XeF_4 is square planar whereas CCl_4 is tetra | vedclass

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Question

(C) $XeF_4$ has $Xe$ as the central atom with $4$ bond pairs and $2$ lone pairs,resulting in $sp^3 d^2$ hybridisation. The presence of $2$ lone pairs

Vedclass · Accepted Answer

in $XeF_4$,'$Xe$' is $sp^3 d^2$ 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

Vedclass · Answer

(C) $XeF_4$ has $Xe$ as the central atom with $4$ bond pairs and $2$ lone pairs,resulting in $sp^3 d^2$ hybridisation. The presence of $2$ lone pairs leads to a square planar geometry.$CCl_4$ has $C$ as the central atom with $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridisation and a tetrahedral geometry.

$XeF_4$ is square planar whereas $CCl_4$ is tetrahedral because

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