Question

$XeF_4$ is square planar where as $CCl_4$ is tetrahedral because

• A. In $XeF_4, �Xe�$ is $sp^2$ hybridised and in $CCl_4 \, �C'$ is $sp^3$ hybridised
• B. In both $XeF_4$ and $CCl_4$ the central atom is $sp^3$ hybridised
• C. In $XeF_4, �Xe'$ is $sp^3d^2$ hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in $CCl_4 \, 'C'$ is $sp^3$ hybridised
• D. Xe is a noble gas. whereas C is a non-metal

Answer 1

Answer: (C)

In $XeF_4, �Xe'$ is $sp^3d^2$ hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in $CCl_4 \, 'C'$ is $sp^3$ hybridised

Answer 2

$XeF _{4}$ is $s p^{3} d^{2}$ hybridised due to $4 \sigma+2 l p$ and shape is square planar.

$CCl _{4}$ is $s p^{3}$ hybridised due to $4 \sigma$ -bond.