JEE Main 2020 · 8 January · evening · Single correct · +4 / −1
A metal (A) on heating in nitrogen gas gives compound B. B on treatment with H₂O gives a colourless gas which when passed through CuSO₄ solution gives a dark blue-violet coloured solution. A and B respectively, are :
- (a) Na and NaNO₃
- (b) Na and Na₃N
- (c) Mg and Mg₃N₂
- (d) Mg and Mg(NO₃)₂
Answer and solution
Answer (c) Mg and Mg₃N₂
Solution
Chemistry ma’am · 4 steps
Deep blue with CuSO₄ means ammonia, and ammonia from water means a nitride, so look for Mg.
- Identify the gas. A colourless gas that turns CuSO₄ solution deep blue-violet is ammonia. It forms the tetraammine copper(II) complex.
Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ (deep blue-violet)
- Work back to B. Ammonia released by plain water points to a metal nitride. B must be a nitride, not a nitrate. A nitrate also needs oxygen, and pure N₂ supplies none.
- Pick the metal. Magnesium burns in nitrogen to give 3Mg + N₂ → Mg₃N₂. Water then hydrolyses it: Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃. Sodium does not combine with N₂ on heating. Among the alkali metals only lithium does.
- So. A is Mg and B is Mg₃N₂, which is option (c).
The idea to keep. Li and the alkaline earth metals (Mg onwards) form nitrides with N₂, and every nitride gives NH₃ with water. A deep blue colour with Cu²⁺ is your ammonia test.
Answer from the NTA final answer key. Solution by PrepWiser.