Question

The hybridization of $Xe$ and the number of lone pairs of electrons on it in $Xe{F}_6$ are:

• A. $s{p}^3{d}^2,0$
• B. $s{p}^3{d}^2,2$
• C. $s{p}^3{d}^2,1$
• D. $s{p}^3{d}^3,1$

Answer 1

Answer: (D)

$s{p}^3{d}^3,1$

As the Xe is attached to the 6 F atoms by the 6 sigma bonds, and it has 1 one pair, on Xe, so total hybrid orbitals are $7$.

Hence, hybridisation = $s{p}^3{d}^3$,

Lone pair on $Xe=1$.

Hence, option $D$ is correct.