$X e F_{4}$ square planar whereas $C C l_{4}$ is tetrahedral because

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In $X e F_{4}$ , ‘Xe’ is $s p^{3} d^{2}$ hybridized but due to the presence of two lone pairs of Electrons shape is square planar where as in $C C l_{4}$ ‘ $C$ ’ is $s p^{3}$ hybridized

$X e$ is noble gas. whereas $C$ is non-metal

In both $X e F_{4}$ and $C C l_{4}$ the central atom is $s p^{3}$ hybridized

In $X e F_{4}$ . ‘Xe’ is $s p^{2}$ hybridised and in $C C l_{4}$ ‘ $C$ ’ is $s p^{3}$ hybridized

answer is C.

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Detailed Solution

In Xenon there are 8 outermost electrons present and hence can able to form 8 bonds with other atoms, whereas in carbon there are 4 outermost electrons Present and hence can able to form 4 bonds.
In ${XeF}_{4}$ , Xenon forms four bonds with fluorine and left two lone pair of electrons. Hence ${XeF}_{4}$ is sp³d² hybridized and exists in square planar.
In ${CCl}_{4}$ , carbon forms four bonds with chlorine and left zero lone pair of electrons. Hence ${CCl}_{4}$ is sp³ hybridized and exists in tetrahedral.