Consider the following species: BrF _5 , XeF _5^- , BF _4^- , ICl...

The steric number $H$ is calculated as $H = \dfrac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms, $C$ is the cationic charge, and $A$ is the anionic charge.

For $sp^3d$ hybridization, $H = 5$.

BrF$_5$: $H = \dfrac{1}{2}(7 + 5) = 6$ ($sp^3d^2$)

XeF$_5^-$: $H = \dfrac{1}{2}(8 + 5 + 1) = 7$ ($sp^3d^3$)

BF$_4^-$: $H = \dfrac{1}{2}(3 + 4 + 1) = 4$ ($sp^3$)

ICl$_4^-$: $H = \dfrac{1}{2}(7 + 4 + 1) = 6$ ($sp^3d^2$)

XeF$_4$: $H = \dfrac{1}{2}(8 + 4) = 6$ ($sp^3d^2$)

SF$_4$: $H = \dfrac{1}{2}(6 + 4) = 5$ ($sp^3d$)

NH$_4^+$: $H = \dfrac{1}{2}(5 + 4 - 1) = 4$ ($sp^3$)

ClF$_3$: $H = \dfrac{1}{2}(7 + 3) = 5$ ($sp^3d$)

XeF$_2$: $H = \dfrac{1}{2}(8 + 2) = 5$ ($sp^3d$)

ICl$_2^-$: $H = \dfrac{1}{2}(7 + 2 + 1) = 5$ ($sp^3d$)

The species with $sp^3d$ hybridization are SF$_4$, ClF$_3$, XeF$_2$, and ICl$_2^-$.

The number of such species is $4$.

Answer: $4$