Hybridization

Xenon has 8 valence electrons. In ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ , it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs $=5$ .
As per the VSEPR Theory  the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.

Following have the sp³d² hybridisation