Give the empirical formulas and names of the compounds formed from the following pairs of ions: (a) \(\mathrm{Rb}^{+}\) and \(\mathrm{I}^{-},\) (b) \(\mathrm{Cs}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{Sr}^{2+}\) and \(\mathrm{N}^{3-},(\mathrm{d}) \mathrm{Al}^{3+}\) and \(\mathrm{S}^{2-}\).

We know that when ions combine to form a compound, they do so in a way that makes the total charge of the compound neutral. This means the positive and negative charges must balance each other out. For (a), the ions are \(\mathrm{Rb}^{+}\) and \(\mathrm{I}^{-}\), both with a charge of 1, so they can combine in a 1:1 ratio. For (b), the ions are \(\mathrm{Cs}^{+}\) and \(\mathrm{SO}_{4}^{2-}\); Cs has a charge of +1 and SO4 a charge of -2, so they will combine in a 2:1 ratio. For (c), the ions are \(\mathrm{Sr}^{2+}\) and \(\mathrm{N}^{3-}\); Sr has a charge of +2 and N a charge of -3, so they will combine in a 3:2 ratio. For (d), the ions are \(\mathrm{Al}^{3+}\) and \(\mathrm{S}^{2-}\); Al has a charge of +3 and S a charge of -2, so they will combine in a 2:3 ratio.

Now that we have the ratios, we can write out the empirical formulas. For (a), \(\mathrm{Rb}^{+}\) and \(\mathrm{I}^{-}\) will combine to give \(\mathrm{RbI}\). For (b), \(\mathrm{Cs}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) combine to give \(\mathrm{Cs}_{2}\mathrm{SO}_{4}\). For (c), \(\mathrm{Sr}^{2+}\) and \(\mathrm{N}^{3-}\) combine to give \(\mathrm{Sr}_{3}\mathrm{N}_{2}\). For (d), \(\mathrm{Al}^{3+}\) and \(\mathrm{S}^{2-}\) combine to give \(\mathrm{Al}_{2}\mathrm{S}_{3}\).

We now use the convention for naming simple ionic compounds, which is 'cation + anion'. For anions, we also replace the ending of the element's name with 'ide'. For polyatomic ions like \(\mathrm{SO}_{4}^{2-}\), we use their common names. So, compound (a) is named Rubidium Iodide, compound (b) is named Cesium Sulfate, compound (c) is named Strontium Nitride, and compound (d) is named Aluminum Sulfide.