The correct shape and I-I-I bond angle respec | vedclass

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(D) In $I_{3}^{-}$ ion,the central iodine atom has $7$ valence electrons. It forms $2$ bonds with other iodine atoms and has $3$ lone pairs of electro

Vedclass · Accepted Answer

Linear; $180^{\circ}$

Vedclass · Answer

(D) In $I_{3}^{-}$ ion,the central iodine atom has $7$ valence electrons. It forms $2$ bonds with other iodine atoms and has $3$ lone pairs of electrons.Total electron pairs = $2$ (bond pairs) + $3$ (lone pairs) = $5$.According to $VSEPR$ theory,the hybridization is $sp^{3}d$ and the electron geometry is trigonal bipyramidal.The $3$ lone pairs occupy the equatorial positions to minimize repulsion.Therefore,the $2$ iodine atoms occupy the axial positions,resulting in a linear shape with an $I-I-I$ bond angle of $180^{\circ}$.

The correct shape and $I-I-I$ bond angle respectively in $I_{3}^{-}$ ion are :-

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