Question

Hybridization and shape of $Xe{F}_4$ is:

• A. $s{p}^{3}d$, trigonal bipyramidal
• B. $s{p}^3$, tetrahedral
• C. $s{p}^3{d}^2$, square planar
• D. $s{p}^3{d}^2$, hexagonal

Answer 1

The correct option is C $s{p}^3{d}^2$, square planar

Structure of $Xe{F}_4$ is shown in figure. Xe has 4 bond pair along with 2 lone pairs in the $Xe{F}_4$. Thus, the hybridisation of $Xe$ is ${sp}^3{d}^2$. Now since it contains two lone pairs it will show square planar geometry.