Factors Affecting Lattice Energy

Once you can calculate lattice energies using Born-Haber cycles, the next step is to understand why lattice energies vary between different ionic compounds and what these variations can tell us about the nature of bonding. This lesson covers the two main factors (charge and radius), the theoretical vs. experimental lattice energy comparison, and how polarisation introduces covalent character.

Factors That Determine Lattice Energy

Lattice energy depends on the electrostatic attraction between the oppositely charged ions in the lattice. Two main factors control the strength of this attraction:

1. Ionic Charge

The greater the charge on the ions, the stronger the electrostatic attraction and the more exothermic (more negative) the lattice energy.

This follows from Coulomb's law. The force of attraction between two charged particles is proportional to the product of their charges:

F ∝ (q⁺ × q⁻) / r²

For example:

NaCl (Na⁺ and Cl⁻, charges 1+ and 1−): ΔH°latt = −787 kJ mol⁻¹
MgO (Mg²⁺ and O²⁻, charges 2+ and 2−): ΔH°latt = −3850 kJ mol⁻¹

The product of charges in MgO is 2 × 2 = 4, compared to 1 × 1 = 1 for NaCl. The lattice energy of MgO is roughly 4–5 times larger, consistent with the charge effect (and helped by smaller ionic radii).

2. Ionic Radius

The smaller the ionic radii, the closer the ions can get to each other, which increases the electrostatic attraction and makes the lattice energy more exothermic.

For example, comparing sodium halides:

NaF: ΔH°latt = −930 kJ mol⁻¹ (F⁻ radius = 133 pm)
NaCl: ΔH°latt = −787 kJ mol⁻¹ (Cl⁻ radius = 181 pm)
NaBr: ΔH°latt = −747 kJ mol⁻¹ (Br⁻ radius = 196 pm)
NaI: ΔH°latt = −704 kJ mol⁻¹ (I⁻ radius = 220 pm)

As the halide ion gets larger (F⁻ → I⁻), the lattice energy becomes less exothermic because the ions are further apart and the electrostatic attraction is weaker.

Comparing Lattice Energies: Comprehensive Table

When asked to compare lattice energies, always consider both charge and radius:

Compound	Ion charges	Cation radius / pm	Anion radius / pm	Lattice energy / kJ mol⁻¹
LiF	1+, 1−	76	133	−1037
LiCl	1+, 1−	76	181	−853
NaF	1+, 1−	102	133	−930
NaCl	1+, 1−	102	181	−787
NaBr	1+, 1−	102	196	−747
NaI	1+, 1−	102	220	−704
KCl	1+, 1−	138	181	−711
KBr	1+, 1−	138	196	−679
MgO	2+, 2−	72	140	−3850
CaO	2+, 2−	100	140	−3401
MgCl₂	2+, 1−	72	181	−2523
CaCl₂	2+, 1−	100	181	−2258

Key patterns from the table:

LiF has a more exothermic lattice energy than NaF because Li⁺ is smaller than Na⁺.
MgO has a much more exothermic lattice energy than NaCl because both charges are doubled and both ions are smaller.
MgCl₂ vs CaCl₂: Mg²⁺ is smaller than Ca²⁺, so MgCl₂ has the more exothermic lattice energy.

Theoretical vs. Experimental Lattice Energies

Lattice energies can be calculated in two ways:

Experimental (Born-Haber cycle): Uses measured thermochemical data (formation enthalpies, ionisation energies, etc.) in a Hess's law cycle. This gives the actual lattice energy.
Theoretical (ionic model): Uses Coulomb's law calculations assuming the ions are perfect spheres with their charges concentrated at the centre. This model assumes purely ionic bonding with no covalent character.

What the Comparison Tells Us

Compound	Theoretical / kJ mol⁻¹	Experimental / kJ mol⁻¹	Difference	Interpretation
NaCl	−770	−787	Small	Good agreement — bonding is essentially ionic
NaF	−912	−930	Small	Good agreement — ionic bonding
AgCl	−770	−905	Large	Significant covalent character
AgI	−736	−876	Very large	Substantial covalent character

When the experimental value is significantly more exothermic than the theoretical value, it means the ionic model underestimates the bonding. The extra stabilisation comes from covalent character — the sharing of electron density between the ions.

Polarisation and Covalent Character

The discrepancy between theoretical and experimental lattice energies is explained by ionic polarisation.

A small, highly charged cation (like Ag⁺ or Li⁺) has a high charge density. It can distort (polarise) the electron cloud of the anion, pulling electron density back towards itself. This partial sharing of electrons introduces covalent character into the ionic bond, which strengthens it beyond what pure electrostatic attraction would predict.

Factors that increase polarisation:

graph LR
    A["High polarising power of cation"] --> C["Greater covalent character"]
    B["High polarisability of anion"] --> C
    C --> D["Experimental lattice energy more exothermic than theoretical"]

High polarising power (of the cation):

Small ionic radius
High charge (2+ or 3+)
Electron configuration that does not shield the nuclear charge effectively (e.g. d-block ions like Ag⁺)

High polarisability (of the anion):

Large ionic radius (big electron cloud is easier to distort)
High negative charge

This is why AgI shows such a large discrepancy: Ag⁺ has a high polarising power (poor shielding from its d electrons), and I⁻ is a very large, polarisable anion.

