A) \[{{B}_{2}}{{H}_{6}}.2N{{H}_{3}}\]
B) \[{{B}_{12}}{{H}_{12}}\]
C) \[{{B}_{3}}{{N}_{3}}{{H}_{6}}\]
D) \[{{(BN)}_{n}}\]
Correct Answer: B
Solution :
Diborane reacts with ammonia and gives different products under different reaction conditions as \[{{B}_{2}}{{H}_{6}}+\underset{(excess)}{\mathop{N{{H}_{3}}}}\,\xrightarrow[\text{temperature}]{\text{At}\,\text{low}}\underset{\text{addition}\,\text{product}}{\mathop{{{B}_{2}}{{H}_{6}}.2N{{H}_{3}}}}\,\] \[{{B}_{2}}{{H}_{6}}+\underset{(excess)}{\mathop{N{{H}_{3}}}}\,\xrightarrow{\text{High}\,\text{temperature}}\] \[\underset{\begin{smallmatrix} \text{boron}\,\text{nitride} \\ \text{(resembles}\,\text{with}\,\text{graphite)} \end{smallmatrix}}{\mathop{{{\text{(BN)}}_{\text{n}}}}}\,\] \[\underset{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,2}{\mathop{{{\text{B}}_{2}}{{H}_{6}}+N{{H}_{3}}}}\,\xrightarrow{\text{High}\,\text{temperature}}\] \[\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{borzine} \\ \text{(inorganic}\,\text{benzene)} \end{smallmatrix}}{\mathop{{{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}}}\,\] Thus, \[{{\text{B}}_{12}}{{H}_{12}}\]is not the product of reaction between diborane and ammonia.
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