Edexcel International A Level Chemistry: Lattice energy (YCH11) – Revision Notes

These notes squeeze part 12B, Lattice energy (outcomes 12.12 to 12.19), of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, into quick-recall form. This is IA2 material belonging to Unit 4: Rates, Equilibria and Further Organic Chemistry. For the full teaching and longer worked examples, go to the lattice energy study guide.

When the notes feel secure, move on to the lattice energy practice questions. Record progress on the Edexcel A Level Chemistry checklist, find other topics on the course hub, and use the free diagnostics to decide what to revise next.

Definitions (12.12 and 12.16)

All of these apply under standard conditions (100 kPa, normally at 298 K).

Term	Meaning	Example equation	Sign
Standard enthalpy change of atomisation, ΔatH	Enthalpy change for producing 1 mol of free gaseous atoms from an element in its standard state	½I₂(s) → I(g)	Always +
First electron affinity	Enthalpy change when 1 mol of gaseous atoms each gain one electron, giving 1 mol of gaseous 1− ions	Br(g) + e⁻ → Br⁻(g)	Usually −
Second electron affinity	Enthalpy change when 1 mol of gaseous 1− ions each gain one more electron	S⁻(g) + e⁻ → S²⁻(g)	+
Lattice energy	Enthalpy change when gaseous ions come together to make 1 mol of ionic solid	Ca²⁺(g) + 2I⁻(g) → CaI₂(s)	Always −
Enthalpy change of solution, ΔsolH	Enthalpy change when 1 mol of ionic solid dissolves to infinite dilution	CaI₂(s) + aq → Ca²⁺(aq) + 2I⁻(aq)	+ or −
Enthalpy change of hydration, ΔhydH, of an ion	Enthalpy change when 1 mol of gaseous ions becomes 1 mol of aqueous ions	I⁻(g) + aq → I⁻(aq)	Always −

Key point: in this specification lattice energy is the exothermic making of the lattice, never its break-up.

Relationships to know

Use	Relationship
Born-Haber cycle (12.13)	ΔfH = ΔatH(metal) + ionisation energies + ΔatH(non-metal) + electron affinities + lattice energy
Solution cycle (12.17)	ΔsolH = −(lattice energy) + Σ ΔhydH(ions)
Feasibility of dissolving (12.19)	ΔStotal = ΔSsystem − ΔsolH / T, with ΔsolH in J mol⁻¹
Covalent character (12.14)	Compare experimental (Born-Haber) and theoretical (electrostatic) lattice energies

Method: any Born-Haber calculation

1. Read the formula and count the ions. A 1:2 compound such as CaI₂ needs two moles of I atoms and two electron affinities.
2. List each step from elements to gaseous ions: atomise the metal, ionise it (both ionisation energies for a 2+ ion), atomise the non-metal, add electrons.
3. Write ΔfH = (sum of those steps) + lattice energy.
4. Rearrange for the unknown, keeping every sign exactly as given.
5. Sense-check the sign of your answer.

Worked reminder. Use these values (kJ mol⁻¹): ΔfH(LiCl) = −409; ΔatH(Li) = +159; first ionisation energy of Li = +520; ΔatH(Cl) = +122; first electron affinity of Cl = −349.

Steps to gaseous ions: 159 + 520 + 122 − 349 = +452

Lattice energy = −409 − 452 = −861 kJ mol⁻¹

Method: drawing the Born-Haber diagram

• Elements in standard states go on the zero line, for example Li(s) + ½Cl₂(g).
• Atomisation and ionisation arrows point up.
• The first electron affinity arrow points down; a second electron affinity points up.
• Lattice energy is a long downward arrow from the gaseous ions to the solid.
• ΔfH links the elements to the solid (downward when exothermic).
• Write species and state symbols on every level, including free electrons, for example Li⁺(g) + e⁻ + Cl(g).

Method: enthalpy change of solution

1. Reverse the lattice energy so it becomes positive.
2. Add each ion’s hydration enthalpy, multiplied by how many of that ion the formula contains.
3. Total the terms.

Worked reminder. Use these values (kJ mol⁻¹): lattice energy of CaI₂ = −2074; ΔhydH(Ca²⁺) = −1616; ΔhydH(I⁻) = −297.

ΔsolH = 2074 + (−1616) + 2(−297) = −136 kJ mol⁻¹

As an energy level diagram: the gaseous ions sit at the top, the solid 2074 below them, and the aqueous ions 2210 below them. The solution therefore ends 136 below the solid.

Covalent character and polarisation (12.14, 12.15)

• Theoretical value: ions treated as perfect spheres held only by electrostatic forces.
• Experimental value: from a Born-Haber cycle using measured data.
• Values close together: bonding is close to purely ionic.
• Experimental clearly more exothermic: extra attraction from partial covalent bonding.
• Cause: the cation polarises the anion, drawing some of its electron density into the region between the nuclei.
• Biggest effect: small, highly charged cation with a large, highly charged anion.

