The oxidation number of oxygen in `KO_(3),Na_(2)O_(2` respectively are:

Options

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

Answer

To determine the oxidation number of oxygen in KO₃ and Na₂O₂, we can follow these steps:

### Step 1: Determine the oxidation state of potassium in KO₃
In KO₃, potassium (K) is an alkali metal, and its oxidation state is always +1.

### Step 2: Set up the equation for KO₃
Let the oxidation state of oxygen be \( x \). Since there are three oxygen atoms in KO₃, we can express the total oxidation state of oxygen as \( 3x \). The overall charge of the compound is neutral (0).

The equation can be set up as follows:
\[ \text{Oxidation state of K} + \text{Oxidation state of O} = 0 \]
\[ +1 + 3x = 0 \]

### Step 3: Solve for \( x \) in KO₃
Rearranging the equation gives us:
\[ 3x = -1 \]
\[ x = -\frac{1}{3} \]

Thus, the oxidation number of oxygen in KO₃ is \(-\frac{1}{3}\).

### Step 4: Determine the oxidation state of sodium in Na₂O₂
In Na₂O₂, sodium (Na) also has an oxidation state of +1.

### Step 5: Set up the equation for Na₂O₂
Let the oxidation state of oxygen again be \( y \). There are two oxygen atoms in Na₂O₂, so the total oxidation state of oxygen can be expressed as \( 2y \). The overall charge of the compound is neutral (0).

The equation can be set up as follows:
\[ \text{Oxidation state of Na} + \text{Oxidation state of O} = 0 \]
\[ 2(+1) + 2y = 0 \]

### Step 6: Solve for \( y \) in Na₂O₂
Rearranging the equation gives us:
\[ 2 + 2y = 0 \]
\[ 2y = -2 \]
\[ y = -1 \]

Thus, the oxidation number of oxygen in Na₂O₂ is \(-1\).

### Final Answers:
- The oxidation number of oxygen in KO₃ is \(-\frac{1}{3}\).
- The oxidation number of oxygen in Na₂O₂ is \(-1\).