Question

The ionic radii of $N{a}^{+},F^{-},O^{2-},N^{3-}$ follow the order

• A. $O^{2-}>F^{-}>N{a}^{+}>N^{3-}$
• B. $N^{3-}>N{a}^{+}>F^{-}>O^{2-}$
• C. $N^{3-}>O^{2-}>F^{-}>N{a}^{+}$
• D. $N{a}^{+}>F^{-}>O^{2-}>N^{3-}$

Answer 1

Answer: (C)

The given species are isoelectronic (same number of electrons) to each other. In case of isoelectronic species, cation with greater positive charge will have a smaller radius because of the greater attraction of the electrons to nucleus. Anion with greater negative charge will have the largest radius. This is because the net repulsion of electrons will out weight the nuclear charge and the ion will expand in size. Thus, the order of ionic radii will be, $N^{3-}>O^{2-}>F^{-}>N{a}^{+}$