The inversion barrier in NHX3 is approximately 5 kcal mol−1 and that of PHX3 is 35 kcal mol−1. This has well-known stereochemical consequences in that amines are not chiral whereas phosphines can be. How can the larger inversion barrier be explained?
Answer
Ammonia is the classic system for spX3 hybridisation save methane. The lone pair (and each of the \unicode[Times]{x3C3}-bonds) has almost 25~\% s-character which corresponds nicely to \ce{sp^3}. However, the whole system can also swing around, changing its hybridisation to \ce{sp^2} and back; a process during which the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.
Elements outside of the second period show a much smaller tendency to involve the s-orbital in bonding. The bonding orbitals only have an s-character of approximately 16~\%. This also means that the phosphorous lone pair has a much higher s-character of approximately 50~\%, while the bonding \unicode[Times]{x3C3}-orbitals have a larger p-character. (This also explains the much smaller \ce{H-P-H} bonding angle of approximately 90^\circ.) An orbital of high s-character has a long way to go to turn into a p-type orbital, and the three largely p-type \unicode[Times]{x3C3}-orbitals have an equally long way to go to give \ce{sp^2}-type orbitals so the interconversion and thereby the inversion of \ce{PH3} is strongly hindered.
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