The ion that has sp^3d^2 hybridization for th | vedclass

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(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central at

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$[ICl_4]^-$

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(B) To determine the hybridization,we use 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.For $[IF_6]^-$: $H = \frac{1}{2}(7 + 6 + 1) = 7$,which corresponds to $sp^3d^3$ hybridization.For $[ICl_4]^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^3d^2$ hybridization.For $[ICl_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.For $[BrF_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.Therefore,the ion with $sp^3d^2$ hybridization is $[ICl_4]^-$,which corresponds to option $B$.

The ion that has $sp^3d^2$ hybridization for the central atom is

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