Among the isoelectronic species the correct order of ionic size is

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${Mg}^{+ 2} < {Na}^{+} < O^{- 2} < N^{- 3}$

$N^{- 3} < {Na}^{+} < O^{- 2} < {Mg}^{+ 2}$

$O^{- 2} < {Mg}^{+ 2} < N^{- 3} < {Na}^{+}$

${Mg}^{+ 2} < {Na}^{+} < N^{- 3} < O^{- 2}$

answer is A.

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Detailed Solution

isoelectronic species contain same no. of electrons with different charges. The size of ion depends on total no. of electrons to proton ratio.

For $For {Mg}^{+ 2} ion, e / p = 10 / 12 = 0.833; For {Na}^{+} ion, e / p = 10 / 11 = 0.909$

$For O^{- 2} ion, e / p = 10 / 8 = 1.25; For N^{- 3} ion, e / p = 10 / 7 = 1.428$

Hence, the increasing order of radii of isoelectronic species is

${Mg}^{+ 2} < {Na}^{+} < O^{- 2} < N^{- 3}$

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