Question

Incorrectly matched pair is:

• A. $Xe{O}_3$ - pyramidal
• B. $Xe{F}_4$ - tetrahedral
• C. $Xe{F}_6$ - disorted octahedral
• D. $XeO{F}_4$ - square pyramidal

Answer 1

Answer: (B)

In $Xe{F}_6$ there are $6$ bond pairs and one lone pair. Thus the shape of $Xe{F}_6$ is distorted octahedral.

In $XeO{F}_4,Xe$ is $s{p}^{3}d$ hybridized and the shape is square pyramidal.

In $Xe{O}_3$, $Xe$ is $s{p}^3$ hybridized and the shape is trigonal pyramidal.

In $Xe{F}_4,Xe$ is $s{p}^3{d}^2$ hybridized. It has $4$ bond pairs and two lone pairs. Thus, the shape is square planar.