Thermal Decomposition Products of Ammonium Nitrite and Ammonium Nitrate

To find the identity of the nitrogen-containing compounds $\mathbf{X}$ and $\mathbf{Y}$, we examine the thermal decomposition reactions of ammonium nitrite and ammonium nitrate:

1. Thermal Decomposition of Ammonium Nitrite ($\text{NH}_4\text{NO}_2$): When ammonium nitrite is gently heated at around $60\text{--}70\ ^\circ\text{C}$, it undergoes an intramolecular redox (comproportionation) reaction to produce dinitrogen gas ($\text{N}_2$) and water: $\text{NH}_4\text{NO}_2 \xrightarrow{60\text{--}70\ ^\circ\text{C}} \text{N}_2\text{(g)} + 2\text{H}_2\text{O}\text{(g)}$ In this reaction, nitrogen in $\text{NH}_4^+$ (oxidation state $-3$) and nitrogen in $\text{NO}_2^-$ (oxidation state $+3$) are converted to elemental nitrogen, $\text{N}_2$ (oxidation state $0$). Therefore, $\mathbf{X} = \text{N}_2$.

2. Thermal Decomposition of Ammonium Nitrate ($\text{NH}_4\text{NO}_3$): When ammonium nitrate is carefully heated at $200\text{--}250\ ^\circ\text{C}$, it undergoes thermal decomposition to form nitrous oxide ($\text{N}_2\text{O}$, also known as dinitrogen monoxide or laughing gas) and water: $\text{NH}_4\text{NO}_3 \xrightarrow{200\text{--}250\ ^\circ\text{C}} \text{N}_2\text{O}\text{(g)} + 2\text{H}_2\text{O}\text{(g)}$ In this reaction, nitrogen in $\text{NH}_4^+$ (oxidation state $-3$) and nitrogen in $\text{NO}_3^-$ (oxidation state $+5$) comproportionate to form $\text{N}_2\text{O}$ (oxidation state $+1$). Therefore, $\mathbf{Y} = \text{N}_2\text{O}$.

Thus, the compounds $\mathbf{X}$ and $\mathbf{Y}$ are $\text{N}_2$ and $\text{N}_2\text{O}$, respectively.

Correct Option: (A)