Oxidation state of O in `KO_(2)` is

Options

A. `-2`
B. 0
C. `+2`
D. None of these

Explanation

To determine the oxidation state of oxygen in potassium superoxide (KO₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**:
We are given KO₂, which is known as potassium superoxide.

2. **Understand the Components**:
In KO₂, potassium (K) is a metal and typically has an oxidation state of +1. The superoxide ion (O₂⁻) consists of two oxygen atoms.

3. **Assign Oxidation State to Potassium**:
Since potassium is in Group 1 of the periodic table, its oxidation state is +1.

4. **Set Up the Equation**:
Let the oxidation state of oxygen be represented as \( x \). In KO₂, we have one potassium atom and two oxygen atoms. The overall charge of the compound is neutral (0). Therefore, we can set up the equation:
\[ +1 + 2x = 0 \]

5. **Solve for x**:
Rearranging the equation gives:
\[ 2x = -1 \]
Dividing both sides by 2:
\[ x = -\frac{1}{2} \]

6. **Interpret the Result**:
The oxidation state of each oxygen atom in KO₂ is -1/2. Since there are two oxygen atoms in the superoxide ion, the total contribution from the oxygen atoms is -1.

7. **Conclusion**:
Therefore, the oxidation state of oxygen in KO₂ is -1/2.

### Final Answer:
The oxidation state of oxygen in KO₂ is -1/2.