List these ions in order of increasing ionic radius: \(\mathrm{N}^{3-}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}, .\mathrm{O}^{2-}\)

Short Answer

Expert verified
Thus, the order of increasing ionic radius is: \(Na^{+}, Mg^{2+}, F^{-}, O^{2-}, N^{3-}\).

Step by step solution

01

Identify Key Information

First, understand the charges on each ion. \(N^{3-}\) has gained three electrons, \(Na^{+}\) has lost one, \(F^{-}\) has gained one, \(Mg^{2+}\) has lost two, and \(O^{2-}\) has gained two. In general, gaining electrons (becoming an anion) makes an atom larger because the additional negatively charged electrons increase electron-electron repulsion, causing the electron cloud to expand. Losing electrons (becoming a cation) makes an atom smaller for converse reasons, as the electron cloud shrinks due to the diminished repulsion.
02

Consider Periodic and Group Positions

Next, consider where each atom originally resides in the periodic table. All five are in Periods 2 or 3, and so all originally have two energy levels. However, species in higher-numbered groups have more protons and thus a greater effective nuclear charge pulling the remaining electrons inward, which tends to decrease atomic size.
03

Arrange in Increasing Radius

Considering both the charge and original position (number of protons) for each ion, arrange them in increasing ionic radius. The ion from the highest-numbered group with the greatest positive charge will be smallest, and the ion from the lowest-numbered group with the greatest negative charge will be largest.

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