JEE Main On 8 April 2017 Question 30
Question: A metal ‘M’ reacts with nitrogen gas to afford $ ‘M _3N’.‘M _3N’ $ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution of $ CuSO _4 $ to form a deep blue compound. ‘M’ and ‘B’ respectively are : [JEE Online 08-04-2017]
Options:
A) Li and $ NH _3 $
B) Na and $ NH _3 $
C) Ba and $ N _2 $
D) Al and $ N _2 $
Show Answer
Answer:
Correct Answer: A
Solution:
- A metal ’ $M$ ’ reacts with nitrogen gas to form $\mathrm{M}_3 \mathrm{~N}$. This suggests that ’ $M$ ’ is likely a metal that can form a nitride.
- $\mathrm{M}_3 \mathrm{~N}$ on heating gives back ’ M ‘. This indicates that the nitride decomposes back to the metal and nitrogen gas upon heating.
- On reaction with water, it produces a gas ’ $B$ ‘. This suggests the nitride reacts with water, likely producing ammonia $\mathrm{NH}_3$.
- Gas ‘B’ reacts with aqueous $\mathrm{CuSO}_4$ to form a deep blue compound. Ammonia $\mathrm{NH}_3$ is known to form a deep blue complex with $\mathrm{CuSO}_4$, specifically $\mathrm{Cu}\left(\mathrm{NH}_3\right)_4 \mathrm{SO}_4$.
Based on these clues, let’s look at the options:
- Option A: Li and $\mathrm{NH}_3$
Lithium nitride $\mathrm{Li}_3 \mathrm{~N}$ reacts with water to produce $\mathrm{NH}_3$, which matches the description of gas ‘B’.
- Option B: Na and $\mathrm{NH}_3$
Sodium does not typically form a nitride that would react to produce $\mathrm{NH}_3$.
- Option C: Ba and $N_2$
Barium nitride $\mathrm{Ba}_3 \mathrm{~N}_2$ reacts with water to produce ammonia, not nitrogen gas.
- Option D: Al and $N_2$
Aluminum nitride AlN reacts with water to produce ammonia, not nitrogen gas.
The correct answer is Option $\mathrm{A}: \mathrm{Li}$ and $\mathrm{NH}_3$. This is because lithium nitride reacts with water to produce ammonia, which forms a deep blue complex with $\mathrm{CuSO}_4$.
BETA
