Question

Among the oxyacids of chloride, the strongest oxidizing agent is:
A. ${\text{HCl}}{{\text{O}}_{\text{4}}}$
B. ${\text{HCl}}{{\text{O}}_3}$
C. ${\text{HCl}}{{\text{O}}_2}$
D. ${\text{HClO}}$

Answer 1

An element having higher oxidation number is considered to be a good oxidizing agent. On the other hand, an element having lower oxidation number is considered to be a good reducing agent.

Complete step by step answer:
We know that the oxidising power of a compound increases with the increase in the oxidation number of the atom.
We will calculate the oxidation number of the chlorine molecule in each of the given compounds.
In ${\text{HCl}}{{\text{O}}_{\text{4}}}$,
The oxidation number of ${\text{Cl}}\,\,{\text{ = }}\,\,{\text{1 + x + ( - 2)}}\, \times 4{\text{ = }}\, + 7$
In ${\text{HCl}}{{\text{O}}_3}$,
The oxidation number of ${\text{Cl}}\,\,{\text{ = }}\,\,{\text{1 + x + ( - 2)}}\, \times 3{\text{ = }}\, + 5$
In ${\text{HCl}}{{\text{O}}_2}$,
The oxidation number of ${\text{Cl}}\,\,{\text{ = }}\,\,{\text{1 + x + ( - 2)}}\, \times 2{\text{ = }}\, + 3$
In ${\text{HClO}}$,
The oxidation number of ${\text{Cl}}\,\,{\text{ = }}\,\,{\text{1 + x + ( - 2)}}\,{\text{ = }}\,{\text{ + 1}}$
From here we see that, oxidation number of the chlorine atom in ${\text{HCl}}{{\text{O}}_{\text{4}}}$ is the highest, that is +7 and the oxidation number of the chlorine atom in ${\text{HClO}}$ is the lowest, that is +1. The order of the oxidising power is,
${\text{HCl}}{{\text{O}}_{\text{4}}} > {\text{HCl}}{{\text{O}}_3} > {\text{HCl}}{{\text{O}}_2} > {\text{HClO}}$
So, we can conclude that ${\text{HCl}}{{\text{O}}_{\text{4}}}$ is the strongest oxidising agent

So, the correct answer is Option A .

Additional Information:
Perchloric acid $\left( {{\text{HCl}}{{\text{O}}_{\text{4}}}} \right)$ is usually found as an aqueous solution. It is a colourless compound and it is a stronger acid than nitric acid and sulfuric acid.

We can prepare at industries by two methods. In the traditional method we use the high aqueous solubility of sodium perchlorate $\left( {{\text{NaCl}}{{\text{O}}_{\text{4}}}} \right)$. Treating this solution with the hydrochloric acid $\left( {{\text{HCl}}} \right)$ produces perchloric acid by the precipitation of solid sodium chloride. We can write the reaction of this preparation follows:

${\text{NaCl}}{{\text{O}}_{\text{4}}} + {\text{HCl}} \to {\text{NaCl}} + {\text{HCl}}{{\text{O}}_{\text{4}}}$
Perchloric acid has various uses. It is used in making explosives, in the separation of sodium and potassium, it is used as an oxidizer, it is used in the rocket fuel ,used in plating of metals, used as catalyst, used for electropolishing of molybdenum, used as a reagent for determining the 1H-Benzotriazole etc.

Note:
In a redox reaction, there exists one reducing agent and one oxidizing agent. The reducing agent is the one that loses an electron and the oxidizing agent is marked by the gaining of an electron.