I have done a Hartree-Fock (HF) calculation on a dioxetane molecule with one methyl group on one of the C-atoms (Fig 1), with minimal basis (using Molcas). I want to identify the active space that will be used in later CASSCF calculations on the same molecule. The interesting molecular orbitals (MO's) for this active space are the following ones: $\sigma_{CC}$, $\sigma_{CO}$, $\sigma_{CO}^{'}$, $\sigma_{OO}$, $\sigma_{CC}^*$, $\sigma_{CO}^*$, $\sigma_{CO}^{*'}$, $\sigma_{OO}^*$, $n_{OO}$, and $n_{OO}^{'}$. So basically all the ring bonding and anti-bonding MO's, plus the two non-bonding MO's on the oxygen atoms.

The CO MO's are not localized for just one CO pair, but rather they both exist at the same time in an "out-of-phase" and "in-phase" version. These are given in Figs 2-5. The same goes for the non-bonding MO's; they both exist in an out-of-phase and in-phase version, given in Figs 6 and 7. Then there are only the CC and OO MO's left. In the same calculation for 1,2-dioxetane (zero methyl groups, only H atoms on the carbons), I find separate CC and OO MO's. However, for this one-methyl dioxetane, it looks like the CC and OO MO's are combined, giving out-of-phase and in-phase version of both the bonding and anti-bonding. My issue is that the MO's look a bit ambiguous, in that some of the lobes are not where I expected them to be. These MO's are given Figs 8-10.

For it to be a $\sigma$ orbital, the electron density should be in the bond axis between the atoms. However, in Figure 8 this is not the case. It's like the same-phase lobes attract each other and overlap instead inside the ring. Is it correct to interpret this as the CC/OO in-phase bonding MO? Figure 10 looks more like the $\sigma_{OO}^*$ than the in-phase $\sigma_{CC/OO}^*$, as the orientation of the lobes on the C atoms seem strange to me. Especially the MO in Figure 11 is not obviously classified, in my opinion. The OO contribution seems very small, and the lobe orientations on the oxygens are not what I would expect for an anti-bonding $\sigma$ orbital.

However, there are no other candidates for these MO's, so I'm left with the current classification, even if it feels a bit incorrect. Perhaps this is a common issue in analyzing molecular orbitals in computational chemistry.

P.S. I apologize for the bad figure qualities, I don't know how to visualize molden files in any other way.


Figure 1. The molecule

CO in phase

Figure 2. CO bonding in phase

CO out of phase

Figure 3. CO bonding out of phase

CO anti in phase

Figure 4. CO anti-bonding in phase

CO anti out of phase

Figure 5. CO anti-bonding out of phase

n in phase

Figure 6. Non-bonding in phase

n out of phase

Figure 7. Non-bonding out of phase

CC OO in phase

Figure 8. CC/OO bonding in phase

CC OO out of phase

Figure 9. CC/OO bonding out of phase

CC OO anti in phase

Figure 10. CC/OO anti-bonding in phase

CC OO anti out of phase

Figure 11. CC/OO anti-bonding out of phase

  • 2
    $\begingroup$ Generating neat orbitals/surfaces from molden/wfn-files $\endgroup$ Commented Oct 27, 2016 at 17:53
  • $\begingroup$ Avogadro does not render it correctly. The molecule is invisible, and weird layers appear. The orbitals themselves seem okay, though. $\endgroup$
    – Yoda
    Commented Oct 27, 2016 at 18:37
  • $\begingroup$ I know, that's also my problem with Avogadro. But there are other programs than avogadro. :) $\endgroup$ Commented Oct 27, 2016 at 18:46
  • 1
    $\begingroup$ I found out that wxMacMolPlt worked quite well. But the software part is kind of beside the point of the question :P $\endgroup$
    – Yoda
    Commented Oct 31, 2016 at 6:11
  • 1
    $\begingroup$ @Martin-マーチン Oh Molden... Overlooked by more sophisticated programs but it will forever be a useful workhorse. $\endgroup$ Commented Apr 12, 2018 at 1:23

1 Answer 1


Natural bond orbital (NBO) methods may help you to localized the orbitals of your system after a post-HF calculation. I'm not familiar with Molcas but I know that several computational chemistry softwares have NBO methods implemented. Take a look at this simple explanation that uses MolPro:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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