Question

In the $Xe{F}_4$ molecule, the $Xe$ atom is in the:

• A. $s{p}^2$-hybridisation state
• B. $s{p}^3$-hybridisation state
• C. $s{p}^{2}d$-hybridized state
• D. $s{p}^3{d}^2$-hybridized state

Answer 1

The correct option is D $s{p}^3{d}^2$-hybridized state

$Xe{F}_4$ molecule has 4 bond pairs and 2 lone pairs in the molecule. So all the electron pairs surrounding the atom are 6 i.e lone pairs and bond pairs. Hence hybridization is $s{p}^3{d}^2$

The geometry of the molecule is octahydral.

But the shape of the molecule is square planar shape, due to lone pair- bond pair arrangement.

Hence option $D$ is correct.