13
$\begingroup$

I calculated some orbitals and I get quite strange results with orbital geometries I wouldn't expect from MO theory. A more experienced colleague just said: "That's always the case with big basis sets, just ignore those and search for the ones that make sense" which is a very unsatisfying answer and he couldn't explain what's the cause or what they mean. I also noticed that these occur more often/with lower energy when using HF but they also occur using DFT.

Can anyone explain what happens here and why?

Here is one example with the first two looking like I would expec.

The first 5 unoccupied orbitals using HF (6-311+G(d,p)) including energies in eV:

enter image description here

$\endgroup$
5
  • 6
    $\begingroup$ Don't put too much thought into virtual orbitals. They do not contribute to the Fock matrix, hence they don't have any effect on occupied orbitals or the ground state energy. $\endgroup$ – Feodoran Feb 9 '19 at 8:17
  • 2
    $\begingroup$ @Feodoran but they are used in (F)MO theory besides other applications, that's why we look at them after all. And that's why I would like to understand what's going on here and not just ignore it or don't think much about it. $\endgroup$ – DSVA Feb 9 '19 at 9:09
  • 3
    $\begingroup$ @DSVA That is why I said "don't put too much thought them". They are valid eigenfunctions of the Fock matrix, and can be used for interpretaions, e.g. the LUMO energy in Koopmans' theorem. But they are still virtual, as in not really there. Can you try what happens when you remove the diffuse basis functions? $\endgroup$ – Feodoran Feb 9 '19 at 13:01
  • $\begingroup$ Maybe they do exist. But if they do, we can ask: (1) how much of the electron density do these far-out lobes carry? (Probably less than it looks like in the drawing) and (2) do they affect the observedelectronic structure of the molrcule? (If they require antibonding interactions to form, then no). We may be observing a tempest in a teapot. $\endgroup$ – Oscar Lanzi Feb 9 '19 at 13:52
  • 1
    $\begingroup$ @OscarLanzi Virtual orbitals don't carry any of the electron density in HF and DFT calculations, and AFAIK they have no effect whatsoever on the ground-state electronic structure of the molecule. $\endgroup$ – hBy2Py Feb 12 '19 at 18:10
6
$\begingroup$

In addition to PAEP's answer, I want to shed some light on how much meaning virtual orbitals may have.

First of all, there is the whole discussion on whether orbital in general are meaningful or not. Without going into details here: they are an approximation and as such they may give meaningful interpretations occasionally. If those interpretations do not reflect reality, we need to go to more sophisticated theories.

Why do we even have virtual orbitals?

As already mentioned, they show up because in linear algebra (which we use to numerically solve the Hartree-Fock-Roothan-Hall equation), we get as many eigenfunction (MOs) as basis functions (AOs) are supplied. Since usually more AOs are supplied, than electrons are in the systems, we get unoccupied or virtual orbitals. And if we want to improve accuracy by adding more basis function (e.g. diffuse/augmented or polarization functions) we do get more virtual orbitals. (At this point one could ask what the additional virtual orbitals even mean. Probably not much, that is why I suggested to compare your result to a calculation without diffuse basis functions.)

Are the virtual optimized?

What happens during the HF algorithm is, that we iteratively optimize the MOs based on the current (approximate) wave function (or electron density, if that helps imagining it). In HF this corresponds to the Slater-determinant, which is constructed from all occupied MOs. So all MOs (occupied and virtual) are optimized, but the optimization is with respect to the occupied MOs only.

Another perspective would be to say that the occupied MOs are coupled together via the electron-electron interaction. Thus they somehow need to "work together" to minimize the ground state energy. The virtual orbitals however, do not contribute to that energy and are not coupled by the electron-electron interaction (because they are empty). Maybe one could say, that the virtual orbitals are not optimized with respect to each other. But maybe this statement does not really make sense, since due to the missing coupling there is nothing they could be optimized for. However, they are still valid eigenfunctions of the Fock operator and are therefore part of the orthonormal basis set of molecular orbitals.

Summary

So there is something well defined about virtual orbitals which may justify interpretations based on them. But this is more vague/not as rigid as for occupied orbitals. Therefore a meaningful interpretation of virtual orbitals may not work as well as for occupied orbitals (and even those are limited).

$\endgroup$
4
$\begingroup$

As already mentioned in the comments you have to be careful when interpreting the meaning of the virtual orbitals in an ab initio or DFT calculation. In the calculation the only orbitals that are optimized are the occupied orbitals while the virtual (unoccupied) orbitals give you a guess of the shape you expect from them if they were occupied. To look at it from a simple perspective, somehow you are using the LCAO method, you obtain as many MO as AO you are using in your basis set, but you only optimize the MO that are occupied.

There is a second point that may be relevant to your question. In many programs the default orbitals you get in your calculation are the canonical orbitals (delocalized orbitals) that even if occupied may not have an evident chemical interpretation on its own. For those purposes it may be better to use localized orbitals (e.g. NBO) that may be easier to interpret.

Two book recommendations:

  1. C. J. Cramer, Essentials of Computational Chemistry. Theories and Models, Wiley (2004).
  2. Frank Weinhold, Clark R. Landis, Discovering chemistry with natural bond orbitals, Wiley (2012).
$\endgroup$
4
  • 1
    $\begingroup$ @IanBush purposes vs. porpoises. I'm quite certain dolphins have little to do with these types of orbitals. || OP: Since you have overruled I'll leave it to you to fix it. $\endgroup$ – Martin - マーチン Jun 14 at 21:45
  • 1
    $\begingroup$ @Martin-マーチン In case you don't know it's a fairly common UK piece of word play - and I had assumed that that's what the OP was aiming at with the misspelling $\endgroup$ – Ian Bush Jun 15 at 7:36
  • 2
    $\begingroup$ @IanBush It might be my inner German talking, but I don't think this is the right place for word play; especially since a large part of our community does not speak English as their first language. That kind of confusing language let me to reject your edit in the first place. $\endgroup$ – Martin - マーチン Jun 15 at 18:21
  • $\begingroup$ Thanks for the comments. It is a little embarassing. Guess I was watching a David Attenborough program around the time I wrote the piece... $\endgroup$ – PAEP Jun 20 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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