Hydrolysis of `XeF_2` gives 

Options

A. `Xe and O_2` only
B. `Xe, HF and O_2`
C. `O_2` only
D. `XeF_6` and `O_2` only

Answer

To solve the question regarding the hydrolysis of `XeF_2`, we will follow these steps:

### Step 1: Write the hydrolysis reaction
When xenon difluoride (`XeF_2`) undergoes hydrolysis, it reacts with water (`H_2O`). The general reaction can be written as:

\[ \text{XeF}_2 + \text{H}_2O \rightarrow \text{Xe} + \text{HF} + \text{O}_2 \]

### Step 2: Balance the reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The unbalanced reaction is:

\[ \text{XeF}_2 + \text{H}_2O \rightarrow \text{Xe} + \text{HF} + \text{O}_2 \]

To balance it, we need to account for the fluorine and oxygen atoms. The balanced reaction will be:

\[ 2 \text{XeF}_2 + 2 \text{H}_2O \rightarrow 2 \text{Xe} + 4 \text{HF} + \text{O}_2 \]

### Step 3: Identify the products
From the balanced equation, we can see that the products of the hydrolysis of `XeF_2` are:
- Xenon (`Xe`)
- Hydrogen fluoride (`HF`)
- Oxygen (`O_2`)

### Step 4: Determine the correct answer
Based on the products identified, the correct answer to the question "Hydrolysis of `XeF_2` gives" is:

**Xenon, Hydrogen Fluoride, and Oxygen.**

### Final Answer:
The hydrolysis of `XeF_2` gives xenon (`Xe`), hydrogen fluoride (`HF`), and oxygen (`O_2`).

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