The hybridization of the central atom in ICl^ | vedclass

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(D) To determine the hybridization of the central atom $I$ in $ICl^{+}_2$,we use the formula for steric number $(SN)$:$SN = \frac{1}{2} [V + M - C + A

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$sp^3$

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(D) To determine the hybridization of the central atom $I$ in $ICl^{+}_2$,we use the formula for steric number $(SN)$:$SN = \frac{1}{2} [V + M - C + A]$Where:$V$ = Number of valence electrons of the central atom $(I = 7)$$M$ = Number of monovalent atoms attached $(Cl = 2)$$C$ = Cationic charge $(+1)$$A$ = Anionic charge $(0)$$SN = \frac{1}{2} [7 + 2 - 1 + 0] = \frac{8}{2} = 4$$A$ steric number of $4$ corresponds to $sp^3$ hybridization.Therefore,the central iodine atom in $ICl^{+}_2$ is $sp^3$ hybridized.

The hybridization of the central atom in $ICl^{+}_2$ is

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