2023 H2 Chem P1 Q10 — Chemical bonding Solution

The answer is A (CaCl, \(\text{CaCl}_2\), \(\text{MgCl}_2\)).
The magnitude of lattice energy is proportional to \(\frac{q_+ \times q_-}{r_+ + r_-}\), so it is larger for ions of higher charge and smaller radius. The anion is \(\text{Cl}^-\) in all three, so only the cation differs.
• CaCl and \(\text{CaCl}_2\): \(\text{Ca}^{2+}\) has a higher charge and a smaller radius than \(\text{Ca}^+\), so the magnitude of the lattice energy of \(\text{CaCl}_2\) is greater than that of CaCl.
• \(\text{CaCl}_2\) and \(\text{MgCl}_2\): both cations are 2+, but \(\text{Mg}^{2+}\) has a smaller radius than \(\text{Ca}^{2+}\) (one fewer electron shell), so the magnitude of the lattice energy of \(\text{MgCl}_2\) is greater than that of \(\text{CaCl}_2\).
Increasing magnitude: CaCl, then \(\text{CaCl}_2\), then \(\text{MgCl}_2\).
• B is wrong: it puts CaCl as the largest, but \(\text{Ca}^+\) has the lowest charge and the largest radius of the three cations, so CaCl has the smallest lattice energy.
• C is wrong: it puts \(\text{MgCl}_2\) below \(\text{CaCl}_2\), but \(\text{Mg}^{2+}\) is smaller than \(\text{Ca}^{2+}\) with the same charge, so \(\text{MgCl}_2\) has the larger lattice energy.
• D is wrong: this is the order of decreasing magnitude, the reverse of what is asked.