Benzene Quantum Systems

7/23/2019 Benzene Quantum Systems

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Benzene:

1

2

3

4

5

6

We will solve Schodinger equation for this molecule by consider ing only p-orbitals of six carbons under the Huckel

approximation. Huckel approximation, though quite crude, provides very useful results.

Since a-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,

a-bonding is much stronger than ^-bonding, ^- and ^ D molecular orbitals lie within the a ? a D gap. Thus HOMO and

LUMO orbitals are either of ^,^D-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 a-bonding, i.e. molecule’s shape.

Huckel approximation can be summarized with following statements:

1) < p i | p j >= N ij - all overlap integrals are zero

2) < p i | H | p i >= J - diagonal elements equal atomic energies of carbon p z electrons

3) < p i | H | p j >= KÝneighborsÞ - the interaction energy is nonzero only for neighboring carbons.

Both, J and K, are negative and can be paraterized for a typical conjugated carbon and carbon-carbon bond. Such

a parameterization can be a lso introduced for hetero atoms, S, O, N

If we write the secular equation for benzene

det

J ? E K 0 0 0 K

K J ? E K 0 0 0

0 K J ? E K 0 0

0 0 K J ? E K 0

0 0 0 K J ? E K

K 0 0 0 K J ? E

= 0, : ?4K6+ 9K4ÝJ ? E Þ 2 ? 6K2ÝJ ? E Þ 4

+ ÝJ ? E Þ 6= 0.

The solutions can be found E = J + K, E = J ? 2K, E = J ? K, E = J + 2K. For an arbitrary cyclic molecule with identical

conjugated bonds, the solutions can be written as E = J + 2Kcos 2^n k and graphically represented as enegries

horizontally leveled at the corners of the appropriate polygon symmetrically placed on one of its corners:

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"Graphical solution" to cyclic aromatics

E = α + 2β cos(2πk/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 @ ^ , The character matrix is also shown on the top. C 2v ’s and a v ’s were chosen to go through

carbon atoms and C vv

2 ’s and a d ’s go in between.

D6h E 2C 6 2C 3 C 2 3C 2v 3C vv

2 i 2S 3 2S 6 ah 3ad 3av

A1 g 1 1 1 1 1 1 1 1 1 1 1 1

A2 g 1 1 1 1 ?1 ?1 1 1 1 1 ?1 ?1

B1 g 1 ?1 1 ?1 1 ?1 1 ?1 1 ?1 1 ?1

B2 g 1 ?1 1 ?1 ?1 1 1 ?1 1 ?1 ?1 1

E 1 g 2 1 ?1 ?2 0 0 2 1 ?1 ?2 0 0

E 2 g 2 ?1 ?1 2 0 0 2 ?1 ?1 2 0 0

A1u 1 1 1 1 1 1 ?1 ?1 ?1 ?1 ?1 ?1

A2u 1 1 1 1 ?1 ?1 ?1 ?1 ?1 ?1 1 1

B1u 1 ?1 1 ?1 1 ?1 ?1 1 ?1 1 ?1 1

B2u 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

@^ 6 0 0 0 ?2 0 0 0 0 ?6 0 2

@3 N 36 0 0 0 ?4 0 0 0 0 12 0 4

That representation can be reduced into irreducible for this point group:

@^ = B 2 g + E 1 g + A2u + E 2u

A bit faster approach is to recognize that D6h =D6 × C i and thus use a smaller group, D6, instead:

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D6 E 2C 6 2C 3 C 2 3C 2v 3C vv

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

E 1 2 1 ?1 ?2 0 0

E 2 2 ?1 ?1 2 0 0

@^ 6 0 0 0 ?2 0

That representation can be reduced into irreducible for this point group:

@^ = B 2 + E 1 + A 2 + E 2 which can be easily ass igned g and u subscr ipt based on the required negative character

with respect tooperation a h in D 6h. Thus @ ^ = B2 g + E 1 g + 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, a2u with energy J + 2K is

the lowest because both, J and K, are negative. The approach is handy but might be confusing when trying to find

those c1 and c2 coefficients to finish the construction.

b2g α − 2β

e2u α − β

e1g α + β

a2u α + 2β

c1

c1

-c1

-c1

c1

c1

-c1

-c1

c1

c2

-c1

-c2

-c1

c1

+

+ +

+

+

+

+ +

+

-

-

-

--

-

-c2

- -

c1

c1

c1

c1

c1 c1

c1

-c1-c1

-c1

-c2

b) More straight forward (also a bi t more tedious) approach is by using the projection operator:

P W => R

e WÝ RÞ DO R

Again, we’l l resort to a smaller group, D6, instead of D6h. Then projection operators would look shorter :0). For

example: P A 2 = 1O E + 1OC 6 + 1OC 6?1 + 1OC 3

?1 + 1OC 3 + 1OC 2 ? 1OC 2v Ý1Þ ? 1OC 2

v Ý2Þ ? 1OC 2v Ý3Þ ? 1OC 2

vv Ý1Þ ? 1OC 2vv Ý2Þ ? 1OC 2

vv Ý3Þ.

