Lattice energy and Born-Haber cycles — practice questions

1. State the Born-Haber cycle equation based on Hess's Law: $\Delta H_f^\circ = \Delta H_{at(Na)}^\circ + IE_1(Na) + \Delta H_{at(Cl)}^\circ + EA_1(Cl) + \Delta H_{latt}^\circ$
2. Rearrange the equation to make lattice energy the subject: $\Delta H_{latt}^\circ = \Delta H_f^\circ - (\Delta H_{at(Na)}^\circ + IE_1(Na) + \Delta H_{at(Cl)}^\circ + EA_1(Cl))$
3. Substitute the given values into the rearranged equation: $\Delta H_{latt}^\circ = (-411) - ((+107) + (+496) + (+122) + (-349))$
4. Calculate the value of the bracket: $(107 + 496 + 122 - 349) = +376$ kJ mol⁻¹
5. Complete the final calculation: $\Delta H_{latt}^\circ = -411 - 376 = -787$ kJ mol⁻¹

Answer: The lattice energy of sodium chloride is -787 kJ mol⁻¹.

There are two main factors contributing to the difference in lattice energy:

1. Ionic Charge: The primary reason is the magnitude of the ionic charges. MgO is formed from Mg²⁺ and O²⁻ ions. NaF is formed from Na⁺ and F⁻ ions. The product of the charges in MgO is (2+) × (2-) = 4, whereas in NaF it is (1+) × (1-) = 1. The much larger product of charges in MgO leads to far stronger electrostatic forces of attraction between the ions.
2. Ionic Radii: The sum of ionic radii for MgO (0.072 + 0.140 = 0.212 nm) is slightly smaller than for NaF (0.102 + 0.133 = 0.235 nm). This smaller internuclear distance in MgO also contributes to a stronger attraction.

Conclusion: While the smaller ionic radii in MgO play a role, the dominant factor is the doubling of charge on both the cation and anion, which dramatically increases the electrostatic attraction and results in a significantly more exothermic lattice energy for MgO compared to NaF.