## Sunday 7 January 2007

### Benzene's MOs

I am planning to take a few posts for covering the principles of quantum mechanics (with a slight mathematical touch). Today's post is some kind of introduction to get your attention (or tell you that you won't have to check this blog for the next days). I am going to outline the principles of MO theory on the example of benzene.

The images show a cross section through the π-MOs of benzene 75 pm above the molecules plane. When you look at them you have to consider that the wave function changes it's sign if you move to the other side of the plane.

The MOs were calculated according to the Hückel-theory. The approximation in this case is very good. There is no other possible way for LCAO because of symmetry considerations. The character table for the group D6h can be seen here.

b1g

 e1u e1u e2g e2g

a2u

The molecular orbitals are arranged according to their energy. The three lower ones are populated with two electrons each. The first striking observation you make is that the energy scheme is a hexagon just like the molecule. This is actually true for all annullenes and the basis of Hückel's rule. I will talk about that later.

The next thing you notice is that you get an extra nodal plane each time you move up a step. This can be understood with the fact that wave functions have to be orthogonal. In general the energy is higher the more nodal planes you have. The first explanation would be that nodal planes mean more curvature and more curvature means higher kinetic energy. This is actually not true because this would violate the virial law, stating that
-T = V/2 = E
(T ... kinetic energy, V ... potential energy, E ... resulting energy, E = T + V).
The virial law states that higher kinetic energy corresponds to lower energy (just like a satellite travelling close to the earth will need high kinetic energy not to crash). In fact the orbitals are contracted but you don't get that from LCAO.

The next thing to consider is that each set of degenerate orbitals has to be a representation of the symmetry group. For non degenerate orbitals this means that they have to comply with all symmetry operations (real orbitals are symmetrical or antisymmetrical). For degenerate orbitals it means that you don't leave the subvectorspace when you apply a symmetry operator.

Any linear combination of degenerate orbitals is an orbital with the same energy eigenvalue. Therefore any orthogonal set of orbitals can be chosen to represent the eigenvalue. It makes sense to use the most symmetrical orbitals. For degenerate orbitals it is only possible that they comply with any commutative subgroup of the molecules symmetry group. In this case this is D2h. C6h would be commutative, too. But it has complex eigenvectors.

Maybe that was too much already for an easy reading blog. It just amazes me how much you know from a little bit of symmetry group theory and linear algebra without having solved any differential equation.

Ψ*Ψ said...

We spent a nice long while on this in a phys org class I took last semester. What you're posting is already more interesting, more in depth, and I can pretty well assume you won't screw up the math. (It was a pretty awful class.)

Felix said...

thanks, nice to hear that

there is no possibility to srew up math since everything is perfectly logical ;)

Ψ*Ψ said...

Well, you'd think so...but after a while, lots of organikers seem to lose it. I think it has to do with solvent inhalation, maybe?

Felix said...

poor organikers

but I have to admit that I am able to perform that impossible task of screwing up math often enough, too

Anonymous said...

1) you have ordered the energies of the 2 degenerate states incorrectly (E1g<E2u)

2) you have the wrong inversion symmetry label on the E1 orbitals (they should be E1g, not E1u).

3) you have the wrong symmetry label on the highest energy orbital (should be B2g not B1u)

Look it up. this is all over the net.

A quick simple search gives the following energy ordering:

B2g
0.3480 Ha
E2u
0.1322 Ha
E1g
–0.3396 Ha
A2u
–0.5073 Ha

Felix said...

thanks, it is true that the lower energy pair of orbitals should have g parity. I changed that.the question about "1" or "2" probably depends on the coordinate system. there is no strict way to choose which one is the C_2' and C_2'' and then the sigma_d and sigma_v are adjusted accordingly.