Ionic Bonding

Ionic bonding is one of the three main types of chemical bonding you need to understand for A-Level Chemistry. It occurs between metals and non-metals and involves the complete transfer of electrons from one atom to another, producing oppositely charged ions that are held together by strong electrostatic attraction.

Formation of Ions

Atoms are electrically neutral — they contain equal numbers of protons and electrons. When a metal atom reacts with a non-metal atom, the metal loses one or more electrons from its outer shell, forming a positive ion (cation). The non-metal gains those electrons, forming a negative ion (anion). Both ions achieve a stable noble gas electron configuration.

For example, sodium (2,8,1) loses one electron to become Na⁺ (2,8), and chlorine (2,8,7) gains that electron to become Cl⁻ (2,8,8). Each ion now has a full outer shell.

The driving force behind this electron transfer is the difference in ionisation energy and electron affinity between the two atoms. Metals have low ionisation energies (they lose electrons easily), while non-metals have high electron affinities (they gain electrons readily).

Worked Example: Predicting Ions from Electron Configuration

Calcium (1s² 2s² 2p⁶ 3s² 3p⁶ 4s²) has 2 electrons in its outermost shell. It loses both to form Ca²⁺ (achieving the argon configuration). Oxygen (1s² 2s² 2p⁴) needs 2 electrons to fill its outer shell, forming O²⁻ (achieving the neon configuration). The compound formed is CaO, with ions in a 1:1 ratio because the charges balance.

Aluminium (2,8,3) loses 3 electrons to form Al³⁺. Fluorine (2,7) gains 1 electron to form F⁻. Three fluoride ions are needed per aluminium ion to balance charges, giving the formula AlF₃.

Exam tip: Always check that the total positive charge equals the total negative charge in the formula. If Al is 3+ and F is 1−, you need three F⁻ ions: Al³⁺(F⁻)₃ → AlF₃.

The Ionic Bond

The ionic bond itself is the electrostatic attraction between oppositely charged ions. This is a crucial definition to learn. Note that an ionic bond is not between two specific atoms — it acts in all directions. Each positive ion attracts every surrounding negative ion, and vice versa.

This non-directional nature of ionic bonding means that ionic compounds do not exist as discrete molecules. Instead, they form giant ionic lattices — extended three-dimensional structures containing billions of ions arranged in a regular, repeating pattern.

Sodium Chloride (NaCl) Structure

The sodium chloride lattice is the most common example you will encounter. In this structure:

Each Na⁺ ion is surrounded by 6 Cl⁻ ions (octahedral coordination)
Each Cl⁻ ion is surrounded by 6 Na⁺ ions
The coordination number is therefore 6:6

The ions are packed in a face-centred cubic arrangement. If you imagine a cube, Na⁺ and Cl⁻ ions alternate at every corner and along every edge. This maximises the attractive forces between oppositely charged ions while minimising repulsion between like charges.

Magnesium Oxide (MgO) Structure

Magnesium oxide has the same crystal structure as NaCl but with some important differences. Magnesium loses two electrons to form Mg²⁺, and oxygen gains two electrons to form O²⁻. The higher charges on the ions mean that the electrostatic attraction is much stronger.

This is reflected in the melting points: NaCl melts at 801°C, while MgO melts at 2852°C. Both have the same structure, but the doubled charges in MgO produce dramatically stronger ionic bonds.

Lattice Energy

Lattice energy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. It is always exothermic (negative) because forming a lattice releases energy as the ions come together from an infinite separation.

For NaCl: Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH_lattice = −787 kJ mol⁻¹

The more negative (more exothermic) the lattice energy, the more stable the ionic lattice.

Factors Affecting Lattice Energy

Two key factors determine the magnitude of lattice energy:

1. Ionic charge: Higher charges produce stronger electrostatic attraction. The force between ions is proportional to the product of the charges (q⁺ × q⁻). Doubling one charge doubles the attraction; doubling both quadruples it. This is why MgO (2+ and 2−) has a much more exothermic lattice energy than NaCl (1+ and 1−).

2. Ionic radius: Smaller ions allow the charges to get closer together, increasing the electrostatic attraction. Lattice energy is inversely proportional to the sum of the ionic radii. LiF, where both ions are very small, has a more exothermic lattice energy than KBr, where both ions are larger.

These two factors together give the concept of charge density — the ratio of charge to size. High charge density (small, highly charged ions) leads to stronger lattice energies.

