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