A) Inorganic benzene \[({{B}_{3}}{{N}_{3}}{{H}_{6}})\]
B) \[{{[{{H}_{3}}N\to B{{H}_{2}}\leftarrow N{{H}_{3}}]}^{+}}{{[B{{H}_{4}}]}^{-}}\]
C) Boron nitride \[{{(BN)}_{x}}\]
D) \[[B{{H}_{2}}^{+}{{[{{H}_{3}}N\to B{{H}_{4}}\leftarrow N{{H}_{3}}]}^{-}}\]
Correct Answer: B
Solution :
\[\Rightarrow \,\,\frac{{{x}_{1}}\,+{{x}_{2}}+{{x}_{3}}+.....{{x}_{9}}}{9}=15\] (Excess) \[\sum\limits_{i=1}^{9}{{{x}_{i}}=135}\] \[{{x}_{10}}\] (Excess) \[\sum\limits_{i=1}^{9}{{{x}_{i}}=135}+{{x}_{10}}\] (Inorganic graphite) \[\Rightarrow \,\,{{x}_{0}}=160\,-135\,=25\] \[\tilde{\ }p\vee q\] (Borazine) (Inorganic benzene)
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