Using which on p 1 results in

P A 2 p1 = p 1 + p 2 + p 6 + p 3 + p 5 + p 4 + p 1 + p 5 + p 3 + p 2 + p 4 + p 6 = 2Ý p1 + p 2 + p 3 + p 4 + p 5 + p 6Þ

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After normalizing we obtain:S 4Ýa2uÞ = 1

6

Ý p1 + p 2 + p 3 + p 4 + p 5 + p 6Þ. The same way it can be dobe with others.

Some attention is required for doubly degenerate species. The projection operator for E 2 :

P E 2 = 2O E ? 1OC 6 ? 1OC 6?1 ? 1OC 3

?1 ? 1OC 3 + 2OC 2 .when applied to say p 1 gives:

P E 2 p1 = 2 p1 ? p 2 ? p 6 ? p 5 ? p 3 + 2 p4 .That gives only one symmetry adapted orbital,

S 5 = 1

2 3

Ý2 p1 ? p 2 ? p 6 ? p 5 ? p 3 + 2 p4Þ

Another has to be found by generating a similar one after applying P E 2 say to p2:

P E 2 p2 = 2 p2 ? p 3 ? p 1 ? p 6 ? p 4 + 2 p5

and making it or thogonal to S 5:

S 6 = 2 p2 ? p 3 ? p 1 ? p 6 ? p 4 + 2 p5 ? S 5 < S 5 |2 p2 ? p 3 ? p 1 ? p 6 ? p 4 + 2 p5 >=

= 2 p2 ? p 3 ? p 1 ? p 6 ? p 4 + 2 p5 ? 1

2 3

2

Ý2 p1 ? p 2 ? p 6 ? p 5 ? p 3 + 2 p4Þ ?6 =

= 3

2 Ý p2 ? p 3 + p 5 ? p 6 Þ

Analogously: P E 1 = 2O E + 1OC 6 + 1OC 6?1 ? 1OC 3

?1 ? 1OC 3 ? 2OC 2 and

P E 1 p1 = 2 p1 + p 2 + p 6 ? p 5 ? p 3 ? 2 p4 ; =>S 2 = 1

2 3

Ý2 p1 + p 2 ? p 3 ? 2 p4 ? p 5 + p 6Þ

P E 1 p2 = 2 p2 + p 3 + p 1 ? p 6 ? p 4 ? 2 p5 and

S 3 = 2 p2 + p 3 + p 1 ? p 6 ? p 4 ? 2 p5 ? 1

2 3

2

Ý2 p1 + p 2 ? p 3 ? 2 p4 ? p 5 + p 6ÞÝ6Þ =

= 3

2 p2 + 3

2 p3 ?

3

2 p5 ?

3

2 p6 = 3

2 Ý p2 + p 3 ? p 5 ? p 6 Þ

B2 g S 1 = 1

6

Ý p1 ? p 2 + p 3 ? p 4 + p 5 ? p 6Þ

E 1 g S 2 = 1

2 3

Ý2 p1 + p 2 ? p 3 ? 2 p4 ? p 5 + p 6Þ

S 3 = 1

2 Ý p2 + p 3 ? p 5 ? p 6 Þ

A2u S 4 = 1

6

Ý p1 + p 2 + p 3 + p 4 + p 5 + p 6Þ

E 2u S 5 = 1

2 3

Ý2 p1 ? p 2 ? p 3 + 2 p4 ? p 5 ? p 6Þ

S 6 = 1

2 Ý p2 ? p 3 + p 5 ? p 6 Þ

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 resembl ing the value of coefficient in the eigen function.

