`underset((1:20))(Xe+F_(2)) overset(573K,60-70" bar")to?`
The compound formed in above reaction is

Options

A. `XeF_(2)`
B. `XeF_(4)`
C. `XeF_(6)`
D. `XeOF_(4)`

Answer

To solve the question regarding the reaction of xenon (Xe) with fluorine (F₂) at a specific temperature and pressure, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Conditions**:
- The reactants are xenon (Xe) and fluorine (F₂).
- The ratio of the reactants is 1:20 (Xe:F₂).
- The reaction occurs at a temperature of 573 K and a pressure of 60-70 bar.

2. **Determine the Expected Product**:
- Under these conditions, xenon reacts with fluorine to form a fluorinated compound. The high pressure and temperature favor the formation of a more complex fluorine compound.

3. **Analyze the Possible Products**:
- The possible products from the reaction of xenon with fluorine include:
- Xenon difluoride (XeF₂)
- Xenon tetrafluoride (XeF₄)
- Xenon hexafluoride (XeF₆)
- Xenon oxo tetrafluoride (XeOF₄)

4. **Select the Correct Product Based on Conditions**:
- Given the high ratio of fluorine (1:20), the conditions (573 K and 60-70 bar) are conducive to the formation of xenon hexafluoride (XeF₆), which is a stable product formed under these conditions.

5. **Conclusion**:
- Therefore, the compound formed in the reaction of xenon with fluorine at the specified conditions is **Xenon hexafluoride (XeF₆)**.

### Final Answer:
The compound formed in the reaction of Xe + F₂ at 573 K and 60-70 bar is **XeF₆** (xenon hexafluoride).

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