Did you know that adding the second electron to a gaseous oxygen ion requires energy, yet magnesium oxide is a remarkably stable ionic solid that forms exothermically? These facts are not contradictory. The second electron affinity of oxygen is endothermic because we are forcing an electron onto an already negatively charged O⁻ ion. But when Mg²⁺ and O²⁻ ions form a solid ionic lattice, the strong attractions release a large amount of energy. The overall energy balance can favour magnesium oxide formation even though one of the individual steps is uphill.
For JC1 and JC2 students searching for an H2 Chemistry tutor in Hougang, second electron affinity of oxygen, MgO Born–Haber cycles, lattice formation enthalpy, magnesium oxide energetics, electron affinity signs or A-Level Chemistry 9476 Hess’s Law practice, this is a powerful test of whether a student actually understands the steps in an energy cycle.
I ask students to explain the two electron-affinity equations separately. The first electron is accepted by a neutral oxygen atom; the second electron is accepted by a negative oxygen ion. The second step has a very different electrostatic cost. Then I ask why the final ionic solid can still be stable. If the learner can explain the strong lattice-energy contribution, the overall cycle begins to make sense.
This guide follows the 2027 Singapore H2 Chemistry 9476 treatment of lattice enthalpy, electron affinity, ionisation energy, Hess’s Law and Born–Haber cycles. The numerical values used below are an explicitly stated internally consistent illustrative dataset. Different reliable data books may use slightly different values, particularly for oxygen’s difficult-to-determine second electron affinity, so use the exact data given by the examination question.
Hougang Chemistry learning library · Official 2027 H2 Chemistry syllabus listing
What does an electron affinity describe?
An electron affinity enthalpy change describes adding an electron to a mole of gaseous atoms or ions in a specified state.
For oxygen’s first electron affinity:
O(g) + e⁻ → O⁻(g).
This step has a negative enthalpy change under the familiar data convention: energy is released when a neutral gaseous oxygen atom accepts the electron to form a gaseous oxide(I)-charged anion, O⁻.
For the second electron affinity:
O⁻(g) + e⁻ → O²⁻(g).
This step has a positive enthalpy change, meaning energy must be supplied.
The two equations differ in the charge of the species receiving the electron. That charge changes the electrostatic interaction profoundly.
A student who says “oxygen is electronegative, so every extra electron must release energy” has overlooked the negative charge already on O⁻.
Why the first electron affinity is usually exothermic
Start with a neutral isolated oxygen atom in the gaseous state.
Its nucleus attracts electrons, and the arrival of an electron can lead to a more favourable bound electron arrangement for the resulting singly negative species.
For oxygen:
O(g) + e⁻ → O⁻(g), ΔHₑₐ₁ < 0.
Representative data values put the first electron-affinity enthalpy near −141 kJ mol⁻¹.
That negative sign corresponds to heat released for the process as written.
A negative electron-affinity enthalpy is not the same as a negative electron charge. It describes a change in enthalpy, not the sign of the particle.
Also, the value is for the isolated gaseous atom and ion, not oxygen sitting in a solid oxide lattice or hydrated in water.
The physical state specified in the definition is essential.
Why the second electron affinity is different
Now the starting species is O⁻, not neutral O.
The incoming electron has negative charge and the existing ion already carries overall negative charge.
There is therefore a strong electrostatic repulsion against forcing an additional electron onto the isolated gaseous anion.
The attraction from the oxygen nucleus remains, but the overall electron-addition process under this gaseous-ion definition requires an energy input.
So:
O⁻(g) + e⁻ → O²⁻(g), ΔHₑₐ₂ > 0.
That is the meaning of an endothermic second electron affinity.
This is not claiming the entire formation of MgO must be endothermic. It only describes one hypothetical thermochemical stage of forming gaseous ions.
The next key question is what happens when those ions assemble into an ionic solid.
A second electron affinity is not a second ionisation energy
The vocabulary is easy to confuse.
Ionisation energy involves removing an electron from a gaseous atom or cation.
For example:
Mg⁺(g) → Mg²⁺(g) + e⁻.
That is magnesium’s second ionisation energy.
