A metal X on heating in nitrogen gas gives Y. | vedclass

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Question

(B) The metal $X$ is Magnesium $(Mg)$.When $Mg$ is heated in nitrogen gas,it forms Magnesium nitride $(Y = Mg_3N_2)$:$3Mg + N_2 \xrightarrow{\Delta} M

Vedclass · Accepted Answer

$Mg_3N_2$

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(B) The metal $X$ is Magnesium $(Mg)$.When $Mg$ is heated in nitrogen gas,it forms Magnesium nitride $(Y = Mg_3N_2)$:$3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$When $Mg_3N_2$ reacts with water,it produces ammonia gas $(NH_3)$:$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3(g)$Ammonia gas reacts with $CuSO_4$ solution to form a deep blue complex,tetraamminecopper$(II)$ sulfate:$Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$ (Deep blue colour)Therefore,$Y$ is $Mg_3N_2$.

$A$ metal $X$ on heating in nitrogen gas gives $Y$. $Y$ on reaction with ${H_2}O$ produces a colourless gas. When this gas is passed through $CuSO_4$ solution,it gives a blue colour. What is $Y$?

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