Lattice Energy Data for Key Compounds

Compound	Ions	Lattice energy / kJ mol⁻¹	Melting point / °C
LiF	Li⁺, F⁻	−1037	845
NaCl	Na⁺, Cl⁻	−787	801
KBr	K⁺, Br⁻	−682	734
MgO	Mg²⁺, O²⁻	−3850	2852
CaO	Ca²⁺, O²⁻	−3401	2614
Al₂O₃	Al³⁺, O²⁻	−15916	2072

Notice how the lattice energy broadly correlates with melting point: more exothermic lattice energy → higher melting point.

Properties of Ionic Compounds

The strong electrostatic forces in ionic lattices explain several characteristic properties:

High melting and boiling points: A large amount of energy is required to overcome the strong attractions between ions throughout the lattice. The melting points generally increase with lattice energy.

Brittleness: When a force is applied to an ionic lattice, layers of ions may shift. If like-charged ions are brought next to each other, the resulting repulsion causes the crystal to shatter.

Electrical conductivity: Solid ionic compounds do not conduct electricity because the ions are held in fixed positions and cannot move. When melted or dissolved in water, the ions become free to move and can carry charge, so molten and aqueous ionic compounds are good electrical conductors.

Solubility in water: Many ionic compounds dissolve in water because the polar water molecules can surround and stabilise the individual ions (hydration). The ion-dipole attractions between the water molecules and the ions must be strong enough to compensate for the lattice energy that must be overcome.

Comparison: Ionic vs Covalent vs Metallic Properties

Property	Ionic	Simple covalent	Giant covalent	Metallic
Melting point	High	Low	Very high	Variable (high)
Hardness	Hard, brittle	Soft	Very hard	Malleable
Solid conductivity	No	No	No (except graphite)	Yes
Molten conductivity	Yes	No	N/A	Yes
Solubility in water	Often soluble	Depends on polarity	Insoluble	Insoluble

Predicting Ionic Character

The tendency to form ionic bonds increases with greater difference in electronegativity between the two elements. A rough guideline is that an electronegativity difference greater than about 1.7 typically results in an ionic bond, though this boundary is not sharp — bonding exists on a continuum from purely covalent to predominantly ionic.

Elements from Group 1 and Group 2 bonded with elements from Group 6 and Group 7 typically form ionic compounds. The further apart the elements are on the periodic table (in terms of electronegativity), the more ionic the bond.

Worked Example: Predicting Melting Point Order

Predict the order of melting points: NaF, NaCl, NaBr, NaI.

All four compounds have the NaCl-type structure with 1+ and 1− ions. The cation (Na⁺) is the same in all cases. The anion changes: F⁻ < Cl⁻ < Br⁻ < I⁻ in size. Smaller anions allow closer approach, giving stronger electrostatic attraction and more exothermic lattice energies.

Predicted order: NaI < NaBr < NaCl < NaF (increasing melting point).

Actual values confirm this: NaI (661°C) < NaBr (747°C) < NaCl (801°C) < NaF (993°C).

Common exam mistake: Students sometimes say "NaF has stronger ionic bonds because fluorine is more electronegative." While fluorine is indeed more electronegative, the correct reasoning is about ionic radius — F⁻ is the smallest halide ion, allowing the closest approach and strongest electrostatic attraction. Electronegativity explains why the bond is ionic, but ionic radius explains the relative strength.

A-Level Deep Dive: Ionic Bonding

Spec mapping

Edexcel 9CH0 specification Topic 2 — Bonding and Structure, sub-topic 2.1 covers ionic bonding: how ionic compounds form by electron transfer, the giant ionic lattice, and how ionic radius and ionic charge govern lattice energy and the physical properties that follow (refer to the official specification document for exact wording). Although ionic bonding is introduced in Year 1, it is examined throughout the qualification. Paper 1 (Advanced Inorganic and Physical Chemistry) assesses it directly, both as definitional recall and as a quantitative tool inside Topic 8 Born-Haber cycles and Topic 13 group 2 trends. Paper 3 (General and Practical Principles) uses ionic-compound chemistry in qualitative analysis questions — silver halide precipitation, group 2 sulfate solubility trends, and ion identification all rest on lattice-energy versus hydration-enthalpy reasoning developed here. The Edexcel formula booklet does not list lattice energy values — they must be quoted in the question stem or derived from a Born-Haber cycle using the data provided.

