Xenon tetrafluoride ( XeF_ 4 ) is a square planar molecule. The...

The central atom Xenon (Xe) has 8 valence electrons.

In $XeF_{4}$, Xe forms 4 sigma bonds with 4 Fluorine atoms, which uses 4 of its valence electrons. The remaining 4 valence electrons form 2 lone pairs on the Xe atom.

Steric number = Number of sigma bonds + Number of lone pairs = $4 + 2 = 6$.

A steric number of 6 corresponds to $sp^{3}d^{2}$ hybridization.

The total number of valence electrons around the central Xe atom in $XeF_{4}$ is the sum of the electrons in the 4 sigma bonds and the 2 lone pairs.

Total valence electrons around Xe = $4 \times 2$ (bonding electrons) + $2 \times 2$ (lone pair electrons) = $8 + 4 = 12$.

Thus, the hybridization is $sp^{3}d^{2}$ and the total number of valence electrons around Xe is $12$.

Answer: $sp^{3}d^{2}$ and $12$