Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
In Hückel theory we are only interested in π systems, where $\mathrm p_z$ orbitals overlap. One of the approximations in Hückel theory is that the overlapping $\mathrm p_z$ orbitals are orthonormal:
$$ S_{ij} = \begin{cases} 1\quad i = j \\ 0\quad i \neq j \\ \end{cases} $$
If orbital overlap leads to bond formation, how can the overlap integrals, between $\mathrm p_z$ orbitals, be equal to 0? Maybe I am missing something obvious, but doesn't a π-bond demand non-zero overlap between p-orbitals?
Hückel theory would still work if there would be overlap.
In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system.
There is the opposite view in Extended Huckel Methods whereby the assumption is that the overlap entries depend upon the off-diagonal elements. In ordinary Huckel theory, the diagonal elements are represented as \begin{align} \alpha_{ii} &= \langle \psi_i | \hat{H} | \psi_i \rangle \\ \hat{H} &= -\frac{\hbar^2}{2}\nabla^2 + V \end{align}
But in the extended thoery, the offdiagonal elements are not denoted by $\beta_{ij} = \delta_{ij}$ but more as
\begin{align} \beta_{ij} &= - \frac{C}{2}\Big \langle \psi_i \Big |-\frac{\hbar^2}{2}\nabla^2 + V \Big | \psi_j \Big \rangle\\ &= - \frac{C}{2} \left(\alpha_{ii} + \alpha_{jj} \right)S_{ij} \end{align} Where $S_{ij}$ is the usual notation for the overlap between the corresponding atomic orbitals.
Here $C$ is a constant, usually taken equal to $7/4$ and has units of energy.
