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

Options

A. both give out `O_(2)`
B. `XeF_(6)` gives `O_(2)` and does not
C. `XeF_(2)` along gives `O_(2)`
D. Neither of them gives HF

Explanation

To solve the question 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 reaction of xenon difluoride (\( \text{XeF}_2 \)) with water can be represented as follows:

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

**Products:**
- Xenon (\( \text{Xe} \))
- Hydrogen fluoride (\( \text{HF} \))
- Oxygen (\( \text{O}_2 \))

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

Next, we analyze the hydrolysis of xenon hexafluoride (\( \text{XeF}_6 \)):

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

**Products:**
- Xenon trioxide (\( \text{XeO}_3 \))
- Hydrogen fluoride (\( \text{HF} \))

### Step 3: Summary of Products

From the hydrolysis reactions, we can summarize the products obtained from each reaction:

- From \( \text{XeF}_2 \): \( \text{Xe} + 2 \text{HF} + \frac{1}{2} \text{O}_2 \)
- From \( \text{XeF}_6 \): \( \text{XeO}_3 + 6 \text{HF} \)

### Step 4: Analyzing the Options

Now, we will analyze the options based on the products formed:

1. **Both give \( \text{O}_2 \)**: This is incorrect because \( \text{XeF}_6 \) does not produce \( \text{O}_2 \).
2. **\( \text{XeF}_6 \) gives \( \text{O}_2 \) and \( \text{XeF}_2 \)**: This is also incorrect because \( \text{XeF}_6 \) does not produce \( \text{O}_2 \).
3. **\( \text{XeF}_2 \) gives \( \text{O}_2 \)**: This is correct as \( \text{XeF}_2 \) produces \( \frac{1}{2} \text{O}_2 \).
4. **Neither of them gives \( \text{HF} \)**: This is incorrect because both reactions produce \( \text{HF} \).

### Conclusion

The correct option is that \( \text{XeF}_2 \) gives \( \text{O}_2 \) while \( \text{XeF}_6 \) does not produce \( \text{O}_2 \).