Consider the isoelectronic species, ${Na}^{+}, {Mg}^{2 +}, F^{-}$ and $O^{2 -}$ . The correct order of increasing length of their radii is :

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$O^{2 -} < F^{-} < {Mg}^{2 +} < {Na}^{+}$

$\begin{array}{r} {Mg}^{2 +} < {Na}^{+} < F^{-} < O^{2 -} \end{array}$

$\begin{array}{r} O^{2 -} < F^{-} < {Na}^{+} < {Mg}^{2 +} \end{array}$

$\begin{aligned} F^{-} < O^{2 -} < {Mg}^{2 +} < {Na}^{+} \end{aligned}$

answer is B.

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Detailed Solution

Ionic radii for isoelectronic species decrease as nuclear charge rises (i.e., no. of protons). The cation with the greater +ve charge will therefore have a smaller radius and the anion with the greater -ve charge will have a bigger radius for isoelectronic species.
Ionic radii should therefore increase in the following order: $\begin{array}{r} {Mg}^{2 +} < {Na}^{+} < F^{-} < O^{2 -} \end{array}$ .