The hybrid state of sulphur in SO2 molecule is :

The correct option is sp².

To determine the hybridization of the central sulfur atom in a sulfur dioxide (${\mathrm{SO}}_2$) molecule, we can use the steric number concept:
$\text{Steric Number}=\text{Number of sigma (}\sigma \text{) bonds}+\text{Number of lone pairs on the central atom}$

Let's break down the electronic structure step-by-step:

1. Valence Electrons:
Sulfur (S) belongs to Group 16 of the periodic table and has 6 valence electrons.

2. Bonding and Lone Pairs:
In ${\mathrm{SO}}_2$, the sulfur atom forms two double bonds with two oxygen atoms (O=S=O). Each double bond consists of one sigma ($\sigma$) bond and one pi ($\pi$) bond.
- Total sigma ($\sigma$) bonds = 2
Out of the 6 valence electrons of sulfur, 4 are shared in bonding (2 for each double bond), leaving 2 non-bonding valence electrons.
- Total lone pairs = 1 (since 2 electrons make 1 pair)

3. Calculating Steric Number:
$\text{Steric Number}=2\text{ (sigma bonds)}+1\text{ (lone pair)}=3$

A steric number of 3 corresponds to sp² hybridization, where one s orbital and two p orbitals mix to form three equivalent ${\text{sp}}^2$ hybrid orbitals in a trigonal planar arrangement (bent molecular geometry due to the lone pair).