Which compound is prepared by the following reaction ?
`underset(("2:1 volume ratio"))(Xe+F_(2)) underset(673K)overset(Ni)to`

Options

A. `XeF_(4)`
B. `XeF_(2)`
C. `XeF_(6)`
D. None of these.

Explanation

To determine the compound prepared by the reaction of xenon (Xe) with fluorine (F2) in a 2:1 volume ratio at a temperature of 673 K in the presence of nickel (Ni), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Conditions**:
- We have xenon (Xe) and fluorine (F2) as reactants.
- The reaction takes place at a temperature of 673 K with nickel as a catalyst.
- The volume ratio of Xe to F2 is given as 2:1.

2. **Understanding the Reaction**:
- Xenon reacts with fluorine to form various fluorinated xenon compounds, such as XeF2, XeF4, and XeF6.
- The specific product formed can depend on the stoichiometry of the reactants.

3. **Analyze the Volume Ratio**:
- In the given reaction, the volume ratio of 2:1 indicates that for every 2 volumes of xenon, there is 1 volume of fluorine.
- This suggests that the reaction will favor the formation of a product that corresponds to this ratio.

4. **Determine the Likely Product**:
- When xenon reacts with fluorine in a 2:1 volume ratio, the predominant product formed is xenon difluoride (XeF2).
- If the ratio were different (for example, 1:1 or lower amounts of fluorine), other products like XeF4 or XeF6 could be formed, but with the 2:1 ratio, XeF2 is the main product.

5. **Conclusion**:
- Therefore, the compound prepared by the reaction of xenon with fluorine in a 2:1 volume ratio at 673 K with nickel is **XeF2**.

### Final Answer:
The compound prepared by the reaction is **XeF2**.