JEE Advanced 2024 Paper 1 Q02 Chemistry Inorganic Chemistry p-Block (Nitrogen) Chemistry Medium

JEE Advanced 2024 Paper 1 · Q02 · p-Block (Nitrogen) Chemistry

At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid ($HNO_2$) gives the species

  1. A. $H_3O^+$, $NO_3^-$ and $NO$
  2. B. $H_3O^+$, $NO_3^-$ and $NO_2$
  3. C. $H_3O^+$, $NO^-$ and $NO_2$
  4. D. $H_3O^+$, $NO_3^-$ and $N_2O$
Reveal answer + step-by-step solution

Correct answer:A

Solution

$HNO_2$ is unstable in aqueous solution and undergoes disproportionation. Nitrogen is in the $+3$ oxidation state in $HNO_2$; on disproportionation it goes to a higher oxidation state (+5 in $NO_3^-$) and a lower state (+2 in $NO$). The balanced reaction in water is: $$3HNO_2 \longrightarrow 2NO + NO_3^- + H_3O^+$$ Alternatively written as $3HNO_2 \rightarrow HNO_3 + 2NO + H_2O$. Therefore the species formed are $H_3O^+$, $NO_3^-$ and $NO$.

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