Consider the following species: BrF_5, XeF_5^ | vedclass

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(B) The hybridization $H$ is determined by the formula $H = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons of the central a

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$4$

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(B) The hybridization $H$ is determined by the formula $H = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons of the central atom, $M$ is the number of monovalent atoms, $C$ is the cationic charge, and $A$ is the anionic charge.$sp^3d$ hybridization corresponds to a steric number of $5$.$1$) $SF_4$: $H = \frac{1}{2}(6 + 4) = 5$ $(sp^3d)$$2$) $ClF_3$: $H = \frac{1}{2}(7 + 3) = 5$ $(sp^3d)$$3$) $XeF_2$: $H = \frac{1}{2}(8 + 2) = 5$ $(sp^3d)$$4$) $ICl_2^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$ $(sp^3d)$$5$) $XeF_5^-$: $H = \frac{1}{2}(8 + 5 + 1) = 7$ $(sp^3d^3)$$6$) $ICl_4^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$ $(sp^3d^2)$$7$) $BrF_5$: $H = \frac{1}{2}(7 + 5) = 6$ $(sp^3d^2)$$8$) $XeF_4$: $H = \frac{1}{2}(8 + 4) = 6$ $(sp^3d^2)$$9$) $BF_4^-$: $H = \frac{1}{2}(3 + 4 + 1) = 4$ $(sp^3)$$10$) $NH_4^+$: $H = \frac{1}{2}(5 + 4 - 1) = 4$ $(sp^3)$The species with $sp^3d$ hybridization are $SF_4, ClF_3, XeF_2, ICl_2^-$. Thus, the total count is $4$.

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 . . . . . . .

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