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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Linear

Vedclass · Answer

(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?

Similar Questions

What are the structures and shapes of $SF_4$ and $ClF_3$?

The geometries of $XeF_4$ and $XeOF_4$ respectively are

Provide the structure of the $SO_{2}$ molecule and calculate the formal charge on each atom.

In which of the following covalent compounds is the bond angle maximum?

The correct order of bond angle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module