QJEEAdv24Ch26c4: JEE Advanced 2024 Chemistry - Class 12 JEEAdv24Ch26c4 Solution: At room temperature, disproportionation of an aqueous solution…

To determine the products of the disproportionation of nitrous acid ($${\mathrm{HNO}}_2$$) at room temperature, let's analyze the chemical behavior of $${\mathrm{HNO}}_2$$ in aqueous solution.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced, yielding two different products. In the case of nitrous acid ($${\mathrm{HNO}}_2$$), the reaction can be represented as follows:

$$3{\mathrm{HNO}}_2\to {\mathrm{HNO}}_3+{\mathrm{H}}_2\mathrm{O}+2\mathrm{NO}$$

Breaking this down, we see that nitrous acid disproportionates into nitric acid ($${\mathrm{HNO}}_3$$), water ($${\mathrm{H}}_2\mathrm{O}$$), and nitrogen monoxide ($$\mathrm{NO}$$). Additionally, in an aqueous solution, nitric acid dissociates to form $${{\mathrm{H}}_3\mathrm{O}}^{+}$$ (hydronium ion) and $${{\mathrm{NO}}_3}^{-}$$ (nitrate ion).

Therefore, the species formed from this disproportionation reaction are $${{\mathrm{H}}_3\mathrm{O}}^{+}$$, $${{\mathrm{NO}}_3}^{-}$$, and $$\mathrm{NO}$$.

From the given options, the correct answer is:

Option A

$${{\mathrm{H}}_3\mathrm{O}}^{+}$$, $${{\mathrm{NO}}_3}^{-}$$, and $$\mathrm{NO}$$