Question

$B_2{H}_6+N{H}_3\underset{\text{low temperature}}{\overset{\text{excess of}N{H}_3}{\to }}A$
Compound A is ionic in nature which on strong heating gives a compound which is aromatic in nature. Sum of number of lone pair(s) present in cationic and anionic part of A is

Answer 1

• $B_2{H}_6+N{H}_3\underset{\text{low temperature}}{\overset{\text{excess of}N{H}_3}{\to }}\left[B{H}_2\right(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}$
• On heating A it form Borazine :

$\left[B{H}_2\right(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}\stackrel{△}{\to }\underset{Borazine}{B_3{N}_3{H}_6}$

Since lone pairs of ammonia are involved in bond forming. So product has no lone pair.