Which compound is prepared by the following reaction
`Xe + 2F_(2) overset(Ni vestel)underset(673 K, 5-6 atm)(to)`
`(1:5 volume ratio)`

Options

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

Answer

To determine the compound prepared by the reaction \( \text{Xe} + 2\text{F}_2 \) in a nickel vessel at 673 K and 5-6 atm, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are xenon (Xe) and fluorine (F2). Xenon is a noble gas, which is generally inert but can react with highly electronegative elements like fluorine.

**Hint:** Consider the properties of xenon and its ability to form compounds under specific conditions.

### Step 2: Analyze the Reaction Conditions
The reaction occurs in a nickel vessel at a temperature of 673 K and a pressure of 5-6 atm. These conditions are significant because they provide the energy and pressure needed to overcome the inertness of xenon and facilitate the reaction with fluorine.

**Hint:** High temperatures and pressures can help in breaking the inertness of noble gases.

### Step 3: Determine the Stoichiometry of the Reaction
The stoichiometry given is a 1:5 volume ratio of xenon to fluorine. This means for every 1 mole of xenon, 5 moles of fluorine are used. However, since the reaction shows \( \text{Xe} + 2\text{F}_2 \), we can interpret that 2 moles of \( \text{F}_2 \) (which is equivalent to 4 moles of fluorine atoms) react with 1 mole of xenon.

**Hint:** Understand how to interpret volume ratios in terms of moles, especially when dealing with gases.

### Step 4: Predict the Product
Xenon can form compounds with fluorine. In this reaction, xenon will bond with fluorine atoms. Each fluorine atom requires one electron to complete its octet, and xenon can provide these electrons.

In this case, xenon will form bonds with four fluorine atoms, leading to the formation of \( \text{XeF}_4 \). Although \( \text{XeF}_6 \) can also form in negligible amounts, the primary product in this reaction is \( \text{XeF}_4 \).

**Hint:** Consider the octet rule and how many bonds xenon can form with fluorine.

### Step 5: Conclusion
The compound formed by the reaction \( \text{Xe} + 2\text{F}_2 \) under the given conditions is \( \text{XeF}_4 \).

**Final Answer:** \( \text{XeF}_4 \)

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