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At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid (\({\mathrm{HNO}}_{2}\)…

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At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid (\({\mathrm{HNO}}_{2}\)) gives the species

a

\({H}_{3}{O}^{+},N{{O}_{3}}^{−}andNO\)

b

\({H}_{3}{O}^{+},N{{O}_{3}}^{−}andN{O}_{2}\)

c

\({H}_{3}{O}^{+},N{O}^{−}andN{O}_{2}\)

d

\({H}_{3}{O}^{+},N{{O}_{3}}^{−}and{N}_{2}O\)

✓ Correct answer: a)

\({H}_{3}{O}^{+},N{{O}_{3}}^{−}andNO\)

Explanation

In HNO₂:

Oxidation state of N = +3

In disproportionation, same species undergoes oxidation and reduction simultaneously

So N(+3) can:

Oxidize → +5 (NO₃⁻)
Reduce → +2 (NO)

3 HNO₂ → HNO₃ + 2 NO + H₂O

In aqueous solution: HNO₃ → H₃O⁺ + NO₃⁻

Final species present
H₃O⁺
NO₃⁻
NO

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