Question

The correct order of atomic radii for the following isoelectronic species $F^{-},O^{2-},N{a}^{+},M{g}^{2+}$ is:

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

Answer 1

The correct option is B $O^{2-}>F^{-}>M{g}^{2+}>N{a}^{+}$

Isoelectronic species are those which have the same number of valence electrons.
Ionic radius of the cation is always smaller than that of the parent atom. Ionic radius of the anion is always greater than that of the parent atom. Greater the effective nuclear charge, smaller the ionic radii. Greater the negative charge greater the size of the anion. Greater the postive charge, smaller the size of the cation.

Among the ions, $O^{2-},F^{-},N{a}^{+},M{g}^{2+}$ the effective nuclear charge and the positive charge increases and the ionic radius decreases. Hence the correct answer is option (b) i.e,
$O^{2-}>F^{-}>M{g}^{2+}>N{a}^{+}$