Edexcel International A Level Chemistry: Lattice energy (YCH11)

This guide teaches part 12B, Lattice energy, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, taking outcomes 12.12 to 12.19 in order. All of it is International A2 (IA2) content, assessed in the specification’s Unit 4: Rates, Equilibria and Further Organic Chemistry. It builds on Hess’s law and Group 2 chemistry (Unit 2) and on ionic bonding (Unit 1).

Once you have read it, shorten it into memory aids with the lattice energy revision notes, then try the lattice energy practice questions. Tick each outcome off on the Edexcel A Level Chemistry checklist, and browse the other topics on the course hub. If charge density and polarising power feel rusty, revisit ionic and covalent bonding before section 12.15.

Coverage at a glance

Spec ref	You need to…
12.12	Define standard enthalpy change of atomisation, electron affinity and lattice energy
12.13	Build Born-Haber cycles and calculate any missing step
12.14	Compare experimental and theoretical lattice energies to judge the degree of covalent bonding
12.15	Explain that covalency in an ionic bond comes from polarisation of anions by cations
12.16	Define enthalpy change of solution and enthalpy change of hydration of an ion
12.17	Calculate ΔsolH with energy cycles and energy level diagrams
12.18	Explain how ionic charge and ionic radius affect ΔhydH and lattice energy
12.19	Predict solubility from entropy and enthalpy changes of solution; explain the Unit 2 solubility trends

12.12 Three definitions to learn word for word

All three refer to standard conditions: a pressure of 100 kPa, normally quoted at 298 K.

Standard enthalpy change of atomisation, ΔatH. The enthalpy change for producing 1 mol of separate gaseous atoms from an element in its standard state. Breaking bonds and separating particles always needs energy, so ΔatH is always positive.

• Cs(s) → Cs(g)
• ½Cl₂(g) → Cl(g)
• ½I₂(s) → I(g) (iodine must be sublimed as well as split)

ΔatH(Cl) is half the Cl–Cl bond enthalpy, because it refers to 1 mol of atoms.

Electron affinity. The first electron affinity is the enthalpy change when 1 mol of gaseous atoms each pick up one electron, giving 1 mol of gaseous 1− ions.

• Cl(g) + e⁻ → Cl⁻(g) (exothermic: the incoming electron is pulled in by the nucleus)
• O⁻(g) + e⁻ → O²⁻(g) (second electron affinity: endothermic)

The second value is positive because the new electron approaches a particle that is already negatively charged. Work has to be done against that repulsion.

Lattice energy. This specification defines lattice energy as an exothermic quantity: the enthalpy change when gaseous ions come together to make 1 mol of an ionic solid.

• Cs⁺(g) + Cl⁻(g) → CsCl(s)
• Mg²⁺(g) + 2F⁻(g) → MgF₂(s)

It cannot be measured directly, so it is found indirectly with a Born-Haber cycle.

12.13 Building and using Born-Haber cycles

A Born-Haber cycle applies Hess’s law to the formation of an ionic solid. The direct route is ΔfH: elements in their standard states turn straight into the solid. The indirect route turns the elements into gaseous atoms, then gaseous ions, and finally packs the ions into the lattice. Both routes start and finish at the same place, so their enthalpy changes are equal:

ΔfH = ΔatH(metal) + ionisation energies + ΔatH(non-metal) + electron affinities + lattice energy

To draw the cycle as an energy level diagram, put the elements on a baseline at zero. Endothermic steps (atomisation, ionisation, second electron affinity) go up; exothermic steps (first electron affinity, lattice energy, usually ΔfH) go down. Label every level with its species and state symbols, and keep free electrons on the levels where they exist, for example Cs⁺(g) + e⁻ + Cl(g).

Worked example 1: lattice energy of caesium chloride

Use these values (kJ mol⁻¹): ΔfH(CsCl) = −443; first ionisation energy of Cs = +376; ΔatH(Cs) = +76; ΔatH(Cl) = +122; first electron affinity of Cl = −349.

Step 1. Total the steps that convert the elements into gaseous ions:

76 + 376 + 122 + (−349) = +225 kJ mol⁻¹

Step 2. Hess’s law gives −443 = +225 + lattice energy, so

lattice energy = −443 − 225 = −668 kJ mol⁻¹

Ions attracting each other release energy, so a negative answer makes sense.

