A) \[C{{l}^{-}}\]
B) \[{{S}^{2-}}\]
C) \[N{{a}^{+}}\]
D) \[{{F}^{-}}\]
Correct Answer: B
Solution :
\[C{{l}^{-}}\]and\[{{S}^{2-}}\]are isoelectronic species (containing equal number of electrons, i.e., 18). Their radii would be different because of their different nuclear charge. Anion with the greater negative charge will have the larger radius hence \[{{S}^{2-}}\]has larger size than\[C{{l}^{-}}.\] For isoelectronic species, radii\[\propto \frac{1}{Z}\] Further\[\text{N}{{\text{a}}^{\text{+}}}\]and\[{{\text{F}}^{-}}\]are also isoelectronic species but they have 10 electrons. Hence, their sizes are smaller than \[{{\text{S}}^{2-}}\]ion.
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