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 give a dark purple solution. X is:

Options

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

Answer

To solve the question, we need to identify the element 'X' in the compound `XO2` based on the given reactions and products formed. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Reaction
The question states that when `XO2` is fused with an alkali metal hydroxide (like KOH) in the presence of an oxidizing agent (like KNO3), a dark green product is formed.

### Step 2: Identifying the Dark Green Product
The dark green product formed in the reaction is likely to be a manganese compound, specifically `K2MnO4`. This is known to be formed when manganese dioxide (`MnO2`) reacts with potassium hydroxide in the presence of an oxidizing agent.

### Step 3: Disproportionation Reaction
The question mentions that this dark green product disproportionates in an acidic solution to yield a dark purple solution. The dark purple solution is identified as potassium permanganate (`KMnO4`), which is a well-known compound.

### Step 4: Writing the Balanced Equation
The balanced equation for the disproportionation of `K2MnO4` in acidic solution can be written as:
\[ K2MnO4 + 4H^+ \rightarrow 2KMnO4 + MnO2 + 2H2O \]

### Step 5: Conclusion
From the above steps, we can conclude that the element 'X' in the compound `XO2` is manganese (Mn). The other options (chromium, vanadium, and titanium) do not yield the same products under the given conditions.

### Final Answer
Thus, the answer is:
**X is Manganese (Mn).**

---