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Energetics

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Summary — What This Chapter Covers

Lattice energy (Δ_lattH) — the energy released when gaseous ions come together to form 1 mole of ionic solid. Always exothermic, always huge and negative.
Born–Haber cycles — a Hess's Law diagram that breaks the formation of an ionic compound into small steps (atomisation, ionisation, electron affinity, lattice formation) so we can calculate the one step we can't measure directly.
Key enthalpy definitions — atomisation, first/second electron affinity, ionisation energy — each with its own energy "sign" (endo or exo) that you must know cold.
Polarisation — real ionic bonds aren't 100% ionic. Small, highly-charged cations distort ("polarise") large anions, adding covalent character — which is why experimental and theoretical lattice energies disagree.
Enthalpy of solution & hydration — what happens energetically when an ionic solid dissolves in water: breaking the lattice apart (endothermic) and hydrating the free ions (exothermic).
Trends — how ionic radius and ionic charge affect the size of lattice energy and hydration enthalpy (charge density is the master variable).
Predicting solubility — using entropy (ΔS_total) alongside enthalpy of solution to explain why some ionic solids dissolve and others don't.

1. Born–Haber Key Terms

1.1 Lattice Energy (Δ_lattH)

Think of an ionic solid like NaCl as a crowd of positive and negative ions that have "fallen into place" next to each other, held tight by electrostatic attraction. Lattice energy is the enthalpy change when that crowd forms from scratch — starting from ions floating freely as a gas, with no attraction between them at all.

Because oppositely-charged ions want to be near each other, snapping them together into a solid lattice releases a huge amount of energy. That's why lattice energy is always exothermic (a big negative number, often in the hundreds or thousands of kJ mol⁻¹).

Definition Lattice energy (formation) = the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions, under standard conditions.

Na⁺(g) + Cl⁻(g) → NaCl(s) Δ_lattH = −776 kJ mol⁻¹

Why so negative? The bigger (more negative) the lattice energy, the stronger the ionic bonding — because more energy had to be released to pull the ions together, which means the electrostatic attraction between them was very strong.

Watch out You cannot measure lattice energy directly in the lab with one experiment — there's no way to physically "watch" gaseous ions collapse into a solid. That's exactly why we need Born–Haber cycles: they let us calculate it indirectly from values we can measure.

Practice Question

Write the equation (with state symbols) that represents the lattice energy of magnesium oxide.

1.2 Enthalpy of Atomisation (Δ_atH)

Before you can turn an element into ions, you first have to get it into single, separate gas atoms. That's atomisation — and it always costs energy, because you're breaking whatever bonds or forces hold the element together (metallic bonds in a solid metal, or covalent bonds in a diatomic gas like Cl₂).

Definition The enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state. Always endothermic (positive) — you're breaking bonds, never forming them.

Element's standard state	Example equation
Solid metal (e.g. Na, K)	`Na(s) → Na(g) Δ_atH = +108 kJ mol⁻¹`
Diatomic gas (e.g. Cl₂)	`½Cl₂(g) → Cl(g) Δ_atH = +121 kJ mol⁻¹`
Liquid (e.g. Hg)	`Hg(l) → Hg(g)`

Why the "½" for chlorine? The definition demands exactly 1 mole of gaseous atoms as the product — not 1 mole of the element in its normal form. Since Cl₂ contains 2 atoms, you need only ½ a mole of Cl₂ molecules to make 1 mole of Cl atoms. This trips up a lot of students — always check your equation produces exactly 1 mole of atoms.

Practice Question

Write the equation for the standard enthalpy change of atomisation of bromine, given that bromine's standard state is a liquid, Br₂(l).

1.3 Electron Affinity (Δ_eaH)

Electron affinity is about an atom (or ion) gaining an electron to become more negative. The first electron affinity is almost always exothermic — an incoming negative electron is attracted to the positive nucleus, so energy is released. But the second electron affinity is a completely different story.

Definition The enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms (or ions) to form 1 mole of gaseous ions, under standard conditions.

Step	Equation	Sign
1st electron affinity of Cl	`Cl(g) + e⁻ → Cl⁻(g) Δ_eaH = −364 kJ mol⁻¹`	Exothermic
2nd electron affinity of O	`O⁻(g) + e⁻ → O²⁻(g) Δ_eaH = +844 kJ mol⁻¹`	Endothermic

Why does the second one flip sign? Once an atom has already gained one electron, it's now a negatively charged ion (e.g. O⁻). Trying to force a second negative electron onto an already-negative ion means fighting against strong repulsion between like charges. That repulsion has to be overcome with an input of energy — hence endothermic.

