Question

What is the value of the smallest bond angle in icl4−? enter the smallest bond angle of the molecule?

Answer 1

This compound has a square planar molecular geometry. Therefore the bond angles is 90 everywhere. So the smallest is 90 degrees.

See the figure attached.

Answer 2

Answer : The value of the smallest bond angle in $ICl_4^-$ is, $90^o$

Explanation :

Formula used  :

$\text{Number of electrons}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

First we have to determine the hybridization of the given molecule, $ICl_4^-$.

$\text{Number of electrons}=\frac{1}{2}\times [7+4+1]=6$

The number of electrons is 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central atom iodine, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar and the bond angle between the lone pair and the lone pair will be, $180^o$  and the bond angle between the lone pair and the bond pair will be, $90^o$.

The structure of the given molecule is shown below.

Hence, the value of the smallest bond angle in $ICl_4^-$ is, $90^o$