The hybridisation of any compound can be calculated as,

$\mathrm{X}=\frac{1}{2}$

[Number of Valence electrons at central atom

+ Number of mono valent atoms attached to central atom+total negative charge on the molecule-total positive charge on the molecule]

X = number of hybrid orbitals

$\therefore$ For ${\mathrm{XeF}}_4\left(\mathrm{X}\right)=\frac{1}{2}(8+4-0)=6$

$\therefore$ The hybridisation is $s{p}^3{d}^2$