Question

A metal A on heating in nitrogen gas gives compound B. B on treatment with ${\mathrm{H}}_2\mathrm{O}$ gives a colourless gas which when passed through ${\mathrm{CuSO}}_4$ solution gives a dark blue-violet coloured solution A and B respectively, are :

• A. $\mathrm{Na}$ and ${\mathrm{Na}}_3\mathrm{N}$
• B. $\mathrm{Mg}$ and ${\mathrm{Mg}}_3{\mathrm{N}}_2$
• C. $\mathrm{Mg}$ and $\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2$
• D. $\mathrm{Na}$ and ${\mathrm{NaNO}}_3$

Answer 1

Answer: (A), (B)

$\mathrm{Mg}$ and ${\mathrm{Mg}}_3{\mathrm{N}}_2$

Explanation for the correct options:

A) $\mathbf{Na}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{NaN}}_{\mathbf{3}}$ and B) $\mathbf{Mg}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{Mg}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$:

When metal Mg or Na is heated with nitrogen gas it give ${\mathrm{Mg}}_3{\mathrm{N}}_2$ or ${\mathrm{NaN}}_3$.

$3\mathrm{Mg}\left(\mathrm{s}\right)+{\mathrm{N}}_2\left(\mathrm{s}\right)\to {\mathrm{Mg}}_3{\mathrm{N}}_2\left(\mathrm{s}\right)$

$3\mathrm{Na}\left(\mathrm{s}\right)+\left(1/2\right){\mathrm{N}}_2\left(\mathrm{g}\right)\to {\mathrm{Na}}_3\mathrm{N}\left(\mathrm{s}\right)$

${\mathrm{Mg}}_3{\mathrm{N}}_2$ or ${\mathrm{Na}}_3\mathrm{N}$ on treatment with water gives colorless ammonia.

${\mathrm{Mg}}_3{\mathrm{N}}_2\left(\mathrm{s}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 3\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

${\mathrm{Na}}_3\mathrm{N}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 3\mathrm{NaOH}\left(\mathrm{s}\right)+{\mathrm{NH}}_3\left(aq\right)$

Ammonia reacts with ${\mathrm{CuSO}}_4$ to form blue complex $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4$.

${\mathrm{CuSO}}_4\left(\mathrm{s}\right)+4{\mathrm{NH}}_3\left(\mathrm{g}\right)\to \left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4\left(\mathrm{s}\right)$

Explanation for the incorrect options:

C) $\mathbf{Mg}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Mg}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$:

This option is not applicable because when $\mathrm{Mg}$ react with nitrogen gas it do not form $\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2$.

So the compound $\mathrm{B}$ cannot be $\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2$.

D) $\mathbf{Na}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{NaNO}}_{\mathbf{3}}$:

This option is not applicable because when $\mathrm{Na}$ react with nitrogen gas it do not form ${\mathrm{NaNO}}_3$.

So the compound $\mathrm{B}$ cannot be ${\mathrm{NaNO}}_3$.

Hence options A and B are correct, solutions $\mathrm{A}$ and $\mathrm{B}$ obtained after the reactions are $\mathbf{Na}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{NaN}}_{\mathbf{3}}$and $\mathbf{Mg}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{Mg}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$.