Question

The oxidation number of oxygen in KO\(_3\), Na\(_2\)O\(_2\) is

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

Hint:

In peroxides, oxygen is always \(-1\). In superoxides (like KO\(_2\)), oxygen is \(-\frac{1}{2}\).

Answer 1

The Correct Option is D

Step 1: Oxidation state of oxygen in KO\(_3\).
In KO\(_3\), potassium has oxidation state \(+1\).
Let oxidation state of oxygen be \(x\).
\[ +1 + 3x = 0 \Rightarrow 3x = -1 \Rightarrow x = -\frac{1}{3} = -0.33 \]
Step 2: Oxidation state of oxygen in Na\(_2\)O\(_2\).
Na\(_2\)O\(_2\) is a peroxide.
In peroxides, oxygen has oxidation state \(-1\).
Step 3: Final conclusion.
So oxidation numbers are \(-0.33\) and \(-1\).
Final Answer:
\[ \boxed{\text{(D) } -0.33,\,-1} \]