A) \[Si{{F}_{4}}\]and \[S{{F}_{4}}\]
B) \[I{{O}^{-}}_{3}\]and\[Xe{{O}_{3}}\]
C) \[B{{H}^{+}}_{4}\]and\[N{{H}^{+}}_{4}\]
D) \[P{{F}^{-}}_{6}\]and \[S{{F}_{6}}\]
Correct Answer: A
Solution :
Shape (structure) of a species can be ascertained on the basis of hybridisation of its central atom, which in turn can be known by determining the number of hybrid bonds (H). Thus, \[H=\frac{1}{2}(V+X-C+A)\] For \[Si{{F}_{4}}:H=\frac{1}{2}(4+4-0+0)=4;s{{p}^{3}}\] (Tetrahedral structure) For \[S{{F}_{4}}:H=\frac{1}{2}(6+4-0+0)=5;s{{p}^{3}}d\] (See-saw structure) Hence the two compounds have different structures.
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