Question

The correct structure of $Xe{F}_4$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$Xe{F}_4$ has $s{p}^3{d}^2$ hybridisation. Hence, it will have octahedral geometry. When we arrange all the four $F$ atoms along with the 2 lone pairs in that geometry, it is found to have a square planar shape. There are four pairs of bonding electrons and two lone pairs in the molecule. The two lone pairs will be oriented at ${180}^{\circ }$ to each other to minimize lone pair-lone pair repulsions.