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

We already have the ionic compound formed by the combination of Na⁺ and O⁻², which is Na₂O. Now let's complete the row by combining the rest of the cations with O⁻² ions. 1. Ca²⁺ and O²⁻ - We need two O²⁻ ions to balance the charge of Ca²⁺ ion, so the formula becomes: CaO₂ → CaO (simplified). 2. Fe²⁺ and O²⁻ - We need two O²⁻ ions to balance the charge of Fe²⁺ ion, so the formula becomes: FeO₂ → FeO (simplified). 3. Al³⁺ and O²⁻ - We will need three O²⁻ ions and two Al³⁺ ions to balance the charges, so the formula becomes: Al₂O₃

Now, let's complete the second row by combining the cations with NO₃⁻ ions. 1. Na⁺ and NO₃⁻ - We need one Na⁺ ion to balance the charge of NO₃⁻ ion, so the formula becomes: NaNO₃ 2. Ca²⁺ and NO₃⁻ - We need two NO₃⁻ ions to balance the charge of Ca²⁺ ion, so the formula becomes: Ca(NO₃)₂ 3. Fe²⁺ and NO₃⁻ - We need two NO₃⁻ ions to balance the charge of Fe²⁺ ion, so the formula becomes: Fe(NO₃)₂ 4. Al³⁺ and NO₃⁻ - We need three NO₃⁻ ions to balance the charge of Al³⁺ ion, so the formula becomes: Al(NO₃)₃

Now, let's complete the third row by combining the cations with SO₄²⁻ ions. 1. Na⁺ and SO₄²⁻ - We need two Na⁺ ions to balance the charge of SO₄²⁻ ion, so the formula becomes: Na₂SO₄ 2. Ca²⁺ and SO₄²⁻ - We need one Ca²⁺ ion to balance the charge of SO₄²⁻ ion, so the formula becomes: CaSO₄ 3. Fe²⁺ and SO₄²⁻ - We need one Fe²⁺ ion to balance the charge of SO₄²⁻ ion, so the formula becomes: FeSO₄ 4. Al³⁺ and SO₄²⁻ - We will need three Al³⁺ ions and two SO₄²⁻ ions to balance the charges, so the formula becomes: Al₂(SO₄)₃

Finally, Let's complete the last row by combining the cations with AsO₄³⁻ ions. 1. Na⁺ and AsO₄³⁻ - We need three Na⁺ ions to balance the charge of AsO₄³⁻ ion, so the formula becomes: Na₃AsO₄ 2. Ca²⁺ and AsO₄³⁻ - We need three Ca²⁺ ions and two AsO₄³⁻ ions to balance the charges, so the formula becomes: Ca₃(AsO₄)₂ 3. Fe²⁺ and AsO₄³⁻ - We need three Fe²⁺ ions and two AsO₄³⁻ ions to balance the charges, so the formula becomes: Fe₃(AsO₄)₂ 4. Al³⁺ and AsO₄³⁻ - We need one Al³⁺ ion to balance the charge of AsO₄³⁻ ion, so the formula becomes: AlAsO₄ Now, with all the steps completed, the fully filled table should look like this: $$ \begin{array}{c|c|c|c|c} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Ca}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Al}^{3+} \\ \hline \mathrm{O}^{2-} & \mathrm{Na}_{2} \mathrm{O} & \mathrm{CaO} & \mathrm{FeO} & \mathrm{Al}_{2}\mathrm{O}_{3} \\ \mathrm{NO}_{3}^{-} & \mathrm{NaNO}_{3} & \mathrm{Ca(NO}_{3})_{2} & \mathrm{Fe(NO}_{3})_{2} & \mathrm{Al(NO}_{3})_{3} \\ \mathrm{SO}_{4}^{2-} & \mathrm{Na}_{2}\mathrm{SO}_{4} & \mathrm{CaSO}_{4} & \mathrm{FeSO}_{4} & \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \\ \mathrm{AsO}_{4}{ }^{3-} & \mathrm{Na}_{3}\mathrm{AsO}_{4} & \mathrm{Ca}_{3}(\mathrm{AsO}_{4})_{2} & \mathrm{Fe}_{3}(\mathrm{AsO}_{4})_{2} & \mathrm{AlAsO}_{4} \\ \hline \end{array} $$