Chapter 3 · Bonding & Lattice Energy

Bonding & Lattice Energy

What holds a crystal together — and how strongly? This chapter covers the five types of bonding in solids, the meaning of lattice energy, the Born–Lande equation, the Born–Haber cycle and the Madelung constant. The Born–Haber numerical appears almost every year — learn it by heart.

🔗 5 Bond Types⚡ Lattice Energy🔄 Born–Haber Cycle✎ PYQ Corner

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Five Types of Bonding in Solids

Based on the forces holding the particles together, solids are classified into five types:

Ionic solids: held by electrostatic attraction between positive and negative ions. Examples: NaCl, MgO, CaF₂.
Covalent (network) solids: atoms joined by a continuous network of covalent bonds in all directions. Examples: diamond, graphite, SiO₂ (quartz), SiC.
Metallic solids: positive metal ions (kernels) in a "sea" of delocalised (mobile) electrons. Examples: Fe, Cu, Na, Ag.
Molecular (van der Waals) solids: discrete molecules held together by weak van der Waals (dispersion) forces. Examples: solid Ar, I₂, naphthalene, dry ice (CO₂).
Hydrogen-bonded solids: molecules linked by directional hydrogen bonds — stronger than van der Waals forces. Examples: ice (H₂O), solid HF, solid NH₃.

Property	Ionic	Covalent (network)	Metallic	Molecular (van der Waals)	Hydrogen-bonded
Bonding force	Electrostatic (ion–ion)	Covalent bonds	Metallic (electron sea)	Weak dispersion forces	Hydrogen bonds
Melting point	High	Very high	Variable (low to high)	Low	Moderate
Hardness	Hard but brittle	Very hard	Hard to soft, malleable	Soft	Soft
Electrical conductivity	Conduct when molten or in solution, not as solid	Poor (graphite is an exception)	Good in solid state	Poor	Poor
Examples	NaCl, MgO	Diamond, SiO₂	Cu, Fe, Na	I₂, solid Ar	Ice, solid HF

💡 Memory trick: Order the bond strengths as Ionic > Covalent ≈ Metallic > H-bond > van der Waals. Then the melting points follow the same order — one trend answers many one-mark questions.

Exam tip: "Why are ionic solids hard but brittle?" — Answer: shifting one layer brings like charges face to face; the strong repulsion then cracks the crystal along the plane.

Five Types of Bonding in Solids

Ionic (electron transfer), covalent network (shared pairs), metallic (electron sea), molecular (weak van der Waals forces), and hydrogen-bonded solids.

⚡

Lattice Energy — Definition and Factors

Lattice energy (U) is the energy released when one mole of a solid ionic crystal is formed from its gaseous ions placed at infinite separation. For NaCl:

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH = −U (U is taken as a positive number)

Equivalently, it is the energy required to separate one mole of the solid into gaseous ions. A larger lattice energy means a more stable crystal.

Factors affecting lattice energy

Charge on ions: lattice energy increases sharply with charge, because U ∝ z₊ × z₋. Example: MgO (2 × 2) has a much larger lattice energy than NaCl (1 × 1).
Size of ions: lattice energy decreases as the interionic distance r₀ increases, because U ∝ 1/r₀. Example: NaF > NaCl > NaBr > NaI in lattice energy.
Crystal structure: enters through the Madelung constant — different structures (NaCl-type vs CsCl-type) have slightly different lattice energies.

Sign convention (important!): Lattice energy U is written as a positive number (energy needed to break the lattice), while the formation of the lattice from gaseous ions releases energy, so ΔH_{lattice formation} = −U. Examiners check this sign carefully in Born–Haber numericals.

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Born–Lande Equation

Born and Lande derived a theoretical expression for the lattice energy of an ionic crystal by adding the attraction between oppositely charged ions and the repulsion when electron clouds overlap:

U = − ( N_A · M · z₊ · z₋ · e² / 4πε₀r₀ ) × ( 1 − 1/n )

Meaning of each term:

N_A — Avogadro number (we want energy per mole).
M — Madelung constant, which accounts for the attractions and repulsions of all ions in the lattice, not just one pair.
z₊, z₋ — charges (valencies) of cation and anion.
e — charge on the electron; ε₀ — permittivity of free space.
r₀ — equilibrium interionic distance (r₊ + r₋).
n — Born exponent, a measure of the repulsion (typically 5–12; larger ions are more compressible, so n is larger). The term (1 − 1/n) corrects for repulsion.

