Question

Central $ \text{O} $ atom in $ {{\text{O}}_{\text{3}}} $ is _____________ hybridized.
(A) $ \text{sp} $
(B) $ \text{s}{{\text{p}}^{\text{2}}} $
(C) $ \text{s}{{\text{p}}^{3}} $
(D) $ \text{ds}{{\text{p}}^{\text{2}}} $

Answer 1

Hint: Hybridization is the phenomenon of intermixing of different atomic orbitals to get hybrid orbitals that have properties of all the participating atomic orbitals. This in turn influences the molecular geometry and bonding properties of the orbitals.

Formula Used: $ \text{H =}\dfrac{\text{V + X - C + A}}{\text{2}} $
Where H is the hybridization of the central element, V is the number of the electrons present in the valence shell of the element, x is the number of monovalent groups present in the molecule, C is the charge if the species is cationic and A is the charge if the species is anionic.

Complete step by step solution:
For ozone, the central element is oxygen, the number of electrons in the valence shell of oxygen is 6, there are no monovalent groups, and the compound is electrically neutral, so there is no cationic or anionic charge. Therefore,
 $ \text{H =}\dfrac{\text{6 + 0 - 0 + 0}}{\text{2}}=\dfrac{6}{2}=3 $
As there are three hybrid orbitals that are involved in bonding, the hybridization is $ \text{s}{{\text{p}}^{\text{2}}} $ and the structure of ozone is “triangular”.
So the correct answer is option B.

Note:
The shape of any molecule is dependent on the hybridization of the central atom of that compound and the hybridization of the orbitals is defined as the intermixing of the atomic orbitals to give hybrid orbitals that have properties of the atomic orbitals involved.
The Valence Shell Electron Pair Repulsion Theory of the VSEPR theory guides the bonding pattern and the shape of the molecules according to the number of the lone pairs and the bond pairs present in the molecule and this is because, the order of repulsion among the electron pairs is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.