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

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

The hybridization of $Xe{F}_4$ = number of lone pairs + number of sigma bonds around $Xe$ atom $=2+4=6$. The steric number 6 implies that hybridization is $s{p}^3{d}^2$ and hence the geometry is octahedral.