Worked example with full mark scheme

Question (8 marks): Use the data below to construct a Born-Haber cycle and calculate the lattice energy of magnesium oxide, ΔH°_lat[MgO(s)].

Step	Enthalpy change	Value / kJ mol⁻¹
Enthalpy of formation of MgO(s)	ΔH°_f	−602
Enthalpy of atomisation of Mg(s)	ΔH°_at(Mg)	+148
1st ionisation energy of Mg	ΔH°_ie1	+738
2nd ionisation energy of Mg	ΔH°_ie2	+1451
Enthalpy of atomisation of O₂(g)	ΔH°_at(O)	+249 (per mole of O atoms)
1st electron affinity of O	ΔH°_ea1	−141
2nd electron affinity of O	ΔH°_ea2	+798

Solution with mark scheme:

Step 1 — write the cycle. By Hess's Law, the route from elements in standard states to MgO(s) via gaseous ions equals the direct route (ΔH°_f):

$\Delta H^{\circ}_f = \Delta H^{\circ}_{\text{at}}(Mg) + \Delta H^{\circ}_{\text{ie}1} + \Delta H^{\circ}_{\text{ie}2} + \Delta H^{\circ}_{\text{at}}(O) + \Delta H^{\circ}_{\text{ea}1} + \Delta H^{\circ}_{\text{ea}2} + \Delta H^{\circ}_{\text{lat}}$

B1 — correct cycle (or labelled diagram with arrows in the right direction). Common error: drawing the lattice-energy arrow upwards (lattice dissociation) and then forgetting to flip the sign at the end.

Step 2 — substitute values.

$-602 = 148 + 738 + 1451 + 249 + (-141) + 798 + \Delta H^{\circ}_{\text{lat}}$

M1 — substituting all seven enthalpies with the correct signs. Common error: using +141 for the first electron affinity (electron affinities are exothermic for the first electron — they should be negative). Another common error: forgetting that O₂(g) → 2O(g) is +498 in total but the table already gives the per-mole-of-atoms value (+249), so no doubling is needed here. Always read the units line of the data table.

Step 3 — sum the right-hand side excluding lattice energy.

$148 + 738 + 1451 + 249 - 141 + 798 = 3243 \text{ kJ mol}^{-1}$

M1 — correct arithmetic sum.

Step 4 — solve for ΔH°_lat.

$\Delta H^{\circ}_{\text{lat}} = -602 - 3243 = -3845 \text{ kJ mol}^{-1}$

A1 — correct numerical answer. A1 — correct sign (negative, i.e. exothermic) and units (kJ mol⁻¹). B1 — final value rounded sensibly (3 s.f., consistent with input data). A1 — explicit statement that the lattice energy is for the formation route Mg²⁺(g) + O²⁻(g) → MgO(s), distinguishing it from lattice dissociation.

Total: 8 marks (B2 M2 A4).

The doubly-positive and doubly-negative ions account for the unusually large magnitude — roughly four to five times that of NaCl — which previews the synoptic argument used in the next section. Common errors at the global level: (i) flipping the sign of the first electron affinity; (ii) doubling ΔH°_at(O) when the table already gives the per-atom value; (iii) treating the second electron affinity as negative — for oxide it is endothermic (+798) because adding a second electron to an already negative O⁻ ion requires energy to overcome electrostatic repulsion.

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): Magnesium chloride (MgCl₂) is significantly more soluble in water than magnesium fluoride (MgF₂), despite MgF₂ having the more exothermic lattice energy. Using lattice energy and hydration enthalpy reasoning, explain this observation.

Mark scheme decomposition by AO:

