# The hybridisation of $'Xe'$ atom in $XeF_4$ molecule is

$\begin{array}{1 1}(a)\;dsp^2&(b)\; sp^3\\(c)\;sp^3d&(d)\;sp^3d^2\end{array}$

Square planar $(XeF_4)$ with $sp^3d^2$ hybridisation.
Hence (d) is the correct answer.