The hybridization of Xe in ${\text{XeF}}_6$ is

The formal for calculating hybridization at central atom is

 $\text{H}\text{.O}\text{=}\frac{1}{2}\left[V+M-C-A\right]$

Here, V is valency , M is number of mono-valency atoms attached to the central atom,

C means charge of cation and A means charge of anion. Then,

$\text{H}\text{.O}\text{=}\frac{1}{2}[8+6-0-0]=7\left(s{p}^{3}d32\right)$

In${\text{XeF}}_6$ , the hybridization of Xe is$s{p}^3{d}^3$  . There are 1 lone pair and 6 bond pairs present around the central atom, Xenon.