Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "X" with 6-equivalents of "Y". "X" and "Y", respectively are :

Solution

Chemistry ma’am · 4 steps

Borazine needs B–H and N–H bonds, so take diborane for boron and ammonia for nitrogen.

1. Look at borazine. Borazine, B₃N₃H₆, is a six-membered ring of alternating B and N atoms. Every B carries one H and every N carries one H. So the boron source must already have B–H bonds, and the nitrogen source must have N–H bonds.
2. Pick the reagents. Diborane, B₂H₆, supplies boron with B–H bonds. Ammonia, NH₃, supplies nitrogen with N–H bonds. They first form an adduct, [BH₂(NH₃)₂]⁺[BH₄]⁻, which loses H₂ on heating.

   3B₂H₆ + 6NH₃ → 3[BH₂(NH₃)₂]⁺[BH₄]⁻ →(Δ) 2B₃N₃H₆ + 12H₂

3. Check the count. Left side: 6 B, 6 N and 18 + 18 = 36 H. Right side: 6 B, 6 N and 12 + 24 = 36 H. The ratio is 3 of B₂H₆ to 6 of NH₃, as the question says.
4. So. X is B₂H₆ and Y is NH₃, which is option (b).

The idea to keep. For borazine, remember the pair diborane plus ammonia in a 1 : 2 ratio, heated so that H₂ leaves. The B–H and N–H bonds in the ring come straight from these two reagents.

Answer from the NTA final answer key. Solution by PrepWiser.