Question

The compound which do not react with \[{\text{KMn}}{{\text{O}}_4}\] :
A. Perdisulphuric acid
B. Sulphurous acid
C. Hydrogen sulphide
D. Hydrogen peroxide

Answer 1

Hint: \[{\text{KMn}}{{\text{O}}_4}\] is an oxidising agent. Elements with maximum oxidation state cannot be oxidized further and elements with minimum oxidation state cannot be reduced further.

Complete step by step solution:
In a redox reaction, oxidation and reduction both occur simultaneously. Increase in oxidation state is known as oxidation and decrease in oxidation state is known as reduction. Reducing agents are those chemical compounds which can reduce other and oxidized itself, for example lithium aluminium hydride. Oxidizing agents are those chemical compounds which oxidize other at the cost of being reduced, for example nitric acid and potassium permanganate or \[{\text{KMn}}{{\text{O}}_4}\] .
The chemical compounds which are already in maximum oxidation state do not undergo further oxidation and can only behave as oxidizing agents. This is because these species have already donated possible pair of electron and now can only gain electrons for example in \[{\text{KMn}}{{\text{O}}_4}\] the manganese is in maximum oxidation state that is +7. Similarly elements with minimum oxidation state possible do not undergo reduction and can only act as reducing agents in order to oxidize themselves.
In sulphurous acid \[{{\text{H}}_2}{\text{S}}{{\text{O}}_3}\] , Sulphur is in +4 oxidation state and it can be oxidized easily by \[{\text{KMn}}{{\text{O}}_4}\] . In hydrogen sulphide \[{{\text{H}}_2}{\text{S}}\] , sulphur is in -2 oxidation state, it can also be easily oxidised by \[{\text{KMn}}{{\text{O}}_4}\] . In hydrogen peroxide \[{{\text{H}}_2}{{\text{O}}_2}\] , oxygen is in -1 oxidation state and can easily be oxidized by \[{\text{KMn}}{{\text{O}}_4}\]. But in perdisulphuric acid \[{{\text{H}}_2}{{\text{S}}_2}{{\text{O}}_7}\] , sulphur is in maximum oxidation state that is +6 and therefore its further oxidation is not possible because its oxidation number cannot be increase any further.

Thus, the correct option is A.

Note:
Acidified \[{\text{KMn}}{{\text{O}}_4}\] acts as an oxidizing agent. In \[{\text{KMn}}{{\text{O}}_4}\] . \[{\text{Mn}}\] is in +7 oxidation state and oxidation state change from +7 to +2. \[{\text{KMn}}{{\text{O}}_4}\] is acidified by treating it with dilute sulphuric acid.