Oxidation state of Oxygen in `O_(2)F_(2)` is __________.

Answer

To find the oxidation state of oxygen in the compound \( O_2F_2 \), we can follow these steps:

### Step 1: Assign oxidation states
- Let the oxidation state of oxygen be \( x \).
- The oxidation state of fluorine (F) is known to be \(-1\) because it is the most electronegative element in the periodic table.

### Step 2: Write the equation based on the compound
- In the compound \( O_2F_2 \), there are 2 oxygen atoms and 2 fluorine atoms.
- The total oxidation states can be expressed as:
\[ 2x + 2(-1) = 0 \]

### Step 3: Simplify the equation
- This simplifies to:
\[ 2x - 2 = 0 \]

### Step 4: Solve for \( x \)
- Rearranging the equation gives:
\[ 2x = 2 \]
- Dividing both sides by 2 results in:
\[ x = 1 \]

### Conclusion
- Thus, the oxidation state of oxygen in \( O_2F_2 \) is \( +1 \).

### Final Answer
The oxidation state of oxygen in \( O_2F_2 \) is **+1**.

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