Why does trisilylphosphine $\ce{P(SiH3)3}$ (ICSD Entry: 72676, [1]) have a pyramidal structure while trisilylamine $\ce{N(SiH3)3}$ (ICSD Entry: 201428, [2]) is planar?

Crystal structures of trisilylamine and trisilylphosphine

I think since silicon has a vacant d-orbital, it would form a backbond, wherein phosphorus shares its lone pair, as is the case with trisilylamine. This would result in $\ce{sp^2}$ hybridization, i.e. a trigonal planar structure.

However, the crystal structure confirms the pyramidal $\ce{P}$ atom, which is also preserved in the gas phase [1]:

Although IR and Raman spectroscopic results (Davidson, Ebsworth, Sheldrick & Woodward, 1966) were consistent with trisilylphosphine being similar to trisilylamine in that the central atom was surrounded by a trigonal planar arrangement of $\ce{Si}$ atoms, electron diffraction studies (Beagley, Robiette & Sheldrick, 1968) established pyramidal geometry in the gas phase.


The crystal structure of trisilylphosphine may be contrasted with those of trisilylamine and trimethylamine. As $\ce{N(SiH3)3}$ has a planar heavy-atom skeleton and concomitantly lacks a stereochemically active lone pair (Barrow & Ebsworth, 1984), the absence of discrete intermolecular $\ce{Si\bond{...}N}$ contacts — seen in solid dimethylsilylamine (Blake, Ebsworth & Welch, 1984a) and chlorosilyl-N,N-dimethylamine (Anderson, Blake, Cradock, Ebsworth, Rankin & Welch, 1986) — is unsurprising. Like $\ce{P(SiH3)3},$ trimethylamine is clearly pyramidal $[\angle \ce{CNC} = 110.40(7)^\circ]$ but the absence of vacant d orbitals on the $\ce{C}$ atoms removes the possibility of these acting as acceptors and the crystal comprises isolated molecules (Blake, Ebsworth & Welch, 1984b).

J.D. Lee Concise Inorganic Chemistry [3, p. 77] also specifically mentions that $\ce{P(SiH3)3}$ has a trigonal pyramidal structure:

In case of $\ce{P(SiH3)3},$ the internuclear distance is large and $\ce{P}$ atom has its own vacant d orbital. Hence the tendency to donate lone pair is very less and it adopts pyramidal structure and $\ce{P}$ atom is $\ce{sp^3}$ hybridized.


  1. Blake, A. J.; Ebsworth, E. A. V.; Henderson, S. G. D. Structure of Trisilylphosphine, $\ce{P(SiH3)3},$ at $\pu{100 K}.$ Acta Crystallogr C Cryst Struct Commun 1991, 47 (3), 486–489. DOI: 10.1107/S0108270190009593.
  2. Barrow, M. J.; Ebsworth, E. A. V. Crystal and Molecular Structure of Trisilylamine at $\pu{115 K}.$ J. Chem. Soc., Dalton Trans. 1984, No. 4, 563. DOI: 10.1039/dt9840000563.
  3. Sudarsan Guha. J.D. Lee Concise Inorganic Chemistry, for JEE (Main & Advanced), 4th ed.; Wiley India Pvt. Ltd.: India, 2017 (reprint 2020). ISBN 978-81-265-9114-5.
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    $\begingroup$ also wikipedia explains the reasons for the geometry in P(CH3)3...en.wikipedia.org/wiki/Trimethylphosphine. Could possibly extrapolate to the trisilyl $\endgroup$
    – user129490
    Aug 1 at 17:01
  • $\begingroup$ @lazarus98 but since Carbon doesn't have any vacant d orbitals being a second period element and nor any lone pairs while forming trimethylphosphine I am unable to explain the reason why backbonding doesn't take place in trisilyl since that does have a vacant d orbital and phosphorus has a lone pair that it can use up $\endgroup$ Aug 1 at 17:05
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    $\begingroup$ Just read chemistry.stackexchange.com/questions/69568/… instead of spawning some "vacant d orbital" nonsense. $\endgroup$
    – Mithoron
    Aug 1 at 17:33
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    $\begingroup$ chemistry.stackexchange.com/questions/152243/… $\endgroup$
    – Mithoron
    Aug 1 at 17:39
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    $\begingroup$ @Lazarus98 Just read the linked answers. I dunno why anyone would even know some trisilylphosphine exist at all, and still buy into stuff deprecated at least 30 years ago. $\endgroup$
    – Mithoron
    Aug 1 at 17:46

1 Answer 1


trisilyl boron nitrogen and phosphorus

Let us analyze the three cases:

  1. $\ce{B(SiH3)3}$
  2. $\ce{N(SiH3)3}$
  3. $\ce{P(SiH3)3}$

We are going to look at the back-bonding aspect. It is expected that the structures will be more planer as the extent of back-bonding increases. In this case the $\ce{N(SiH3)3}$, $\ce{Si}$-$\ce{N}$-$\ce{Si}$ bond angle is 119.8 degrees. which is just 120 degrees meaning that the structure is, more or less, triangular planer.

First Observations


Boron has no lone-pairs to back-donate, or according to VSEPR theory, we expect it to be triangular planer, as predicted by ab-initio calculations.


Nitrogen has a lone pair of electrons that can back-bond with silicon. In this case the $\ce{Si}$-$\ce{N}$-$\ce{Si}$ bond angle is 119.8 degrees.


Phosphorous also has a lone pair of electrons that can back-bond with silicon. In this case the $\ce{Si}$-$\ce{P}$-$\ce{Si}$ bond angle is 101.8 degrees.

At RHF level of theory, not only is there more back-bonding in trisilylamine, it is close to triangular planer trisilylboron, or that the electorn pair is completely shared between $\ce{N}$ and silyl groups.

Back-bonding or absence of isolated p-orbitals on the central atom can lead to planarity.


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