B1 (AO1.1) — state that solubility depends on the balance between lattice energy (energy to break apart the ionic lattice into gaseous ions) and the sum of the hydration enthalpies of the cation and anion (energy released when gaseous ions are surrounded by polar water molecules).
M1 (AO2.1) — recognise that F⁻ is smaller than Cl⁻, so MgF₂ has a more exothermic (more negative) lattice energy. Therefore the lattice is harder to break apart.
M1 (AO2.1) — recognise that F⁻, being smaller, also has a more exothermic hydration enthalpy than Cl⁻ (smaller ion = higher charge density = stronger ion–dipole attraction with water).
A1 (AO2.2) — explain that for MgF₂ the increase in lattice energy on going from Cl⁻ to F⁻ outweighs the corresponding increase in hydration enthalpy. The energy needed to break the lattice is not fully recovered by hydration, so the dissolution is less energetically favourable.
B1 (AO3.1) — quantitative or comparative framing: ΔH_sol = −ΔH_lat + ΔH_hyd(Mg²⁺) + 2·ΔH_hyd(X⁻). For MgF₂ this sum is more endothermic than for MgCl₂.
B1 (AO1.2) — note that entropy and kinetics also influence solubility, but the lattice-vs-hydration enthalpy balance is the dominant thermodynamic factor at A-Level.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Edexcel Paper 1 questions of this type reward candidates who explicitly compare the two changes — the marks are not for stating values, but for identifying which factor "wins" and why.

Synoptic links

Ionic bonding is one of the most heavily synoptic topics in 9CH0. The lattice-energy reasoning developed here surfaces in at least five other parts of the specification:

Topic 4 (Inorganic — Group 2): the trend of decreasing solubility for group 2 sulfates (BaSO₄ < CaSO₄ < MgSO₄) and increasing solubility for group 2 hydroxides (Mg(OH)₂ < Ca(OH)₂ < Ba(OH)₂) is explained directly by the lattice-vs-hydration enthalpy competition, with the controlling factor switching depending on whether the anion or the cation has the dominant size-change.
Topic 8 (Energetics II): Born-Haber cycles formalise the lattice-energy calculation done above. The mean bond enthalpy versus actual lattice energy comparison reveals partial covalent character in highly polarising compounds — a Fajans-rules application that hinges on ionic-radius and charge density data introduced here.
Topic 13 (Energetics, Entropy and Equilibria): ΔG = ΔH − TΔS for dissolution makes ionic-compound solubility a free-energy calculation. Lattice and hydration enthalpies provide the ΔH input.
Topic 11 (Acid-base equilibria): the acidity of hydrated metal-ion complexes such as [Al(H₂O)₆]³⁺ is a direct consequence of the high charge density of small, highly charged cations polarising the bound water molecules — the same charge-density reasoning used here for lattice energy.
Topic 1 (Atomic structure): ionic radii correlate with electron configuration. Isoelectronic series (e.g. O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺) demonstrate how nuclear charge alone, with constant electron count, controls radius — a Topic 1 idea required for the lattice-energy comparisons here.

Mark-scheme literacy

Ionic-bonding questions on 9CH0 distribute AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / recall)	30–40%	Defining ionic bonding, lattice energy, hydration enthalpy; stating structure of NaCl-type lattice; quoting the formation-route convention for ΔH°_lat
AO2 (application)	40–50%	Applying Born-Haber cycles to numerical data; predicting and explaining trends in melting point, solubility and thermal stability across a series of compounds; using charge-density reasoning to compare lattice energies
AO3 (analysis / evaluation)	15–25%	Evaluating limitations of the purely ionic model (e.g. polarisation, partial covalent character, deviations between theoretical Madelung values and Born-Haber experimental values)

Examiner-rewarded phrasing includes: "electrostatic attraction between oppositely charged ions throughout the lattice" (the full definition — partial credit only if any element is missing); "more exothermic lattice energy" (preferable to "stronger lattice energy" because it ties the comparison to the energy axis); "smaller ionic radius gives higher charge density and therefore stronger electrostatic attraction"; and "the lattice is overcome and the individual ions are hydrated by polar water molecules". Phrases that lose marks: "ionic bonds break when dissolved" — the lattice is overcome by hydration, but the bonding model itself is still ionic; "the molecule of NaCl" — NaCl does not exist as discrete molecules in the solid state; "ionic compounds conduct because electrons move" — they conduct because ions move when molten or aqueous.

A specific 9CH0 pattern: questions phrased "explain in terms of structure and bonding" demand both — naming the giant ionic lattice (structure) and the electrostatic attraction (bonding). Stating only one element typically caps the mark at half.

Grade-band model answers

3-mark question

Question: Define the term lattice energy and state whether it is exothermic or endothermic, giving a reason.

Grade A response (~210 words):*

The lattice energy (ΔH°_lat) of an ionic compound is the standard enthalpy change when one mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions of 100 kPa and a stated temperature, usually 298 K. For sodium chloride, this is the enthalpy change for the process Na⁺(g) + Cl⁻(g) → NaCl(s).

