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