Chemistry · Redox Reactions

JEE Advanced 2024 — Paper 1 — Question 35

At room temperature, disproportionation of aqueous in situ generated $\mathrm{HNO_2}$ gives the species:

Option A:
$\mathrm{H_3O^+, NO_3^-, NO}$
Correct
Option B:
$\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NO}_{3}^{-}$ and $\mathrm{NO}_{2}$
Option C:
$\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NO}^{-}$ and $\mathrm{NO}_{2}$
Option D:
$\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NO}_{3}{ }^{-}$ and $\mathrm{N}_{2} \mathrm{O}$

Answer: A

At room temperature, $\mathrm{HNO_2}$ undergoes disproportionation:

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

In aqueous medium: $\mathrm{HNO_3 \rightarrow H_3O^+ + NO_3^-}$

Thus the species formed are $\mathrm{H_3O^+, NO_3^-, NO}$ .

Thus, the correct option is A.

Answer key and solution verified before publishing.

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