Lattice Energy Explained (H2 Chemistry)

Quick answer: Lattice energy is the enthalpy change when one mole of an ionic solid forms from its gaseous ions under standard conditions. It is negative because lattice formation releases energy. Its magnitude increases with higher ionic charge and smaller ionic radius, and you find it from a Born-Haber cycle using Hess’s law.

Lattice energy is one of the most heavily tested ideas in H2 Chemistry energetics. It appears in definition questions, trend comparisons, Born-Haber calculations, and solubility explanations. This guide explains what it means, what controls its size, how to use it in a cycle, and where students lose marks. It sits in Topic 7 of the H2 Chemistry Syllabus Explained and builds on ideas from the A Level Periodic Table guide.

Table of Contents

What Is Lattice Energy?

An ionic solid contains positive and negative ions arranged in a regular three-dimensional lattice. The oppositely charged ions attract one another through strong electrostatic forces. When gaseous ions come together to form the lattice, energy is released because the ions move to a lower-energy state.

For example, the formation of one mole of solid sodium chloride is written as:

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH°latt ≈ −787 kJ mol⁻¹

The negative sign is important. It shows that lattice formation is exothermic. Some textbooks define lattice energy in the reverse direction, as the energy needed to separate one mole of an ionic solid into gaseous ions, which makes the value positive. In H2 Chemistry questions, always check which direction and sign convention is being used before you start a calculation.

Factors Affecting Lattice Energy

The magnitude of the lattice energy depends mainly on two factors.

• Ionic charge: higher charges produce stronger electrostatic attraction and a larger magnitude of lattice energy.
• Ionic radius: smaller ions can approach each other more closely, giving stronger attraction and a larger magnitude.

A useful qualitative relationship for trend questions is:

Magnitude of lattice energy ∝ |z⁺ × z⁻| / (r₊ + r₋)

Here z⁺ and z⁻ are the ionic charges and r₊ + r₋ is the sum of the ionic radii, which is the distance between the centres of neighbouring ions. This explains trends. It is not the equation you use to calculate an exact value in H2 Chemistry; for that you use a Born-Haber cycle.

• Ionic Charge

Charge has the bigger effect. MgO contains Mg²⁺ and O²⁻, while NaCl contains Na⁺ and Cl⁻. Both are ionic solids, but MgO has a far larger magnitude because the attraction between 2+ and 2− ions is much stronger.

• Ionic Radius

When the charges are the same, smaller ions give a larger magnitude. LiF has a larger magnitude than KCl: both contain 1+ and 1− ions, but Li⁺ and F⁻ are smaller than K⁺ and Cl⁻, so the ions get closer together. To understand why ions differ in size, see Atomic Radius Explained, Effective Nuclear Charge and the Shielding Effect guide.

Lattice Energy Values: Common Examples

The table uses the formation convention (gaseous ions → solid lattice). Values vary slightly between data sources, so in an exam always use the value given in the question or the H2 Chemistry Data Booklet.

Ionic solid	Ions in lattice	Representative ΔH°latt / kJ mol⁻¹	Trend point
LiF	Li⁺ and F⁻	−1046	Small 1+ / 1− ions give a large magnitude
NaCl	Na⁺ and Cl⁻	−787	Reference example in many Born-Haber cycles
KCl	K⁺ and Cl⁻	−715	Larger ions than NaCl give a smaller magnitude
NaBr	Na⁺ and Br⁻	about −747	Larger halide ion gives a smaller magnitude
NaI	Na⁺ and I⁻	about −704	Even larger halide ion, smaller magnitude
MgO	Mg²⁺ and O²⁻	−3791	Higher charges give a very large magnitude
CaO	Ca²⁺ and O²⁻	about −3414	Larger Ca²⁺ gives a smaller magnitude than MgO

Lattice energy magnitude for KCl, NaCl, LiF, CaO and MgO: 2+/2− compounds are several times larger than 1+/1− compounds, and within each group smaller ions give a larger magnitude.

Notice how large the gap is between the two groups. Going from 1+/1− to 2+/2− ions multiplies the magnitude by roughly three to four, while changing the ion size within a group moves it by tens or a few hundred kJ mol⁻¹. That is why charge is the first thing to compare.

