A) \[{{B}_{2}}{{H}_{6}}\]
B) LiH
C) \[B{{H}_{4}}\]
D) \[IB{{H}_{2}}\]
Correct Answer: A
Solution :
\[LiAI{{H}_{4}} + {{I}_{2}} \xrightarrow{{}}\,\,{{B}_{2}}{{H}_{6}}\] \[{{\operatorname{B}}_{2}}{{H}_{6}}\,+\,\,3{{O}_{2}}\,\,\xrightarrow{{}}\,\,\underset{Boric\,\,anhydride}{\mathop{{{B}_{2}}{{O}_{3}}}}\,\] \[{{\operatorname{B}}_{2}}{{H}_{6}} +N{{H}_{3}} \xrightarrow{{}}\,\,B{{H}_{3}}\cdot N{{H}_{3}}\]You need to login to perform this action.
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