$\mathrm{KO}_2$, reacts with water to form $A, B$ and $C$. $B$ forms $C$ when it reacts with iodine in basic medium. What are $B$ and $C$ respectively?

• A

  $\mathrm{KOH}, \mathrm{H}_2 \mathrm{O}_2$
• B

  $\mathrm{K}_2 \mathrm{O}_2, \mathrm{H}_2 \mathrm{O}_2$
• C

  $\mathrm{KOH}, \mathrm{O}_2$
• Correct answer
  D

  $\mathrm{H}_2 \mathrm{O}_2, \mathrm{O}_2$

Solution

(i) When $\mathrm{KO}_2$ reacts with water, it gives $(A)$, $(B)$ and $(C)$ as follows:
(ii) When $(B)$, i.e. $\mathrm{H}_2 \mathrm{O}_2$ reacts with iodine in basic medium, it gives $(C)$, i.e. $\mathrm{O}_2$, as shown below :
$\because \mathrm{KOH}$ can not reduce $\mathrm{I}_2$, thus option (d) is the correct answer.

Hence, the correct answer is option: 'D'.