Electron affinity, in the relevant convention, concerns adding an electron.
For oxygen:
O⁻(g) + e⁻ → O²⁻(g).
That is oxygen’s second electron-affinity enthalpy change.
Both can have positive enthalpy changes, but the electron is travelling in opposite directions in the formal equations.
An H2 student should draw the electron on the correct side before looking at any numerical value. It prevents many sign errors.
Why magnesium contributes two electrons
A magnesium atom has the ground-state configuration:
Mg: [Ne]3s².
It commonly forms Mg²⁺ through two successive electron-removal processes:
Mg(g) → Mg⁺(g) + e⁻.
Mg⁺(g) → Mg²⁺(g) + e⁻.
Both ionisation enthalpies are positive. Magnesium’s first and second electrons must be removed from the atom or ion at an energy cost.
Oxygen accepts those two electrons conceptually to reach O²⁻, but the two electron-affinity steps have different signs.
A tempting mistake is to pair magnesium’s two positive ionisations with two negative oxygen electron affinities and assume the charges must balance energetically.
Electrical charge conservation is not the same as energy conservation with equal and opposite numerical enthalpies. The electrons cancel in the overall chemical equations, but their energy changes are independent.
The lattice formation step provides another major energy contribution.
Why the crystal lattice releases so much energy
The gaseous ions Mg²⁺ and O²⁻ carry charges of +2 and −2.
When these oppositely charged ions assemble into the solid MgO ionic lattice, there are strong electrostatic attractions between them.
The corresponding conceptual lattice-formation equation is:
Mg²⁺(g) + O²⁻(g) → MgO(s).
Under the lattice formation enthalpy convention, this step is exothermic, so its enthalpy is negative.
A useful representative magnitude for MgO is several thousand kilojoules per mole of formula units. The attraction is especially substantial because the ions have high charge magnitudes and the ionic separation in the crystal is relatively small.
That large negative lattice contribution can more than compensate for the positive ionisation energies and the positive second electron affinity of oxygen.
This is how a stable solid can result from a cycle containing several costly isolated-ion steps.
Formation and dissociation conventions must be separated
There are two common ways to define a lattice enthalpy.
Lattice formation enthalpy:
Mg²⁺(g) + O²⁻(g) → MgO(s).
Its sign is negative.
Lattice dissociation enthalpy:
MgO(s) → Mg²⁺(g) + O²⁻(g).
Its sign is positive.
The numerical magnitudes are the same for the exact reverse processes under a consistent state convention, but the signs are opposite.
The 9476 syllabus uses lattice energy with the gaseous-ions-to-solid formation convention. Always check the wording of the actual exam question.
The mistake is not using either convention. The mistake is using a value from one convention with an equation drawn in the other direction without reversing its sign.
A complete Born–Haber cycle for MgO
The direct standard formation equation is:
Mg(s) + ½O₂(g) → MgO(s).
An alternative Hess-law path takes the elements to gaseous atoms, makes gaseous ions and then forms the solid lattice.
The conceptual steps are:
- Mg(s) → Mg(g): atomisation or sublimation of magnesium.
- Mg(g) → Mg⁺(g) + e⁻: first magnesium ionisation.
- Mg⁺(g) → Mg²⁺(g) + e⁻: second magnesium ionisation.
- ½O₂(g) → O(g): formation of one mole of gaseous oxygen atoms.
- O(g) + e⁻ → O⁻(g): oxygen’s first electron affinity.
- O⁻(g) + e⁻ → O²⁻(g): oxygen’s second electron affinity.
- Mg²⁺(g) + O²⁻(g) → MgO(s): lattice formation.
When the equations are added, intermediate gaseous atoms, ions and electrons cancel. The net equation is the direct formation of one mole of MgO from its elements in their standard states.
Hess’s Law says that the enthalpy change depends on the initial and final thermodynamic states, not the imaginary route used to calculate it.
The Born–Haber cycle therefore lets us add measurable and modelled energy changes in a consistent way.
