Question

Out of $N{a}^{+},M{g}^{+2},O^{-2}and{N}^{-3},$ the pair of species showing minimum and maximum IP would be :

• A. $N{a}^{+},M{g}^{+2}$
• B. $M{g}^{+2},N^{-3}$
• C. $N^{-3},M{g}^{+2}$
• D. $O^{-2},N^{-3}$

Answer 1

Answer: (C)

$N^{-3},M{g}^{+2}$

$N{a}^{+},M{g}^{+2},O^{-2}and{N}^{-3},$ are isoelectronic species. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. Species with smallest size will have maximum IP and the one with biggest size will have minimum IP.