Question

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

• A. $F^{-}$ < $O^{2-}$ < $M{g}^{2+}$ < $N{a}^{+}$
• B. $M{g}^{2+}$ < $N{a}^{+}$ < $F^{-}$ < $O^{2-}$
• C. $O^{2-}$ < $F^{-}$ < $N{a}^{+}$ < $M{g}^{2+}$
• D. $O^{2-}$ < $F^{-}$ < $M{g}^{2+}$ < $N{a}^{+}$

Answer 1

Answer: (B)

$M{g}^{2+}$ < $N{a}^{+}$ < $F^{-}$ < $O^{2-}$

For isoelectronic species, ionic radii decrease with an increase in nuclear charge (i.e., no. of protons). Thus, for isoelectronic species, the cation with the greater +ve charge will have a smaller radius and the anion with greater -ve charge will have a larger radius.

Thus, the correct order of increasing ionic radii is $M{g}^{2+}$ < $N{a}^{+}$ <$F^{-}$< $O^{2-}$