The ion that has sp^{3}d^{2} hybridization fo | vedclass

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(A) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence ele

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$[ICl_{4}]^{-}$

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(A) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.						Species			Hybridization							$[ICl_{2}]^{-}$			$sp^{3}d$							$[ICl_{4}]^{-}$			$sp^{3}d^{2}$							$[BrF_{2}]^{-}$			$sp^{3}d$							$[IF_{6}]^{-}$			$sp^{3}d^{3}$			For $[ICl_{4}]^{-}$,the central atom $I$ has $7$ valence electrons,$4$ monovalent $Cl$ atoms,and $1$ negative charge. Thus,$H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^{3}d^{2}$ hybridization.

Column-$I$	Column-$II$
$A$. Boron in $[B(OH)_4]^-$	$1$. $sp^2$
$B$. Aluminum in $[Al(H_2O)_6]^{3+}$	$2$. $sp^3$
$C$. Boron in $B_2H_6$	$3$. $sp^3d^2$
$D$. Carbon in Buckminsterfullerene
$E$. $Si$ in $SiO_4^{4-}$
$F$. $Ge$ in $[GeCl_6]^{2-}$

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

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