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. $$ \begin{array}{|l|c|c|c|c|} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Ca}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Al}^{3+} \\ \hline \mathrm{O}^{2-} & \mathrm{Na}_{2} \mathrm{O} & & & \\ \hline \mathrm{NO}_{3}^{-} & & & & \\ \hline \mathrm{SO}_{4}{ }^{2-} & & & & \\ \hline \mathrm{AsO}_{4}{ }^{3-} & & & & \\ \hline \end{array} $$

Ionic compounds form when a metal (cation) transfers one or more electrons to a non-metal (anion), resulting in a neutral compound. The charges between cations and anions must balance each other, meaning the overall charge of the ionic compound must be zero. With this in mind, we can complete the given table by finding the appropriate ratio of ions to achieve a neutral compound and writing its chemical formula.

1. Pair Ca^2+ with O^2-: They both have a charge of 2, but with opposite signs, so they will combine in a 1:1 ratio to form CaO. 2. Pair Ca^2+ with NO_3^-: Ca has a +2 charge, while NO_3^- has a -1 charge. The compound needs two NO_3^- ions to balance the +2 charge of Ca. The formula is Ca(NO_3)_2. 3. Pair Ca^2+ with SO_4^2-: They both have a charge of 2, but with opposite signs, so they will combine in a 1:1 ratio to form CaSO_4. 4. Pair Ca^2+ with AsO_4^3-: Ca has a +2 charge, while AsO_4 has a -3 charge. The ion charges can be balanced by combining three Ca^2+ ions with two AsO_4^3- ions, resulting in a formula of Ca_3(AsO_4)_2.

1. Pair Fe^2+ with O^2-: They both have a charge of 2, but with opposite signs, so they will combine in a 1:1 ratio to form FeO. 2. Pair Fe^2+ with NO_3^-: Fe has a +2 charge, while NO_3^- has a -1 charge. The compound needs two NO_3^- ions to balance the +2 charge of Fe. The formula is Fe(NO_3)_2. 3. Pair Fe^2+ with SO_4^2-: They both have a charge of 2, but with opposite signs, so they will combine in a 1:1 ratio to form FeSO_4. 4. Pair Fe^2+ with AsO_4^3-: Fe has a +2 charge, while AsO_4 has a -3 charge. The ion charges can be balanced by combining three Fe^2+ ions with two AsO_4^3- ions, resulting in a formula of Fe_3(AsO_4)_2.

1. Pair Al^3+ with O^2-: Al has a +3 charge, while O has a -2 charge. The ion charges can be balanced by combining two Al^3+ ions with three O^2- ions, resulting in a formula of Al_2O_3. 2. Pair Al^3+ with NO_3^-: Al has a +3 charge, while NO_3^- has a -1 charge. The compound needs three NO_3^- ions to balance the +3 charge of Al. The formula is Al(NO_3)_3. 3. Pair Al^3+ with SO_4^2-: Al has a +3 charge, while SO_4^2- has a -2 charge. The ion charges can be balanced by combining two Al^3+ ions with three SO_4^2- ions, resulting in a formula of Al_2(SO_4)_3. 4. Pair Al^3+ with AsO_4^3-: They both have a charge of 3, but with opposite signs, so they will combine in a 1:1 ratio to form AlAsO_4. Now the table is complete: $$ \begin{array}{|l|c|c|c|c|} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Ca}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Al}^{3+} \\ \hline \mathrm{O}^{2-} & \mathrm{Na}_{2} \mathrm{O} & \mathrm{CaO} & \mathrm{FeO} & \mathrm{Al}_{2} \mathrm{O}_{3} \\ \hline \mathrm{NO}_{3}^{-} & \mathrm{NaNO}_{3} & \mathrm{Ca(NO}_{3})_{2} & \mathrm{Fe(NO}_{3})_{2} & \mathrm{Al(NO}_{3})_{3} \\ \hline \mathrm{SO}_{4}{ }^{2-}& \mathrm{Na}_{2} \mathrm{SO}_{4} & \mathrm{CaSO}_{4} & \mathrm{FeSO}_{4} & \mathrm{Al}_{2} \mathrm{(SO}_{4})_{3} \\ \hline \mathrm{AsO}_{4}{ }^{3-}& \mathrm{Na}_{3} \mathrm{AsO}_{4} & \mathrm{Ca}_{3} \mathrm{(AsO}_{4})_{2} & \mathrm{Fe}_{3} \mathrm{(AsO}_{4})_{2} & \mathrm{AlAsO}_{4} \\ \hline \end{array} $$