A) \[N{{H}_{3}}\]to \[NH_{4}^{+}\]
B) \[{{H}_{2}}O\]to \[{{H}_{3}}{{O}^{+}}\]
C) \[C{{H}_{4}}\]to\[{{C}_{2}}{{H}_{6}}~\]
D) \[B{{F}_{3}}\]to \[BF_{4}^{-}\]
Correct Answer: D
Solution :
| \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization}{\mathop{N{{H}_{3}}}}\,}}\,\] | \[\xrightarrow{{}}\] | \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization}{\mathop{NH_{4}^{+}}}\,}}\,\] |
| \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,C}{\mathop{C{{H}_{4}}}}\,\,}}\,\] | \[\xrightarrow{{}}\] | \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,C}{\mathop{{{C}_{2}}{{H}_{6}}}}\,\,}}\,\] |
| \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,}{\mathop{{{H}_{2}}O}}\,\,}}\,\] | \[\xrightarrow{{}}\] | \[\underset{Tetrahedral\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,}{\mathop{{{H}_{3}}{{O}^{+}}}}\,\,}}\,\] |
| \[\underset{Tetrahedral\,planar\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,}{\mathop{B{{F}_{3}}}}\,\,}}\,\] | \[\xrightarrow{{}}\] | \[\underset{Tetrahedral\,\,geometry}{\mathop{\underset{s{{p}^{3}}\,\,hybridization\,\,}{\mathop{BF_{4}^{-}}}\,\,}}\,\] |
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