Question

The correct geometry and hybridization for $Xe{F}_4$ are

• 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

Answer: (D)

$Xe{F}_4$ is $A{B}_4{L}_2$ 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.