`XeF_(2)` on hydrolysis (in the presence of alkali) yield :

Options

A. `XeOF_(2)`
B. `XeO_(2)`
C. `XeO_(2)F_(2)`
D. `Xe`

Answer

To solve the question regarding the hydrolysis of xenon difluoride (XeF₂) in the presence of alkali, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant in this case is xenon difluoride (XeF₂). We will also need water (H₂O) and a strong alkali, such as sodium hydroxide (NaOH), for the hydrolysis reaction.

2. **Write the Hydrolysis Reaction**:
When xenon difluoride undergoes hydrolysis in the presence of an alkali, it reacts with water. The general reaction can be represented as:
\[ \text{XeF}_2 + \text{H}_2\text{O} \xrightarrow{\text{NaOH}} \text{Xe} + \text{HF} + \text{NaF} \]

3. **Analyze the Products**:
The products of this reaction include:
- Xenon (Xe)
- Hydrofluoric acid (HF)
- Sodium fluoride (NaF)

4. **Identify the Main Product**:
The main gaseous product formed in this reaction is xenon (Xe), which is a noble gas.

5. **Conclusion**:
Therefore, upon hydrolysis of xenon difluoride in the presence of alkali, the primary product is xenon.

6. **Choose the Correct Option**:
Based on the analysis, the correct answer from the options provided is option D, which states that xenon is formed.

### Final Answer:
The hydrolysis of XeF₂ in the presence of alkali yields xenon (Xe), hydrofluoric acid (HF), and sodium fluoride (NaF). The correct option is D.

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