QJEEM24Ch262e: JEE Main 2024 Chemistry - Class 12 JEEM24Ch262e Solution: Given below are two statements : Statement (I) : Fusion of with …

Statement I is indeed true. When $\mathrm{MnO}_2$ (manganese dioxide) is fused with $\mathrm{KOH}$ (potassium hydroxide) and an oxidising agent such as $\mathrm{KNO}_3$ (potassium nitrate), it forms dark green potassium manganate ( $\mathrm{K}_2\mathrm{MnO}_4$ ). The reaction can be represented as follows:

$\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}$

This process involves the oxidation of manganese dioxide to manganate ion ( $\mathrm{MnO}_4^{2-}$ ) in an alkaline medium.

Statement II is also true. The manganate ion ( $\mathrm{MnO}_4^{2-}$ ), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion ( $\mathrm{MnO}_4^-$ ), which is characterized by a deep purple color. This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell. The reaction can be represented as follows:

$\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-$

Thus, both statements I and II are accurate, making the correct answer:

Option D: Both Statement I and Statement II are true.