Question

A compound of boron X reacts at ${200}^{\circ }C$ temperature with $N{H}_3$ to give another compound $Y$ which is called as inorganic benzene. The compound $Y$ is a colourless liquid and is highly sensitive. Its melting point is $-{57}^{\circ }C$. The compound $X$ with excess $N{H}_3$ and at a still higher temperature gives boron nitride ${\left(BN\right)}_n$. The compounds $X$ and $Y$ respectively are:

• A. $B{H}_3\text{and}{B}_2{H}_6$
• B. $NaB{H}_4\text{and}{C}_6{H}_6$
• C. $B_2{H}_6\text{and}{B}_3{N}_3{H}_6$
• D. $B_4{C}_3\text{and}{C}_6{H}_6$

Answer 1

The correct option is C $B_2{H}_6\text{and}{B}_3{N}_3{H}_6$

The reactions involved are
$\underset{\left(X\right)}{3B_2{H}_6}+6N{H}_3⟶\underset{\left(Y\right)}{2B_3{N}_3{H}_6}+12H_2$
$B_2{H}_6+N{H}_3\left(excess\right)⟶(BN{)}_n+H_2$

Y is borazole, which is isosteric and isoelectronic with benzene and hence called as Inorganic benzene.