In compound, `OF_(2)` the oxidation state for 'O' is:

Options

A. `+2`
B. `-2`
C. `+4`
D. `+6`

Explanation

To determine the oxidation state of oxygen in the compound OF₂ (oxygen difluoride), we can follow these steps:

### Step 1: Understand the compound
The compound OF₂ consists of one oxygen atom and two fluorine atoms. We need to find the oxidation state of oxygen (let's denote it as 'x').

### Step 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 OF₂, their total contribution to the oxidation state will be:
\[ 2 \times (-1) = -2 \]

### Step 3: Set up the equation
The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can set up the following equation:
\[ x + (-2) = 0 \]

### Step 4: Solve for 'x'
Now, we can solve for 'x':
\[ x - 2 = 0 \]
\[ x = +2 \]

### Conclusion
The oxidation state of oxygen in OF₂ is +2.