Question

A compound o f boron $X$ reacts at $200^{\circ}C$ temperature with $NH_{3}$ to give another compound $Y$ which is called as inorganic benzene. The compound $Y$ is a colourless liquid and is. highly light sensitive. Its melting point is $-57^{\circ}C$ . The compound $X$ with excess of $NH_{3}$ and at a still higher temperature gives boron nitride $(BN)_{n}$ The compounds $X$ and $Y$ are respectively :

• A. $BH_{3}$ and $B_{2}H_{6}$
• B. $NaBH_{4}$ and $C_{6}H_{6}$
• C. $B_{2}H_{6}$ and $B_{3}N_{3}H_{6}$
• D. $B_{4}C_{3}$ and $C_{6}H_{6}$

Answer 1

Answer: (C)

The reactions involved are
$\underset{(X)}{3B_{2}H_{6}}+6NH_{3} \to \underset{(Y)}{2B_{3}N_{3}H_{6}}+12H_{2}$;
$B_{2}H_{6}+NH_{3}$ (excess) $\xrightarrow{\text{at higher temperature} } (BN)_{n}+H_{2}$
$Y$ is borazole which is isosteric with benzene