💡 How to remember: The equation says "lattice energy = electrostatic attraction of the whole lattice (Madelung) minus a small repulsion correction". Attraction dominates, so U comes out positive (energy needed to break the crystal).

Kapustinskii equation: When the exact crystal structure (and hence the Madelung constant) is not known, Kapustinskii's simpler formula is used: U ≈ −(120200 · ν · z₊z₋ / d) × (1 − 34.5/d) kJ mol⁻¹, where d = r₊ + r₋ in pm and ν = number of ions per formula unit. It assumes a rock-salt type structure and works well for most ionic solids.

Born–Landé Potential Energy Curve

Attraction pulls ions together (−a/r) while short-range repulsion (+b/rⁿ) pushes back; the total energy is minimum at the equilibrium distance r₀.

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Born–Haber Cycle for NaCl

The Born–Haber cycle applies Hess's law: the enthalpy of formation of an ionic solid can be reached directly from the elements, or step-by-step through gaseous atoms and ions. Since enthalpy is a state function, both paths give the same total — so we can calculate the lattice energy from measurable quantities.

The five steps (with correct signs)

Step	Process	ΔH (kJ mol⁻¹)
1. Sublimation	Na(s) → Na(g)	+108
2. Dissociation	½Cl₂(g) → Cl(g)	+121 (half of the Cl–Cl bond energy)
3. Ionisation	Na(g) → Na⁺(g) + e⁻	+496 (first ionisation energy)
4. Electron affinity	Cl(g) + e⁻ → Cl⁻(g)	−349 (energy is released)
5. Lattice formation	Na⁺(g) + Cl⁻(g) → NaCl(s)	−U = −788 (energy released)

By Hess's law:
ΔH_f° = ΔH_sub + ½D + IE + EA + (−U)

So U = ΔH_sub + ½D + IE + EA − ΔH_f°

= 108 + 121 + 496 − 349 − (−411) = 787 ≈ 788 kJ mol⁻¹

Exam tip: This numerical is asked almost every year. Write all five steps with correct signs first, then apply Hess's law. The two places students lose marks: forgetting the ½ before the dissociation energy, and writing electron affinity with the wrong sign (it is negative — energy is released when Cl gains an electron).

Born–Haber Cycle for NaCl (kJ/mol)

The lattice energy (−788 kJ/mol) is found by Hess’s law: the direct route (ΔHf = −411) must equal the long route through gaseous atoms and ions.

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Madelung Constant

In a crystal, one ion is attracted by all oppositely charged neighbours and repelled by all similarly charged ones — not just its nearest neighbours. The Madelung constant (M) is a number that adds up all these attractions and repulsions for the whole lattice. It depends only on the geometry of the crystal structure.

Crystal structure	Madelung constant (M)
NaCl (rock-salt) type	1.7476
CsCl type	1.7627
Zinc blende (ZnS) type	1.6381

Why does it matter? In the Born–Lande equation, U ∝ M. A larger Madelung constant means stronger net attraction and a larger lattice energy for the same ions. This is why the CsCl structure (M = 1.7627) is slightly favoured over the NaCl structure for large ions like Cs⁺.

💡 Memory trick: The three values rise and fall as 1.74 – 1.76 – 1.63 for NaCl, CsCl, zinc blende. Remember: "zinc blende is the smallest (1.63), CsCl the largest (1.76)".

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Crystal-Field Considerations in Solids

Transition-metal ions in a crystal feel the electric field of the surrounding anions — just like ligands in a complex. This crystal-field stabilisation energy (CFSE) adds extra stability on top of the simple ionic (Born–Lande) lattice energy.

