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Chapter 8: Problem 76

Arrange the following isoelectronic species in order of (a) increasing ionic radius and (b) increasing ionization energy: \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\).

Short Answer

Expert verified
The order of increasing ionic radius is \( \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-} \). The order of increasing ionization energy is \( \mathrm{O}^{2-} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{2+} \).

Step by step solution

01

Arrange Isoelectronic Species in Order of Increasing Ionic Radius

In isoelectronic species, the species with the most positive charge will have the smallest ionic radius because of the higher attractive force acting on the same number of electrons. Hence, the order is: \( \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-} \)
02

Arrange Isoelectronic Species in Order of Increasing Ionization energy

Ionization energy increases with an increase in effective nuclear charge. In the case of isoelectronic species, the species with the most positive charge will have a higher effective nuclear charge and therefore the highest ionization energy. Hence, the order is: \( \mathrm{O}^{2-} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{2+} \)

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