Worked example 2: a compound with two anions

Use these values (kJ mol⁻¹) to find ΔfH(MgF₂): ΔatH(Mg) = +148; first and second ionisation energies of Mg = +738 and +1451; ΔatH(F) = +79; first electron affinity of F = −328; lattice energy of MgF₂ = −2957.

Magnesium steps: 148 + 738 + 1451 = +2337

Fluorine steps: the formula needs 2 mol of F atoms and 2 mol of F⁻ ions, so 2(79) + 2(−328) = −498

ΔfH = 2337 + (−498) + (−2957) = −1118 kJ mol⁻¹

12.14 and 12.15 Experimental and theoretical values

You can arrive at a lattice energy in two independent ways.

• Experimental: worked out from a Born-Haber cycle built on measured enthalpy changes. It therefore includes whatever bonding is really present.
• Theoretical: calculated from electrostatic theory, treating the ions as perfect spheres with evenly spread charge that interact only through electrostatic attraction and repulsion. This is the purely ionic model.

When the two numbers are close, the ionic model is a good description. When the experimental value is clearly more exothermic, the ions are bonded more strongly than pure electrostatics allows. The extra bonding is partial covalent character.

Worked example 3: reading the comparison

Use these values (kJ mol⁻¹):

Compound	Experimental	Theoretical	Gap	Gap as % of theoretical
CsCl	−668	−652	16	2.5%
AgCl	−905	−833	72	8.6%

CsCl fits the ionic model well. AgCl does not: its experimental value is 72 kJ mol⁻¹ more exothermic than theory predicts, so its bonding has significant covalent character. Percentages help when lattice energies differ greatly in size.

The cause: polarisation

A cation pulls on the electron cloud of a neighbouring anion and distorts it. The distortion is called polarisation. Some anion electron density is drawn into the space between the two nuclei, where it is shared. Shared electron density is a partial covalent bond, and it strengthens the lattice beyond the ionic prediction.

Polarisation is greatest when:

• the cation is small and highly charged, so its charge density and polarising power are high;
• the anion is large and highly charged, so its outer electrons are far from its nucleus and easily pulled out of shape.

You would therefore predict much covalent character in MgI₂ and very little in CsF. The Born-Haber comparison in 12.14 is the evidence that the effect is real.

12.16 Solution and hydration: two more definitions

Enthalpy change of solution, ΔsolH. The enthalpy change when 1 mol of an ionic compound dissolves in so much water that its ions no longer influence one another (infinite dilution), under standard conditions.

CsI(s) + aq → Cs⁺(aq) + I⁻(aq)

Enthalpy change of hydration, ΔhydH, of an ion. The enthalpy change when 1 mol of a gaseous ion becomes surrounded by water to give 1 mol of that ion in aqueous solution, under standard conditions.

Cs⁺(g) + aq → Cs⁺(aq)

Hydration always releases energy. Water is polar: cations attract the partially negative oxygen end of water molecules, while anions attract the partially positive hydrogen ends.

12.17 Finding ΔsolH from a cycle

Split dissolving into two imagined stages. First, separate the solid into gaseous ions; this is the reverse of lattice energy, so it is endothermic. Second, hydrate those gaseous ions, which is exothermic.

ΔsolH = −(lattice energy) + Σ ΔhydH(ions)

Worked example 4: caesium iodide

Use these values (kJ mol⁻¹): lattice energy of CsI = −604; ΔhydH(Cs⁺) = −276; ΔhydH(I⁻) = −297.

Step 1. Separate the lattice: +604

Step 2. Hydrate both ions: −276 + (−297) = −573

Step 3. ΔsolH = 604 + (−573) = +31 kJ mol⁻¹

For the energy level diagram, put CsI(s) low down and Cs⁺(g) + I⁻(g) well above it, joined by an upward arrow of +604 labelled as the reverse of the lattice energy. A downward arrow of 573 from the gaseous ions reaches Cs⁺(aq) + I⁻(aq). That level finishes 31 kJ mol⁻¹ above the solid, which shows the dissolving is endothermic.

