Which compound is prepared by the following reaction?
`Xe+2F_(2) underset(673K, 5-6 atm)overset(Ni" vessel ")(to)`

Options

A. `XeF_2`
B. `XeF_6`
C. `XeF_4`
D. `XeOF_2`

Answer

To determine which compound is prepared from the reaction of xenon (Xe) with fluorine (F2) under the given conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Conditions**:
- The reactants are xenon (Xe) and fluorine (F2).
- The reaction occurs at a temperature of 673 K and a pressure of 5-6 atm in the presence of a nickel vessel.

2. **Understand the Reactivity**:
- Xenon is a noble gas that can form compounds with highly electronegative elements like fluorine.
- Fluorine has a high affinity for xenon, which allows for the formation of xenon fluorides.

3. **Determine the Possible Products**:
- The possible products from the reaction of xenon with fluorine include:
- Xenon difluoride (XeF2)
- Xenon tetrafluoride (XeF4)
- Xenon hexafluoride (XeF6)
- Xenon oxytetrafluoride (XeOF2)

4. **Analyze the Reaction Conditions**:
- The high temperature (673 K) and high pressure (5-6 atm) favor the formation of more stable and higher fluorinated products.
- Under these conditions, xenon tetrafluoride (XeF4) is likely to be formed, as it is a stable product of xenon reacting with fluorine.

5. **Confirm the Product**:
- The balanced chemical reaction can be represented as:
\[ \text{Xe} + 2 \text{F}_2 \rightarrow \text{XeF}_4 \]
- This indicates that the product formed is indeed xenon tetrafluoride (XeF4).

6. **Conclusion**:
- Based on the analysis, the compound prepared from the reaction of xenon with fluorine under the specified conditions is **xenon tetrafluoride (XeF4)**.

### Final Answer:
The compound prepared is **Xenon tetrafluoride (XeF4)**.