When `MnO_(2)`, is fused with KOH in the presence of `KNO_(3)` as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidises Kl to compound (C)whereas an acidified solution of compound (B) oxidises KI to (D), Identity (A), (B), (C), and (D).

Explanation

To solve the given question step by step, we will identify the compounds A, B, C, and D based on the reactions described.

### Step 1: Identify Compound (A)
When `MnO2` is fused with `KOH` in the presence of `KNO3`, it forms a dark green compound. The reaction can be represented as:
\[ \text{MnO}_2 + 4 \text{KOH} + 2 \text{KNO}_3 \rightarrow 2 \text{K}_2\text{MnO}_4 + 2 \text{H}_2\text{O} \]
The dark green compound formed is **Potassium Manganate (K2MnO4)**. Thus,
**Compound (A) = K2MnO4**.

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