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a2u

e1g

e2u

b2g

The energies of these MOcan be found by evaluating appropriate < S i | H |S i >:

E(b2 g ) = < S 1 | H |S 1 >=< 1

6

Ý p1 ? p 2 + p 3 ? p 4 + p 5 ? p 6Þ| H | 1

6

Ý p1 ? p 2 + p 3 ? p 4 + p 5 ? p 6Þ >=

= 1

6

< p1 | H |Ý p1 ? p 2 + p 3-p 4 +p 5 ? p 6Þ > ? < p2 | H |Ý p1 ? p 2 + p 3 -p 4 +p 5-p 6

Þ > +

+ < p3 | H |Ý p 1 ? p 2 + p 3 ? p 4 + p 5-p 6

Þ > ? < p4 | H |Ý p 1-p 2 + p 3 ? p 4 + p 5 ? p 6

Þ > +

< p5 | H |Ý p 1-p 2 +p 3 ? p 4 + p 5 ? p 6Þ > ? < p6 | H |Ý p1 -p 2 +p 3-p 4

+ p 5 ? p 6Þ >

=

matrix element for highlighted orbitals are zero

= 1

6

< p1 | H |Ý p1 ? p 2 ? p 6Þ > ? < p2 | H |Ý p1 ? p 2 + p 3Þ > + < p3 | H |Ý? p2 + p 3 ? p 4Þ > ?

? < p4 | H |Ý p3 ? p 4 + p 5Þ > + < p5 | H |Ý? p4 + p 5 ? p 6Þ > ? < p6 | H |Ý p1 + p 5 ? p 6Þ >

=

= 1

66 J ? 2K = J ? 2K

Energies of other orbitals are calculated the same way:

E Ýb2 g Þ S 1 = 1

6

Ý p1 ? p 2 + p 3 ? p 4 + p 5 ? p 6Þ

E Ýe1 g Þ S 2 = 1

2 3

Ý2 p1 + p 2 ? p 3 ? 2 p4 ? p 5 + p 6Þ

S 3 = 1

2 Ý p2 + p 3 ? p 5 ? p 6 Þ

E Ýa2uÞ = S 4 =

1

6 Ý p1 + p 2 + p 3 + p 4 + p 5 + p 6Þ

E Ýe2uÞ S 5 = 1

2 3

Ý2 p1 ? p 2 ? p 3 + 2 p4 ? p 5 ? p 6Þ

S 6 = 1

2 Ý p2 ? p 3 + p 5 ? p 6 Þ

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a1u

2β : a1u

= 0.408(p1+ p

2+p

3+p

4+p

5+p

6)

α

e1g

β : e1g

(1) = 0.289(2p1+ p

2−p

3−2p

4−p

5+p

6); e

1g

(2) = 0.5( p2+p

3−p

5−p

6)

b2g

-2β : b2g

= 0.408(p1− p

2+p

3−p

4+p

5−p

6)

e2u

-β : e2u

(1) = 0.289(2p1− p

2−p

3+2p

4−p

5−p

6); e

2u

(2) = 0.5( p2−p

3+p

5−p

6)

We see that MOenergies are symmetric with respect tovalue of J. All planar conjugated hydrocarbons with suchproperty 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.

The ground state configuration is Ýa2uÞ2Ýe1 g Þ

4=

æ X

1 A1 g . 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 2K.

The lowest excited state configuration is Ýa2uÞ2Ýe1 g Þ

3Ýe2uÞ1 (with energy 2K above the ground state) from which

following states can be constructed:

Ýe1 g Þ × Ýe2uÞ = B1u + B2u + E 1u .each of which can be either a singlet or a triplet, i.e.:

b2g

e2u

e1g

a1u

b2g

e2u

e1g

a1u

b2g

e2u

e1g

a1u

X:1 A1g

1B2u

, 1B1u

, 1E1u

3B2u

, 3B1u

, 3E1u

At this level of comlexity we cannot choose which excited state is the lowest energy, 1 B1u , 1 B2u , or 1 E 1u (all we know

is that they each have corresponding triplet of lower energy 3 B1u < 1 B1u , 3 B2u < 1 B2u , and 3 E 1u < 1 E 1u). Wehave toanalyse other higher energy one-electron excited state configurations and find if there are any of the sam e

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 3K) are:Ýa2uÞ2Ýe1 g Þ

3Ýb2 g Þ1 and Ýa2uÞ

1Ýe1 g Þ4Ýe2uÞ

1 .

