Huckel method for cyclooctatetraene (C8H8)

[JorisL]

Hi

I'm studying for my course of quantum chemistry and I have some issues with applying the Hückel or thight binding method.

I can calculate the energies with ease using Frost-Musulin diagrams.
The problem is finding expressions for the molecular orbitals.
I can use symmetry considerations and the pairing theorem to find some of the orbitals.

This is an overview of what I got until now.

The energies are determined using a frost-musulin diagram.
They are $$E_1 = \alpha +2\beta\, ,E_2 = E_3 = \alpha +\sqrt{2}\beta\, ,E_4 = E_5=\alpha\, ,E_6=E_7 = \alpha -\sqrt{2}\beta\, ,E_8=\alpha -2\beta$$ where $$E_1,\, E_2\, and\, E_3$$ correspond to bondig orbitals.
$$E_4\, and\, E_5$$ correspond with nonbonding orbitals and the other energies are anti-bonding.

The molecular orbitals (MO's) are denoted by $$\phi_i$$ where the i corresponds with the energy. They consist of the atomic orbitals (AO's) denoted by $$\chi_i$$.

Because the lowest and highest energy orbitals must contain all symmetry, all Atomic orbitals should be included. The ring is perfectly symmetrical so they all have the same sign and coefficient for the lowest energy.

$$\phi_1 = \frac{1}{2\sqrt{2}}\left(\chi_1 +\chi_2 +\chi_3 +\chi_4 +\chi_5 +\chi_6 +\chi_7 +\chi_8\right)$$

Because of the pairing theorem I know the highest energy orbital has a similar form only with alternating signs.

$$\phi_8 = \frac{1}{2\sqrt{2}}\left(\chi_1 -\chi_2 +\chi_3 -\chi_4 +\chi_5 -\chi_6 +\chi_7 -\chi_8\right)$$

Know I will try to use some symmetry, the pairing theorem and some simple calculus to determine the other orbitals.

The image in the attachment shows how I named the atoms.

I assume for orbital $$\phi_2$$ that it has a nodal plane through atom 1 and 5. For $$\phi_3$$ I'll assume a nodal plane perpendicular to that i.e. through 3 and 7.

I know that $$E_{tot} = 2\cdot E_1 + 4\cdot E_2 + 2\cdot E_4 = 8\alpha +2\beta (2+2\sqrt{2})$$
Hence the sum $$\sum_{k,l}p_kl = 2+2\sqrt{2}$$.
Because of symmetry I state that the bond order is equal for all neighbours i.e. $$p_{12} = \frac{2+2\sqrt{2}}{8} = \frac{1+\sqrt{2}}{4}$$

But I can't get any further with this.
I can use normality of the orbital to say that
$$1 = \int dV \phi^*_2\phi_2 = c_1^2+c_2^2+c_3^2+c_4^2+c_5^2+c_6^2+c_7^2+c_8^2$$

Furthermore $$c_1 = c_5 = 0$$ because of the nodal plane.
Also $$c_2=c_4=-c_6=-c_8$$ and $$c_3=-c_7$$ from symmetry.

When I use this information about the coefficients and the bond order between 1 and 2, I find found that $$c_4 = 0$$ and consequently that $$c_3 = \infty$$

I have no idea what else I can use because when I want to use the fact that the electron density is uniform as well, I need to include the 2 non-bonding orbitals as well.

Am I looking in the right direction? Or should I start over in another way?

Joris

 

[cgk]

This may be a stupid question, but can't you just write down the whole tight binding Hamiltonian as a N x N matrix and then diagonalize it? Using Python/scipy or Matlab or your favorite numerical scripting language this could probably be done in about 20 lines of code.

 

[JorisL]

Of course I can do that. But the exercise is made in such a way that we have to use symmetry etc. And I can use the practice. I don't always find the correct symmetry element/operation I have to use.