Consider \(\mathrm{S}, \mathrm{Cl}\), and \(\mathrm{K}\) and their most common ions. (a) List the atoms in order of increasing size. (b) List the ions in order of increasing size. (c) Explain any differences in the orders of the atomic and ionic sizes.

To determine the atomic sizes of \(\mathrm{S}\), \(\mathrm{Cl}\), and \(\mathrm{K}\), let's first find their atomic numbers: - \(\mathrm{S}\) (sulfur) has the atomic number 16. - \(\mathrm{Cl}\) (chlorine) has the atomic number 17. - \(\mathrm{K}\) (potassium) has the atomic number 19. These elements are all in the same period (row) of the periodic table, which means they have the same number of electron shells. As we move from left to right across a period, the atomic size generally decreases, primarily due to the increased effective nuclear charge experienced by the electrons, which pulls them closer to the nucleus. Therefore, we can determine that the order of increasing atomic size should be: $$\mathrm{K} < \mathrm{S} < \mathrm{Cl}$$

Now, let's identify the common ions for each element and determine their sizes: - \(\mathrm{S}\) typically forms the \(\mathrm{S^{2-}}\) ion. - \(\mathrm{Cl}\) typically forms the \(\mathrm{Cl^{-}}\) ion. - \(\mathrm{K}\) typically forms the \(\mathrm{K^{+}}\) ion. When an atom gains or loses electrons to form an ion, its size will change. For instance, when an atom loses an electron to create a positively charged ion (like \(\mathrm{K^{+}}\)), the overall electron-electron repulsion decreases, causing the remaining electrons to be pulled closer to the nucleus, resulting in a smaller ionic radius. Conversely, when an atom gains electrons to form a negatively charged ion (like \(\mathrm{S^{2-}}\) and \(\mathrm{Cl^{-}}\)), the electron-electron repulsion increases, causing the electrons to spread out and the ionic radius to increase. By following this logic, we can determine the order of increasing ionic size: $$\mathrm{K^{+}} < \mathrm{Cl^{-}} < \mathrm{S^{2-}}$$

The differences in the orders of atomic and ionic sizes can be attributed to the changes in electron-electron repulsion and effective nuclear charge experienced by the electrons when these atoms form their respective ions. Since potassium loses an electron to form the \(\mathrm{K^{+}}\) ion, its electron-electron repulsion decreases, causing the ion to have a smaller radius than its parent atom. Sulfur and chlorine gain electrons to form \(\mathrm{S^{2-}}\) and \(\mathrm{Cl^{-}}\) ions, respectively, leading to increased electron-electron repulsion and a corresponding increase in ionic radius. Thus, the atomic size order is: $$\mathrm{K} < \mathrm{S} < \mathrm{Cl}$$ while the ionic size order is: $$\mathrm{K^{+}} < \mathrm{Cl^{-}} < \mathrm{S^{2-}}$$