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An explosion takes place when conc. ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is added to ${\text{KMn}}{{\text{O}}_4}$ which of the following is formed:

Last updated date: 13th Jun 2024
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Hint:To solve this question, it is required to have knowledge about the chemical reaction of potassium permanganate in an acidic medium. In ${\text{KMn}}{{\text{O}}_4}$ , the manganese ion is present in its highest oxidation state, i.e. $ + 7$. Reaction of ${\text{KMn}}{{\text{O}}_4}$ with acids reduces the manganese ion to a lower oxidation state which differs according to the pH of the solution. The reaction of sulphuric acid with potassium permanganate is the only reaction in which the oxidation number of manganese does not change.

Complete step by step answer:
As we know, the Mn containing products from redox reactions depends on the pH. Acidic solutions of permanganate are reduced to the faintly pink ${\text{M}}{{\text{n}}^{2 + }}$ ion and water. In neutral solutions, permanganate is reduced by only three electrons to give ${\text{Mn}}{{\text{O}}_2}$ . The manganese ion is present in $ + 4$ oxidation state as ${\text{M}}{{\text{n}}^{4 + }}$ . In an alkaline solution, ${\text{KMn}}{{\text{O}}_4}$ reduces into a green ${{\text{K}}_2}{\text{Mn}}{{\text{O}}_4}$ .
On reaction with concentrated ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ , ${\text{KMn}}{{\text{O}}_4}$ gives ${\text{M}}{{\text{n}}_2}{{\text{O}}_7}$ in which manganese ion is present in the oxidation state of $ + 7$. The compound is unstable and in the presence of sunlight can also be explosive. The vapour can ignite paper dipped with alcohol. It is a dark green compound and is also highly volatile.
$\therefore $ The correct option is option A, i.e. ${\text{M}}{{\text{n}}_2}{{\text{O}}_7}$

Note: Potassium permanganate reacts with most compounds and reduces itself while oxidizing the other compound. But only in the case of sulphuric acid, potassium permanganate does not reduce itself and remains the same. It is an acidic anhydride of permanganic acid. The reaction goes as follows:
$2{\text{KMn}}{{\text{O}}_4} + 2{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{M}}{{\text{n}}_2}{{\text{O}}_7} + {{\text{H}}_2}{\text{O + 2KHS}}{{\text{O}}_4}$