Question

The central atom in the iodite ion, 102, is surrounded by 1 [2] [3] [4] [5] single bond, I double bond and 2 lone pairs of electrons. 3 bonding pairs and I unshared pair of electrons. 1 bonding pair and 3 unshared pairs of electrons. 2 double bonds and no unshared pairs of electrons. 4 bonding pairs and 4 lone pairs of electrons.

Answer 1

Answer : The correct option is 1 single bond, 1 double bond and 2 lone pairs of electrons.

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

Now we have to determine the hybridization of the iodite ion $IO_2^-$ molecules.

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

The number of electrons are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 2 atoms around the central iodine atom, the third and fourth 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 bent or angular.

From the molecular geometry of iodite ion we conclude that, there are two lone pairs of electrons, 1 single bond and 1 double bond are present on central atom of iodine.

Hence, the correct option is 1 single bond, 1 double bond and 2 lone pairs of electrons.