The decomposition of \(\text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7\) on heating liberates nitrogen gas (N2) along with water vapor and chromium(III) oxide \(\text{Cr}_2\text{O}_3\). Therefore, the gas liberated from this reaction is N2.
Among the given options:
(A) Treating \(\text{H}_2\text{O}_2\) with \(\text{NaNO}_2\) does not produce nitrogen gas. It results in the formation of water and oxygen gas.
(B) Heating \(\text{NH}_4\text{NO}_3\) decomposes it into nitrogen gas, water, and oxygen gas. However, this is not the same gas obtained from the decomposition of \(\text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7\), which produces N2.
(C) Treating \(\text{Mg}_3\text{N}_2\) with H2O produces ammonia gas (NH3), not nitrogen gas.
(D) Heating \(\text{NH}_4\text{NO}_2\) decomposes it into nitrogen gas, water, and oxygen gas. This is the correct option.
Therefore, the gas obtained by heating \(\text{NH}_4\text{NO}_2\) is the same gas liberated from the decomposition of \(\text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7\), which is nitrogen gas (N2). So, the correct answer is option (D) Heating \(\text{NH}_4\text{NO}_2\)
Correct answer: Heating \( NH_4NO_2 \)
On heating, \((NH_4)_2Cr_2O_7\) decomposes as:
$$ (NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4H_2O + N_2 \uparrow $$
It liberates nitrogen gas (N₂).
Similarly, when ammonium nitrite (NH₄NO₂) is heated, it decomposes to give:
$$ NH_4NO_2 \rightarrow N_2 \uparrow + 2H_2O $$
Thus, the same gas, nitrogen (N₂), is evolved in both reactions.
When \((NH_4)_2Cr_2O_7\) is heated, it decomposes to produce nitrogen gas (N2), chromium(III) oxide (Cr2O3), and water (H2O). The reaction is:
\((NH_4)_2Cr_2O_7(s) \xrightarrow{\Delta} N_2(g) + Cr_2O_3(s) + 4H_2O(g)\)
We need to find which of the given reactions also produces nitrogen gas.
Therefore, the correct answer is:
Heating \(NH_4NO_2\)