Born–Haber CycleJEE Advanced

Concept

Why does Na⁺Cl⁻ exist when ionising sodium costs 496 kJ/mol? The Born–Haber cycle itemises the energy budget: sublimation, bond-breaking and ionisation are expensive, but electron affinity and — above all — the enormous lattice energy repay it. Hess's law closes the loop and lets us extract U, which no experiment measures directly.

Key formula

Derivation

Build NaCl(s) two ways: directly from elements (ΔH_f), or via gaseous atoms → gaseous ions → solid. Hess's law equates them.

Everything except U is independently measurable, so $U = \Delta H_f - (\Delta H_{sub} + \tfrac12 D + IE + EA)$. For NaCl: −411 − (108+121+496−349) = −787 kJ/mol.

Scenarios to explore

• Born–Haber Cycle — Lattice energy from the thermodynamic staircase of NaCl.

Real-world applications

• Ranking ionic compound stabilities (MgO's U ≈ −3800 kJ/mol!).
• Explaining why NaCl₂ or MgCl don't exist (second IE vs extra lattice payback).
• Testing ionic models: covalency shows as U_experimental > U_calculated.

JEE exam tips

• Lattice energy ∝ q₊q₋/(r₊+r₋) — higher charges & smaller ions ⇒ bigger U (Kapustinskii intuition).
• MgO vs NaCl: 2+/2− charges ≈ 4× the lattice energy.
• The cycle explains why noble-gas salts don't form — no lattice payback beats the IE cost.

Common mistakes

• Forgetting the ½ on the dissociation energy of Cl₂.
• Sign chaos — EA and U are typically negative (energy released).
• Using electron GAIN enthalpy sign conventions inconsistently.

Exam traps to avoid

• ΔH_f measures overall stability, but a POSITIVE single step (IE) is fine if the cycle closes negative.
• Second EA of oxygen is POSITIVE (O⁻ + e⁻ → O²⁻ costs energy) — lattice energy still wins in MgO.