The shape and type of hybridization of $XeOF_4$ according to VSEPR theory is:

square pyramidal, $sp^3d^2$

The correct answer is Option (4) → square pyramidal, $sp^3d^2$.

The central atom is Xenon, \(Xe\) and its atomic number is \(54\), so its electronic configuration is

\([_{36}Kr] 4d^{10}5s^25p^6\)

So, \(Xe\) has 8 valence elecetrons.

\(\text{Hybridization = } \frac{1}{2}[8+4-0+0]\)

\(\text{Hybridization = } \frac{1}{2}[12]\)

\(\text{Hybridization = } 6\)

The hybridization number of \(XeOF_4\) is 6. Thus, the hybridization of \(Xe\) in \(XeOF_4\) is \(sp^3d^2\). Due to the presence of one lone pair the structure is square pyramidal.