(Cross-posting from Chemistry.SE due to recommendation that this be posted here; see comment section)

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?

  • $\begingroup$ Since you cross posted, you should also include a link to this post on the Chem SE one. This will ensure that if someone comes across that post later on, they will see it got answered here. $\endgroup$
    – Tyberius
    Commented Sep 8, 2021 at 2:00

1 Answer 1


Note that there is no absolute distinction between non-bonding orbitals and bonding (or antibonding) orbitals, especially in multiatomic molecules, because the orbitals that are generally considered as non-bonding still more or less mix with the orbitals of neighboring atoms. For example, the oxygen lone pairs of water are usually regarded as perfectly non-bonding orbitals, but they are not, since they mix with the hydrogen 2s and 2p orbitals.

As NBO is designed for arbitrary basis sets (not just minimal ones), to meaningfully allow for non-bonding orbitals, one must define non-bonding orbitals as orbitals whose bonding and antibonding characters are below a certain non-zero threshold. Thus the lone pair of trimethylamine is classified as non-bonding, because its antibonding character is below the default threshold.

  • $\begingroup$ So the LUMO of trimethylborane is also nonbonding, may I ask? $\endgroup$ Commented Sep 7, 2021 at 4:46
  • 1
    $\begingroup$ Yes, your results indicate that the antibonding character of the LUMO of trimethylborane must be higher than the default threshold of the NBO program, while the HOMO of trimethylamine does not make it to the threshold $\endgroup$
    – wzkchem5
    Commented Sep 7, 2021 at 8:22

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