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This question already has an answer here:

Phosphorus uses its $\ce{3s}$, three $\ce{3p}$ and one $\ce{3d}$ orbitals for its hybridization in $\ce{PCl5}$ and similar compounds.

However, shouldn't it use its $\ce{4s}$ rather than $\ce{3d}$ orbitals as the former is of lower energy?

I initially thought that this was not the case because hybridization cannot take place between orbitals of different principal quantum numbers, but then learnt that this indeed was possible, for example in coordination compounds with $\ce{dsp^2}$ hybridization such as $\ce{[Pt(NH3)2Cl2]}$.

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marked as duplicate by Mithoron, Jan, Tyberius, pentavalentcarbon, airhuff Sep 18 '17 at 18:49

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