question_answer

Among the following, the pair in which the two species are not isostructural, is:        

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: While \[\text{S}{{\text{F}}_{\text{4}}}\]is not tetrahedral but it is distorted tetrahedral because in it S is \[s{{p}^{3}}\]d hybrid. \[_{16}S=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3p_{x}^{2}3p_{y}^{1}3p_{z}^{1}\] (In ground state) \[=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3p_{x}^{1}3p_{y}^{1}3p_{z}^{1,}3d_{xy}^{1}\] \[s{{p}^{3}}\]d-hybridisation (In first excitation state) Hence, five \[\text{s}{{\text{p}}^{\text{3}}}\]d hybrid orbitals are obtained. One orbital is already paired and rest four are overlapped with four p-orbitals of four fluorine atoms on their axis in trigonal bipyramidal form. This structure is distorted from trigonal bi-pyramidal to tetrahedral due to involvement of repulsion between lone pair and bond pair.