Question

Which is having a higher bond angle and why? $Cl{O_2}$ or $ClO_2^ - $?

Answer 1

Hint: The angle formed by three atoms over at least two bonds is known as a bond angle. The rotational angle is the angle created by the first three atoms and the plane formed by the last three atoms with four atoms bonded together in a chain.

Complete answer:
The central atom of chlorine has two lone pairs and the chlorine atom is also attached to the oxygen by two sigma bonds. Hence it has $s{p^3}$ hybridization. But due to the two lone pairs present on chlorine the lone pair $ - $ lone pair repulsion is greater, due to which the bond angle decreases from ${109.5^ \circ }$ to ${105^ \circ }$.
Since chlorine has $7$ valence electrons and oxygen has $6$ valence electrons. Therefore, $Cl{O_2}$ has $7 + 6 + 6 = 19$ total valence electrons. Since oxygen atom is larger as compared to the chlorine atom, that also contributes to the substantially larger bond angle as compared to $ClO_2^ - $.
Since chlorine has $7$ valence electrons, oxygen has $6$ valence electrons and has one electron in excess due to the negative charged ion. Therefore, $ClO_2^ - $ has $7 + 6 + 6 + 1 = 20$ total valence electrons to distribute. The bond order of each $Cl - O$ bond in the resonance hybrid structure is lower than the bond order in each $Cl = O$ bond in $Cl{O_2}$.
Therefore, $Cl{O_2}$ has a higher bond angle as compared to $ClO_2^ - $

Note:
A resonance hybrid is a chemical, atom, ion, or radical that exhibits resonance and has a structure written as the average of two or more structural formulas, each divided by a double-headed arrow.