Question

Calcium imide on hydrolysis gives gas (P) which on oxidation by bleaching powder gives gas (Q), gas (Q) on reaction with magnesium gives compound (R) which on hydrolysis again gives gas (P). Identity (P), (Q), and (R)

• A. ${\mathrm{N}}_2,{\mathrm{N}}_2{\mathrm{O}}_5,\mathrm{Mg}\left({\mathrm{NO}}_3\right)$
• B. ${\mathrm{N}}_2,{\mathrm{NH}}_3,{\mathrm{MgNH}}_4$
• C. ${\mathrm{NH}}_3,{\mathrm{NO}}_2,\mathrm{Mg}{\left({\mathrm{NO}}_2\right)}_2$
• D. ${\mathrm{NH}}_3,{\mathrm{N}}_2,{\mathrm{Mg}}_3{\mathrm{N}}_2$

Answer 1

The correct option is D

${\mathrm{NH}}_3,{\mathrm{N}}_2,{\mathrm{Mg}}_3{\mathrm{N}}_2$

The explanation for the correct answer

Calcium imide ($\mathrm{CaNH}$) on hydrolysis gives gas ${\mathrm{NH}}_3$ and ${\mathrm{CaOH}}_2$.

$\underset{\mathrm{Calcium}\mathrm{imide}}{\mathrm{CaNH}}\left(\mathrm{s}\right)+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\left(\mathrm{l}\right)\to \underset{\mathrm{Calcium}\mathrm{hydroxide}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}\left(\mathrm{s}\right)+\underset{\mathrm{Ammon}\mathrm{ia}}{{\mathrm{NH}}_3}\left(\mathrm{g}\right)$

So P is ${\mathrm{NH}}_3$

$\underset{\mathrm{Calcium}\mathrm{imide}}{\mathrm{CaNH}}\left(\mathrm{s}\right)+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\left(\mathrm{l}\right)\to \underset{\mathrm{Calcium}\mathrm{hydroxide}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}\left(\mathrm{s}\right)+\underset{\mathrm{Ammon}\mathrm{ia}}{{\mathrm{NH}}_3}\left(\mathrm{g}\right)$

${\mathrm{NH}}_3$ on oxidation by bleaching powder gives gas ${\mathrm{N}}_2$ as byproducts.

So Q is ${\mathrm{N}}_2$

$3\underset{\mathrm{Calcium}\mathrm{hypochlorite}}{{\mathrm{CaOCl}}_2\left(\mathrm{s}\right)}+2\underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{g}\right)}\to 3\underset{\mathrm{Calcium}\mathrm{chloride}}{{\mathrm{CaCl}}_2\left(\mathrm{s}\right)}+3\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Nitrogen}}{{\mathrm{N}}_2\left(\mathrm{g}\right)}$

$N_2$ on reaction with Magnesium gives compound $M{g}_3{N}_2$

$\underset{\mathrm{Nitrogen}}{{\mathrm{N}}_2}\left(\mathrm{g}\right)+3\underset{\mathrm{Magnesium}}{\mathrm{Mg}}\left(\mathrm{s}\right)\to \underset{\mathrm{Magnesium}\mathrm{nitride}}{{\mathrm{Mg}}_3{\mathrm{N}}_2}\left(\mathrm{s}\right)$

$M{g}_3{N}_2$ on hydrolysis gives $N{H}_3$ which is P.

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

Therefore,

P = $N{H}_3$, Q = $N_2$, R = $M{g}_3{N}_2$

the correct option is (d)

The explanation for the incorrect answer

options (a), (b), and (c) do not match with the above explanation.

Hence option (d) is the correct option