What is the geometry of the IBr^{-}_{2} ion? | vedclass

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(A) To determine the geometry of the $IBr^{-}_{2}$ ion,we calculate the steric number $(X)$:$X = \frac{1}{2} [V + M - C + A]$Where $V$ is the number o

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(A) To determine the geometry of the $IBr^{-}_{2}$ ion,we calculate the steric number $(X)$:$X = \frac{1}{2} [V + M - C + A]$Where $V$ is the number of valence electrons of the central atom $(I = 7)$,$M$ is the number of monovalent atoms attached $(Br = 2)$,$C$ is the cationic charge $(0)$,$A$ is the anionic charge $(1)$.$X = \frac{1}{2} [7 + 2 + 1] = \frac{10}{2} = 5$.$A$ steric number of $5$ corresponds to $sp^{3}d$ hybridization,which has a trigonal bipyramidal electron geometry.For $IBr^{-}_{2}$,there are $2$ bond pairs and $3$ lone pairs.According to $VSEPR$ theory,the $3$ lone pairs occupy the equatorial positions to minimize repulsion,resulting in a linear molecular geometry.

What is the geometry of the $IBr^{-}_{2}$ ion?

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