Across the 3d series, the experimental lattice energies of M²⁺ halides and oxides do not rise smoothly with decreasing ionic size. Instead they show a "double-hump" curve with maxima around d³ and d⁸ configurations.
This is because CFSE is maximum for d³ (t_2g³) and d⁸ (t_2g⁶e_g²) in an octahedral field, and zero for d⁰, d⁵ (high-spin) and d¹⁰.
Ions like Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) lie on the smooth "ionic only" line, while others (e.g. Ni²⁺, V²⁺) lie above it by exactly their CFSE.

Exam tip: Only a qualitative answer is expected: "Explain the double-hump curve in lattice energies of 3d metal(II) halides." — Answer: the humps are due to extra crystal-field stabilisation energy in octahedral sites; maxima at d³ and d⁸, minima (on the ionic line) at d⁰, d⁵ and d¹⁰.

Double-Hump Curve: CFSE Effect on Lattice Energy

A purely ionic model predicts a smooth rise, but real lattice energies hump upward at d³ and d⁸ — the extra stability comes from crystal field stabilisation energy.

✎ PYQ Zone — Chapter 3

Every previous-year question from this chapter's topics, with year, paper, marks and a full exam-ready answer. Tap a question to open its detailed solution.

2022B.Sc. CC-64 marks

Q3(b)(i) — “What do you mean by lattice energy of an ionic crystal? Calculate the lattice energy of NaCl using the following data: Madelung Constant (A) = 1·748, Equilibrium ionic distance = 2·79 Å, Born Exponent = 8·0, Electronic charge = 4·8×10⁻¹⁰ esu.” ⭐⭐⭐

📖 Detailed answer ▼

Lattice energy (U): the energy released when one mole of a solid ionic compound is formed from its gaseous ions at infinite separation — Na⁺(g) + Cl⁻(g) → NaCl(s). It is the quantitative measure of the strength of the ionic bond; a larger value means a more stable crystal.

Born–Landé equation (in cgs/esu units, the form the paper's data uses):

U = −(N_A·A·z⁺z⁻·e² / r₀)·(1 − 1/n)

N_A = 6.022×10²³ mol⁻¹ — Avogadro number (energy per mole).
A = 1.748 — Madelung constant for the rock-salt structure; it sums the attraction/repulsion of the whole lattice.
z⁺ = z⁻ = 1 for NaCl; e = 4.8×10⁻¹⁰ esu — electronic charge.
r₀ = 2.79 Å = 2.79×10⁻⁸ cm — equilibrium interionic distance.
n = 8.0 — Born exponent; (1 − 1/n) corrects for short-range repulsion of electron clouds.

Calculation:

N_A·A·e²/r₀ = (6.022×10²³ × 1.748 × (4.8×10⁻¹⁰)²) ÷ (2.79×10⁻⁸) = (6.022×10²³ × 1.748 × 23.04×10⁻²⁰) ÷ (2.79×10⁻⁸) ≈ 8.69×10¹² erg mol⁻¹ U = −8.69×10¹² × (1 − ⅛) = −8.69×10¹² × 0.875 ≈ −7.61×10¹² erg mol⁻¹

Converting (1 erg = 10⁻⁷ J): U ≈ −761 kJ mol⁻¹.

Result: lattice energy of NaCl ≈ −761 kJ mol⁻¹ (the negative sign means 761 kJ is released per mole when the lattice forms). The experimental value from Born–Haber data is −787 kJ mol⁻¹ — the theoretical Born–Landé value agrees within ~3%, which is why this equation is trusted.

2018B.Sc. CC-63 marks

Q2(d)(i) — “What is Madelung's constant? What is its significance?” ⭐⭐⭐

📖 Detailed answer ▼

Madelung constant (A): in a crystal, one ion is attracted by all neighbouring ions of opposite charge and repelled by all ions of the same charge — not just its nearest neighbours, but the whole infinite lattice. Adding up all these Coulomb terms, the electrostatic energy per ion pair can be written in the compact form

E = −A·z⁺z⁻e² / (4πε₀r₀)

where A is the Madelung constant — a pure number that depends only on the geometry of the crystal structure, not on the sizes or charges of the ions.

