• # question_answer 57) A compound X of boron reacts with ammonia on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating $B{{F}_{3}}$with lithium aluminium hydride. The compound X and Y are represented by the formulae : (a) ${{B}_{2}}{{H}_{6}},{{B}_{3}}{{N}_{3}}{{H}_{6}}$       (b) ${{B}_{2}}{{O}_{3}},{{B}_{3}}{{N}_{3}}{{H}_{6}}$                (c)$B{{F}_{3}},{{B}_{3}}{{N}_{3}}{{H}_{6}}$                        (d) ${{B}_{3}}{{N}_{3}}{{H}_{6}},{{B}_{2}}{{H}_{6}}$

(a) ${{B}_{2}}{{H}_{6}}+2N{{H}_{3}}\xrightarrow{Low\,temp.}{{B}_{2}}{{H}_{6}}\cdot 2N{{H}_{3}}$ $3{{B}_{2}}{{H}_{6}}\cdot 2N{{H}_{3}}\xrightarrow{200\,{}^\circ C}\underset{\begin{smallmatrix} Borazole(y) \\ (lnorganic\,benzene) \end{smallmatrix}}{\mathop{2{{B}_{3}}{{N}_{3}}{{H}_{6}}}}\,+12{{H}_{2}}$ $4B{{F}_{3}}+3LiAl{{H}_{4}}\xrightarrow{Ether}\underset{(X)}{\mathop{2{{B}_{2}}{{H}_{6}}}}\,+3LiCl+3AlC{{l}_{3}}$