Oxidation state of oxygen in `F_2O` is

Options

A. `+1 `
B. `-1 `
C. `+2 `
D. `-2 `

Explanation

To determine the oxidation state of oxygen in \( F_2O \) (oxygen difluoride), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**: The compound in question is \( F_2O \), which is known as oxygen difluoride.

2. **Know the Oxidation State of Fluorine**: Fluorine is the most electronegative element and typically has an oxidation state of \(-1\) in compounds. Since there are two fluorine atoms in \( F_2O \), the total contribution from fluorine to the oxidation state is:
\[ 2 \times (-1) = -2 \]

3. **Set Up the Equation**: Let the oxidation state of oxygen be \( x \). The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can set up the equation:
\[ x + (-2) = 0 \]

4. **Solve for \( x \)**: Rearranging the equation gives:
\[ x - 2 = 0 \\ x = +2 \]

5. **Conclusion**: The oxidation state of oxygen in \( F_2O \) is \( +2 \).

### Final Answer:
The oxidation state of oxygen in \( F_2O \) is \( +2 \).

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