Question

The structure of $S{{O}_{3}}$ molecule in the gaseous phase contains:
(A) Only $\sigma$- bond between sulphur and oxygen
(B) $\sigma$ bonds and a (p$\pi$-p$\pi$) bonds between sulphur and oxygen
(C) $\sigma$bonds and a d$\pi$- p$\pi$) bonds between sulphur and oxygen
(D) $\sigma$ bonds and a (p$\pi$-p$\pi$) bonds and a ( d$\pi$- p$\pi$) bonds between sulphur and oxygen

Answer 1

To answer this question we should be aware when $\sigma$bonds, (p$\pi$-p$\pi$) bonds and ( d$\pi$- p$\pi$) bonds are formed. Valence electrons are the electrons that are present in the outermost shell of an atom upon this the covalency of the atom depends.

Complete answer:
Sulphur has six electrons in its outermost shell that is its covalency is six. Covalency of an atom is the number bonds an atom can make. In solid state the $S{{O}_{3}}$molecule exists in a cluster that is the number of$S{{O}_{3}}$ molecules bonded together to form a solid structure. In the gas phase the $S{{O}_{3}}$ molecule exists as a single molecule which is also said to exist in the free State.
$\sigma$- bond is formed by overlapping of atomic orbitals or hybrid orbitals in their axis.
For example: overlap of s-s, s-p,$s{{p}^{2}}$-$s{{p}^{2}}$ overlap along their axis.
$\pi$- bond is formed by the side to side overlap of molecular orbital or atomic orbital along a plane perpendicular to the plane. If the donating orbital is p and the orbital to which the electron pair is donated is p then it is known as P$\pi$-p$\pi$.
Similarly, if the donating orbital is d and the orbital to which the electron pair is donated is p then it is known as d$\pi$- p$\pi$
Firstly, let's draw the Lewis structure of$S{{O}_{3}}$:

There are three delocalised $\pi$ bonds, out of which one of them is P$\pi$-p$\pi$and other two are d$\pi$- p$\pi$. The sulphur is $d\pi -p\pi$hybridized.
The electronic configuration of sulphur is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$.
Sulphur in ground state:

In excited state:

Oxygen has 6 electrons. So, the valence electrons of oxygen make bond with unpaired electrons.

The 6 valence electrons in oxygen are of 2p orbital.

The red coloured electrons in the above mentioned diagram are the electrons of oxygen. The $s{{p}^{2}}$-$s{{p}^{2}}$ overlap forms $\sigma$ bond as sulphur in $S{{O}_{3}}$ is $s{{p}^{2}}$ hybridized.

From the above diagram it is evident that option D is the correct answer.

Note: Knowing the valance electrons of a particular element may help us to figure out their structure. Always whenever, a double bond is formed. Here one of the bonds will be $\sigma$ bond and the other will be $\pi$ bond. Now, the $\pi$ bond can be p$\pi$-p$\pi$bond, d$\pi$- p$\pi$bond or d$\pi$-d$\pi$bond.