Identification of Colourless Ammonium Salt Producing Gas with Zinc

To determine the identity of the colourless salt(s) $\mathbf{H}$, we analyze the reactions step by step:

1. Initial Reaction with excess $\text{NaOH}$:
   The colourless salt $\mathbf{H}$ contains the ammonium ion ($\text{NH}_4^+$). Upon boiling with excess $\text{NaOH}$, ammonia gas ($\text{NH}_3$) is evolved:
   $\text{NH}_4^+ + \text{OH}^- \xrightarrow{\Delta} \text{NH}_3\uparrow + \text{H}_2\text{O}$
   Ammonia ($\text{NH}_3$) is a colourless, pungent-smelling, and non-flammable gas under standard atmospheric conditions. After prolonged boiling, all ammonium ions are converted to $\text{NH}_3$ and expelled, causing the gas evolution to cease.

2. Reaction upon addition of $\text{Zn}$ dust:
   The remaining solution contains unreacted, excess $\text{NaOH}$ along with the anion of the salt $\mathbf{H}$. When $\text{Zn}$ dust is added to this strongly alkaline solution, the gas evolution restarts (producing the same non-flammable gas, $\text{NH}_3$), which indicates the reduction of nitrogen-containing oxyanions to ammonia:

   • If the anion is nitrate ($\text{NO}_3^-$):
     $\text{NO}_3^- + 4\text{Zn} + 7\text{OH}^- + 6\text{H}_2\text{O} \longrightarrow \text{NH}_3\uparrow + 4[\text{Zn}(\text{OH})_4]^{2-}$
   • If the anion is nitrite ($\text{NO}_2^-$):
     $\text{NO}_2^- + 3\text{Zn} + 5\text{OH}^- + 5\text{H}_2\text{O} \longrightarrow \text{NH}_3\uparrow + 3[\text{Zn}(\text{OH})_4]^{2-}$

3. Comparison with other options:

   • For $\text{NH}_4\text{Cl}$ and $(\text{NH}_4)_2\text{SO}_4$, the anions $\text{Cl}^-$ and $\text{SO}_4^{2-}$ are not reduced to $\text{NH}_3$ by $\text{Zn}/\text{NaOH}$. The addition of $\text{Zn}$ to excess $\text{NaOH}$ in these cases would yield hydrogen gas ($\text{H}_2$), which is highly flammable, contradicting the condition that the evolution of the non-flammable gas restarts.

Therefore, the colourless salt(s) $\mathbf{H}$ can be $\text{NH}_4\text{NO}_3$ and $\text{NH}_4\text{NO}_2$.

Correct Options:
(A) and (B)