Diborane reacts with ammonia on heating to give borazine, B3N3H6 (a planar six-membered ring), with release of hydrogen: 3B2H6 + 6NH3 -> 2B3N3H6 + 12H2. The sum of the number of moles of H2 released per mole of borazine and the number of sigma bonds in one molecule of borazine is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pb042

Diborane reacts with ammonia on heating to give borazine, $B X_{3} N X_{3} H X_{6}$ (a planar six-membered ring), with release of hydrogen: $3 B X_{2} H X_{6} + 6 NH X_{3} 2 B X_{3} N X_{3} H X_{6} + 12 H X_{2}$ . The sum of the number of moles of $H X_{2}$ released per mole of borazine and the number of $σ$ bonds in one molecule of borazine is

Per mole of borazine,

12/2 = 6 mol

H X_{2}

are released. Borazine has 6 ring

B - N

σ

bonds, 3

B - H

and 3

N - H

bonds, i.e.

6 + 3 + 3 = 12

σ

bonds (the

π

system is delocalised and has no extra

σ

bonds). The sum is

6 + 12 = 18