CUET · CHEMISTRY · PYQ PAPER 2025
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\(KMnO _4\) 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 disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
On heating \(K _2 MnO _4\) decomposes at 513 K to give \(K _2 MnO _4\). Permanganate ion is tetrahedral and diamagnetic
Acidified \(KMnO _4\) acts 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
- A One-third of its molecular weight
- B One-fifth of its molecular weight
- C half of its molecular weight
- D Equal to its molecular weight
Answer & Solution
Correct Answer
(B) One-fifth of its molecular weight
Step-by-step Solution
Detailed explanation
Oxidation state of Mn in \(KMnO _4\) is \(+7\). In acidic medium, \(Mn ^{7+}\) is reduced to \(Mn ^{2+}\). Change in oxidation state \((\Delta x) = 7 - 2 = 5\). Equivalent weight \( = \frac{\text{Molecular weight}}{ \Delta x} = \frac{\text{Molecular weight}}{5}\).
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