Question

The type of hybrid orbitals used by chlorine atom in $Cl{O_2}$ is:
(A) $s{p^3}$
(B) $s{p^2}$
(C) $sp$
(D) $s{p^3}d$

Answer 1

Hybridisation is the data that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. Hybridisation is also an expansion of the valence bond theory.

Complete step by step answer:
We know that $ClO_2^ -$ is having the $s{p^3}$ hybridization.
In the ground state electronic configuration of $Cl$ is $3{s^2}3{p^5}$ and the excited electron configuration of $Cl$ is $3{s^2}3{p^4}3{d^1}$. So, the electronic configuration of $O$ will be $2{s^2}2{p^4}$
From above we get that the chlorine is having $s{p^3}$ hybridization. The lone electrons are in s and p orbitals while the two singly filled p – orbitals will be forming the sigma bonds. While the pi bond is being formed after that by the overlap of the half filled $3d$ orbitals with the half filled $2p$ orbital of the $O$ atom.
So, the answer is (A) $s{p^3}$hybridization and hybrid orbitals.

Additional information:
Uses of $Cl{O_2}$:
1. It is used as disinfectant in treatment of water.
2. Used as antimicrobial agents in water that are used in processing poultry, washing vegetables and fruits.
3. Used for processing wood pulp that is used for making paper.
4. Used for sterilizing medical equipment.

Note:
$ClO_2^ -$ has tetrahedral geometry and molecular geometry is bent. Since there are $2$ lone pairs on chlorine the electron pair repulsion will result in a bond angle that will be slightly less than ${109.5^\circ }$.