Consider the species XeF4, XeOF4, ICl4^-, BrF5, SF6, XeO3, XeF2, ClF3. Among these, take those whose central atom is sp^3d^2 hybridised and which…

Hybridisation from steric number: $\mathrm{XeF_4}$ (4 bond pairs + 2 lone pairs = 6): $sp^3d^2$, square planar, nonpolar. $\mathrm{XeOF_4}$ (5 bonds incl. Xe=O, 1 lone pair): $sp^3d^2$, square pyramidal, polar. $\mathrm{ICl_4^-}$ (4 + 2 lone pairs): $sp^3d^2$, square planar, nonpolar. $\mathrm{BrF_5}$ (5 + 1): $sp^3d^2$, square pyramidal, polar. $\mathrm{SF_6}$ (6 + 0): $sp^3d^2$, octahedral, nonpolar. $\mathrm{XeO_3}$: $sp^3$. $\mathrm{XeF_2}$: $sp^3d$ (nonpolar, but not $sp^3d^2$). $\mathrm{ClF_3}$: $sp^3d$, T-shaped, polar. Qualifying species: $\mathrm{XeF_4}$ (2 lone pairs), $\mathrm{ICl_4^-}$ (2), $\mathrm{SF_6}$ (0). Total $=2+2+0=4$.