Which is the right structure of ${\mathrm{XeF}}_4$?

${\mathrm{XeF}}_4$ has ${\mathrm{sp}}^3{\mathrm{d}}^2$hybridization with 2 lone pairs so, it has square planar structure.

${\mathrm{XeF}}_4$ has a steric number of 6. There are two lone pairings and four bond pairs.
Fluoride's four atoms form a connection with its four electrons. All four of them are in the same plane. To reduce the repulsion between lone and bond pairs of electrons, the two non-bonded pairs of electrons (lone pair) are positioned along the perpendicular axis, resulting in a square planar molecular shape.

Hence, option C is correct.