Question

What is the Hybridisition of ${\mathrm{SO}}_2$?

Answer 1

Hybridisation: The process of combining of the atomic orbitals to produce new orbitals (hybrid orbitals) of varying energies and shapes than the initial orbitals, is called hybridisation.

VSEPR theory:-

• VSEPR stands for, Valence Shell Electron Pair Repulsion theory.
• In 1940, Sidgwick and Powell proposed this theory on the basis of the repulsion interaction of the electron pairs in the valence shell of the atoms.
• Basically from this theory we can calculate the number of electrons pairs present in a molecule with the help of this formula : $\frac{1}{2}\times [V+N-C+A]$
• Here, V= the number of valence electrons present in the central atom, N= total number of monovalent atoms, C= cationic charge, and A= anionic charge.

Hybridisation of Sulphur dioxide using VSEPR theory:-

• In Sulphur dioxide ${\mathrm{SO}}_2$, there are 2 Oxygen atoms that are linked with 1 Sulphur atom.
• Electronic configuration of Sulphur is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^4$, hence it has 6 valence electrons in its outermost shell.
• Here no monovalent atom is present.
• So according to the formula, for calculating electron pair, the values of V= 6 ($\because$Sulphur has 6 valence electrons), N= 0($\because$no monovalent atom is present), and C= 0 and A=0 ($\because$No cations or anions are present in Sulphur dioxide, hence charge is 0).
• Putting the values in the formula we get the number of electron pair

$$\begin{gathered} =\frac{1}{2}\times [V+N-C+A] \\ =\frac{1}{2}\times [6+0-0+0] \\ =\frac{1}{2}\times 6 \\ =3 \end{gathered}$$

• Here total of 3 electron pairs are present, out of them 2 are bond pairs $(\mathrm{S}=\mathrm{O})$ and 1 is a lone pair present in Sulphur.
• Now since there are 3 electron pairs present, the hybridization will be ${\mathrm{sp}}^2$, so its shape will be V-shape or Bent.

Therefore, the hybridisation of ${\mathrm{SO}}_2$ is ${\mathrm{sp}}^2$.