Question

$Xe{F}_4$ is square planar where as $CC{l}_4$ is tetrahedral because

• A. In $Xe{F}_4,‘Xe‘$ is $s{p}^2$ hybridised and in $CC{l}_4‘C^{\prime }$ is $s{p}^3$ hybridised
• B. In both $Xe{F}_4$ and $CC{l}_4$ the central atom is $s{p}^3$ hybridised
• C. In $Xe{F}_4,‘X{e}^{\prime }$ is $s{p}^3{d}^2$ hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in $CC{l}_4{}^{\prime }{C}^{\prime }$ is $s{p}^3$ hybridised
• D. Xe is a noble gas. whereas C is a non-metal

Answer 1

Answer: (C)

$Xe{F}_4$ is $s{p}^3{d}^2$ hybridised due to $4\sigma +2lp$ and shape is square planar.

$CC{l}_4$ is $s{p}^3$ hybridised due to $4\sigma$ -bond.