A) \[B{{H}_{3}}\] and \[{{B}_{2}}{{H}_{6}}\]
B) \[NaB{{H}_{4}}\]and\[{{C}_{6}}{{H}_{6}}\]
C) \[{{B}_{2}}{{H}_{6}}\]and \[{{B}_{3}}{{N}_{3}}{{H}_{6}}\]
D) \[{{B}_{4}}{{C}_{3}}\]and \[{{C}_{6}}{{H}_{6}}\]
Correct Answer: C
Solution :
[c] The reactions involved are \[3{{B}_{2}}{{H}_{6}}+6N{{H}_{3}}\to 2{{B}_{3}}{{N}_{3}}{{H}_{6}}+12{{H}_{2}};\] (X) (Y) \[{{B}_{2}}{{H}_{6}}+N{{H}_{3}}(excess)\xrightarrow{at\,higher\,temperature}{{(BN)}_{n}}+{{H}_{2}}\]Y is borazole which is isosteric with benzene.You need to login to perform this action.
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