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Q. A compound $X$, of boron reacts with $NH_3$ on heating to give another compound $Y$ which is called inorganic benzene. The compound $X$ can be prepared by treating $BF_3$ with Lithium aluminium hydride. The compounds $X$ and $Y$ are represented by the formulas

The p-Block Elements

Solution:

A compound $X$, of boron, reacts with $NH_3$ on heating to give another compound $Y$ which is called inorganic benzene.

$\underset{X\text{(Diborane)}}{3B_2H_6} + 6NH_3 \rightarrow 3[BH_2(NH_3)_2]^+[BH_4]^- $
$\xrightarrow{\text{heat}} \underset{Y\text{(Borazole/Inorganic Benzene}}{2B_3N_3H_6} + 12H_2$

$4BF_3 + 3LiAlH_4 \rightarrow \underset{X}{2B_2H_6} + 3LiF + 3AlF_3$