Complete the table by filling in the formula for the ionic compound formed by each pair of cations and anions, as shown for the first pair. $$ \begin{array}{|l|c|c|c|c|} \hline \text { Ion } & \mathrm{K}^{+} & \mathrm{NH}_{4}^{+} & \mathrm{Mg}^{2+} & \mathrm{Fe}^{3+} \\ \hline \mathrm{Cl}^{-} & \mathrm{KCl} & & & \\ \hline \mathrm{OH}^{-} & & & & \\ \hline \mathrm{CO}_{3}^{2-} & & & & \\ \hline \mathrm{PO}_{4}{ }^{3-} & & & & \\ \hline \end{array} $$

For each pair of ions, first determine the charge of the cation and anion. This information is provided in the superscript of each ion.

With the charges identified, you can now determine the correct ratio of cations and anions needed to form a neutral compound. The simplest way to do this is by crossing the charges (i.e., the top number of the cation becomes the subscript of the anion, and vice versa). Note that if the subscripts remain the same, you only need to use one of the ions.

Now, write the chemical formulas for the ionic compounds formed by each pair of cations and anions. Don't forget the subscripts from the previous step. $$ \begin{array}{|l|c|c|c|c|} \hline \text { Ion } & \mathrm{K}^{+} & \mathrm{NH}_{4}^{+} & \mathrm{Mg}^{2+} & \mathrm{Fe}^{3+} \\\ \hline \mathrm{Cl}^{-} & \mathrm{KCl} & \mathrm{NH}_{4}\mathrm{Cl} & \mathrm{MgCl}_{2} & \mathrm{FeCl}_{3} \\\ \hline \mathrm{OH}^{-} & \mathrm{KOH} & \mathrm{NH}_{4}\mathrm{OH} & \mathrm{Mg(OH)}_{2} & \mathrm{Fe(OH)}_{3} \\\ \hline \mathrm{CO}_{3}^{2-} & \mathrm{K}_{2}\mathrm{CO}_{3} & \mathrm{(NH}_{4}\mathrm{)_2}\mathrm{CO}_{3} & \mathrm{MgCO}_{3} & \mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3} \\\ \hline \mathrm{PO}_{4}{ }^{3-} & \mathrm{K}_{3}\mathrm{PO}_{4} & \mathrm{(NH}_{4}\mathrm{)_3}\mathrm{PO}_{4} & \mathrm{Mg}_{3}(\mathrm{PO}_{4})_{2} & \mathrm{FePO}_{4} \\\ \hline \end{array} $$