 What is the hybridisation of the central P atom in $\mathrm{PCl_5}$?
- $sp^2$
- $sp^3d^2$
- $sp$
- $sp^3d$
Show answer and 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 $\mathrm{PCl_5}$.