Determine the steric number (bond pairs + lone pairs) on the central atom in each species. $s{p}^{3}d$ hybridisation corresponds to a steric number of 5.

${\text{BrF}}_5$: 5 bonds + 1 lp = 6, $s{p}^3{d}^2$.

${\text{XeF}}_5^{-}$: 5 bonds + 2 lp = 7, $s{p}^3{d}^3$.

${\text{BF}}_4^{-}$: 4 bonds + 0 lp = 4, $s{p}^3$.

${\text{ICl}}_4^{-}$: 4 bonds + 2 lp = 6, $s{p}^3{d}^2$.

${\text{XeF}}_4$: 4 bonds + 2 lp = 6, $s{p}^3{d}^2$.

${\text{SF}}_4$: 4 bonds + 1 lp = 5, $s{p}^{3}d$ ✓.

${\text{NH}}_4^{+}$: 4 bonds + 0 lp = 4, $s{p}^3$.

${\text{ClF}}_3$: 3 bonds + 2 lp = 5, $s{p}^{3}d$ ✓.

${\text{XeF}}_2$: 2 bonds + 3 lp = 5, $s{p}^{3}d$ ✓.

${\text{ICl}}_2^{-}$: 2 bonds + 3 lp = 5, $s{p}^{3}d$ ✓.

Species with $s{p}^{3}d$: SF${}_4$, ClF${}_3$, XeF${}_2$, ICl${}_2^{-}$. Count $=4$.