Explanation

(c)
In $XeF_{4},$ the central atom, Xe, has eight electrons in its outermost shell. Out of these four are used for forming four $\sigma$-bonds with F and four remain as lone pairs.
$\therefore\ {XeF}_{4}\ \Longrightarrow 4\ \sigma\ bonds + 2\ lone\ pairs$
$\Longrightarrow 6\ bybridised\ orbitals,\ i.e.,\ {sp}^{3}d^{2}$ hybridisation
Since, two lone pairs of electrons are present, the geometry of $XeF_{4}$ becomes square planar from octahedral.
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