Disproportionation Species of Nitrous Acid in Aqueous Solution

To determine the species formed during the disproportionation of an aqueous solution of in situ generated nitrous acid ($\text{HNO}_2$), we analyze the stability and redox behavior of $\text{HNO}_2$.

1. Oxidation State Analysis: In nitrous acid ($\text{HNO}_2$), the oxidation state of nitrogen is $+3$. Since $+3$ is an intermediate oxidation state for nitrogen (which ranges from $-3$ to $+5$), $\text{HNO}_2$ is thermodynamically unstable in aqueous solution at room temperature and undergoes a disproportionation reaction.

2. Disproportionation Reaction: Nitrous acid disproportionates into nitric acid ($\text{HNO}_3$, where nitrogen has an oxidation state of $+5$) and nitric oxide ($\text{NO}$, where nitrogen has an oxidation state of $+2$): $3\text{HNO}_2(\text{aq}) \longrightarrow \text{HNO}_3(\text{aq}) + 2\text{NO}(\text{g}) + \text{H}_2\text{O}(\text{l})$

3. Dissociation in Aqueous Solution: In an aqueous medium, the strong acid $\text{HNO}_3$ completely ionizes into hydronium ions ($\text{H}_3\text{O}^+$) and nitrate ions ($\text{NO}_3^-$): $\text{HNO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \longrightarrow \text{H}_3\text{O}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$

   Combining the overall chemical process: $3\text{HNO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \longrightarrow \text{H}_3\text{O}^+(\text{aq}) + \text{NO}_3^-(\text{aq}) + 2\text{NO}(\text{g})$

Thus, the species present in the solution after disproportionation are $\text{H}_3\text{O}^+$, $\text{NO}_3^-$, and $\text{NO}$.

Correct Option: (A)