Common Mistake Students often assume "electron affinity = always exothermic." It's true for the first electron affinity of virtually every element, but the second (and any further) electron affinity is endothermic due to electron–electron repulsion. Always check which electron affinity you're being asked about.

Practice Question

Explain why the first electron affinity of chlorine is exothermic, but the second electron affinity of oxygen is endothermic.

2. Constructing Born–Haber Cycles

A Born–Haber cycle is just Hess's Law applied to ionic compounds. The whole idea: there are two different "routes" to get from the elements in their standard states to the finished ionic solid — a short direct route (enthalpy of formation) and a long indirect route (atomisation → ionisation → electron affinity → lattice formation). Since both routes start and end at the same place, they must release/absorb the same total energy.

Golden rule for drawing the diagram Energy increases upward on the page.
• Endothermic steps → arrow points up (energy of substances increases)
• Exothermic steps → arrow points down (energy of substances decreases)

Step-by-step: Building the NaCl cycle

Na⁺(g) + Cl(g) + e⁻ ─┐ ↑ +500 (IE₁) │ −364 (EA₁) Na(g) + Cl(g) │ Na⁺(g) + Cl⁻(g) ↑ +121 (atom. Cl) │ Na(g) + ½Cl₂(g) │ Δlatt H (what we want) ↑ +108 (atom. Na) │ Na(s) + ½Cl₂(g) ────── −411 (ΔfH) ─┴──→ NaCl(s)

Reading this: starting from the elements at the bottom, you can go up and around (atomise → ionise → electron affinity → lattice energy) or go straight down via enthalpy of formation. Both arrive at the same final state, so their total energy changes must be equal.

Building tip Start by drawing the elements in their standard states roughly a third of the way up the page (this is your "base line"). Then build the two-step "atomise, then ionise" pathway upward on one side, and drop the enthalpy of formation arrow down to the ionic solid on the other. Finally connect the gaseous ions to the solid with the lattice energy arrow — that's usually the missing piece you're solving for.

Worked Example: KCl Cycle

Step	Equation	Enthalpy term
Atomise K(s)	K(s) → K(g)	Δ_atH(K)
Ionise K(g)	K(g) → K⁺(g) + e⁻	IE₁
Atomise ½Cl₂(g)	½Cl₂(g) → Cl(g)	Δ_atH(Cl)
Electron affinity of Cl	Cl(g) + e⁻ → Cl⁻(g)	EA₁
Sum of the above	—	Δ₁H
Lattice formation	K⁺(g) + Cl⁻(g) → KCl(s)	Δ_lattH

The "direct route" (K(s) + ½Cl₂(g) → KCl(s)) is the enthalpy of formation. The "indirect route" is the sum of every step listed above. By Hess's Law, they're equal — which is exactly how the calculation section works.

Practice Question

Sketch (in words/steps) the six stages needed for a Born–Haber cycle for magnesium oxide, MgO. (Hint: Mg loses 2 electrons, O gains 2 electrons.)

3. Born–Haber Calculations

Once the cycle is drawn, the maths is just Hess's Law rearranged. The full version of the equation lists every single step:

Full equation Δ_fH = Δ_atH + Δ_atH + IE + EA + Δ_lattH

It's much easier to think of it in three simplified chunks:

Simplified equation Δ_fH = Δ₁H + Δ_lattH

where Δ₁H = the sum of everything needed to turn the elements into gaseous ions (atomisation + ionisation + electron affinity).

Rearranged to solve for the value you usually need — lattice energy:

Rearranged for lattice energy Δ_lattH = Δ_fH − Δ₁H

Worked Example: Lattice Energy of KCl

	Δ_atH / kJ mol⁻¹	IE / EA / kJ mol⁻¹
K	+90	+418
Cl	+122	−349

Δ_fH(KCl) = −437 kJ mol⁻¹

Δlatt H = ΔfH − [Δat H(K) + Δat H(Cl) + IE₁(K) + EA₁(Cl)] Δlatt H = (−437) − [(+90) + (+122) + (+418) + (−349)] Δlatt H = (−437) − (281) Δlatt H = −718 kJ mol⁻¹

Remember: double or halve when needed If a compound has 2 moles of a single-charge ion (e.g. MgCl₂), you must double the relevant atomisation and electron affinity values for chlorine, because you're now forming 2 moles of Cl⁻ from 2 moles of Cl atoms — not just 1.

