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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?

1 Answers1

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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.

Ivo Filot
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  • 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? –  Dec 30 '15 at 09:44
  • 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: https://en.wikipedia.org/wiki/H%C3%BCckel_method#H.C3.BCckel_solution_for_ethylene. – Ivo Filot Dec 30 '15 at 10:11
  • so, what does resonance integral being zero imply? –  Dec 30 '15 at 10:16
  • The resonance integrals are the (off-diagonal) integrals of neighbouring atoms. If those are zero as well, then you were correct that you cannot find any MO because every MO will then simply be an AO and all these MO/AO will have the same energy. (and your system is no longer conjugated) – Ivo Filot Dec 30 '15 at 10:27
  • In those examples, $S_{12}= S_{21}= 0$ meaning there is no overlap between atom 1 and 2; still $H_{12}= H_{21}\ne 0= \beta$; I'm not getting that- if there is no overlap which is evident from $s_{12}$ being zero, how can there be non-zero $H_{12}$ as it would mean there would be overlapping? I'm really not understanding how $S_{12}= 0$ and $H_{12}\ne 0$ go with each other. –  Dec 30 '15 at 13:20
  • Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij} = \delta_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle \phi_{i}|\hat{H}|\phi_{j} \rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a nonzero outcome. – Ivo Filot Dec 30 '15 at 13:33
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    Use \langle \rangle to get $\langle ;\rangle$ rather than <> to get $<;>;.$ –  Dec 30 '15 at 13:35
  • (thanks for mentioning about the mathjax commands) To give an example of my previous comments; have a look at this website: http://www.ivofilot.nl/posts/view/29/Introduction+to+Electronic+Structure+Calculations%3A+The+variational+principle. If you look at example 2, you see that $\psi_{1}$ and $\psi_{2}$ are orthogonal to each other so they have zero overlap. Yet the off-diagonal elements of the Hamiltonian matrix are nonzero. – Ivo Filot Dec 30 '15 at 13:47
  • You are right at all your points but still I'm not getting the physical insight; I'll post it as a different question where I can reflect my problem more clearly. Thanks, nevertheless for the answer; +1. –  Dec 30 '15 at 14:04
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    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. – Ivo Filot Dec 30 '15 at 14:14
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    It's probably worth mentioning that there is an extended Hückel method that includes both $\sigma$ and $\pi$ electrons. – Geoff Hutchison Dec 30 '15 at 14:18