Home › JEE Advanced › Chemistry › Chemical Bonding ›  What is the hybridisation of the central P atom in $\mathrm{PCl_5}$?
A$sp^2$
B$sp^3d^2$
C$sp$
D$sp^3d$
Answer & Solution
Correct answer: D. $sp^3d$
P in $\mathrm{PCl_5}$ has five bonded chlorines and no lone pair. That gives five electron domains. The hybridisation needed is $sp^3d$ (one $s$ + three $p$ + one $d$ orbital = five hybrid orbitals). The molecular geometry is trigonal bipyramidal.
$sp^3d^2$ would give six domains and an octahedral geometry, which is what you see in $\mathrm{SF_6}$, not $\mathrm{PCl_5}$.
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