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 ${\mathrm{B}}_{2}{\mathrm{H}}_{6}·2{\mathrm{NH}}_{3}$ which is formulated as $\left[{\mathrm{BH}}_{2}{\left({\mathrm{NH}}_{3}\right)}_{2}\right]+\left[{\mathrm{BH}}_{4}\right]$ further heating gives borazine, B3N3H6 also called borazole.

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

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.

$4B{F}_{3}+3{\mathrm{LiAlH}}_{4}\to \underset{\left(\mathrm{X}\right)}{2{\mathrm{B}}_{2}{\mathrm{H}}_{6}}+3{\mathrm{AlF}}_{3}+3\mathrm{LiF}$