A) \[Si{{F}_{4}}\]and\[S{{F}_{4}}\]
B) \[IO_{3}^{-}\]and\[Xe{{O}_{3}}\]
C) \[\text{B}{{\text{H}}_{\text{4}}}\]and\[\text{NH}_{4}^{+}\]
D) \[\text{PF}_{6}^{-}\]and\[\text{SF}_{6}^{{}}\]
Correct Answer: A
Solution :
\[\text{Si}{{\text{F}}_{\text{4}}}\]and\[\text{S}{{\text{F}}_{\text{4}}}\]are not is structural because \[\text{Si}{{\text{F}}_{\text{4}}}\]is tetrahedral due to \[\text{s}{{\text{p}}^{3}}\]hybridisation of Si. \[_{14}Si=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}}3{{p}^{2}}\] (In ground state) \[_{14}Si=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}}3{{p}^{2}}\] (In excited state) \[_{14}Si=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{1}}3{{p}^{3}}\](In excited state) Hence, four equivalent\[s{{p}^{3}}\]-hybrid orbitals are obtained and they are overlapped by four p-orbitals of four fluorine atoms on their axes. Thus it shows following structure:You need to login to perform this action.
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