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Chemistry Question on trends in periodic table

The correct order of increasing radii of the ions Br−,F−,O2− {Br^{-}, F^{-} , O^{2-}} and S2− {S^{2-}} is as follows:

A

\ceBr−<F−<O2−<S2−\ce{Br^{-} < F^{-} < O^{2-} < S^{2-}}

B

\ceS2−<O2−<F−<Br−\ce{S^{2-} < O^{2-} < F^{-} < Br^{-}}

C

\ceF−<O2−<S2−<Br−\ce{F^{-} < O^{2-} < S^{2-} < Br^{-}}

D

\ceF−<Br−<O2−<S2−\ce{F^{- } < Br^{-} < O^{2-} < S^{2-}}

Answer

\ceF−<O2−<S2−<Br−\ce{F^{-} < O^{2-} < S^{2-} < Br^{-}}

Explanation

Solution

F− {F^{-}} and O2− {O^{2-}} are isoelectronic but nuclear charge of F−>O2− {F^{-} > O^{2-}}. Hence, ionic radius of F−<O2− { F^{-} < O^{2-}}