Formation of Boron Nitride from Various Boron Compounds

To determine which of the given species react with ammonia ($\text{NH}_3$) to yield boron nitride ($\text{BN}$), we analyze the chemical properties and high-temperature reactions of each:

1. Reaction with Boron ($\text{B}$): When elemental boron is heated strongly (at red to white heat, $\sim 1000^\circ\text{C}-1200^\circ\text{C}$) in an atmosphere of ammonia gas, it directly reacts to form boron nitride: $2\text{B}(s) + 2\text{NH}_3(g) \xrightarrow{\Delta} 2\text{BN}(s) + 3\text{H}_2(g)$

2. Reaction with Diborane ($\text{B}_2\text{H}_6$): The reaction of diborane with excess ammonia at elevated temperatures leads to the ultimate formation of polymeric boron nitride (also known as inorganic graphite): $\text{B}_2\text{H}_6(g) + 2\text{NH}_3(g) \xrightarrow{\text{high } T} 2\text{BN}(s) + 6\text{H}_2(g)$

3. Reaction with Boron Trioxide ($\text{B}_2\text{O}_3$): Boron trioxide reacts with ammonia at high temperatures (around $800^\circ\text{C}-1000^\circ\text{C}$) to form boron nitride and water vapor: $\text{B}_2\text{O}_3(s) + 2\text{NH}_3(g) \xrightarrow{\Delta} 2\text{BN}(s) + 3\text{H}_2\text{O}(g)$

4. Reaction with Fluoroboric Acid ($\text{HBF}_4$): Fluoroboric acid acts as an acid and reacts with ammonia in a simple acid-base neutralization reaction to yield ammonium tetrafluoroborate: $\text{HBF}_4(aq) + \text{NH}_3(g) \rightarrow \text{NH}_4\text{BF}_4(aq)$ It does not yield boron nitride ($\text{BN}$).

Hence, (A), (B), and (C) [or (B) and (C) considering strictly compounds] are correct.

Correct Options: (A), (B), (C) (or (B), (C))