$\mathrm{XeF}_4$ is square planar where as $\mathrm{CCl}_4$ is tetrahedral because

1. in $\mathrm{XeF}_4$, ' $\mathrm{Xe}$ ' is $s p^2$ hybridised and in $\mathrm{CCl}_4$ ' $\mathrm{C}$ ' is $s p^3$ hybridised
2. in both $\mathrm{XeF}_4$ and $\mathrm{CCl}_4$ the central atom is $s p^3$ hybridised
3. in $\mathrm{XeF}_4$, ' $\mathrm{Xe}$ ' is $s p^3 d^2$ hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in $\mathrm{CCl}_4$ 'C' is $s p^3$ hybridised
4. $X e$ is noble gas, whereas $C$ is a non-metal

Solution

$\mathrm{XeF}_4$ is $s p^3 d^2$ hybridised due to $4 \sigma+2 l p$ and shape is square planar. $\mathrm{CCl}_4$ is $s p^3$ hybridised due to $4 \sigma$-bond.

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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