Answer:
In the reaction:
\[\underset{(s{{p}^{2}})}{\mathop{B{{F}_{3}}}}\,+\,\underset{(s{{p}^{3}})}{\mathop{_{\centerdot
}^{\centerdot }N{{H}_{3}}}}\,\to
H-\overset{\overset{H}{\mathop{|}}\,}{\mathop{\underset{\underset{\text{H}}{\mathop{\left|
s{{p}^{3}} \right.}}\,}{\mathop{N_{\centerdot }^{\centerdot }}}\,}}\,\to
\overset{\overset{\text{F}}{\mathop{\text{ }\!\!|\!\!\text{
}}}\,}{\mathop{\underset{\underset{\text{F}}{\mathop{\left|
\text{s}{{\text{p}}^{\text{3}}} \right.}}\,}{\mathop{\text{B}}}\,}}\,-F\]
According to Lewis Acid-Base
theory:
\[N{{H}_{3}}\to \] Lewis Base
(Electron pair donor)
\[B{{F}_{3}}\to \] Lewis Acid
(Electron pair acceptor)
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