Question

Which of the conditions correctly represents product formation in the reaction between $ B_2H_6 $ and $ NH_3 $ ?

• A. When the ratio between $ B_2H_6 $ and $ NH_3 $ at high temperature is $ 1 : 1 $ , borazine is produced
• B. When the ratio between $ B_2H_6 $ and $ NH_3 $ at high temperature is $ 1 : 2 $ , borazine is produced
• C. Excess $ NH_3 $ with $ B_2H_6 $ at low temperature produces boron nitride
• D. Excess $ NH_3 $ with $ B_2H_6 $ at high temperature produces $ B_2H_6 \cdot 2NH_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 $473\,K$ decomposes to give borazine
$\underset{\text{Diborane}}{{3B_{2}H_{6}}}+ \underset{\text{Ammonia}}{{6NH_{3}}}\xrightarrow{ \text{Low temp.} }\underset{\text{Addition product}}{{3[BH_{2}(NH_{3})_{2}]^{+}}}+[BH_{4}]^{-}\xrightarrow{ 473\, k }\underset{\text{Borazine}}{2B_{3}N_{3}H_{6}}+12H_{2}$
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)} \xrightarrow[\text{temp.}]{\text{High}} 2BN_{\left(s\right)}$