A) \[C{{H}_{4}}\xrightarrow{{}}{{C}_{2}}{{H}_{6}}\]
B) \[N{{H}_{3}}\xrightarrow{{}}NH_{4}^{+}\]
C) \[B{{F}_{3}}\xrightarrow{{}}BF_{4}^{-}\]
D) \[{{H}_{2}}O\xrightarrow{{}}{{H}_{3}}{{O}^{+}}\]
Correct Answer: C
Solution :
[a] \[\underset{\begin{smallmatrix} 4bp+0lp \\ Hybridisation\,\,\,s{{p}^{3}} \\ Structure\text{ }tetrahedral \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,\xrightarrow{{}}\underset{tetrahedral}{\mathop{\underset{\begin{smallmatrix} 4\,bp \\ s{{p}^{3}} \end{smallmatrix}}{\mathop{C{{H}_{3}}}}\,-\underset{\begin{smallmatrix} 4\,bp \\ s{{p}^{3}} \end{smallmatrix}}{\mathop{C{{H}_{3}}}}\,}}\,\] [b] \[\underset{\begin{smallmatrix} 3bp+1lp \\ Hybridisation\,\,s{{p}^{3}} \\ Structure\text{ }pyramidal \end{smallmatrix}}{\mathop{N{{H}_{3}}}}\,\xrightarrow{{}}\underset{tetrahedral}{\mathop{\underset{\begin{smallmatrix} 4\,bp \\ s{{p}^{3}} \end{smallmatrix}}{\mathop{NH_{4}^{+}}}\,}}\,\] [c] \[\underset{\begin{smallmatrix} 3bp \\ Hybridisation\,\,s{{p}^{2}} \\ Structure\text{ }trigonal\text{ }planar \end{smallmatrix}}{\mathop{B{{F}_{3}}}}\,\xrightarrow{{}}\underset{tetrahedral}{\mathop{\underset{\begin{smallmatrix} 4\,bp \\ s{{p}^{3}} \end{smallmatrix}}{\mathop{BF_{4}^{-}}}\,}}\,\] [d] \[\underset{\begin{smallmatrix} 2bp+2lp \\ Hybridisation\,\,s{{p}^{3}} \\ Structure\text{ }angular \end{smallmatrix}}{\mathop{{{H}_{2}}O}}\,\xrightarrow{{}}\underset{pyramidal}{\mathop{\underset{\begin{smallmatrix} 3\,bp1lp \\ s{{p}^{3}} \end{smallmatrix}}{\mathop{{{H}_{3}}{{O}^{+}}}}\,}}\,\] Thus, conversation of \[B{{F}_{3}}\] into \[BF_{4}^{-}\] involves change in both hybridisation and shape.
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