Find the number of $s{p}^{3}d$ hybridized species:

${\left[{\mathrm{XeF}}_4\right]}^{2+}, \left[{\mathrm{CH}}_4\right], {\left[{\mathrm{ClF}}_4\right]}^{+}, {\left[{\mathrm{SF}}_5\right]}^{+}, \left[{\mathrm{BCl}}_3\right], \left[{\mathrm{XeF}}_6\right], {\left[{\mathrm{IF}}_2\right]}^{-}$

Hybridization is calculated using the following formula:

$\mathrm{Hybridization}=\frac{\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}}{2}$

V=valence electrons of central metal. M=monovalent atom, C is the cationic charge and A is the anionic charge.

$$\begin{gathered} \mathrm{For} {\left[{\mathrm{XeF}}_4\right]}^{2+}=\frac{8+4-2}{2}=\frac{10}{2}=5\left({\mathrm{sp}}^3\mathrm{d}\right) \\ \mathrm{For} \left[{\mathrm{CH}}_4\right]=\frac{4+4}{2}=4 \left({\mathrm{sp}}^3\right) \\ \mathrm{For} {\left[{\mathrm{ClF}}_4\right]}^{+}=\frac{7+4-1}{2}=\frac{10}{2}=5\left({\mathrm{sp}}^3\mathrm{d}\right) \\ \mathrm{For} {\left[{\mathrm{SF}}_5\right]}^{+}=\frac{6+5-1}{2}=\frac{10}{2}=5\left({\mathrm{sp}}^3\mathrm{d}\right) \\ \mathrm{For} \left[{\mathrm{BCl}}_3\right]=\frac{3+3}{2}=3 \left({\mathrm{sp}}^2\right) \\ \mathrm{For} \left[{\mathrm{XeF}}_6\right]=\frac{8+6}{2}=7 \left({\mathrm{sp}}^3{\mathrm{d}}^3\right) \\ \mathrm{For} {\left[{\mathrm{IF}}_2\right]}^{-}=\frac{7+2+1}{2}=\frac{10}{2}=5\left({\mathrm{sp}}^3\mathrm{d}\right) \end{gathered}$$