What is the oxidation number of `Xe` in `XeOF_(2)`?

Explanation

To determine the oxidation number of xenon (Xe) in the compound xenon oxydifluoride (XeOF₂), we can follow these steps:

### Step 1: Assign variables
Let the oxidation number of xenon (Xe) be represented as \( X \).

### Step 2: Assign known oxidation numbers
- The oxidation number of oxygen (O) is typically \(-2\).
- The oxidation number of fluorine (F) is \(-1\) since it is the most electronegative element.

### Step 3: Set up the equation
In the compound XeOF₂, we have:
- 1 oxygen atom contributing \(-2\) to the total oxidation state.
- 2 fluorine atoms contributing \(2 \times (-1) = -2\) to the total oxidation state.

The overall charge of the compound is neutral (0), so we can set up the equation:
\[ X + (-2) + (-2) = 0 \]

### Step 4: Simplify the equation
This simplifies to:
\[ X - 4 = 0 \]

### Step 5: Solve for \( X \)
Now, we can solve for \( X \):
\[ X = 4 \]

### Conclusion
The oxidation number of xenon in XeOF₂ is \( +4 \).

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