Alkaline oxidative fusion of \(MnO_2\) gives “A” which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:
To solve this problem, we need to understand the chemical transformations taking place. The process described involves two main steps: the alkaline oxidative fusion of \( \text{MnO}_2 \) and subsequent electrolytic oxidation.
Based on these reactions, we can determine that "A" is \( \text{MnO}_4^{2-} \), and "B" is \( \text{MnO}_4^- \).
Thus, the correct answer is the option that identifies A and B as \( \text{MnO}_4^{2-} \) and \( \text{MnO}_4^- \) respectively.
Correct Option: \( \text{MnO}_4^{2-} \) and \( \text{MnO}_4^- \).
The process described involves the oxidative reactions of manganese compounds in alkaline solutions.
Alkaline Oxidative Fusion of \(MnO_2\): When \(MnO_2\) is subjected to oxidative fusion in an alkaline medium, it can form \(MnO_4^{2-}\) (manganate ion) under alkaline conditions. The reaction can be represented as:
\(2MnO_2 + 2OH^- + O_2 \rightarrow 2MnO_4^{2-} + H_2O\).
Electrolytic Oxidation of Product A: The product A formed from the oxidative fusion of \(MnO_2\) can be oxidized electrolytically in alkaline solution to form \(MnO_4^-\) (permanganate ion). The overall reaction is:
\(MnO_2 + 2e^- \rightarrow MnO_4^-\).
Thus, the products A and B formed in the given reaction sequence are: A = \(MnO_4^{2-}\) and B = \(MnO_4^-\).