Worked Example: Comparing NaCl and AgCl

Both NaCl and AgCl have the same crystal structure and similar ionic radii for the cation (Na⁺ = 102 pm, Ag⁺ = 115 pm). Yet:

NaCl: theoretical = −770, experimental = −787 (difference = 17 kJ mol⁻¹)
AgCl: theoretical = −770, experimental = −905 (difference = 135 kJ mol⁻¹)

The much larger discrepancy for AgCl is because Ag⁺ has a [Kr]4d¹⁰ electron configuration. The d electrons are poor at shielding the nuclear charge, giving Ag⁺ a higher effective nuclear charge and greater polarising power than Na⁺. This distorts the Cl⁻ electron cloud, introducing covalent character and additional stabilisation.

Fajan's Rules: A Summary

Fajan's rules predict when ionic compounds will have significant covalent character:

Factor	More ionic character	More covalent character
Cation size	Large cation (low charge density)	Small cation (high charge density)
Cation charge	Low charge (+1)	High charge (+2, +3)
Anion size	Small anion	Large anion (easily polarised)
Anion charge	Low charge (−1)	High charge (−2, −3)
Cation electron config	Noble gas config (e.g. Na⁺)	Non-noble gas config (e.g. Ag⁺, Pb²⁺)

Real-World Application: Why MgO Is So Refractory

Magnesium oxide has one of the most exothermic lattice energies of any binary compound (−3850 kJ mol⁻¹). This is due to the high charges (2+ and 2−) and small ionic radii. As a result, MgO has an extremely high melting point (2852 °C) and is used as a refractory lining material in furnaces and kilns in the steel industry. Understanding lattice energy directly explains why certain materials can withstand extreme temperatures.

Summary

Lattice energy depends on ionic charge (higher charge = more exothermic) and ionic radius (smaller ions = more exothermic).
Compare experimental (Born-Haber) and theoretical (ionic model) lattice energies to assess covalent character.
Large discrepancies indicate significant polarisation and covalent character.
Polarisation increases with small, highly charged cations and large, easily distorted anions.
Fajan's rules provide a systematic framework for predicting covalent character in ionic compounds.

A-Level Deep Dive: Factors Affecting Lattice Energy

Spec mapping

Edexcel 9CH0 specification Topic 13 requires explanation of how lattice energy varies with ionic charge and ionic radius — fundamentally, with charge density — and use of these trends to predict relative thermal stabilities, solubilities and other properties of ionic compounds (refer to the official specification document for exact wording). The relationship ΔH°_lat ∝ −|z⁺z⁻|/(r⁺ + r⁻) underpins comparison of NaCl vs MgO, MgO vs CaO, NaF vs NaCl, and the trend in thermal stability of group 2 carbonates and nitrates. These questions appear on Paper 1 as predictive items and on Paper 3 as multi-strand synoptic items combining lattice energy with group 2 chemistry. The Edexcel data booklet provides ionic radii for common cations and anions, which candidates must use to rank lattice energies.

Worked example with full mark scheme

Question (8 marks):

(a) The lattice energies of NaCl and MgO are −787 and −3791 kJ mol⁻¹. Explain why MgO has a more exothermic lattice energy. (3)

(b) The lattice energies of MgCO₃ and CaCO₃ are −3122 and −2811 kJ mol⁻¹. The lattice energies of MgO and CaO are −3791 and −3401 kJ mol⁻¹. Predict and explain which carbonate decomposes at the lower temperature. (5)

Solution with mark scheme:

(a) B1 — MgO has higher charge product: |z⁺z⁻| = 2 × 2 = 4 vs 1 × 1 = 1 for NaCl.

B1 — MgO has smaller ionic radii (Mg²⁺ < Na⁺; O²⁻ < Cl⁻), so smaller r⁺ + r⁻.

B1 — by ΔH°_lat ∝ −|z⁺z⁻|/(r⁺ + r⁻), the larger numerator (charge product) and smaller denominator (sum of radii) for MgO give a more exothermic lattice energy. The factor-of-five difference (3791/787 ≈ 4.8) reflects that the charge effect (factor 4) dominates over the radius effect.

(b) Carbonate decomposition: MCO₃(s) → MO(s) + CO₂(g).

M1 — ΔH°_decomp = ΔH°_lat(carbonate)_diss − ΔH°_lat(oxide)_diss + (small) ΔH for CO₃²⁻ → O²⁻ + CO₂.

For MgCO₃: ΔH°_lat(diss): +3122; for MgO: +3791. Difference: +3791 − 3122 = +669 kJ mol⁻¹ "extra" stability for the oxide product.

For CaCO₃: ΔH°_lat(diss): +2811; for CaO: +3401. Difference: +3401 − 2811 = +590 kJ mol⁻¹.

M1 — comparison shows that for MgO the carbonate-to-oxide transformation gains more lattice stability (+669) than for CaO (+590).

B1 — therefore MgCO₃ decomposes at a lower temperature than CaCO₃.

B1 — explanation: the small Mg²⁺ ion polarises the large CO₃²⁻ ion strongly, weakening the C–O bonds and lowering the activation energy for decomposition.

B1 — equivalently, the lattice energy of MgO (relative to MgCO₃) is more strongly stabilising than CaO (relative to CaCO₃) because Mg²⁺ benefits more from the smaller anion (O²⁻ < CO₃²⁻).

Total: 8 marks (M2 B6).

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):