Question:

Given below are two statements:
Statement (I) : Fusion of MnO$_2$ with KOH and an oxidising agent gives dark green K$_2$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.

Updated On: Nov 4, 2025
  • Both Statement I and Statement II is true
  • Both Statement I and Statement II is false
  • Statement I is true but Statement II is false
  • Statement I is false but Statement II is true
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The Correct Option is A

Approach Solution - 1

The question provides two statements related to the chemistry of manganese compounds, and we need to determine the veracity of these statements based on chemical reactions and principles.

Statement I: Fusion of \(MnO_2\) with \(KOH\) and an oxidising agent gives dark green \(K_2MnO_4\).

Explanation: When manganese dioxide (\(MnO_2\)) is fused with potassium hydroxide (\(KOH\)) in the presence of an oxidizing agent such as chlorine (\(Cl_2\)) or potassium nitrate (\(KNO_3\)), it forms potassium manganate (\(K_2MnO_4\)), which is dark green.

This reaction can be represented as: \(2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O\)

Therefore, Statement I is true.

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

Explanation: In alkaline medium, the manganate ion (\(MnO_4^{2-}\)) can undergo further oxidation to form permanganate ion (\(MnO_4^{-}\)), especially under electrolytic conditions. This transformation occurs due to the gain of an electron by \(MnO_4^{2-}\):

\(MnO_4^{2-} \ + \ e^- + \ H_2O \rightarrow \ MnO_4^- + 2OH^-\)

This shows an increase in oxidation state from Mn (+6 in manganate) to Mn (+7 in permanganate).

Hence, Statement II is also true.

Conclusion: Based on the explanations, both Statement I and Statement II are true. Therefore, the correct answer is: Both Statement I and Statement II is true.

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Approach Solution -2

Statement I: True. Fusion of MnO$_2$ with KOH in the presence of oxygen gives K$_2$MnO$_4$ (potassium manganate), which is dark green in color: \[ \text{MnO}_2 + 4\text{KOH} + \text{O}_2 \xrightarrow{\text{fusion}} 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}. \]
Statement II: True. In alkaline medium, the manganate ion (MnO$_4^{2-}$) undergoes electrolytic oxidation to form the permanganate ion (MnO$_4^-$): \[ \text{MnO}_4^{2-} \rightarrow \text{MnO}_4^- + e^-. \]
Both statements are correct as they describe valid chemical processes.

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