`XeF_(2)` and `XeF_(6)` are separately hydrolysed then:

Options

A. both give out `O_2`
B. `XeF_6` gives `O_2 and XeF2` does not
C. `XeF_2` alone gives `O_2`
D. Neither of them gives HF

Answer

To solve the problem regarding the hydrolysis of \( \text{XeF}_2 \) and \( \text{XeF}_6 \), we will analyze the reactions step by step.

### Step 1: Hydrolysis of \( \text{XeF}_2 \)

The hydrolysis of \( \text{XeF}_2 \) can be represented by the following chemical equation:

\[ \text{XeF}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Xe} + 2 \text{HF} + \frac{1}{2} \text{O}_2 \]

**Hint:** Remember that hydrolysis involves the reaction of a compound with water, leading to the formation of new products.

### Step 2: Hydrolysis of \( \text{XeF}_6 \)

Now, let's consider the hydrolysis of \( \text{XeF}_6 \):

\[ \text{XeF}_6 + 3 \text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6 \text{HF} \]

**Hint:** In this case, \( \text{XeF}_6 \) reacts with water to form xenon trioxide and hydrofluoric acid.

### Step 3: Analyzing the Products

From the reactions above, we can summarize the products:

- **For \( \text{XeF}_2 \)**: The products are \( \text{Xe} \), \( \text{HF} \), and \( \frac{1}{2} \text{O}_2 \).
- **For \( \text{XeF}_6 \)**: The products are \( \text{XeO}_3 \) and \( \text{HF} \).

**Hint:** Pay attention to the formation of oxygen gas (\( \text{O}_2 \)) in the products.

### Step 4: Conclusion

From the analysis, we can conclude:

- \( \text{XeF}_2 \) produces \( \frac{1}{2} \text{O}_2 \) upon hydrolysis.
- \( \text{XeF}_6 \) does not produce \( \text{O}_2 \) upon hydrolysis.

Thus, the correct answer is that \( \text{XeF}_2 \) gives \( \text{O}_2 \) while \( \text{XeF}_6 \) does not.

**Final Answer:** The correct option is that \( \text{XeF}_2 \) alone gives \( \text{O}_2 \).