Significance:

It compresses the effect of the entire infinite lattice into a single number — without it we could not calculate lattice energies theoretically.
It is the key structural input of the Born–Landé equation (U ∝ A), so two compounds of the same ions in different structures have different lattice energies.
It distinguishes structure types energetically: NaCl-type A = 1.748, CsCl-type A = 1.763, zinc-blende-type A = 1.638. The slightly larger A for CsCl is one reason large cations like Cs⁺ prefer 8-coordination.

One line: the Madelung constant is the lattice's "geometry factor" — same ions, different structure, different A, different lattice energy.

2021B.Sc. CC-65 marks🔁 3 years

Q7 — “What is Born-Haber Cycle? Calculate the lattice energy of NaCl crystal from the following data by use of Born-Haber Cycle. Sublimation energy (S) = 108.7 kJ mol⁻¹, Dissociation energy for Cl₂ (D) = 225.9 kJ mol⁻¹, Ionization potential of Na(g) (I) = 489.5 kJ mol⁻¹, Electron affinity of Cl(g) (E) = −351.4 kJ mol⁻¹, Enthalpy of formation of NaCl (ΔH_f) = −414.2 kJ mol⁻¹.” ⭐⭐⭐⭐

Born–Haber theme repeated: 2020 · B.Sc. CC-6 · Q7 (MgS) · 2018 · B.Sc. CC-6 · Q3(a)(ii) (KF)

📖 Detailed answer ▼

Born–Haber cycle: an application of Hess's law. The standard enthalpy of formation of an ionic solid can be reached by two routes — directly from the elements, or step-by-step through gaseous atoms and ions. Since enthalpy is a state function, both routes must give the same total, so we can calculate the lattice energy (which cannot be measured directly) from measurable quantities like sublimation and ionisation energies.

The five steps (write these with correct signs — this is where marks are won or lost):

1. Sublimation: Na(s) → Na(g); ΔH = S = +108.7 kJ mol⁻¹ (energy absorbed).
2. Dissociation: ½Cl₂(g) → Cl(g); ΔH = ½D = 225.9/2 = +112.95 kJ mol⁻¹ (only half the Cl–Cl bond energy — we need one Cl atom).
3. Ionisation: Na(g) → Na⁺(g) + e⁻; ΔH = I = +489.5 kJ mol⁻¹.
4. Electron affinity: Cl(g) + e⁻ → Cl⁻(g); ΔH = E = −351.4 kJ mol⁻¹ (energy is released, so negative).
5. Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s); ΔH = U (unknown — this is what we find).

Hess's law: direct route = long route

ΔH_f = S + ½D + I + E + U U = ΔH_f − S − ½D − I − E U = (−414.2) − (108.7) − (112.95) − (489.5) − (−351.4) U = −414.2 − 108.7 − 112.95 − 489.5 + 351.4 ≈ −774 kJ mol⁻¹

Lattice energy of NaCl ≈ −774 kJ mol⁻¹ — i.e. 774 kJ is released per mole when gaseous Na⁺ and Cl⁻ form the solid. This agrees well with the Born–Landé theoretical value (−761 kJ mol⁻¹). Exam traps: (1) always take half the Cl₂ dissociation energy; (2) electron affinity is negative (energy released); (3) draw the box cycle with each arrow labelled — the diagram alone earns step marks.

2020B.Sc. CC-65 marks

Q7 — “Define lattice energy. Establish Born-Haber cycle for the formation of MgS(s) starting from Mg(s) and S₈(s), and hence calculate the electron affinity of S(g) for the S(g) + 2e⁻ ⟶ S²⁻(g) reaction using the thermochemical data given below: Enthalpy of formation = 345 kJmol⁻¹, Enthalpy of sublimation of Mg(s) = 153 kJmol⁻¹, Sum of 1ˢᵗ and 2ⁿᵈ ionization potentials of Mg(g) = 2187 kJmol⁻¹, Enthalpy of atomization of S₈(s) = 559 kJmol⁻¹, Lattice energy of MgS(s) = 2948 kJmol⁻¹.” ⭐⭐⭐

📖 Detailed answer ▼

Lattice energy (U): the energy released when one mole of a solid ionic compound is formed from its gaseous ions at infinite separation — here, Mg²⁺(g) + S²⁻(g) → MgS(s). Because the ions are doubly charged, MgS has a very large lattice energy.

