VKC Notes · Ionic bonding · Civil Services P-I · IFoS P-I

3.4 Ionic Bonding: Lattice Energy, Born–Haber Cycle, Born–Landé and Kapustinskii Equations, Fajans’ Rules

★★★ 2021–2025: 6 parts, 46 marks (Civil Services 3 parts / 20 m · IFoS 3 / 26 m). Almost always a number: build the cycle, put in the data, read the sign.

Syllabus lines. Civil Services: “Ionic bond, characteristics of ionic compounds, lattice energy, Born–Haber cycle”. IFoS: the same, plus “factors affecting stability of ionic compounds”. Radius ratio and crystal structures are in 3.1; ionisation enthalpy and electron gain in 14.1.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

3.4.1 The ionic bond and the properties of ionic compounds ★★

An ionic bond is the non-directional electrostatic attraction between oppositely charged ions in a lattice. It forms when the metal has a low ionisation enthalpy, the non-metal a large negative electron-gain enthalpy, and above all when the lattice enthalpy released pays for making the gaseous ions; electron transfer alone is always uphill (Na → Na⁺ costs 496, Cl → Cl⁻ returns only 349 kJ mol⁻¹).SN
Properties follow from strong, non-directional Coulomb forces in a 3-D lattice: high melting and boiling points (NaCl 801 °C, MgO 2852 °C); hard but brittle (a shear brings like charges face to face); insulators as solids, conductors when molten or dissolved; soluble in polar solvents of high dielectric constant (water, ε = 78), insoluble in non-polar ones; reactions in solution are fast (ions already exist).SN
The structure taken (rock salt, CsCl, fluorite …) is set by the radius ratio r₊/r₋ and the stoichiometry; see 3.1. A larger Madelung constant goes with a higher coordination number.

3.4.2 Lattice energy: the Born–Landé equation and the Madelung constant ★★★

Lattice enthalpy ΔH_L: the enthalpy change for MX(s) → M⁺(g) + X⁻(g), always positive (endothermic); quote the lattice formation enthalpy with a minus sign. Lattice energy U (at 0 K) and ΔH_L (298 K) differ by only a few kJ mol⁻¹.SN
Coulomb term for one ion: around Na⁺ in rock salt sit 6 Cl⁻ at r, 12 Na⁺ at √2 r, 8 Cl⁻ at √3 r, 6 Na⁺ at 2r … The series 6 − 12/√2 + 8/√3 − 6/2 + … converges (slowly) to the Madelung constant A = 1.748: it depends only on the geometry of the lattice.SN
Born repulsion B/rⁿ between closed shells; setting dU/dr = 0 at the equilibrium distance r₀ fixes B and gives the Born–Landé equation. The Born exponent n (from compressibility) is 5 (He-type ion), 7 (Ne), 9 (Ar), 10 (Kr), 12 (Xe); use the average for the two ions.SN

(3.4.1)

Born–Mayer form (exponential repulsion): replace (1 − 1/n) by (1 − d*/r₀) with d* = 34.5 pm. Both say the same thing: U ∝ z₊z₋/r₀: high charges and small ions give large lattice energies.SN

Structure	Rock salt	CsCl	Fluorite	Rutile	Zinc blende	Wurtzite
Madelung constant A	1.748	1.763	2.519	2.408	1.638	1.641

Worked example: lattice energy of NaCl from eq. 3.4.1 (r₀ = r(Na⁺) + r(Cl⁻) = 102 + 181 = 283 pm, n = ½(7 + 9) = 8)

U = −858 × (1 − ⅛) = −751 kJ mol⁻¹. The Born–Haber value is −787: the ionic model is within 5 % for NaCl.SN

3.4.3 The Born–Haber cycle ★★★ asked 4 times

Lattice energy cannot be measured directly. Hess’s law closes a cycle: forming MX(s) directly from the elements equals forming it through gaseous atoms and ions. Energy demanding: sublimation, dissociation, ionisation. Energy evolving: electron gain (first), lattice formation CSE 2025 Q1(b) 5m.SN

(3.4.2)

Reading the cycle: making the gaseous ions costs +376 kJ mol⁻¹ (108 + 121 + 496 − 349); the lattice returns 787. The lattice, not the electron transfer, makes NaCl stable.SN

Solved PYQ IFoS 2023 Q1(b) 8m

ΔH(sub) Li = 161, ΔH(dis) = 79, IE = 531, EA = −328, ΔH(xtal) = −1239 kJ mol⁻¹. Find ΔfH°(LiF).

Li(s) + ½F₂(g) → LiF(s). Take 79 as the atomisation enthalpy of ½F₂ (one mole of F atoms).
ΔfH° = 161 + 79 + 531 − 328 − 1239 = −796 kJ mol⁻¹.SN

Solved PYQ CSE 2021 Q1(b) 10m

Using U(NaCl) = −757, U(NaCl₂) = −2155, I₁ = 495, I₂ = 4561, EA = −348, ½D = 121, ΔH(sub) = 108 kJ mol⁻¹, can NaCl₂ form?

NaCl: ΔfH = 108 + 121 + 495 − 348 − 757 = −381 kJ mol⁻¹.
NaCl₂: ΔfH = 108 + 2(121) + 495 + 4561 + 2(−348) − 2155 = +2555 kJ mol⁻¹.
The second electron comes from the closed 2p⁶ core (I₂ ≈ 9 I₁); the larger lattice energy of a doubly charged cation cannot repay it. Entropy cannot rescue it (ΔfG ≈ ΔfH here): NaCl₂ does not form. The same cycle explains why MgCl does not exist (it disproportionates to Mg + MgCl₂, whose lattice energy repays I₂ of Mg).SN

Fig. 3.4.1 Born–Haber cycle for NaCl, to scale.

Draw it in 60 s: zero line for the elements; step up 108, 121, 496; step down 349; big drop 787 to NaCl(s) at −411; a direct arrow 0 → −411.

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