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The decreasing order of the ionic radii of the elements

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${Al}^{+ 3} > {Mg}^{+ 2} > {Na}^{+} > F^{-} > O^{- 2}$

${Na}^{+} > F^{-} > {Mg}^{+ 2} > O^{- 2} > {Al}^{+ 3}$

$O^{- 2} > F^{-} > {Na}^{+} > {Mg}^{+ 2} > {Al}^{+ 3}$

${Na}^{+} > {Mg}^{+ 2} > {Al}^{+ 3} > O^{- 2} > F^{-}$

answer is C.

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Detailed Solution

O²⁻,F⁻,Na⁺,Mg²⁺ and Al³⁺are isoelectronic species, for isoelectronic species, r∝1/Z

Thus, the larger the atomic number (effective nuclear charge) the smaller is the ionic radii.

So decreasing order of radii is: O²⁻>F⁻>Na⁺>Mg²⁺>Al³⁺

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