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Step-by-Step Solution
Step 1: Identify the Metal (A)
The question states that a metal (A) reacts with nitrogen gas upon heating to give a compound (B). Among the provided options, magnesium (Mg) is known to react with nitrogen to form magnesium nitride (Mg3N2) when heated. This hints that metal (A) is magnesium, Mg.
Step 2: Form the Nitride (Compound B)
When magnesium is heated in the presence of nitrogen, it forms magnesium nitride. The balanced chemical equation is:
$3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$
This identifies compound (B) as magnesium nitride, Mg3N2.
Step 3: Reaction of (B) with Water
Next, compound (B) = Mg3N2 is treated with water. Magnesium nitride reacts with water to form magnesium hydroxide and ammonia gas:
$Mg_3N_2 + 6H_2O \longrightarrow 3Mg(OH)_2 + 2NH_3$
The gas produced in this step is ammonia (NH3), which is colorless.
Step 4: Test with Copper Sulfate (CuSO4)
The question also mentions that when the colorless gas (ammonia) is passed through a CuSO4 solution, it acquires a dark blue or violet-blue color. This color change is attributed to the formation of the tetraammine copper(II) complex:
$[Cu(H_2O)_4]^{2+} + 4NH_3 \longrightarrow [Cu(NH_3)_4]^{2+} + 4H_2O$
In solution form, it is written as:
$CuSO_4 + 4NH_3 \longrightarrow [Cu(NH_3)_4]SO_4$
This complex, tetraammine copper(II) sulfate, is responsible for the deep blue color observed.
Step 5: Final Answer
From the above steps, metal (A) is magnesium (Mg) and compound (B) is magnesium nitride (Mg3N2).