Comparing Lattice Energies in Exam Questions

Comparison questions ask which compound has the larger lattice energy and why. The strongest answers link ionic charge, ionic radius, and electrostatic attraction directly to the magnitude. Charge first, radius second.

Which has the greater magnitude of lattice energy, MgO or NaCl? MgO has the greater magnitude because Mg²⁺ and O²⁻ have higher ionic charges than Na⁺ and Cl⁻. The stronger electrostatic attraction between the oppositely charged ions means more energy is released when the lattice forms.

Which has the greater magnitude of lattice energy, LiF or KCl? LiF has the greater magnitude because Li⁺ and F⁻ are smaller than K⁺ and Cl⁻, so the ions approach more closely and the electrostatic attraction is stronger.

When comparing ions with the same charge, ionic radius increases down a group, so the magnitude falls: LiCl > NaCl > KCl. See Ionisation Energy Explained for why cation size changes the way it does.

Lattice Energy and Born-Haber Cycles

A Born-Haber cycle is a Hess’s law cycle for an enthalpy change that cannot be measured directly. It links the enthalpy change of formation of an ionic solid to the steps needed to make gaseous ions: atomisation, ionisation energy, bond dissociation, electron affinity and lattice formation. Hess’s law says the overall enthalpy change is the same whichever route you take, so you can rearrange the terms to find the unknown.

Step	Example for NaCl	Typical sign
1. Atomise sodium	Na(s) → Na(g)	Positive
2. Ionise sodium	Na(g) → Na⁺(g) + e⁻	Positive
3. Dissociate chlorine	½Cl₂(g) → Cl(g)	Positive
4. Add an electron to chlorine	Cl(g) + e⁻ → Cl⁻(g)	Negative
5. Form the lattice	Na⁺(g) + Cl⁻(g) → NaCl(s)	Negative

Born-Haber energy-level diagram for NaCl: atomisation +108, half bond dissociation +121, first ionisation energy +496, electron affinity −349, and lattice energy −787 kJ per mole, adding up to the enthalpy of formation of −411.

Read the diagram from the left. Each upward arrow is endothermic, and each downward arrow is exothermic. The long downward arrow on the left (ΔH°f) is the route that is measured directly; the stepped route on the right is the one that contains the unknown lattice energy.

Worked Example 1: Lattice Energy of Sodium Chloride

Data for the formation of NaCl(s): – ΔH°f[NaCl(s)] = −411 kJ mol⁻¹ – Atomisation of Na = +108 kJ mol⁻¹ – First ionisation energy of Na = +496 kJ mol⁻¹ – ½ bond dissociation enthrall of Cl₂ = +121 kJ mol⁻¹ – Electron affinity of Cl = −349 kJ mol⁻¹

The unknown is the lattice energy for Na⁺(g) + Cl⁻(g) → NaCl(s).

1. Write the equations for every step, with state symbols, before substituting any numbers.
2. Apply Hess’s law: ΔH°f = sum of all steps.
3. Set up: −411 = 108 + 496 + 121 − 349 + ΔH°latt
4. Add the known terms: 108 + 496 + 121 − 349 = +376
5. Rearrange: ΔH°latt = −411 − 376
6. ΔH°latt = −787 kJ mol⁻¹

The result is negative because forming the lattice from gaseous ions is exothermic. If a question defines lattice energy as the energy needed to separate the solid into gaseous ions, the value is +787 kJ mol⁻¹.

Worked Example 2: Lattice Energy of Magnesium Chloride

Exam questions often move to Mg because it forms a 2+ ion and needs two chloride ions. Formula: MgCl₂, so everything involving chlorine is doubled.

Data (textbook values; use the data given in your question): – ΔH°f[MgCl₂(s)] = −641 kJ mol⁻¹ – Atomisation of Mg = +148 kJ mol⁻¹ – First ionisation energy of Mg = +738 kJ mol⁻¹ – Second ionisation energy of Mg = +1451 kJ mol⁻¹ – ½ bond dissociation enthrall of Cl₂ = +121 kJ mol⁻¹ (per Cl atom) – Electron affinity of Cl = −349 kJ mol⁻¹ (per Cl atom)

The lattice energy is for Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s).

