JEE Advanced 2024 — Paper 1 — CHEM-I solution

JEE Advanced2024Paper 1CHEM-I

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

A
$H_{3} O^{+}$ , $N O_{3}^{-}$ and NO
B
$H_{3} O^{+}$ , $N O_{3}^{-}$ and $N O_{2}$
C
$H_{3} O^{+}$ , $N O^{-}$ and $N O_{2}$
D
$H_{3} O^{+}$ , $N O_{3}^{-}$ and $N_{2} O$

Solution

Nitrogen in HNO $_{2}$ has oxidation state $+ 3$ . In disproportionation, part of it is oxidised (to a higher state) and part is reduced (to a lower state).

The standard disproportionation in aqueous medium:

$3 HN O_{2} ⟶ HN O_{3} + 2 NO + H_{2} O$

Here N goes from $+ 3$ to $+ 5$ (in $N O_{3}^{-}$ , oxidation) and from $+ 3$ to $+ 2$ (in NO, reduction). In aqueous solution, HNO $_{3}$ ionises completely to give $H_{3} O^{+}$ and $N O_{3}^{-}$ .

$∴$ The species present are $H_{3} O^{+}$ , $N O_{3}^{-}$ and NO. Option (A).

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