Animated Solution for Chemistry - s and p-Block Elements: The compound(s) which react(s) with ${\text{NH}}_3$ to give boron nitride (BN) is(are)

Welcome to a fascinating journey through the chemistry of the p-block! This problem from JEE Advanced tests not only your knowledge of inorganic reactions but also your attention to detail. Let's break down the interactions of various boron species with ammonia to see which ones yield the elusive inorganic graphite, Boron Nitride ($\text{BN}$).

The question explicitly asks for compound(s) that react with ${\text{NH}}_3$ to give boron nitride. Option (A) presents us with elemental Boron ($\text{B}$).

It is a chemical fact that when elemental boron is heated with ammonia at high temperatures, it does indeed form boron nitride:

However, there is a massive catch here! Boron is an element, not a compound. The examiners deliberately set a linguistic trap. Because the question strictly specifies "compound(s)", we must ruthlessly eliminate Option (A).

Let's move to Option (B), Diborane (${\text{B}}_2{\text{H}}_6$). The reaction between diborane and ammonia is one of the most celebrated sequences in inorganic chemistry. Initially, at lower temperatures, ammonia (a hard Lewis base) attacks diborane, causing an unsymmetrical cleavage to form an ionic adduct:

When this adduct is heated to around $200^{\circ }\text{C}$, it undergoes a condensation reaction, releasing hydrogen gas and forming Borazine (${\text{B}}_3{\text{N}}_3{\text{H}}_6$), famously known as "inorganic benzene" due to its isoelectronic and isostructural relationship with benzene.

But the story doesn't end there. If we continue to heat borazine above $200^{\circ }\text{C}$, it polymerizes and cross-links, ultimately yielding polymeric Boron Nitride:

Thus, diborane is a valid compound that eventually leads to boron nitride. Option (B) is correct.

Option (C) gives us Boron trioxide (${\text{B}}_2{\text{O}}_3$). This is actually the standard industrial precursor for manufacturing boron nitride.

When liquid boron trioxide is reacted with ammonia gas at blistering temperatures (around $1200^{\circ }\text{C}$), a direct replacement of oxygen by nitrogen occurs:

The resulting boron nitride adopts a layered structure remarkably similar to graphite, earning it the moniker "inorganic graphite". It is an excellent high-temperature lubricant. Option (C) is absolutely correct.

Finally, we evaluate Option (D), Fluoroboric acid (${\text{HBF}}_4$). This is a very strong Brønsted acid. Ammonia, on the other hand, is a classic Brønsted base.

When these two meet, they don't engage in complex structural rearrangements or polymerizations. Instead, they undergo a straightforward, rapid acid-base neutralization. The ammonia molecule simply accepts a proton from the acid:

The product is ammonium fluoroborate, a stable salt. No boron nitride is formed here, making Option (D) incorrect.

In conclusion, by carefully navigating the chemical reactions and the precise wording of the question, we find that Diborane and Boron trioxide are the correct compounds.