Given borane's ($\ce{BH3}$) LUMO, which of these Lewis bases would have a more stabilizing interaction? In acid bases that means the higher HOMO and lower LUMO equals a stronger dative bond I believe.

So which would have a stronger interaction (greater net stabilization) with borane, trimethylphosphine or trimethylamine:

$$\ce{P(CH3)3 ~vs~ N(CH3)3}$$

Phosphorus is less EN and it's electrons reside in a higher energy 3p orbital, but doesn't it compensate for this by placing the electron pair an orbital with more "s" character? Also carbon is more EN than phosphorous drawing some strain.

I would say trimethylphosphine has the HOMO just because of it's higher orbitals and lower overall EN making donating a pair of electrons more favorable.

Am I right?

  • $\begingroup$ @Dave You mean an orbital with more s character, not a character, right? $\endgroup$
    – Dissenter
    Mar 4, 2015 at 1:58
  • $\begingroup$ If so I suppose you could check using bond angles ... although there is another issue at hand -the relative sizes of P and N $\endgroup$
    – Dissenter
    Mar 4, 2015 at 1:59

1 Answer 1


OK, first off, let's comment on the stability question. When making a Lewis adduct (and rationalizing with the frontier orbital approach), you want the resulting complex to be stable. That requires two possibilities:

  1. The HOMO of the Lewis Base ($\ce{NH3}$ or $\ce{PH3}$ here) is low (i.e., hard to remove an electron).
  2. The coupling between the acid and base orbitals is large.

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I want those electrons to be stabilized. Either (or both) approaches can work. So to correct your statement, you don't want a "high HOMO" or "low LUMO."

The catch, of course, is that a large splitting between the Lewis acid LUMO and Lewis base HOMO can lead to low coupling (poor orbital overlap).

Now, I went and ran some quick B3LYP/6-31G(d) calculations on both $\ce{N(CH3)3}$ and $\ce{P(CH3)3}$ because I couldn't find any quick comparisons in the textbooks I use.

The amine has a computed orbital energy of -5.64 eV, and the phosphine has a computed orbital energy of -5.98 eV. (Standard disclaimers about orbital energies being non-physical apply.)

This would suggest that the phosphine would form a stronger bond with a borane.

There's supporting evidence here: "Ab initio molecular orbital study of the substituent effect on ammonia and phosphine–borane complexes" Journal of Molecular Structure: THEOCHEM, (2004) 709(1–3), pp 103-107.

The complexation energies of $\ce{H3BXH_{$3−n$}F_{$n$}}~(\ce{X~$=$~N, P}; n=0–3)$ and the proton affinities of $\ce{XH_{$3–n$}F_{$n$}}$ compounds have been investigated at the G2(MP2) level of theory. The G2(MP2) results show that the phosphine complexes are more stable than the corresponding ammonia ones. ... The NBO partitioning scheme shows that the stability of the phosphine complexes was related to the hyperconjugation effect.

Basically, the phosphine complexes seem more stable because there is some back-donation of electrons from the borane into the phosphine. It's not just a simple frontier orbital picture - more orbital interactions come into play.

  • 1
    $\begingroup$ For some reason, that I don't know, you have to enclose one character (non-numeric) sub- and superscripts in the mhchem environment, i.e. $\ce{F_n}$ produces $\ce{F_n}$, while $\ce{F_{n}}$ produces $\ce{F_{n}}$. $\endgroup$ Mar 6, 2015 at 3:05

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