Hydrolysis of `XeF_4` with excess water gives :

Options

A. `XeOF_4`
B. `XeO_2F_2`
C. `Xe`
D. `XeO_4`

Explanation

To solve the question regarding the hydrolysis of `XeF_4` with excess water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactant in this case is xenon tetrafluoride (XeF₄) and water (H₂O).

2. **Understand the Reaction**: When XeF₄ reacts with excess water, it undergoes hydrolysis. This reaction is known to be vigorous.

3. **Write the Reaction**: The hydrolysis of XeF₄ can be represented by the following chemical equation:
\[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{HF} + \text{O}_2 \]

4. **Balance the Reaction**: To ensure that the number of atoms of each element is the same on both sides of the equation, we need to balance it. The balanced equation is:
\[ \text{XeF}_4 + 2\text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + 4\text{HF} + \text{O}_2 \]

5. **Identify the Products**: From the balanced equation, the products of the reaction are:
- Xenon (Xe)
- Xenon trioxide (XeO₃)
- Hydrogen fluoride (HF)
- Oxygen (O₂)

6. **Conclusion**: Therefore, the hydrolysis of `XeF₄` with excess water produces xenon, xenon trioxide, hydrogen fluoride, and oxygen.

### Final Answer:
The products of the hydrolysis of `XeF₄` with excess water are:
- Xenon (Xe)
- Xenon trioxide (XeO₃)
- Hydrogen fluoride (HF)
- Oxygen (O₂)