Cycle setup (Hess's law — direct route = route through gaseous ions):

Sublimation: Mg(s) → Mg(g); ΔH = +153 kJ mol⁻¹.
Ionisation (1st + 2nd): Mg(g) → Mg²⁺(g) + 2e⁻; ΔH = +2187 kJ mol⁻¹.
Atomisation of sulphur: ⅛S₈(s) → S(g); ΔH = 559/8 = +69.9 kJ mol⁻¹ (we need only one S atom, so take one-eighth of the S₈ value).
Electron affinity (unknown): S(g) + 2e⁻ → S²⁻(g); ΔH = E.
Lattice formation: Mg²⁺(g) + S²⁻(g) → MgS(s); ΔH = U.

Sign convention — read carefully. The paper prints magnitudes without signs (345, 2948). Formation of MgS from its elements is exothermic, so ΔH_f = −345 kJ mol⁻¹; forming the lattice from gaseous ions releases energy, so U = −2948 kJ mol⁻¹. Sublimation, ionisation and atomisation all absorb energy (positive).

Solving for E:

ΔH_f = ΔH_sub + (I₁+I₂) + ⅛ΔH_at + E + U E = ΔH_f − ΔH_sub − (I₁+I₂) − ⅛ΔH_at − U E = (−345) − (153) − (2187) − (69.9) − (−2948) E = −345 − 153 − 2187 − 69.9 + 2948 ≈ +193 kJ mol⁻¹

Electron affinity S(g) → S²⁻(g) ≈ +193 kJ mol⁻¹ (endothermic overall). Why positive? Adding the first electron releases energy (−200 kJ mol⁻¹), but forcing a second electron onto an already negative S⁻ ion costs a lot of energy (+456 kJ mol⁻¹) — the sum is positive. MgS is stable only because its huge lattice energy (−2948 kJ mol⁻¹) outweighs everything else.

2018B.Sc. CC-63 marks

Q3(a)(ii) — “Calculate the heat of formation (ΔH_f) of KF from its elements from the following data by the use of Born-Haber cycle. Sublimation energy of K(s) = 87·8 kJ mol⁻¹, Dissociation energy of F₂(D) = 158·9 kJ mol⁻¹, Ionization energy of K(g)(I) = 414·2 kJ mol⁻¹, Electron affinity for F(g)(E) = −334·7 kJ mol⁻¹, Lattice energy of KF (U₀) = −807·5 kJ mol⁻¹.” ⭐⭐⭐

📖 Detailed answer ▼

Born–Haber cycle for KF (Hess's law): the heat of formation from the elements equals the sum of the steps through gaseous atoms and ions:

ΔH_f = S + ½D + I + E + U₀

Sublimation: K(s) → K(g); S = +87.8 kJ mol⁻¹.
Dissociation: ½F₂(g) → F(g); ½D = 158.9/2 = +79.45 kJ mol⁻¹ (half — we need one F atom).
Ionisation: K(g) → K⁺(g) + e⁻; I = +414.2 kJ mol⁻¹.
Electron affinity: F(g) + e⁻ → F⁻(g); E = −334.7 kJ mol⁻¹ (released, so negative).
Lattice formation: K⁺(g) + F⁻(g) → KF(s); U₀ = −807.5 kJ mol⁻¹ (already given negative — energy released).

Substitution:

ΔH_f = 87.8 + 79.45 + 414.2 + (−334.7) + (−807.5) ΔH_f = 581.45 − 1142.2 ≈ −561 kJ mol⁻¹

Heat of formation of KF ≈ −561 kJ mol⁻¹ (experimental −567 kJ mol⁻¹ — close). Note the sign logic: the first three steps cost energy, but the huge electron affinity of fluorine and the large lattice energy pay it back with interest — that is why KF forms so readily.