The hybridization states of the central atoms | 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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$sp^3d, sp^3d^2, sp^3d$

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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.$1$. For $I_3^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.$2$. For $ICl_4^-$: Central atom $I$ has $V=7$,$M=4$,$A=1$. $H = \frac{1}{2}(7 + 4 + 1) = 6$. Hybridization is $sp^3d^2$.$3$. For $ICl_2^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.Thus,the hybridization states are $sp^3d, sp^3d^2, sp^3d$.

The hybridization states of the central atoms of the ions $I_3^-$,$ICl_4^-$ and $ICl_2^-$ are respectively

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