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This is an excerpt about Simple Hückel Theory from Elements of Physical Chemistry by Peter Atkins:

The first step that Hückel took was to ignore the $\sigma$-bonding framework and focus solely on the $\pi$ electrons. That is, he assumed that the atoms had taken up the positions they have in the actual molecule, then calculated the properties of the $\pi$ orbitals that matched that framework.

[...] we show in Derivation 14.3 that the Schrödinger equation for the orbitals $\hat H \psi = E\psi\;,$ then becomes the following pair of simultaneous equations for the coefficients: $$(H_{AA}- E)c_A- (H_{AB}- ES)c_B= 0\\ (H_{BA}- ES)c_A+ (H_{BB}- E)c_B= 0$$ ... . These equations are called the secular equations.

Hückel then made further approximations. First he neglected all the overlap integrals and set $S= 0$ wherever it appear. .....

I didn't quite get why he made this approximation. There must be some reason for neglecting the overlap integral, I suppose. Also, Atkins didn't mention why Hückel only made calculations for $\pi$ orbitals only; does his equations not work for $\sigma$-bonding?

So my questions are:

$\bullet$ What is(are) the reason(s) behind making the approximation of putting all the overlap integrals $0\;?$ After all, making $S= 0$ would mean there is no overlap.

$\bullet$ Why did Hückel work only on $\pi$ orbital? Doesn't his calculation work for $\sigma$-bonding?

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What is(are) the reason(s) behind making the approximation of putting all the overlap integrals 0?

I am not 100% sure here, but using such a simplification, the mathematics becomes significantly easier. In principle, 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. You do not have to back-transform your solution as is the case in the Hatree-Fock method.

Why did Hückel work only on π orbital? Doesn't his calculation work for σ-bonding?

Only $\pi$ electron molecular orbitals were included because these determine the properties of the molecules that Hückel was studying (conjugated molecules). The $\sigma$ electrons were ignored. This is known as $\sigma$-$\pi$ separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. As a consequence though, the Hückel method is in principle limited to planar systems.

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  • $\begingroup$ Taking $S=0$ implies an orthogonal basis set of AOs & this indeed makes the maths easy. But if there is no overlap between any of the constituting AOs, then how can MO form? $\endgroup$
    – user5764
    Dec 30, 2015 at 9:44
  • $\begingroup$ Even without overlap, by the diagonalisation of the Hamiltonian matrix you obtain eigenvectors and -values. Those vectors represent the molecular orbitals which are linear combinations of the atomic orbitals. For example, in the ethylene system you obtain $\psi_{0} = \frac{\phi_{0} + \phi_{1}}{\sqrt{2}}$ and $\psi_{1} = \frac{\phi_{0} - \phi_{1}}{\sqrt{2}}$ for the HOMO and LUMO, respectively. This is for instance demonstrated on the Wikipedia page concerning the Hückel method: en.wikipedia.org/wiki/…. $\endgroup$
    – Ivo Filot
    Dec 30, 2015 at 10:11
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    $\begingroup$ Use \langle \rangle to get $\langle \;\rangle$ rather than <> to get $<\;>\;.$ $\endgroup$
    – user5764
    Dec 30, 2015 at 13:35
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    $\begingroup$ It does not really have a physical interpretation. You should see it as something transcendental. Like a decoherence of the wave function if you will. That is the reason why diagonalization of the Hamiltonian matrix results in the correct energies of the electronic states: because you get rid of the off-diagonal elements. It is actually a very good question and deserves a more elaborate answer than what I can give in this comment. $\endgroup$
    – Ivo Filot
    Dec 30, 2015 at 14:14
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    $\begingroup$ It's probably worth mentioning that there is an extended Hückel method that includes both $\sigma$ and $\pi$ electrons. $\endgroup$
    – Geoff Hutchison
    Dec 30, 2015 at 14:18

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