Question

Which of the conditions correctly represents product formation in the reaction between $B_2{H}_6$ and $N{H}_3$ ?

• A. When the ratio between $B_2{H}_6$ and $N{H}_3$ at high temperature is $1:1$ , borazine is produced
• B. When the ratio between $B_2{H}_6$ and $N{H}_3$ at high temperature is $1:2$ , borazine is produced
• C. Excess $N{H}_3$ with $B_2{H}_6$ at low temperature produces boron nitride
• D. Excess $N{H}_3$ with $B_2{H}_6$ at high temperature produces $B_2{H}_6\cdot 2N{H}_3$

Answer 1

Answer: (B)

Diborane combines with ammonia in $1:2$ ratio at low temperature to form an addition product which on heating at $473K$ decomposes to give borazine
$\underset{\text{Diborane}}{3B_2{H}_6}+\underset{\text{Ammonia}}{6N{H}_3}\stackrel{\text{Low temp.}}{\to }\underset{\text{Addition product}}{3[B{H}_2(N{H}_3{)}_2{]}^{+}}+[B{H}_4{]}^{-}\stackrel{473k}{\to }\underset{\text{Borazine}}{2B_3{N}_3{H}_6}+12H_2$
Boron combines with nitrogen in the ratio $2:1$ at high temperatures to give boron nitride
$2B_{\left(s\right)}+N_{2\left(g\right)}\underset{\text{temp.}}{\overset{\text{High}}{\to }}2B{N}_{\left(s\right)}$