JEE Chemistry 2024 Question with Solution

Given: Chlorine undergoes disproportionation in alkaline medium.

Find: The values of $a$, $b$, $c$, and $d$ in

$$a\text{Cl}_2 + b\text{OH}^- \rightarrow c\text{ClO}^- + d\text{Cl}^- + e\text{H}_2\text{O}$$

Disproportionation means the same element undergoes both oxidation and reduction. Here, chlorine in $\text{Cl}_2$ forms $\text{Cl}^-$ and $\text{ClO}^-$.

The half-reactions are:

$$\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$$ $$\text{Cl}_2 + 4\text{OH}^- \rightarrow 2\text{ClO}^- + 2\text{H}_2\text{O} + 2e^-$$

After combining and canceling electrons, the balanced overall reaction is:

$$\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O}$$

So, $a = 1$, $b = 2$, $c = 1$, and $d = 1$.

Therefore, the correct option is A. The listed options list is incomplete, but the available option text and solution both indicate $1, 2, 1, \text{ and } 1$.