Animated Solution for Chemistry - s and p-Block Elements: The compound(s) which generate(s) ${\text{N}}_2$ gas upon thermal decomposition below $300^{\circ }\text{C}$ is (are)

Welcome to a classic exploration of inorganic chemistry! Thermal decomposition is like a stress test for molecules. When we turn up the heat, we force compounds to reveal their true stability.

In this problem, we are on a hunt. We need to identify which of the given compounds will release nitrogen gas, ${\text{N}}_2$, when heated to a moderate temperature of below $300^{\circ }\text{C}$.

This specific temperature constraint is not just a random number; it is a carefully placed trap by the examiners. Let's break down each candidate and see how they react under the flame.

The Temperature Trap

Our first suspect is ammonium nitrate, ${\text{NH}}_4{\text{NO}}_3$. This is a fascinating compound because its decomposition pathway is highly dependent on temperature.

When heated gently, specifically below $300^{\circ }\text{C}$, the ammonium ion and the nitrate ion undergo a comproportionation reaction. The nitrogen in the $-3$ oxidation state and the nitrogen in the $+5$ oxidation state meet in the middle to form nitrous oxide, where nitrogen is in the $+1$ state.

Notice the product! We get nitrous oxide, ${\text{N}}_2\text{O}$, commonly known as laughing gas, not pure nitrogen gas.

Here is the catch: If we were to heat it to much higher temperatures, or detonate it, it would explosively decompose into ${\text{N}}_2$, ${\text{O}}_2$, and ${\text{H}}_2\text{O}$. But under our given conditions, it fails the test.

Ammonium Dichromate

Next, we have ammonium dichromate, $({\text{NH}}_4{)}_2{\text{Cr}}_2{\text{O}}_7$. If you have ever seen the classic "chemical volcano" demonstration in a lab, you know exactly what this compound does.

The dichromate ion, ${\text{Cr}}_2{\text{O}}_7^{2-}$, is a powerful oxidizing agent. When heated, it oxidizes the ammonium ion.

This spectacular reaction produces a massive volume of green chromium(III) oxide powder, steam, and most importantly for us, pure nitrogen gas. This makes option (B) a correct answer.

Barium Azide

Moving on to option (C), we encounter barium azide, $\text{Ba}({\text{N}}_3{)}_2$. Azides are the undisputed champions when it comes to producing nitrogen gas.

In fact, the thermal decomposition of heavy metal azides like barium azide or sodium azide is the standard laboratory method for preparing ultra-pure nitrogen.

The azide ion, ${\text{N}}_3^{-}$, is relatively unstable and eagerly breaks apart to form stable, triple-bonded ${\text{N}}_2$ molecules, leaving behind solid barium metal. This confirms option (C) is also correct.

Magnesium Nitride

Finally, let's look at magnesium nitride, ${\text{Mg}}_3{\text{N}}_2$. Unlike the previous compounds which contained complex, redox-active ions, magnesium nitride is a simple, robust ionic solid.

It consists of highly charged ${\text{Mg}}^{2+}$ and ${\text{N}}^{3-}$ ions locked in a strong crystal lattice.

Because of this immense lattice energy, it is incredibly stable to heat. It will not simply decompose to release nitrogen gas upon moderate heating below $300^{\circ }\text{C}$. It would require extreme conditions to break those bonds.

By systematically testing each compound against the flame of our chemical knowledge, the truth is clear.

Ammonium nitrate falls into the temperature trap, yielding nitrous oxide. Magnesium nitride stands too strong to break. But ammonium dichromate and barium azide both readily decompose to release the nitrogen gas we were looking for.

Therefore, the correct compounds are $({\text{NH}}_4{)}_2{\text{Cr}}_2{\text{O}}_7$ and $\text{Ba}({\text{N}}_3{)}_2$.