When `XO_2` is fused with an alkali metal hydroxide in presence of an oxidizing agent such as `KNO_3` a dark green product is formed which disproportionates in acidic solution to afford a dark purple solution. Xis:

Options

A. Mn
B. Cr
C. V
D. Ti

Explanation

To solve the question, we need to identify the element X in the compound `XO_2` that, when fused with an alkali metal hydroxide in the presence of an oxidizing agent, produces a dark green product that can further disproportionate in acidic solution to yield a dark purple solution.

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- We have the compound `XO_2`, an alkali metal hydroxide (like KOH), and an oxidizing agent (like KNO_3).
- The reaction will involve the conversion of `XO_2` in the presence of these components.

2. **Determine the Possible Elements**:
- The options given are Manganese (Mn), Chromium (Cr), Vanadium (V), and Titanium (Ti).
- We need to consider the oxides of these elements and their behavior in reactions with alkali hydroxides and oxidizing agents.

3. **Analyze the Reaction**:
- When `XO_2` (which we suspect to be MnO2 based on the expected products) is fused with KOH and KNO3, it forms a dark green product, which is K2MnO4 (potassium manganate).
- The reaction can be summarized as:
\[ 2MnO_2 + 4KOH + KNO_3 \rightarrow K_2MnO_4 + KNO_2 + 2H_2O \]

4. **Disproportionation in Acidic Solution**:
- The dark green K2MnO4 can disproportionate in acidic conditions to yield MnO4^- ions, which are known to produce a dark purple solution.
- The reaction can be represented as:
\[ K_2MnO_4 + 4H^+ \rightarrow 2MnO_4^- + 2H_2O \]

5. **Conclusion**:
- The element X that corresponds to the behavior described in the question is Manganese (Mn), as it leads to the formation of the dark green K2MnO4 and subsequently the dark purple MnO4^- in acidic solution.

### Final Answer:
The element X is **Manganese (Mn)**.

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