1. Identify the reaction type

Nitrous acid, $\mathrm{HNO_2}$, is unstable and undergoes disproportionation in aqueous solution.

In $\mathrm{HNO_2}$, the oxidation state of nitrogen is:

$x + 1 + 2(-2) = 0 \Rightarrow x = +3$

So nitrogen is in the $+3$ oxidation state.

In disproportionation, the same species is simultaneously oxidized and reduced.

2. Possible oxidation and reduction products

• Oxidation of $\mathrm{N}(+3)$ gives $\mathrm{N}(+5)$ in $\mathrm{NO_3^-}$
• Reduction of $\mathrm{N}(+3)$ gives $\mathrm{N}(+2)$ in $\mathrm{NO}$

Thus, the likely disproportionation is:

$\mathrm{HNO_2} \rightarrow \mathrm{NO_3^-} + \mathrm{NO}$

Since the reaction is in aqueous acidic medium, $\mathrm{H_3O^+}$ will also appear.

3. Write the balanced reaction

The standard disproportionation of nitrous acid is:

$3\mathrm{HNO_2} \rightarrow \mathrm{HNO_3} + 2\mathrm{NO} + \mathrm{H_2O}$

In aqueous solution, $\mathrm{HNO_3}$ exists as:

$\mathrm{HNO_3} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{NO_3^-}$

So the species present are:

$\mathrm{H_3O^+},\ \mathrm{NO_3^-},\ \mathrm{NO}$

4. Check options

• A: $\mathrm{H_3O^+}, \mathrm{NO_3^-}$ and $\mathrm{NO}$ ✅
• B: contains $\mathrm{NO_2}$, not the usual disproportionation product at room temperature ❌
• C: contains $\mathrm{NO^-}$, which is not formed here ❌
• D: contains $\mathrm{N_2O}$, not the standard disproportionation product ❌

5. Final answer

The correct option is A.