Read the passage carefully and answer the Questions.
\( KMnO_4 \) is prepared by the fusion of \( MnO_2 \) with an alkali metal hydroxide and an oxidizing agent like \( KNO_3 \) to give a dark- green manganate ion which disproportionates to give permanganate as follows.
\( 2MnO_2+ 4KOH + O_2→2K_2MnO_4+2H_2O \)
\( 3K_2MnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O \)
On heating \( KMnO_4 \) decomposes at 513 K to give \( K_2MnO_4 \). Permanganate ion is tetrahedral and diamagnetic. Acidified \( KMnO_4 \) acts as a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
Equivalent weight of \( KMnO_4 \) in its oxidising reactions in acidic medium is taken as
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → One-fifth of its molecular weight
The equivalent weight of an oxidising agent is its molar mass divided by the number of electrons one formula unit accepts in the reaction (its n-factor). The n-factor of \( \ce{KMnO4} \) is the change in the oxidation state of manganese, and that depends on the medium.
In acidic medium permanganate is reduced to \( \ce{Mn^2+} \):
\( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)
Manganese changes from +7 to +2, so each \( \ce{KMnO4} \) accepts 5 electrons:
\( \displaystyle \text{Equivalent weight} = \frac{M}{5} = \frac{158}{5} = 31.6 \)
So in acidic medium the equivalent weight of \( \ce{KMnO4} \) is one-fifth of its molecular weight.
Why the other options are incorrect
Option (1) One-third of its molecular weight — Incorrect. This is the value in neutral or faintly alkaline medium, where permanganate is reduced to \( \ce{MnO2} \) (+7 to +4, 3 electrons).
Option (3) half of its molecular weight — Incorrect. This would need a change of two units in the oxidation state of manganese, which is not one of the usual reactions of permanganate.
Option (4) Equal to its molecular weight — Incorrect. This is the value in strongly alkaline medium, where permanganate is reduced only to manganate, \( \ce{MnO4^2-} \) (+7 to +6, 1 electron).
Note: It seems that there is a mistake in the original NTA paper. The second equation of the passage is printed as 3K2MnO4 + 4H+ → 2KMnO4 + MnO2 + 2H2O, in which the charges do not balance; it should be 3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O.
NCERT (2026-27 edition):
Electrolytic Cells and Electrolysis – “… (ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal / Number of electrons required to reduce the cation).”
Oxides and Oxoanions of Metals – “… If we represent the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt by half-reactions, …
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O (Eo = + 1.52 V)”