A metal (A) on heating in nitrogen gas gives compound B. B on treatment with H_2O gives a colourless gas which when passed through CuSO_4 solution gives a dark blue-violet coloured solution. A and B respectively, are :

P Block Elements PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q19 with Solution

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Correct answer: C

A metal $A$ is heated in nitrogen gas to give compound $B$ .

Among the given metals, magnesium reacts directly with nitrogen on heating to form magnesium nitride:

$3Mg + N_2 \rightarrow Mg_3N_2$

Sodium forming $Na_3N$ is not the standard stable nitride considered here, and $NaNO_3$ or $Mg(NO_3)_2$ cannot form by heating in nitrogen gas alone.

So $B$ is likely a nitride.

Metal nitrides react with water to give ammonia gas.

For magnesium nitride:

$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$

The gas formed is colourless, which matches $NH_3$ .

Ammonia reacts with copper sulphate solution to form a deep/dark blue-violet complex:

$Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$

This is the characteristic tetraammine copper(II) complex, confirming the gas is $NH_3$ .

A: $Na$ and $NaNO_3$
Incorrect. Nitrogen gas does not convert sodium into sodium nitrate.
B: $Na$ and $Na_3N$
Incorrect in this context; sodium nitride is not the expected stable product here.
C: $Mg$ and $Mg_3N_2$
Correct. It forms $NH_3$ with water, which gives deep blue-violet colour with $CuSO_4$ .
D: $Mg$ and $Mg(NO_3)_2$
Incorrect. Nitrogen gas does not form magnesium nitrate.

Thus,

$A = Mg, \qquad B = Mg_3N_2$

So the correct option is C.