Charge and radius (12.18)

Change	Lattice energy	ΔhydH
Higher ionic charge	More exothermic	More exothermic
Smaller ionic radius	More exothermic	More exothermic

Charge normally has the larger effect. Quick comparisons: Ca²⁺ is hydrated more exothermically than K⁺; LiF has a more exothermic lattice energy than LiI.

Solubility (12.19)

• An endothermic ΔsolH does not make a salt insoluble. Work out ΔStotal.
• ΔSsystem for dissolving has two parts: ions leaving the lattice (more disorder) and water ordered round the ions (less disorder). For small, highly charged ions ΔSsystem can be negative.
• Group 2 sulfates: the large anion means lattice energy changes little down the group, while cation hydration becomes much less exothermic. ΔsolH grows more endothermic, so solubility falls.
• Group 2 hydroxides: the small anion means lattice energy becomes much less exothermic down the group, faster than hydration does. ΔsolH grows more favourable, so solubility rises.

Must-know distinctions

• Atomisation vs bond enthalpy: ΔatH(Cl) is per mole of Cl atoms, so it is half the Cl–Cl bond enthalpy.
• Lattice energy vs ΔsolH: lattice energy starts from gaseous ions; ΔsolH starts from the solid.
• ΔhydH vs ΔsolH: hydration starts from one kind of gaseous ion; solution starts from the whole solid compound.
• First vs second electron affinity: the first is usually exothermic; the second is always endothermic.
• Covalent character vs covalent compound: a large experimental–theoretical gap shows partial covalency in an ionic solid, not a molecular substance.

Quick self-test

1. Give the equation that represents ΔatH for bromine.
2. Why is the second electron affinity of oxygen endothermic?
3. Write the equation that represents the lattice energy of strontium chloride.
4. Use these values (kJ mol⁻¹) to find the lattice energy of LiI: ΔhydH(Li⁺) = −522; ΔhydH(I⁻) = −297; ΔsolH(LiI) = −63.
5. Use these values (kJ mol⁻¹) to find ΔfH(NaF): first ionisation energy of Na = +496; ΔatH(Na) = +107; ΔatH(F) = +79; first electron affinity of F = −328; lattice energy of NaF = −930.
6. Which of KCl, CaCl₂ and CaO has the most exothermic lattice energy? Explain briefly.
7. Put Na⁺, Mg²⁺ and K⁺ in order of increasingly exothermic ΔhydH.
8. A salt has experimental lattice energy −850 and theoretical −848 kJ mol⁻¹. What does this suggest?
9. A salt has ΔsolH = +25.0 kJ mol⁻¹ and ΔSsystem = +60.0 J mol⁻¹ K⁻¹. Calculate ΔStotal at 298 K and say whether it dissolves.
10. For the salt in question 9, above what temperature does dissolving become feasible?

Answers

1. ½Br₂(l) → Br(g)
2. O⁻ is already negative, so it repels the incoming electron; energy has to be put in to force the electron on.
3. Sr²⁺(g) + 2Cl⁻(g) → SrCl₂(s)
4. Lattice energy = ΣΔhydH − ΔsolH = (−522 − 297) − (−63) = −756 kJ mol⁻¹
5. ΔfH = 107 + 496 + 79 − 328 − 930 = −576 kJ mol⁻¹
6. CaO: both ions are doubly charged, so the attraction is strongest. CaCl₂ has a 2+ cation but 1− anions; KCl has only single charges.
7. K⁺ < Na⁺ < Mg²⁺ (K⁺ is largest; Mg²⁺ is smallest and doubly charged).
8. The values agree closely, so the bonding is almost purely ionic with very little polarisation.
9. ΔSsurroundings = −25 000 / 298 = −83.9; ΔStotal = 60.0 − 83.9 = −23.9 J mol⁻¹ K⁻¹, so it is not expected to dissolve to give a standard solution at 298 K.
10. ΔStotal = 0 when T = 25 000 / 60.0 = 417 K (assuming ΔsolH and ΔSsystem stay constant).

Where marks are usually lost

• Writing lattice energy as the energy to break the lattice. This specification uses formation of 1 mol of solid from gaseous ions.
• Leaving “gaseous atoms” out of the atomisation definition, or using a whole bond enthalpy for a diatomic element.
• Missing “(g)” on the ions in electron affinity and lattice energy equations.
• Forgetting the second ionisation energy for a 2+ cation.
• Not doubling anion atomisation and electron affinity terms in 1:2 compounds.
• Sign slips while rearranging: write the whole Hess equation first, then substitute.
• Mixing kJ and J when working out ΔSsurroundings for dissolving.
• Saying “the bonding is more covalent” without explaining polarisation of the anion by the cation.
• Explaining a Group 2 trend using only lattice energy or only hydration enthalpy. Compare how both change.

Next steps

• Attempt the lattice energy practice questions and mark them against the worked answers.
• Return to the lattice energy study guide for any self-test answer you missed.
• Use the free diagnostics to pick your next topic.

Official syllabus

Source document: Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Section summarised here: 12B Lattice energy (outcomes 12.12 to 12.19), within Topic 12, Entropy and Energetics.