JEE Main · 2023 · 29 Jan 2023 · Shift 1 · p-Block Elements

"A" obtained by Ostwald's method involving air oxidation of

NH_{3}

, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test

(A) $NO_{2}, N_{2} O_{5}$
(B) $NO_{2}, N_{2} O_{4}$
(C) $NO, NO_{2}$ ✓ Correct
(D) $N_{2} O_{3}, NO_{2}$

Answer: (C)

Solution

In Ostwald's process ammonia is oxidised in air over a catalyst to nitric oxide,

4 N H_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O

, so A is nitric oxide. Further air oxidation converts it to nitrogen dioxide,

2 NO + O_{2} \to 2 N O_{2}

, so B is nitrogen dioxide. On hydration the dioxide disproportionates,

3 N O_{2} + H_{2} O \to 2 HN O_{3} + NO

, giving the oxoacid nitric acid together with regenerated nitric oxide, and the nitrate from nitric acid answers the brown-ring test. Hence A and B are

NO

and

N O_{2}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 29 Jan 2023 · Shift 1 · p-Block Elements

Solution

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