In $PCl _{5}$, number of hybrid orbitals,
$H=\frac{1}{2}[V+X+A-C]$
[where, $V=$ number of valence electrons
$X=$ number of monovalent atoms
$A$ and $C=$ charge of cation or anion]
$H =\frac{1}{2}[5+5+0-0]$
$=5$
Thus, the hybridisation is $s p^{3} d$.