Neither of them has a proper symmetry. The former makes: e1 g × b 2 g = e 2 g , while the latter: a2u × e 2u = e 2 g

6



E2g

E2g

B2u

3β 3β 4β

We shall include even higher one electron excited state configuration with energy 4K. There is only one,

Ýa2uÞ1Ýe1 g Þ

4Ýb2 g Þ1 , which produces a 2u × b 2 g = b 2u. Thus, out of the three states, 1 B1u, 1 B2u , and 1 E 1u, only 1 B2u has

two configurations. Consequently, it will be the lowest energy state.

7



Since in the D 6h group the dipole moment components behave like E1u (x,y) and A2u (z), the only dipole allowed

transition is C 1 E 1u ?

æ X

1 A1 g which indeed shows a high extinction coefficient, while the lower energy transitions, 1 B1u

and 1 B2u, 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 evvve"v" =< fe v v v

D |W|fe vv v vv >=< fe v

D |W|fe vv >< fv v

D |fv vv >= M eve" < fv 1

v

D |fv 1vv >< f

v 2v

D |fv 2vv >.. .

The selection rules in this case are defined by the selection rules for electronic part M eve" and the Franck-Condon

factor, < fv v

D |fv vv >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:

Av = 0, ±1, ±2, ±3,...

For all other vibrations the totally symmetric integrand appears only for even difference in vibrational number, i.e.

when:

Av = 0, ±2, ±4,...

Incases 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:

H e = H e0

+ >i

/ H e/Q i

Q i +.. = H e0

+ H v

The excited state wavefunction f f 0 becomes accordinly mixed with other zero-order electronic states :

fe v = f f 0

+ >k c k fk

0

where the degree of mixing, ck

= <fk

0 | H v |f f 0

>

E f 0? E k 0

, depends on both, the vibronic coupling element < fk

0 | H v|f f

0> and the

energy separation between the states E f 0 ? E k

0 . The electronic transition moment becomes:

M eve" =< fe v |W|fe vv >=< f f 0 |W|f

e vv

0> +>

k c k < fk

0 |W|fe vv

0>

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 B1u

transition, beeing closer to the allowed E1u, is more intense than B2u. The energy factor E f 0 ? E k

0 is not the only

one responsible for nonzero c k , the other part , < fk 0 | H v |f f

0>, enforces additional selection rules:

< fk 0 | H v|f f

0>=

/ H e/Q i Q i =0

< fk 0 |Q i|f f

0>

or Av = ±1Ý±3,±5,...Þ for vibronicaly active mode, the mode for which the symmetry of @ fk 0

× @Q i matches that of the

ground stateæ X . The rules for other vibrational states would follow the rules of ordinary Franck-Condon factor

described above. Incase of benzene, vibronically allowed transition should be a 2u and e1u. For theæ A

1 B2u state it

translates to a reqire for either b 1 g b1 g × B2u = B2u or e2 g e2 g × B2u = E 1u . There are 12×3-6=30 vibrational

modes in benzene, which for the ground state would be realized in 20 fundamental frequencies:2a1 g + a 2 g + a 2u + 2b1u + 2b2 g + 2b2u + e 1 g + 3e1u + 4e2 g + 2e2u. From these modes only e2 g satisfy the selection rules

(X 15Ýe2 g Þ through X 18Ýe2 g Þ) for vibronic transition. By symmetry all four of them can contribute and those that have

the highest value of / H e/Q i

contribute the most. C-H vibrations do not affect the energies of ^-electrons 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, X18Ýe2 g Þ, 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:æ A

1 B2u18 0

1 - transition from vibrational ground state in theæ X

1 A1 g to

æ A

1 B2u at vibrationally excited state with v"=1 at

X 18Ýe2 g Þ.

Another possibility is to observe a hotæ A

1 B2u18 1

0 transition from the vibrationally excited with v’=1 at X 18Ýe2 g Þæ X

1 A1 g to

theæ A

1 B2u and v"=0 at X 18Ýe2 g Þ. The latter transition is very temperature dependent (hot). What about combinational

transitions? Again, the other vibrations have to be involved in r ing distortion vibrations (preferably C-C stretching

since it affects the Hamiltonian the most) and posses appropriate symmetry, which can be achived by combining

X 18Ýe2 g Þ with totally symmetric combinations of the ring distortion vibrations such as X 2Ýa1 g Þ (992 cm?1 in the ground

state and 923 cm ?1 in the electronically first excited state).

As a result , a series of vibronic transitions æ A

1 B2u20

m18 01 appears as a famous benzene spectrum near 260 nm.

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Benzene Quantum Systems