Ncert Exercises & Solutions

A compound X, of boron, reacts with NH₃ on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF₃ with lithium aluminum hydride. The compounds X and Y are represented by the formulas.

1. B₂H₆, B₃N₃H₆

2. B₂O₃, B₃N₃H₆

3. BF₃, B₃N₃H₆

4. B₃N₃H₆, B₂H₆

Hint: Borazine is known as inorganic benzene.

(i) Reaction of ammonia with diborane gives initially

B_{2} H_{6} \cdot 2 {NH}_{3}

which is formulated as

[{BH}_{2} {({NH}_{3})}_{2}] + [{BH}_{4}]

further heating gives borazine, B₃N₃H₆ also called borazole.

\(\underset{\substack{\text { Diborane } \\(\mathbf{x})}}{\mathrm{3B}_{2} \mathrm{H}_{6}}+6 \mathrm{NH}_{3} \rightarrow \underset{\substack{\text { Borazole }}}{2 \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}}+12 \mathrm{H}_{2}\)

Borazole has a cyclic structure and is isoelectronic with benzene and thus called inorganic benzene or borazine.

(ii) Diborane can be prepared by the reduction of BF₃ with lithium aluminum hydride in diethyl ether.

4B F_{3} + 3 {LiAlH}_{4} \to \underset{(X)}{2 B_{2} H_{6}} + 3 {AlF}_{3} + 3 LiF

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions