Question

Out of the following how many have Xe in $s{p}^{3}d$ hybridization? $Xe{F}_2,Xe{F}_4,Xe{F}_6,XeO{F}_2,Xe{O}_3$ ?

Answer 1

$Xe$ has an electronic configuration of $5s^{2}5p^6$
In $Xe{F}_2$, there are two bonding pairs and three lone pairs of electrons. So acc. to VSEPR theory, the structure would be trigonal bypyramidal to avoid electron pair repulsions.$\left(s{p}^{3}d\right)$
In $Xe{F}_4$, the structure would be square planar. There are four pairs of bonding electrons and two lone pairs in the molecule. The hybridisation would be $s{p}^3{d}^2$
$Xe{F}_6$ has a pentagonal bipyramidal geometry due to $s{p}^3{d}^3$ hybridization. One trans position is occupied by a lone pair giving a distorted octahedral shape.
$XeO{F}_2$ has a trigonal bipyramidal geometry due to $s{p}^{3}d$ hybridization. Two equatorial positions are occupied by lone pairs of electrons giving a Tshape to molecule.
$XeO{F}_4$ has a octahedral geometry due to $s{p}^3{d}^2$ hybridization. One trans position is occupied by a lone pair giving pyramid shape to the molecule.
Thus $XeO{F}_2$ & $Xe{F}_2$ have $s{p}^{3}d$ hybridization.