$XeF_4$ is square planar whereas $CCl_4$ is tetrahedral because
- Ain $XeF_4$,'$Xe$' is $sp^2$ hybridised and in $CCl_4$ '$C$' is $sp^3$ hybridised
- Bin both $XeF_4$ and $CCl_4$ the central atom is $sp^3$ hybridised
- Cin $XeF_4$,'$Xe$' 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
- D$Xe$ is noble gas,whereas $C$ is a non-metal
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