13
$\begingroup$

I'm very confused about the differences between these methods. From my textbook, it states that the Hückel method only takes into account the π bonding interactions, while the extended Hückel method takes into account all the valence electrons. But what exactly does this do?

To my knowledge, the s, pz, px, and py orbitals are all orthogonal to one another and wouldn't mix anyways. Are shielding effects the only thing accounted for by extended Hückel theory?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Ever heard about sigma bonds? That's the difference. $\endgroup$
    – Ivan Neretin
    Commented Jan 31, 2018 at 5:03
  • $\begingroup$ s, pz, px, and py on the same site are orthogonal. However, for example, an s on one site is not orthogonal to a px on a different site. $\endgroup$
    – Ian Bush
    Commented Apr 30, 2018 at 6:35

2 Answers 2

Reset to default
11
+50
$\begingroup$

In the simple Hückel method (SHM) the basis set is limited to p orbitals. This set is limited in a great extent to pz orbitals which constraints the molecular plane to be the xy plane. Basically, you are limited to planar molecules.
The inclusion of all valence s and p orbitals in the extended Hückel method (EHM) naturally lifts the spacial constraints and you can work with non-planar molecules.

These are the two first differences between the two methods. The other points are:

SHM:

  • Orbital energies are limited to same-atom interactions, adjacent-atom interactions while all other interactions are 0.
  • Fock matrix elements are not actually calculated.
  • Overlap integrals are limited to 1 or 0.

EHM:

  • Orbital energies are calculated and vary smoothly with geometry.
  • Fock matrix elements are actually calculated.
  • Overlap integrals are actually calculated.

You can look up the derivation and steps for the implementation of these two methods in this book that I used as a reference:

Errol G. Lewars; Computational Chemistry, Introduction to the Theory and Applications of Molecular and Quantum Mechanics, Second Edition; Springer: 2011. DOI: 10.1007/978-90-481-3862-3

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Would you be willing to comment on the Fock matrix elements? More specifically, I assume it's still $h + 2J - K$, but are there simple approximations for J and K? $\endgroup$
    – pentavalentcarbon
    Commented May 3, 2018 at 14:39
  • 1
    $\begingroup$ @pentavalentcarbon that's where the empirical part of the method comes in. We obtain the whole matrix elements basically directly from experimental ionization potentials for their corresponding orbitals. $\endgroup$
    – Tyberius ♦
    Commented May 12, 2018 at 16:21
3
$\begingroup$

Pentavalentcarbon brought up an interesting point that I thought I could elaborate on: How are the matrix elements of this simplified Hamiltonian obtained? I'll basically just summarize the relevant parts of this Chem-Libre section on the Extended Hückel method

This is where the empirical part of this method comes into play. If you have the overlap matrix elements $S_{ij}$, then the elements of the Hamiltonian $H_{ij}$ can obtained using experimentally obtained values of a given orbital's ionization potential. The diagonal elements are simply taken to the negative of the ionization potential of the corresponding atomic orbital, $$H_{ii}=-\mathrm{IP}$$ The off-diagonal elements take only slightly more effort with $$H_{ij}=\frac{K}{2}(H_{ii}+H_{jj})S_{ij} \text{ , } i\not=j$$ $K$ is just a proportionality constant and a commonly used value is $1.75$, based on a study by Hoffman on the orbital energies of ethane.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Interesting. That expression is identical to what Q-Chem would call the Generalized Wolfsberg-Helmholtz guess, though its matrix elements are in the AO basis, and the diagonal elements are simply the diagonal from $T + V$. For other programs that can use EHM (Dalton comes to mind), I wonder what they are actually doing. $\endgroup$
    – pentavalentcarbon
    Commented May 12, 2018 at 21:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.