2022 H2 Chem P1 Q10 — Chemical bonding Solution

The answer is D.

Lattice energy \(\propto \left|\frac{q_+ \times q_-}{r_+ + r_-}\right|\). The cation charge is +2 and the anion charge is -1 in all four salts, so the salt with the smallest sum of ionic radii, \(r_+ + r_-\), has the most exothermic lattice energy.

Using the Data Booklet ionic radii:
• \(\text{CaBr}_2\): 0.099 + 0.195 = 0.294 nm
• \(\text{NiBr}_2\): 0.069 + 0.195 = 0.264 nm
• \(\text{PbCl}_2\): 0.120 + 0.181 = 0.301 nm
• \(\text{ZnCl}_2\): 0.074 + 0.181 = 0.255 nm

\(\text{ZnCl}_2\) has the smallest sum, so it has the most exothermic lattice energy.

• A is wrong: \(\text{CaBr}_2\) has a larger sum (0.294 nm).
• B is wrong: \(\text{Ni}^{2+}\) (0.069 nm) is smaller than \(\text{Zn}^{2+}\) (0.074 nm), but \(\text{Br}^-\) (0.195 nm) is larger than \(\text{Cl}^-\) (0.181 nm), so the sum for \(\text{NiBr}_2\) (0.264 nm) is still larger than that for \(\text{ZnCl}_2\).
• C is wrong: \(\text{PbCl}_2\) has the largest sum (0.301 nm), so its lattice energy is the least exothermic of the four.