Qualitative Analysis - Result Question 62
####65. An orange solid $A$ on heating gave a green residue $B$, a colourless gas $C$ and water vapour. The dry gas $C$ on passing over heated $Mg$ gave a white solid $D . D$ on reaction with water gave a gas $E$ which formed dense white fumes with $HCl$. Identify $A$ to $E$ and give the reaction involved.
(1993, 3M)
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Solution:
- $\underset{\text { Orange solid }}{A} \stackrel{\Delta}{\longrightarrow} \underset{\text { Green }}{B \downarrow}+\underset{\text { Colourless }}{C \uparrow}+H _2 O(v) \uparrow$
$$ C+Mg \longrightarrow D \text { (white solid) } $$
$D+H _2 O \longrightarrow E(g) \xrightarrow{HCl}$ White fumes.
Hence, $E$ must be ammonia gas so $D$ must be $Mg _3 N _2$ and $C$ is $N _2(g)$. This $N _2$ is obtained on strong heating of $\left(NH _4\right) _2 Cr _2 O _7$ because $\left(NH _4\right) _2 Cr _2 O _7$ is orange solid and produces green $Cr _2 O _3$ residue on heating.
$$ \begin{aligned} & \left(NH _4\right) _2 Cr _2 O _7 \stackrel{\Delta}{\longrightarrow} Cr _2 O _3+N _2+4 H _2 O \ & \text { Orange }(A) \quad \text { Green }(B) \quad C \ & N _2+Mg \stackrel{\Delta}{\longrightarrow} \underset{D}{Mg _3 N _2} \xrightarrow[E]{H _2 O} \underset{NH _3}{Ng _2(OH) _2} \ & NH _3+HCl \longrightarrow NH _4 Cl \text { (white fumes) } \ & E \end{aligned} $$
