Question

For each of the following pairs, indicate which is smaller. a. \(\mathrm{Rb}^{+}\) or \(\mathrm{Na}^{+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Al}^{3+}\) c. \(\mathrm{F}^{-}\) or \(\mathrm{I}^{-}\) d. \(\mathrm{Na}^{+}\) or \(\mathrm{K}^{+}\)

Short answer

The smallest ions for each pair are: a. \(Na^{+}\) b. \(Al^{3+}\) c. \(F^{-}\) d. \(Na^{+}\)

Step-by-step solution

Understanding periodic trends

To solve this problem, we need to understand the periodic trends that affect the size of ions. Moving across a period on the periodic table from left to right, the atomic size generally decreases due to an increased effective nuclear charge. Conversely, moving down a group on the periodic table, the atomic size generally increases due to having more energy levels (shells) of electrons.

Comparing Rb+ and Na+

Both Rb+ and Na+ have a positive charge of +1, meaning they have lost one electron. Now we need to look at their positions on the periodic table. Rubidium (Rb) is below sodium (Na) in group 1 (alkali metals). As we move down this group, the atomic size increases, so Rb+ will be larger than Na+. Thus, Na+ is the smaller ion.

Comparing Mg2+ and Al3+

Magnesium (Mg) and aluminum (Al) are in the same period of the periodic table (period 3). Mg2+ has a charge of +2 and Al3+ has a charge of +3. The greater the positive charge, the stronger the attraction between the electrons and the nucleus, which causes a smaller ion size. Therefore, Al3+ is smaller than Mg2+ since it has a greater positive charge.

Comparing F- and I-

Fluorine (F) and iodine (I) are both in group 17 (halogens) on the periodic table. F- and I- both have a charge of -1, meaning they have gained one electron. As we move down the group from F to I, the size of the atoms increases. As a result, F- is smaller than I-.

Comparing Na+ and K+

Sodium (Na) and potassium (K) are found in group 1 (alkali metals) on the periodic table. Both Na+ and K+ have a charge of +1. As we move down the group from Na to K, the size of the atoms increases. Consequently, Na+ is smaller than K+.

Summary of the results

The smallest ions for each pair are: a. Na+ b. Al3+ c. F- d. Na+