Question

The correct geometry and hybridisation for $Xe{F}_4$ is:

• A. Octahedral, $s{p}^3{d}^2$
• B. Trigonal bipyramidal, $s{p}^{3}d$
• C. Planar triangle, $s{p}^3{d}^3$
• D. Square planar, $s{p}^3{d}^2$

Answer 1

The correct option is A Octahedral, $s{p}^3{d}^2$

$Xe{F}_4$ is $A{B}_4{L}_2$ type in which the xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).

So, the hybridisation is $s{p}^3{d}^2$.
Geometry is octahedral and shape is square planar.

Hence, option (a) is the correct answer.