Lattice energy is exothermic (negative). When oppositely charged gaseous ions, initially infinitely separated, come together to form an ordered three-dimensional lattice, strong electrostatic attractions between the oppositely charged ions are established throughout the structure. Forming these attractions releases energy to the surroundings, so ΔH°_lat is negative.

It is important to specify the formation convention: the reverse process, lattice dissociation (NaCl(s) → Na⁺(g) + Cl⁻(g)), is endothermic with the opposite sign. Some textbooks tabulate the dissociation value as a positive number — candidates must read which convention applies before substituting into a Born-Haber cycle.

Examiner commentary: Full marks (3/3). The candidate gives the full definition (1), states "exothermic / negative" (1), and explains the reason via formation of electrostatic attractions (1). The bonus discussion of sign convention is unmarked but signals examiner-aware sophistication that pays off on longer questions. Many candidates lose marks here by giving only "ionic compound forms from ions" and omitting the "gaseous" qualifier — gaseous-ion specification is essential.

6-mark question

Question: Compare and explain the lattice energies of NaCl (−787 kJ mol⁻¹) and MgO (−3850 kJ mol⁻¹). Refer to ionic charges and ionic radii in your answer.

Grade A response (~290 words):*

The lattice energy of MgO is approximately five times more exothermic than that of NaCl, despite both compounds adopting the same NaCl-type face-centred cubic lattice with 6:6 coordination. Two factors combine to produce this difference, and both increase the strength of the electrostatic attraction between the ions.

Charge. In NaCl the ions carry charges of +1 and −1; in MgO the charges are +2 and −2. The electrostatic force between any pair of oppositely charged ions is proportional to the product of the charges (q⁺ × q⁻). Doubling each charge multiplies the product by four, so on charge alone MgO would have a lattice energy roughly four times that of NaCl.

Radius. The Mg²⁺ ion (radius ≈ 72 pm) is smaller than Na⁺ (≈ 102 pm), and O²⁻ (≈ 140 pm) is smaller than Cl⁻ (≈ 181 pm). The electrostatic force is inversely proportional to the square of the inter-ionic distance, so the smaller sum of ionic radii in MgO further amplifies the attraction.

Combining these factors, the charge-density of the MgO ion pair is much higher than that of NaCl, producing the observed five-fold enhancement of the lattice energy and the much higher melting point of MgO (2852 °C) compared with NaCl (801 °C).

Examiner commentary: Full marks (6/6). The answer is structured around the two factors the question explicitly invites (charge and radius), each developed quantitatively. The use of the word "product" for charges and "inverse-square" for distance shows an engagement with Coulomb's law that earns the AO2 marks. Critically, the candidate references the same lattice geometry — without that, the comparison would be confounded by structural differences. The closing connection to melting point closes the loop on physical consequences and would attract the final AO3 mark on a longer question.

9-mark question

Question: Discuss the factors that determine whether an ionic compound dissolves in water, using LiF (insoluble) and NaCl (soluble) as illustrative examples. Refer to lattice energy, hydration enthalpy and entropy.

Grade A response (~390 words):*

The dissolution of an ionic compound in water is governed by the enthalpy of solution, ΔH°_sol, and the entropy change, ΔS°_sol, combined as ΔG°_sol = ΔH°_sol − TΔS°_sol. Dissolution is thermodynamically spontaneous only when ΔG°_sol is negative.

The enthalpy of solution is constructed from two competing terms. The lattice must first be broken apart into gaseous ions, requiring an input of energy equal to the magnitude of the lattice energy (the dissociation sense, i.e. reversing the formation). The gaseous ions are then hydrated by polar water molecules, releasing the cation hydration enthalpy plus the anion hydration enthalpy. Whether ΔH°_sol is exothermic or endothermic depends on which sum is larger.

For LiF, both ions are very small. Li⁺ has a high charge density and gives a strongly exothermic hydration enthalpy. F⁻ is similarly compact and is also strongly hydrated. However, the small ionic radii combine to produce an extremely exothermic lattice energy (−1037 kJ mol⁻¹), and this dominates: the energy needed to break the LiF lattice exceeds the energy released by hydrating the resulting ions, giving an overall endothermic ΔH°_sol that, even at room temperature, is not offset by the modest entropy gain.