If the formula holds two of an ion, count its hydration enthalpy twice.

The specification also suggests a practical: calculate the enthalpy change when a variety of ionic solids dissolve in water.

12.18 How charge and radius matter

Lattice energy and hydration enthalpy both come from electrostatic attraction, so the same two factors control them.

• Greater ionic charge means stronger attraction, so both values become more exothermic. Ca²⁺ is hydrated much more exothermically than K⁺, and an oxide lattice is far stronger than the matching chloride lattice.
• Smaller ionic radius lets the attracting particles get closer, again making both values more exothermic. Hydration enthalpies of Group 1 cations become less exothermic down the group, and a metal fluoride has a more exothermic lattice energy than the iodide of the same metal.

Charge usually wins. Doubling both charges roughly quadruples the attraction, while the radius differences between neighbouring ions are small by comparison.

12.19 Will it dissolve?

ΔsolH alone cannot settle solubility; many salts dissolve endothermically. Use the total entropy change from part 12A (entropy study guide):

ΔStotal = ΔSsystem + ΔSsurroundings, with ΔSsurroundings = −ΔsolH / T

When ΔStotal is positive, dissolving is feasible. Put ΔsolH into J mol⁻¹ before dividing by T.

Two effects compete inside ΔSsystem. Ions escaping from an ordered lattice increase disorder. Water molecules locked into shells around small or highly charged ions reduce disorder. With small, highly charged ions the second effect can dominate, so ΔSsystem may be negative.

Worked example 5: rubidium chloride

Use these values at 298 K: ΔsolH = +17.3 kJ mol⁻¹; S⦵ (J mol⁻¹ K⁻¹): RbCl(s) 95.9, Rb⁺(aq) 121.5, Cl⁻(aq) 56.5.

ΔSsystem = (121.5 + 56.5) − 95.9 = +82.1 J mol⁻¹ K⁻¹

ΔSsurroundings = −17 300 / 298 = −58.1 J mol⁻¹ K⁻¹

ΔStotal = 82.1 + (−58.1) = +24.0 J mol⁻¹ K⁻¹. RbCl dissolves even though the process takes in heat.

Worked example 6: barium sulfate

Use these values at 298 K: ΔsolH = +26.3 kJ mol⁻¹; ΔSsystem = −102.5 J mol⁻¹ K⁻¹.

ΔSsurroundings = −26 300 / 298 = −88.3 J mol⁻¹ K⁻¹

ΔStotal = −102.5 + (−88.3) = −190.8 J mol⁻¹ K⁻¹, so BaSO₄ is predicted to be insoluble. Both terms are negative; ΔSsystem is negative because water is held in order around the doubly charged ions.

Explaining the Group 2 trends from Unit 2

Outcome 8.15 (Groups 1 and 2) asked you to know that Group 2 hydroxides get more soluble down the group while Group 2 sulfates get less soluble. Compare how lattice energy and cation hydration change.

• Sulfates. SO₄²⁻ is a large ion. As the cation grows down the group, the distance between ion centres changes only a little in proportion, so lattice energy becomes only slightly less exothermic. Cation hydration becomes much less exothermic. ΔsolH becomes more endothermic, and solubility falls.
• Hydroxides. OH⁻ is small, so a larger cation makes a big proportional change to the distance between ions. Lattice energy becomes less exothermic faster than hydration enthalpy does, so ΔsolH becomes more favourable and solubility rises.

Common errors

• Describing lattice energy as the energy to break a lattice apart. Here it is the exothermic formation of the solid from gaseous ions.
• Using the full Cl–Cl bond enthalpy for ΔatH(Cl).
• Missing the second ionisation energy of a 2+ ion, or the doubled anion terms in MgF₂-type formulae.
• Omitting state symbols. Lattice energy equations need (g) on the ions and (s) on the product.
• Dividing a ΔsolH in kJ by T and then adding it to entropies in J mol⁻¹ K⁻¹.
• Saying a big experimental–theoretical gap means the compound “is covalent”. It shows partial covalent character in ionic bonding.

Next steps

• Shorten this guide with the lattice energy revision notes.
• Check yourself with the lattice energy practice questions.
• Find your weakest topics with the free diagnostics.

Official syllabus

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