The oxidation state of oxygen in the following reaction changes, `K+O_(2) rarr KO_(2)`

Options

A. From `0 rarr -1`
B. From `0 rarr -2`
C. From `0 rarr -1/2`
D. From `0 rarr +1`

Answer

To determine the change in oxidation state of oxygen in the reaction \( K + O_2 \rightarrow KO_2 \), we will follow these steps:

### Step 1: Identify the oxidation states in the reactants
- In the reactants, potassium (K) is in its elemental form, so its oxidation state is \( 0 \).
- Oxygen (O) in \( O_2 \) is also in its elemental form, so its oxidation state is \( 0 \).

### Step 2: Write down the oxidation states in the products
- In the product \( KO_2 \), potassium (K) typically has an oxidation state of \( +1 \).
- Let’s denote the oxidation state of oxygen in \( KO_2 \) as \( x \). Since the compound is neutral, the sum of the oxidation states must equal \( 0 \).

### Step 3: Set up the equation for the oxidation state of oxygen
- The formula for \( KO_2 \) indicates that there are two oxygen atoms. Therefore, we can write the equation:
\[ +1 + 2x = 0 \]
where \( +1 \) is the oxidation state of potassium, and \( 2x \) accounts for the two oxygen atoms.

### Step 4: Solve for the oxidation state of oxygen
- Rearranging the equation gives:
\[ 2x = -1 \]
\[ x = -\frac{1}{2} \]

### Step 5: Determine the change in oxidation state
- The oxidation state of oxygen changes from \( 0 \) (in \( O_2 \)) to \( -\frac{1}{2} \) (in \( KO_2 \)).
- Therefore, the change in oxidation state of oxygen is from \( 0 \) to \( -\frac{1}{2} \).

### Conclusion
The oxidation state of oxygen in the reaction changes from \( 0 \) to \( -\frac{1}{2} \). Thus, the correct answer is option C: from \( 0 \) to \( -\frac{1}{2} \).

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