Practice Question

Using the data: Δ_atH(Mg) = +148, Δ_atH(O) = +248, IE₁(Mg) = +736, IE₂(Mg) = +1450, EA₁(O) = −142, EA₂(O) = +770, Δ_fH(MgO) = −602 (all kJ mol⁻¹) — calculate Δ_lattH of MgO.

4. Polarisation

Everything so far has assumed a "perfect" ionic model: perfectly spherical ions, purely electrostatic attraction, whole-number charges completely transferred. In reality, bonding is rarely 100% ionic — there's often a bit of covalent character mixed in, and that's what polarisation explains.

Picture a small, densely-charged cation sitting right next to a big, "squishy" anion. The cation's concentrated positive charge pulls on the anion's electron cloud, dragging some electron density toward itself. The anion's shape gets distorted — no longer a perfect sphere — and some electron density is effectively shared between the two ions. That sharing is exactly what covalent bonding is, so the bond now has some covalent character mixed into its ionic character.

Ionic bonding Ionic bonding + covalent character (+) (−) (+) (−) small large small large, distorted sphere sphere sphere "teardrop" shape

Two factors that control polarisation Cations: small + highly charged = large "polarising power" (charge density high)
Anions: large ionic radius = easily distorted (highly "polarisable")

A quick way to estimate charge density (and hence polarising power):

Approximation charge density ≈ charge ÷ r²

This is why theoretical lattice energies (calculated assuming pure ionic bonding) sometimes disagree noticeably with experimental lattice energies (measured via Born–Haber cycles using real Δ_fH data) — the gap is a measure of how much covalent character is really present.

	NaCl	MgCl₂
Charge density of cation	111	556
Experimental Δ_lattH	−780	−2526
Theoretical Δ_lattH	−770	−2326
Extent of covalent character	very little	more than NaCl

Why it matters Mg²⁺ has a much higher charge density than Na⁺ (smaller radius, double the charge). That gives it much stronger polarising power, distorting the Cl⁻ ions more, adding more covalent character — and that's why the theoretical and experimental values for MgCl₂ diverge more than they do for NaCl.

Practice Question

Which cation would you expect to have greater polarising power: Al³⁺ or Na⁺? Explain your reasoning.

5. Enthalpy of Solution & Hydration

5.1 Key Definitions

Enthalpy of solution, Δ_solH The enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form an infinitely dilute solution.

KCl(s) + aq → K⁺(aq) + Cl⁻(aq)
Can be exothermic OR endothermic depending on the compound.

Enthalpy of hydration, Δ_hydH The enthalpy change when 1 mole of a specific gaseous ion dissolves in sufficient water to form an infinitely dilute solution.

Mg²⁺(g) + aq → Mg²⁺(aq)
Always exothermic — new attractions are formed between the ion and water molecules, releasing energy.

Water is a polar molecule — the oxygen atom carries a slight negative charge (δ−) and the hydrogens carry a slight positive charge (δ+). When an ionic solid dissolves, water molecules cluster around each ion: oxygen atoms point toward cations, hydrogen atoms point toward anions. These "ion–dipole attractions" are what make hydration exothermic, and what physically separates and stabilises the free ions in solution.

5.2 The Energy Cycle

There are two routes from gaseous ions to hydrated (aqueous) ions:

Route 2 (direct): Δhyd H Mg²⁺(g)+Cl⁻(g) ───────────────→ Mg²⁺(aq)+Cl⁻(aq) │ ↑ │ Δlatt H (indirect) Δsol H (indirect) ↓ │ MgCl₂(s) ────────────────┘

Hess's Law relationship Δ_hydH = Δ_lattH + Δ_solH

where Δ_hydH here is the total hydration enthalpy — add together the individual Δ_hydH values of the cation and anion.

Worked Example: Hydration Enthalpy of Cl⁻ (from KCl)

Given: Δ_lattH(KCl) = −711, Δ_solH(KCl) = +26, Δ_hydH(K⁺) = −322 (all kJ mol⁻¹).

Δhyd H(K⁺) + Δhyd H(Cl⁻) = Δlatt H(KCl) + Δsol H(KCl) Δhyd H(Cl⁻) = Δlatt H(KCl) + Δsol H(KCl) − Δhyd H(K⁺) Δhyd H(Cl⁻) = (−711) + (+26) − (−322) Δhyd H(Cl⁻) = −363 kJ mol⁻¹

Handling ions with subscripts (e.g. MgCl₂) Since MgCl₂ contains 2 moles of Cl⁻, the hydration term for chloride must be doubled:
Δ_hydH = Δ_hydH(Mg²⁺) + 2 × Δ_hydH(Cl⁻)
Always check the formula for how many moles of each ion are actually present.

