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A metal $M$ on heating in nitrogen gas gives $Y$. $Y$ on treatment with $\mathrm{H}_2 \mathrm{O}$ gives a colourless gas which when passed through $\mathrm{CuSO}_4$ solution gives a blue colour. $Y$ is
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Verified Answer
The correct answer is:
$\mathrm{Mg}_3 \mathrm{~N}_2$
$\mathrm{Mg}_3 \mathrm{~N}_2$
$\underset{M}{3 \mathrm{Mg}}+\mathrm{N}_2 \longrightarrow \underset{Y}{\mathrm{Mg}_3} \mathrm{~N}_2$
$$
\begin{aligned}
& \underset{Y}{\mathrm{Mg}_3 \mathrm{~N}_2}+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+\underset{\text { colourless }}{2 \mathrm{NH}_3 \uparrow} \\
& \mathrm{CuSO}_4+4 \mathrm{NH}_3 \longrightarrow\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 \\
&
\end{aligned}
$$
Blue complex
$$
\begin{aligned}
& \underset{Y}{\mathrm{Mg}_3 \mathrm{~N}_2}+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+\underset{\text { colourless }}{2 \mathrm{NH}_3 \uparrow} \\
& \mathrm{CuSO}_4+4 \mathrm{NH}_3 \longrightarrow\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 \\
&
\end{aligned}
$$
Blue complex
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