Question

$Xe{F}_4$ is square planar while $Xe{F}_6$ has a distorted octahedrall structure. What is the correct explanation for this observation?

• A. Both molecules have one lone pair of electrons
• B. Both molecules have two lone pairs of electrons
• C. $Xe{F}_4$ does not have any lone pair of electrons; $Xe{F}_6$ has one lone pair of electrons on $Xe$
• D. $Xe{F}_4$ has two lone pairs of electrons on $Xe;Xe{F}_6$ has one lone pair of electrons on $Xe$

Answer 1

Answer: (D)

$Xe{F}_4$ is square planar and $s{p}^3{d}^2$ -hybridised.



$4\sigma +2lp=6$ -hybrid orbital

$s{p}^3{d}^2$ -hybridisation

$Xe{F}_6$ has distorted octahedral structure and $s{p}^3{d}^3$ -hybridised.



$6\sigma +1lp=7$ -hybrid orbital

$\left(s{p}^3{d}^3\right)$ (distorted octahedral)