Practice Question

Given Δ_lattH(MgCl₂) = −2592 kJ mol⁻¹, Δ_solH(MgCl₂) = −55 kJ mol⁻¹, and Δ_hydH(Cl⁻) = −363 kJ mol⁻¹, calculate Δ_hydH(Mg²⁺).

5.3 Predicting Solubility — Enthalpy AND Entropy

Dissolving isn't just about enthalpy. Two things happen when a solid dissolves: the ordered lattice breaks apart (increasing disorder — entropy goes up) but the free ions then get "trapped" by orderly shells of water molecules around them (decreasing entropy of the water). To predict whether dissolving actually happens, you need the total entropy change, not just Δ_solH alone.

Combined equation ΔS_total = ΔS_system − (Δ_solH ÷ T)

A process is thermodynamically favourable (spontaneous) when ΔS_total is positive.

Worked example — ammonium nitrate at 298 K: Δ_solH = +25.8 kJ mol⁻¹ (endothermic!), ΔS_system = +108.7 J K⁻¹ mol⁻¹.

ΔS_surroundings = −ΔsolH / T = −(+25800) / 298 = −86.6 J K⁻¹ mol⁻¹ ΔS_total = ΔS_system + ΔS_surroundings = (+108.7) + (−86.6) = +22.1 J K⁻¹ mol⁻¹ → POSITIVE → spontaneous → soluble!

Key insight Ammonium nitrate dissolving is endothermic (absorbs heat — that's why instant cold packs use it!) yet it still dissolves readily. That only makes sense once you look at entropy: the big increase in disorder from breaking up the solid lattice outweighs the small entropy cost of ordering the water molecules. This is exactly why enthalpy of solution alone can't always predict solubility.

6. Enthalpy Trends: Ionic Charge & Radius

Both lattice energy and hydration enthalpy are governed by the same underlying idea: electrostatic attraction gets stronger when ions are smaller and more highly charged (i.e. when charge density is high). Everything in this section is really just that one idea applied twice.

6.1 Ionic Radius

Rule As ionic radius increases, lattice energy becomes less exothermic (weaker).

Why? A bigger ion spreads its charge over a larger volume (lower charge density), and the centres of the ions end up further apart in the lattice. Since electrostatic attraction weakens rapidly with distance, this results in weaker attraction and a less exothermic lattice energy.

Example: Lattice energy of CsF is less exothermic than KF, because Cs⁺ (Period 6) is a much bigger ion than K⁺ (Period 4) — same anion (F⁻), so the difference comes entirely from cation size.

6.2 Ionic Charge

Rule As ionic charge increases, lattice energy becomes more exothermic (stronger).

Why? Higher charge means higher charge density, which means stronger electrostatic pull between the ions. More energy is released when the ions fall together — hence a bigger (more negative) lattice energy.

Example: CaO's lattice energy is far more exothermic than KCl's — Ca²⁺/O²⁻ carry double the charge of K⁺/Cl⁻ and are smaller ions, so both factors stack in the same direction.

6.3 Hydration Enthalpy — Same Logic

Hydration enthalpy follows exactly the same pattern: higher charge density (smaller, more highly charged ions) = stronger attraction to the polar water molecules = more exothermic (more negative) hydration enthalpy.

Comparison	Result	Reason
F⁻ vs Cl⁻	F⁻ has more negative Δ_hydH	F⁻ is smaller → higher charge density
Mg²⁺ vs Ba²⁺	Mg²⁺ has more negative Δ_hydH	Mg²⁺ is smaller (same charge) → higher charge density

The unifying rule to memorise Whenever a question asks you to compare lattice energies or hydration enthalpies, ask yourself: which ion has the higher charge density? (smaller radius and/or bigger charge). That ion will always be involved in the more exothermic (more negative) value.

Practice Question

Predict, with reasoning, which would have a more exothermic lattice energy: LiF or NaCl.

What to Memorise

Lattice energyEnthalpy change when 1 mole of ionic solid forms from its gaseous ions. Always exothermic (large negative value).

Enthalpy of atomisationEnthalpy change to form 1 mole of gaseous atoms from an element's standard state. Always endothermic.

1st electron affinityEnthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms. Always exothermic.

2nd (or later) electron affinityAdding an electron to an already-negative ion. Always endothermic (repulsion must be overcome).

Enthalpy of solutionEnthalpy change when 1 mole of ionic solid dissolves in sufficient water. Can be exo- or endothermic.

Enthalpy of hydrationEnthalpy change when 1 mole of a specific gaseous ion dissolves in water. Always exothermic.

