules ( 15 e 2g  through  18 e 2g ) for vibronic transition. By
symmetry all four of them can contribute and those that have the highest value of
∂H e
contribute the most. C-H vibrations do not affect the energies of -electrons
∂Q i
very much, it that should be a mode that shakes the molecular sceleton associated
with -system. A likely candidate is C-C-C in plane bending,  18 e 2g , with energy ca.
500-600 cm −1 (606 cm −1 in the ground state and 522 cm −1 in the electronically first
excited state). Thus the excited vibronic transitions that should be observed are:
1
1
1
A B 2u 18 10 - transition from vibrational ground state in the X A 1g to A B 2u at vibrationally
excited state with "1 at  18 e 2g .
1
Another possibility is to observe a hot A B 2u 18 01 transition from the vibrationally excited
1
1
with ’1 at  18 e 2g  X A 1g to the A B 2u and "0 at  18 e 2g . The latter transition is very
temperature dependent (hot). What about combinational transitions? Again, the other
vibrations have to be involved in ring distortion vibrations (preferably C-C stretching
since it affects the Hamiltonian the most) and posses appropriate symmetry, which
9
can be achived by combining  18 e 2g  with totally symmetric combinations of the ring
distortion vibrations such as  2 a 1g  (992 cm −1 in the ground state and 923 cm −1 in the
electronically first excited state).
appears as a famous
As a result, a series of vibronic transitions 1
A B 2u 2 m0 18 10
benzene spectrum near 260 nm.
10