Oxidation state of Fluorine in `OF_2` is

Options

A. 1
B. 2
C. (-2)
D. (-1)

Answer

To determine the oxidation state of fluorine in the compound \( OF_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Known Oxidation States**:
- Oxygen typically has an oxidation state of -2 in most compounds.
- Fluorine is the most electronegative element and always has an oxidation state of -1 in its compounds.

2. **Set Up the Equation**:
- Let the oxidation state of fluorine be \( x \). Since there are two fluorine atoms in \( OF_2 \), the contribution from fluorine will be \( 2x \).
- The overall charge of the compound \( OF_2 \) is neutral (0).

3. **Write the Oxidation State Equation**:
- The equation can be set up as follows:
\[ \text{Oxidation state of O} + \text{Total oxidation state of F} = 0 \]
- Substituting the known values:
\[ -2 + 2(-1) = 0 \]

4. **Solve the Equation**:
- Simplifying the equation:
\[ -2 - 2 = -4 \]
- Since this does not satisfy our equation, we need to check our assumptions.

5. **Re-evaluate the Oxidation State of Fluorine**:
- In \( OF_2 \), the oxidation state of oxygen is indeed -2, and since there are two fluorine atoms, we can set the equation as:
\[ -2 + 2x = 0 \]
- Solving for \( x \):
\[ 2x = 2 \implies x = 1 \]
- Therefore, the oxidation state of fluorine in \( OF_2 \) is -1.

### Final Answer:
The oxidation state of fluorine in \( OF_2 \) is -1.