The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2}\), and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order \(-2631,-5359\), and \(-14,774 \mathrm{~kJ} / \mathrm{mol}\). Match the appropriate formula to each lattice energy. Explain.

First, we need to determine the charges of the ions in each compound. This will help us understand their potential lattice energies. For \(\mathrm{FeCl}_{3}\): Iron (Fe) has a charge of +3 and Chloride (Cl) has a charge of -1. For \(\mathrm{FeCl}_{2}\): Iron (Fe) has a charge of +2 and Chloride (Cl) has a charge of -1. For \(\mathrm{Fe}_{2} \mathrm{O}_{3}\): Iron (Fe) has a charge of +3 and Oxygen (O) has a charge of -2.

Recall that lattice energy generally increases with increasing ionic charge. Comparing the charges in the compounds, we can predict the following trend in lattice energies: $$ \mathrm{Fe}_{2} \mathrm{O}_{3} > \mathrm{FeCl}_{3} > \mathrm{FeCl}_{2} $$ This is because in \(\mathrm{Fe}_{2} \mathrm{O}_{3}\), the charges on both the cation and anion are larger than in the other two compounds. In \(\mathrm{FeCl}_{3}\), we have Fe with a larger charge than in \(\mathrm{FeCl}_{2}\), so its lattice energy is expected to be larger than \(\mathrm{FeCl}_{2}\).

Based on our predicted trend in Step 2, we can now match the given lattice energy values to their respective compounds. We are given the following lattice energy values (in kJ/mol): - \(-14,774\) - \(-5359\) - \(-2631\) Since \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is expected to have the largest lattice energy, we match the largest value (\(-14,774\)) to this compound. Similarly, we match the smallest value (\(-2631\)) to \(\mathrm{FeCl}_{2}\), which is expected to have the smallest lattice energy, and the middle value (\(-5359\)) to \(\mathrm{FeCl}_{3}\). So, the matched lattice energies for each compound are: - \(\mathrm{Fe}_{2} \mathrm{O}_{3}: -14,774 \mathrm{~kJ/mol}\) - \(\mathrm{FeCl}_{3}: -5359 \mathrm{~kJ/mol}\) - \(\mathrm{FeCl}_{2}: -2631 \mathrm{~kJ/mol}\)