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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.
Oxidation states of Mn in \(MnO _2, KMnO _4\) and \(K _2 MnO _4\) are

  1. A \(+4,+7\), and +7 respectively
  2. B \(+6,+7\), and +4 respectively
  3. C \(+4,+6\), and +7 respectively
  4. D \(+4,+7\), and +6 respectively
Verified Solution

Answer & Solution

Correct Answer

(D) \(+4,+7\), and +6 respectively

Step-by-step Solution

Detailed explanation

In \(MnO_2\): \(x + 2(-2) = 0 \Rightarrow x = +4\) In \(KMnO_4\): \(1 + x + 4(-2) = 0 \Rightarrow x = +7\) In \(K_2MnO_4\): \(2(1) + x + 4(-2) = 0 \Rightarrow x = +6\) Oxidation states are \(+4, +7\), and \(+6\) respectively.
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