Chemistry · d and f Block Elements

JEE Main 2024 — 8 April, Shift 2 — Question 76

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II.

Statement (I) : Fusion of MnO2\mathrm{MnO}_{2} with KOH and an oxidising agent gives dark green K2MnO4\mathrm{K}_{2} \mathrm{MnO}_{4}.

Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.

In the light of the above statements, choose the correct answer from the options given below:

  1. Option A:

    Both Statement I and Statement II is true

    Correct
  2. Option B:

    Both Statement I and Statement II is false

  3. Option C:

    Statement I is true but Statement II is false

  4. Option D:

    Statement I is false but Statement II is true

Answer: A

Step-by-step solution

Statement I is True: Fusion of MnO2\mathrm{MnO_2} with KOH\mathrm{KOH} in the presence of an oxidising agent (such as KNO3\mathrm{KNO_3} or air) produces dark green potassium manganate, K2MnO4\mathrm{K_2MnO_4}.

MnO2+4KOH+O2→ fused 2 K2MnO4(Dark  green)+2H2O\mathrm{MnO}_{2}+4 \mathrm{KOH}+\mathrm{O}_{2} \xrightarrow{\text { fused }} 2 \mathrm{~K}_{2} \mathrm{MnO}_{4} (Dark\; green )+2 \mathrm{H}_{2} \mathrm{O}

Statement II is True: The manganate ion (MnO42−)\mathrm{(MnO_4^{2-})} undergoes oxidation in alkaline medium, including electrolytic oxidation, to form the permanganate ion (MnO4−)\mathrm{(MnO_4^-)}.

MnO42−→MnO4−+e−\mathrm{MnO}_{4}^{2-} \rightarrow \mathrm{MnO}_{4}^{-}+\mathrm{e}^{-}

Since both statements are correct, the correct answer is option A.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
d and f Block Elements
Topic
Compounds of Manganese
Given below are two statements, one is labelled as Statement I and… | JEE Main 2024 PYQ with Solution · DhiX AI