The hybridisation and shape of I_3^{-} ion,re | vedclass

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(B) To determine the hybridisation of the central $I$ atom in $I_3^{-}$,we use the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number

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$sp^3d$; linear

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(B) To determine the hybridisation of the central $I$ atom in $I_3^{-}$,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 $I_3^{-}$,the central $I$ atom has $V = 7$ valence electrons,$M = 2$ (two other $I$ atoms),$C = 0$,and $A = 1$.$H = \frac{1}{2} (7 + 2 + 1) = \frac{10}{2} = 5$.$A$ value of $5$ corresponds to $sp^3d$ hybridisation.In $sp^3d$ hybridisation,the electron geometry is trigonal bipyramidal. The central $I$ atom has $3$ lone pairs and $2$ bond pairs. According to $VSEPR$ theory,the lone pairs occupy the equatorial positions to minimize repulsion,resulting in a linear shape for the $I_3^{-}$ ion.

The hybridisation and shape of $I_3^{-}$ ion,respectively,are

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