A) \[\underline{C}{{H}_{3}}C{{H}_{3}}\xrightarrow{{}}\underline{C}H_{3}^{-}+CH_{3}^{+}\]
B) \[N{{H}_{3}}+\underline{B}{{F}_{3}}\xrightarrow{{}}[{{H}_{3}}N\to \underline{B}{{F}_{3}}]\]
C) \[{{H}_{2}}\underline{O}+{{H}^{+}}\xrightarrow{{}}{{H}_{3}}{{\underline{O}}^{+}}\]
D) \[\underline{N}{{H}_{3}}\xrightarrow{{}}\underline{N}H_{2}^{-}+{{H}^{+}}\]
Correct Answer: B
Solution :
[b] \[N{{H}_{3}}+\underline{B}{{F}_{3}}\xrightarrow{{}}[{{H}_{3}}N\to B{{F}_{3}}]\,or\,{{H}_{3}}\overset{\oplus }{\mathop{N}}\,-\underline{\overset{\oplus }{\mathop{B}}\,}{{F}_{3}}\] \[\ell P=0\] \[\ell p=0\] \[bp=3\] \[bp=4\] \[\underset{Plane\text{ }Triangle}{\mathop{s{{p}^{2}}}}\,\] \[\underset{Tetrahedral}{\mathop{s{{p}^{3}}}}\,\]
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