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# 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. ${\text{HClO}}$

Last updated date: 20th Jun 2024
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Hint: 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.

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 .

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:
${\text{NaCl}}{{\text{O}}_{\text{4}}} + {\text{HCl}} \to {\text{NaCl}} + {\text{HCl}}{{\text{O}}_{\text{4}}}$