The correct geometry and hybridization for XeF 4 are

Xenon is a noble gas that has 8 valence electrons. There are 4 Fluorine atoms attached or bonded with Xe, each of a single bond.
So, the number of lone pair or non-bonding electrons surrounding of xenon atom is $8–4=4$ (that means: two lone pair electrons).

Therefore,${\mathrm{XeF}}_4$ molecule has 4 bond pairs and 2 lone pairs in the molecule.

Thus, ${\mathrm{XeF}}_4 \left(2\mathrm{lp}+4\mathrm{bp}\right)$, has a square planar geometry, where the hybridization of $\mathrm{Xe}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.