Born-Haber Cycle & Lattice Energy: JEE Main

The Born-Haber cycle is a Hess's law application to ionic compound formation, connecting lattice energy to experimentally measurable quantities. JEE Main tests it through calculation questions where you identify the missing enthalpy step, and through conceptual questions about which factor determines lattice energy. Master both aspects here.

The Steps of Ionic Compound Formation

Formation of NaCl from Na(s) and Cl₂(g): the overall enthalpy ΔH_f is the sum of five steps. (1) Sublimation of Na(s) → Na(g): ΔH_sub (endothermic, ~108 kJ/mol for Na). (2) Ionisation energy of Na(g) → Na⁺(g) + e⁻: IE₁ (endothermic, ~496 kJ/mol). (3) Dissociation of ½Cl₂(g) → Cl(g): ½ × Bond dissociation energy (~121 kJ/mol). (4) Electron affinity of Cl(g) + e⁻ → Cl⁻(g): EA (exothermic for Cl, ~−349 kJ/mol). (5) Lattice energy: Na⁺(g) + Cl⁻(g) → NaCl(s): ΔH_lattice (exothermic, highly negative ~−788 kJ/mol for NaCl).

Hess's law: ΔH_f = ΔH_sub + IE + ½BDE + EA + ΔH_lattice. JEE gives you four of these quantities and asks for the fifth. Always draw the energy level diagram (Born-Haber cycle) with Na(s) + ½Cl₂(g) at the top-left and NaCl(s) at the bottom, to keep track of signs. Endothermic steps go up; exothermic go down. Take a free thermochemistry mock. For Hess's law basics see our Hess's law guide.

Lattice Energy: Factors and Trends

Lattice energy (magnitude) increases with: (1) higher ionic charges (MgO has much larger lattice energy than NaCl because Mg²⁺ and O²⁻ are doubly charged vs Na⁺ and Cl⁻ singly charged); (2) smaller ionic radii (smaller ions allow the charges to get closer, increasing attraction). Kapustinskii equation: U_L ∝ (q₊ × q₋) / (r₊ + r₋). JEE tests the trend: lattice energy order for NaCl, NaBr, NaI (decreases as anion size increases), and NaF, MgF₂, AlF₃ (increases with cation charge). Most important comparison: MgO vs NaF — MgO (2+, 2−) has much higher lattice energy than NaF (1+, 1−) despite similar internuclear distances. This explains why MgO has a much higher melting point than NaF.

Electron Affinity Trends and Born-Haber Calculations

Electron affinity (EA) = energy released when an electron is added to a neutral gaseous atom. EA₁ of Cl is exothermic (−349 kJ/mol). EA₂ of O (O⁻ + e⁻ → O²⁻) is endothermic (+780 kJ/mol) — the second electron is repelled by the already-negative O⁻. Despite the endothermic second electron affinity, oxide compounds (MgO, CaO) are stable because the lattice energy more than compensates. This argument — "why do ionic compounds with O²⁻ form despite the endothermic EA₂?" — is a direct JEE question.

Anomalous Trends and Diagonal Relationships

The diagonal relationship between Li and Mg (and Be and Al) partly arises from lattice energy: Li⁺ (small, singly charged) and Mg²⁺ (larger but doubly charged) produce similar charge/radius ratios (charge density), so their salts have similar lattice energies and hence similar solubility patterns. LiF is sparingly soluble (like MgF₂ and alkaline earth fluorides) while NaF, KF are readily soluble — the high lattice energy of LiF (small cation + small anion) exceeds the hydration energy, so it is less soluble. JEE tests LiF solubility as a direct anomalous behaviour question. For the full thermochemistry coverage see our chemical thermodynamics guide and our thermochemistry guide.