Consider the following species:
$$BrF_5$$, $$XeF_5^-$$, $$BF_4^-$$, $$ICl_4^-$$, $$XeF_4$$, $$SF_4$$, $$NH_4^+$$, $$ClF_3$$, $$XeF_2$$, $$ICl_2^-$$
Number of species having $$sp^3d$$ hybridized central atom is _______.

The hybridization of a central atom can be determined using the Steric Number (SN) formula:

$$(\text{SN}=\frac{1}{2}\left[V+M-C+A\right])$$

Where 

$$V = \text{valence electrons of central atom}$$, 

$$M = \text{monovalent atoms}$$, 

$$C = \text{cationic charge}$$, and 

$$A = \text{anionic charge}$$. 

A steric number of 5 corresponds to $$(sp^{3}d)$$ hybridization.

1. Species with $$(sp^{3}d)$$ Hybridization $$(\text{SN} = 5)$$:

• $$(\text{SF}_{4}):$$ $$(\text{SN} = \frac{1}{2}(6 + 4) = 5)$$ $$(\rightarrow )$$ 4 Bond Pairs + 1 Lone Pair (See-saw shape)
• $$(\text{ClF}_{3}):$$ $$(\text{SN} = \frac{1}{2}(7 + 3) = 5)$$ $$(\rightarrow )$$ 3 Bond Pairs + 2 Lone Pairs (T-shape)
• $$(\text{XeF}_{2}):$$ $$(\text{SN} = \frac{1}{2}(8 + 2) = 5)$$ $$(\rightarrow )$$ 2 Bond Pairs + 3 Lone Pairs (Linear shape)
• $$(\text{ICl}_{2}^{-}):$$ $$(\text{SN} = \frac{1}{2}(7 + 2 + 1) = 5)$$ $$(\rightarrow )$$ 2 Bond Pairs + 3 Lone Pairs (Linear shape)

2. Other Species in the List:

• $$(\text{BrF}_{5}):$$ $$(\text{SN} = \frac{1}{2}(7 + 5) = 6 \rightarrow sp^3d^2)$$ (Square pyramidal)
• $$(\text{XeF}_{5}^{-}):$$ $$(\text{SN} = \frac{1}{2}(8 + 5 + 1) = 7 \rightarrow sp^3d^3)$$ (Pentagonal planar)
• $$(\text{BF}_{4}^{-}):$$ $$(\text{SN} = \frac{1}{2}(3 + 4 + 1) = 4 \rightarrow sp^3)$$ (Tetrahedral)
• $$(\text{ICl}_{4}^{-}):$$ $$(\text{SN} = \frac{1}{2}(7 + 4 + 1) = 6 \rightarrow sp^3d^2)$$ (Square planar)
• $$(\text{XeF}_{4}):$$ $$(\text{SN} = \frac{1}{2}(8 + 4) = 6 \rightarrow sp^3d^2)$$ (Square planar)
• $$(\text{NH}_{4}^{+}):$$ $$(\text{SN} = \frac{1}{2}(5 + 4 - 1) = 4 \rightarrow sp^3)$$ (Tetrahedral)

Species having $$sp^3{}d$$ hybridization are:

$$SF_4,\ ClF_3,\ XeF_2,\ ICl_2^-$$

Total number: 4