For NaCl, both ions are larger. The hydration enthalpies are individually less exothermic (lower charge density), but the lattice energy is also less exothermic (−787 kJ mol⁻¹). The two effects are now closely matched, and ΔH°_sol is only mildly endothermic. The increase in entropy when an ordered crystal disperses into a solution of hydrated ions provides a positive ΔS°_sol contribution; multiplied by T, this is enough to make ΔG°_sol negative, and NaCl dissolves.

The general rule emerges: lattice energy and hydration enthalpy both increase in magnitude as ionic radius decreases, but lattice energy increases faster. Very small ions therefore tend to give insoluble compounds, despite their strong individual hydration. The competition is delicate, which is why empirical solubility tables are needed alongside thermodynamic reasoning.

Examiner commentary: Full marks (9/9). The answer earns its top band by integrating all three required factors (lattice, hydration, entropy) into a single thermodynamic framework via ΔG = ΔH − TΔS, and by explicitly comparing LiF and NaCl rather than treating them in isolation. The final paragraph generalises the specific examples into a transferable rule — an AO3 evaluation move that distinguishes A* work from competent A.

A-Level-depth misconceptions

The errors that distinguish A from A* on ionic-bonding questions:

Confusing "lattice formation" enthalpy with "lattice dissociation" enthalpy. Formation (gaseous ions → solid lattice) is exothermic; dissociation (solid lattice → gaseous ions) is endothermic. They are equal in magnitude, opposite in sign. Examiners specify the convention — read it carefully before substituting.
Treating lattice energy as the only factor in solubility. Hydration enthalpies of both ions and the entropy term must be considered. A compound can have a very exothermic lattice energy and still dissolve (e.g. CaCl₂); conversely, a compound with weak-looking lattice energy can be insoluble for entropic reasons.
Asserting "NaCl conducts when dissolved because covalent bonds break". NaCl is ionic. The lattice is overcome by hydration, freeing the ions to move. There are no covalent bonds in NaCl to break.
Ignoring the second electron affinity sign for divalent anions. The first EA of O is exothermic (−141 kJ mol⁻¹); the second is endothermic (+798) because adding an electron to an already-negative O⁻ ion requires energy to overcome electrostatic repulsion. Treating both as negative inflates the lattice energy by ~1600 kJ mol⁻¹.
"Ionic bonds are between two ions." The ionic bond is non-directional and acts between every pair of oppositely charged ions throughout the lattice. There is no discrete Na⁺–Cl⁻ pair in solid NaCl.
Confusing electronegativity with charge density. Electronegativity predicts whether a bond will be predominantly ionic or covalent; charge density (charge ÷ size) predicts the strength of the resulting electrostatic attractions. The two ideas are related but answer different questions.
Forgetting polarisation in "ionic" compounds. Highly polarising cations (small, highly charged — e.g. Al³⁺) and highly polarisable anions (large, soft — e.g. I⁻) give compounds with significant covalent character. AlI₃ is far more covalent than NaF, even though both could be drawn as "ionic" using simple electron-transfer diagrams. Fajans' rules formalise this.

Common errors and mark-loss patterns on lattice energy questions

Three patterns repeatedly cost candidates marks on Paper 1 lattice-energy questions. They are about presentation and method, not about knowledge.

Forgetting to double the atomisation enthalpy when the atomic source is diatomic. For Cl₂(g) → 2Cl(g), the enthalpy change equals 2 × ΔH°_at(Cl), because ΔH°_at is conventionally defined per mole of atoms produced. Yet the data table sometimes gives the value as "ΔH°_at(½Cl₂) = +122" (per half mole of Cl₂) and sometimes as "Cl₂(g) → 2Cl(g) +244". Failing to read the units line and silently doubling (or not doubling) destroys the cycle. The cure: write out the equation for each enthalpy term explicitly before plugging in a number.
Sign-flipping electron affinities mid-cycle. First electron affinities are exothermic (negative), second electron affinities are endothermic (positive). Candidates often switch the sign halfway through, especially when the cycle is drawn vertically with up-arrows for endothermic and down-arrows for exothermic processes. The cure: write each enthalpy with its sign in a single horizontal sum at the start, then do the arithmetic in one pass.
Mixing formation and dissociation conventions in one cycle. A Born-Haber cycle must use a consistent convention for the lattice energy arrow. If the arrow points from gaseous ions down to the solid lattice, ΔH°_lat is negative; if it points upwards from the solid to the gaseous ions, the same magnitude is positive. Candidates who flip the arrow direction mid-derivation get the right magnitude with the wrong sign and lose the final A1 mark for sign error.