Born–Haber diagram ruleEndothermic arrows point up; exothermic arrows point down. Energy increases going up the page.

Full Born–Haber equationΔ_fH = Δ_atH + Δ_atH + IE + EA + Δ_lattH

Simplified rearrangementΔ_lattH = Δ_fH − Δ₁H

Solution/hydration cycleΔ_hydH = Δ_lattH + Δ_solH

Charge density estimatecharge density ≈ charge ÷ r²

Entropy & solubilityΔS_total = ΔS_system − (Δ_solH ÷ T). Dissolving is favourable when ΔS_total is positive.

Concepts Checklist

I can define lattice energy and explain why it is always exothermic

I can write correct equations (with state symbols) for atomisation, ionisation, electron affinity, and lattice energy

I can explain why the first electron affinity is exothermic but the second is endothermic

I can draw a full Born–Haber cycle from scratch, including correctly placed arrows (up = endo, down = exo)

I can use a Born–Haber cycle to calculate lattice energy, enthalpy of formation, or any missing step given the rest

I know to double/halve values when a compound contains 2 (or more) moles of one ion

I can explain polarisation and why small, highly charged cations distort large anions

I understand why theoretical and experimental lattice energies can disagree

I can distinguish enthalpy of solution from enthalpy of hydration

I can construct and use the Δ_hydH = Δ_lattH + Δ_solH energy cycle, including for ions with subscripts > 1

I can explain how ionic radius and ionic charge affect lattice energy and hydration enthalpy

I can use entropy alongside enthalpy of solution to predict whether an ionic compound is likely to dissolve

Exam Tips & Common Mistakes

Sign errors are the #1 killer. Before substituting numbers, write out the full equation symbolically first (e.g. Δ_lattH = Δ_fH − Δ₁H) and only then plug in values, keeping every negative sign in brackets. Rushing straight to numbers is where most marks are lost.

Forgetting to double/halve. For compounds like MgCl₂ or MgO where one ion appears more than once (or where atomisation starts from a diatomic element), always check the mole ratio before you calculate — a very common way marks are dropped.

Mixing up direct vs indirect routes. The value you're asked to calculate is always the "direct route" arrow in the cycle. Everything else forms the "indirect route." Label both clearly on your diagram before writing any equation — examiners give method marks for a correctly labelled cycle even if the final number is wrong.

Assuming electron affinity is always exothermic. Only the first electron affinity is (almost) always exothermic. Any subsequent electron affinity (adding to an already-negative ion) is endothermic — this catches students out constantly with O²⁻ and S²⁻ compounds.

State symbols matter for marks. Examiners specifically check that atomisation starts from the correct standard state (s), (l), or (g) for the element in question — get this wrong and you lose marks even if your enthalpy value is correct.

"Explain" questions on trends need both radius AND reasoning. Don't just say "the ion is bigger" — explain why that makes the electrostatic attraction weaker (charge spread over a larger volume, greater distance between ion centres). Examiners want the mechanism, not just the observation.

Don't forget entropy in solubility questions. If asked to explain why an ionic compound is or isn't soluble, and Δ_solH alone doesn't obviously explain it (e.g. it's endothermic but the compound still dissolves), you almost certainly need to bring in ΔS_total using the T-dependent formula.

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5. Enthalpy of Solution & Hydration
6. Enthalpy Trends: Ionic Charge & Radius
Exam Tips & Common Mistakes

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Energetics

Summary — What This Chapter Covers

1. Born–Haber Key Terms

1.1 Lattice Energy (ΔlattH)

1.2 Enthalpy of Atomisation (ΔatH)

1.3 Electron Affinity (ΔeaH)

2. Constructing Born–Haber Cycles

Step-by-step: Building the NaCl cycle

Worked Example: KCl Cycle

3. Born–Haber Calculations

Worked Example: Lattice Energy of KCl

4. Polarisation

5. Enthalpy of Solution & Hydration

5.1 Key Definitions

5.2 The Energy Cycle

Worked Example: Hydration Enthalpy of Cl⁻ (from KCl)

5.3 Predicting Solubility — Enthalpy AND Entropy

6. Enthalpy Trends: Ionic Charge & Radius

6.1 Ionic Radius

6.2 Ionic Charge

6.3 Hydration Enthalpy — Same Logic

What to Memorise

Concepts Checklist

Exam Tips & Common Mistakes

1.1 Lattice Energy (Δ_lattH)

1.2 Enthalpy of Atomisation (Δ_atH)

1.3 Electron Affinity (Δ_eaH)