The oxidation number of oxygen in `KO_3,Na_2O_2` is

Options

A. `3,2`
B. `1,0`
C. `0,1`
D. `-0.33,-1`

Explanation

To determine the oxidation number of oxygen in \( KO_3 \) and \( Na_2O_2 \), we will analyze each compound step by step.

### Step 1: Determine the oxidation number of oxygen in \( KO_3 \)

1. **Identify the oxidation states of other elements**:
- Potassium (K) is an alkali metal and typically has an oxidation state of +1.

2. **Set up the equation**:
- Let the oxidation number of oxygen be \( x \).
- In \( KO_3 \), there are 3 oxygen atoms, so the total contribution from oxygen is \( 3x \).
- The overall charge of the compound is neutral (0).
- The equation can be set up as:
\[ +1 + 3x = 0 \]

3. **Solve for \( x \)**:
\[ 3x = -1 \\ x = -\frac{1}{3} \]

### Step 2: Determine the oxidation number of oxygen in \( Na_2O_2 \)

1. **Identify the oxidation states of other elements**:
- Sodium (Na) is also an alkali metal and has an oxidation state of +1.
- There are 2 sodium atoms, contributing a total of +2.

2. **Set up the equation**:
- Let the oxidation number of oxygen be \( y \).
- In \( Na_2O_2 \), there are 2 oxygen atoms, so the total contribution from oxygen is \( 2y \).
- The overall charge of the compound is neutral (0).
- The equation can be set up as:
\[ 2(+1) + 2y = 0 \]

3. **Solve for \( y \)**:
\[ 2 + 2y = 0 \\ 2y = -2 \\ y = -1 \]

### Summary of Results

- The oxidation number of oxygen in \( KO_3 \) is \( -\frac{1}{3} \).
- The oxidation number of oxygen in \( Na_2O_2 \) is \( -1 \).

### Final Answer

- The oxidation number of oxygen in \( KO_3 \) is \( -\frac{1}{3} \).
- The oxidation number of oxygen in \( Na_2O_2 \) is \( -1 \).