These patterns are endemic to Paper 1 cycle questions: candidates know the enthalpy values, lose marks on bookkeeping.

Going further — university and academic signposting

Ionic bonding and lattice energy point directly toward several undergraduate trajectories:

Solid-state chemistry (Year 2): the Madelung constant gives a rigorous theoretical lattice energy by summing the Coulombic contributions over all ion pairs in the lattice, weighted by the geometric series characteristic of the structure type (NaCl, CsCl, fluorite, rutile, etc.). For NaCl, the Madelung constant is 1.7476. The Born-Landé and Born-Mayer equations combine this with a repulsive term to give theoretical lattice energies that match Born-Haber experimental values to within ~5% for "good" ionic compounds — and the deviation itself diagnoses partial covalent character.
Crystallography (Year 3): the coordination numbers stated qualitatively here (6:6 for NaCl-type, 8:8 for CsCl-type, 4:4 for ZnS-type) are determined experimentally by X-ray diffraction. Bragg's law (nλ = 2d sin θ) converts diffraction angles into lattice spacings, from which the full three-dimensional ion arrangement is reconstructed.
Materials science: high-temperature ceramics such as ZrO₂ (zirconia) and Al₂O₃ (alumina) exploit the very-high-charge cations to give extreme lattice energies and consequently very high melting points (>2000 °C), making them suitable for jet-engine thermal barrier coatings and biomedical implants.
Electrochemistry (Year 1/2): standard electrode potentials of metals correlate strongly with the sum of atomisation, ionisation and hydration enthalpies — exactly the same Born-Haber-style construction used for lattice energy, but terminating at hydrated ions in solution rather than a solid lattice.

Oxbridge interview prompt: "Why is the lattice energy of MgF₂ more exothermic than that of CaF₂, but the solubility of MgF₂ in water is lower than that of CaF₂? Discuss the competition between lattice and hydration enthalpies and explain which factor dominates as you move down group 2."

Required practical reference

Ionic bonding underpins two of the Edexcel A-Level Chemistry Core Practicals. CP3 (the preparation and reactions of compounds of group 2 elements) explores the chemistry of ionic compounds across the group: candidates carry out reactions of group 2 metals with water and dilute acids, and investigate the thermal stability of group 2 carbonates and nitrates. The thermal stability trend — MgCO₃ decomposes most readily, BaCO₃ requires the highest temperature — is rationalised by the smaller, more polarising Mg²⁺ cation distorting the carbonate anion's electron density toward itself, weakening the C–O bonds and lowering the decomposition temperature. This is the same charge-density reasoning developed in this lesson, applied to a covalently bonded anion. CP4 (the qualitative analysis of inorganic ions) uses precipitation reactions of halide ions (Cl⁻, Br⁻, I⁻) with silver nitrate, where lattice energy and solubility-product reasoning predict the colours and solubilities in aqueous ammonia. Candidates who understand the charge-density / lattice-energy framework introduced here can predict CP3 and CP4 outcomes from first principles rather than memorising tables.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Paper 1 — Advanced Inorganic and Physical Chemistry, Topic 2: Bonding and Structure (sub-topic 2.1, ionic bonding). For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Ionic compound<br/>e.g. MgO, NaCl, CaF₂"] --> B["Identify the ions<br/>and their charges"]
    B --> C["Look up or estimate<br/>ionic radii"]
    C --> D{"Compare with<br/>another compound?"}
    D -->|"Yes"| E["Higher charge product<br/>q⁺ × q⁻"]
    D -->|"Yes"| F["Smaller sum<br/>of ionic radii"]
    E --> G["More exothermic<br/>lattice energy"]
    F --> G
    G --> H["Predict properties:"]
    H --> I["Higher melting<br/>and boiling points"]
    H --> J["Lower solubility<br/>(if lattice dominates<br/>over hydration)"]
    H --> K["Greater thermal<br/>stability of derived<br/>carbonates / nitrates"]
    H --> L["No solid conductivity;<br/>conducts when molten<br/>or aqueous"]

    style A fill:#8e44ad,color:#fff
    style G fill:#27ae60,color:#fff
    style I fill:#3498db,color:#fff
    style J fill:#3498db,color:#fff
    style K fill:#3498db,color:#fff
    style L fill:#3498db,color:#fff