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It is well known that the tert-butyl cation's empty 2p orbital on the central carbon is stabilised by hyperconjugation from neighboring C-H bonds. This would mean that the central 2p "lone pair" is antibonding in character in the tert-butyl cation; the same would go for the empty boron 2p orbital of trimethylborane and filled nitrogen 2p orbital of trimethylamine.

However, when I did NBO calculations using the minimal STO-3G basis set on the aforementioned three species (note that the lattermost species is "two electrons too rich" of being isoelectronic with the first two) on Gaussian, I got a surprising answer- the empty "lone pair"s of both the tert-butyl cation and the trimethylborane molecule did turn out to be antibonding in character, but the filled "lone pair" of the trimethylamine molecule was not antibonding in character.

Defining electron-preciseness of a molecule being the statement, if true, that "all molecular orbitals of molecule in question, up to (and including) the highest lying orbital(s) that are not antibonding, are doubly filled and all other orbitals are empty", this would mean that both the trimethylborane molecule and the tert-butyl cation are electron-precise, while the trimethylnitrenium cation, being two electrons poor of the electron-precise trimethylamine molecule, would actually be electron-decifient. Since two isoelectronic species must be either both electron-precise, both electron-rich or both electron-deficient, this means that the NBO calculation method I used (Gaussian, STO-3G) does not accurately predict whether a "lone pair" is of antibonding character or not.

My question now proceeds; which is electron-precise (via the above EHM-based definition) of the following species, trimethylborane or trimethylamine?

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  • $\begingroup$ You may also consider asking this in matter modelling SE, which is not to say that it is off-topic here. $\endgroup$
    – S R Maiti
    Jul 17 at 13:44
  • $\begingroup$ Did you account for the difference in geometry between the molecules? The trimethylnitrenium cation is presumably trigonal planar, whereas the trimethylamine is trigonal pyramidal, so the molecular orbitals will be different. Just removing two electrons from the trimethylamine orbital diagram doesn't get you the MO diagram of the cation. $\endgroup$
    – Andrew
    Jul 19 at 15:23
  • $\begingroup$ I don't think that would create changes in the (anti)bondingness of orbitals, since e.g. the trigonal pyramidal trifluoromethanide ion is still 2 electrons rich of being electron-precise. $\endgroup$ Jul 22 at 14:34
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    $\begingroup$ If you say to a quantum chemist that the calculation was STO-3G they'll just laugh - the results are not trustworthy. I suggest, before you leap to any complicated conclusions, that you redo the calculations with something a bit more believable. And what level of theory are you using? Hartree-Fock? DFT? Something else? $\endgroup$
    – Ian Bush
    Jul 22 at 16:47
  • $\begingroup$ Hartree-Fock is what I used. $\endgroup$ Jul 22 at 22:14

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