Number of Species with sp3d Hybridized Central Atom

To determine the hybridization of the central atom in each of the given species, we can use the steric number (SN) formula:

$\text{Steric Number (SN)} = \frac{1}{2} (V + M - C + A)$

where:

• $V =$ Number of valence electrons on the central atom
• $M =$ Number of monovalent atoms bonded to the central atom
• $C =$ Charge on the cation
• $A =$ Charge on the anion

The relationship between the Steric Number and Hybridization is as follows:

• $\text{SN} = 4 \implies \text{sp}^3$
• $\text{SN} = 5 \implies \text{sp}^3\text{d}$
• $\text{SN} = 6 \implies \text{sp}^3\text{d}^2$
• $\text{SN} = 7 \implies \text{sp}^3\text{d}^3$

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Step-by-Step Analysis of Each Species:

1. $\text{BrF}_5$: $\text{SN} = \frac{1}{2}(7 + 5) = 6 \implies \text{sp}^3\text{d}^2$

2. $\text{XeF}_5^-$: $\text{SN} = \frac{1}{2}(8 + 5 + 1) = 7 \implies \text{sp}^3\text{d}^3$

3. $\text{BF}_4^-$: $\text{SN} = \frac{1}{2}(3 + 4 + 1) = 4 \implies \text{sp}^3$

4. $\text{ICl}_4^-$: $\text{SN} = \frac{1}{2}(7 + 4 + 1) = 6 \implies \text{sp}^3\text{d}^2$

5. $\text{XeF}_4$: $\text{SN} = \frac{1}{2}(8 + 4) = 6 \implies \text{sp}^3\text{d}^2$

6. $\text{SF}_4$: $\text{SN} = \frac{1}{2}(6 + 4) = 5 \implies \mathbf{sp^3d}$

7. $\text{NH}_4^+$: $\text{SN} = \frac{1}{2}(5 + 4 - 1) = 4 \implies \text{sp}^3$

8. $\text{ClF}_3$: $\text{SN} = \frac{1}{2}(7 + 3) = 5 \implies \mathbf{sp^3d}$

9. $\text{XeF}_2$: $\text{SN} = \frac{1}{2}(8 + 2) = 5 \implies \mathbf{sp^3d}$

10. $\text{ICl}_2^-$: $\text{SN} = \frac{1}{2}(7 + 2 + 1) = 5 \implies \mathbf{sp^3d}$

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Conclusion:

The species with $\text{sp}^3\text{d}$ hybridized central atoms are: $\text{SF}_4, \text{ClF}_3, \text{XeF}_2, \text{ and } \text{ICl}_2^-$

Thus, the total number of species having $\text{sp}^3\text{d}$ hybridized central atom is 4.