1.  Atomise Mg: +148

2. Remove two electrons from Mg: +738 + 1451 = +2189

3. Form two Cl atoms: 2 × (+121) = +242

4. Add two electrons to two Cl atoms: 2 × (−349) = −698

5. Sum of the known steps: 148 + 2189 + 242 − 698 = +1881

6. Apply Hess’s law: −641 = 1881 + ΔH°latt

7. ΔH°latt = −641 − 1881 = −2522 kJ mol⁻¹

This is much larger in magnitude than NaCl (−787) because Mg²⁺ has a higher charge and is smaller than Na⁺. The commonly quoted figure is about −2526 kJ mol⁻¹, so small differences are just rounding between data sources.

Why Oxides Need Extra Care

Oxides such as MgO need two electrons added to each oxygen atom to make O²⁻. The first electron affinity of oxygen is exothermic, but the second is endothermic, because you are pushing an electron onto an ion that is already negative. Students often write both electron affinities as negative, which gives the wrong answer. Always check the sign on the second electron affinity and use the data supplied. Despite this large energy cost, MgO is stable because the lattice energy released (−3791 kJ mol⁻¹) is so large.

Lattice Energy and Enthalpy of Solution

Lattice energy also helps explain whether an ionic solid dissolves. Dissolving needs two things: the lattice must be broken apart, which costs energy equal to the size of the lattice energy, and the separated gaseous ions must be hydrated by water, which releases energy.

ΔH°sol = −ΔH°latt + ΣΔH°hyd

The minus sign in front of ΔH°latt reverses the formation direction, because dissolving starts by breaking the lattice.

Worked example (illustrative). Using ΔH°latt(NaCl) = −787, ΔH°hyd(Na⁺) = −406 and ΔH°hyd(Cl⁻) = −364 kJ mol⁻¹: ΔH°sol = −(−787) + (−406) + (−364) = +17 kJ mol⁻¹

The experimental value for NaCl is closer to +4 kJ mol⁻¹ because tabulated data differ slightly between sources. In an exam, the question supplies a consistent set of values, so just substitute carefully.

A large lattice energy makes it harder to dissolve, but lattice energy alone does not decide solubility, since hydration enthalpy and entropy also contribute. The full explanation of how the two trade off down Group 2 is in the Periodic Table guide linked above.

Common Born-Haber Mistakes

• Forgetting that only ½Cl₂ is needed to make one Cl atom in the NaCl cycle, or forgetting to double chlorine terms for MgCl₂.
• Using the full Cl–Cl bond dissociation enthalpy when the equation contains ½Cl₂.
• Treating every electron affinity as negative, including the second electron affinity of oxygen.
• Leaving out the second ionisation energy when forming a 2+ ion.
• Mixing up lattice formation with lattice dissociation, which flips the sign.
• Substituting numbers before writing the equations, which hides sign errors.
• Defining r as the radius of one ion instead of the distance between ion centres in a trend explanation.
• Saying “stronger bonding” instead of “stronger electrostatic attraction between oppositely charged ions”.

How to Answer Lattice Energy Questions

1. Read the definition carefully and identify whether the question uses lattice formation or dissociation.
2. For trend questions, compare ionic charge first, then ionic radius.
3. Use the phrase “stronger electrostatic attraction” to explain the change in magnitude.
4. For Born-Haber calculations, write each step with state symbols before substituting numbers.
5. Check that the final equation balances both the chemical species and the overall enthalpy change.
6. Give the correct sign and units: kJ mol⁻¹.

For general exam technique that applies across Chemistry topics, see the H2 Chemistry revision guide.

Quick Revision Summary

Key idea	Remember
Definition	Enthalpy change when one mole of an ionic solid forms from gaseous ions.
Sign	Formation convention is negative because lattice formation is exothermic.
Charge	Higher ionic charge gives a larger magnitude of lattice energy.
Radius	Smaller ions give stronger attraction and a larger magnitude.
Comparison	Charge often outweighs radius when the charges are different.
Born-Haber	Use Hess’s law and keep the sign of every step consistent.
Oxides	The second electron affinity of oxygen is endothermic.
Solution	ΔH°sol = −ΔH°latt + ΣΔH°hyd
Exam wording	Say “stronger electrostatic attraction”, not just “stronger bonding”.

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