UPSC Civil Services (Mains), Chemistry optional · 2024 · Paper I
Q1(a)(ii). Lattice energies of NaBr and NaI above the calculated values
Question
Sodium bromide and sodium iodide have higher lattice energies than expected from theoretical calculations. Justify.
Answer in brief
The calculation treats the crystal as purely ionic; Br⁻ and I⁻ are large and easily polarised, so the real crystals gain extra covalent and dispersion bonding and the Born–Haber value comes out higher.
Key points
- The Born–Landé equation, U = NA M z⁺z⁻e²/(4πε₀r₀)(1 − 1/n), counts only Coulomb attraction between point charges and short-range repulsion.
- Br⁻ and I⁻ are large ions whose outer electrons are loosely held; Na⁺ pulls this cloud towards itself (Fajans' rules), giving partial covalent character.
- Large, electron-rich anions also attract their neighbours by dispersion (London) forces, largest for I⁻.
- The equation gives 717 and 671 kJ mol⁻¹, while the Born–Haber cycle gives 747 and 704 kJ mol⁻¹.
- For NaF, with the small and hard F⁻, the two values differ by less than 3 %.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coverslattice energyBorn–Landé equationBorn–Haber cycleFajans' rulespolarisationNaI
All parts of Question 1
- 1(a)(i) Most probable distance of the 1s electron5 marks
- 1(a)(ii) Lattice energies of NaBr and NaI above the calculated values5 marks
- 1(b)(i) Separation factor for ²³⁵UF₆ and ²³⁸UF₆ in one effusion stage5 marks
- 1(b)(ii) Unit cell and the cubic Bravais lattices5 marks
- 1(c) Normal boiling point of mercury10 marks
- 1(d)(i) Will CCl₂F₂ condense at 18 atm and 20 °C?5 marks
- 1(d)(ii) Overvoltage and its applications5 marks
- 1(e) Temperature at which the uncatalysed reaction matches catalase at 20 °C10 marks