An illustrative dataset for the full calculation
Use this internally consistent examination-style dataset, all in kJ mol⁻¹ for the specified reaction steps:
| Process | Illustrative enthalpy | Sign reason |
|---|---|---|
| Mg(s) → Mg(g) | +148 | Energy required to produce gaseous atoms |
| Mg(g) → Mg⁺(g) + e⁻ | +738 | First electron removal requires energy |
| Mg⁺(g) → Mg²⁺(g) + e⁻ | +1451 | Second electron removal requires energy |
| ½O₂(g) → O(g) | +249 | O–O bond disruption to gaseous atoms requires energy |
| O(g) + e⁻ → O⁻(g) | −141 | First electron addition releases energy |
| O⁻(g) + e⁻ → O²⁻(g) | +744 | Second electron addition is endothermic |
| Mg²⁺(g) + O²⁻(g) → MgO(s) | −3791 | Formation of stable ionic lattice releases much energy |
These values are a rounded teaching set, not a claim that every chemistry reference publishes identical constants. In particular, some published Born–Haber examples use different second-electron-affinity and lattice-energy numbers but are internally consistent within their own source datasets.
For the specific 2027 examination, follow the supplied numerical data and official booklet rather than importing a value from an unrelated textbook.
Calculate the enthalpy of formation
From Hess’s Law:
ΔH°f(MgO) = 148 + 738 + 1451 + 249 − 141 + 744 − 3791.
First add the energy-consuming terms:
148 + 738 + 1451 + 249 + 744 = 3330 kJ mol⁻¹.
Subtract the first electron-affinity release:
3330 − 141 = 3189 kJ mol⁻¹.
Then include the lattice formation:
3189 − 3791 = −602 kJ mol⁻¹.
So the calculated standard formation enthalpy using the illustrative dataset is:
ΔH°f(MgO) = −602 kJ mol⁻¹.
The overall sign is negative even though oxygen’s second electron-affinity contribution was +744.
The reason is that the large negative lattice formation enthalpy outweighs the net cost of preparing the gaseous ions.
That is the central answer to the title’s question.
Why no individual step must match the overall sign
Hess’s Law adds enthalpy changes algebraically.
A reaction path can contain both endothermic and exothermic components.
If the negative contributions have greater total magnitude than the positive contributions, the net enthalpy can be negative.
In this example, the positive magnesium atomisation and ionisation costs are substantial. The oxygen atomisation and second electron addition also require energy. The first electron affinity is mildly exothermic.
The major stabilisation occurs when Mg²⁺ and O²⁻ form the crystal lattice.
So MgO(s) formation can be exothermic, even though producing hypothetical isolated O²⁻(g) was unfavourable.
This is a general lesson about thermochemical cycles: the sign of one intermediate step is not automatically the sign of the complete reaction.
A reversed question: calculate lattice formation enthalpy
Suppose an exam gives the overall enthalpy of formation −602 kJ mol⁻¹ and all the other six values from the table, but asks for the unknown lattice formation enthalpy, L.
Write:
−602 = 148 + 738 + 1451 + 249 − 141 + 744 + L.
The non-lattice terms sum to:
3189 kJ mol⁻¹.
So:
L = −602 − 3189 = −3791 kJ mol⁻¹.
That is the lattice formation enthalpy under the specified model.
If the question instead asks for lattice dissociation, reverse the sign:
+3791 kJ mol⁻¹.
A student can earn marks for a correct energy cycle even when the final arithmetic is wrong. But the direction, states and signs must be consistent.
Another reversal: calculate the second electron affinity
Now let oxygen’s second electron affinity be the unknown x.
With lattice formation −3791 and overall formation −602:
−602 = 148 + 738 + 1451 + 249 − 141 + x − 3791.
First sum the known non-x terms:
148 + 738 + 1451 + 249 − 141 − 3791 = −1346.
Therefore:
−602 = −1346 + x.
So:
x = −602 + 1346 = +744 kJ mol⁻¹.
The positive sign is exactly what we expect for adding an electron to a negative oxygen ion.
This calculation does not imply we can directly measure a long-lived isolated gaseous oxide dianion in a simple laboratory vessel. The Born–Haber method helps infer an energy associated with a formal thermochemical step that is difficult to study independently.
The scientific reason is still electrostatic repulsion when adding the second electron to O⁻.
Why MgO’s lattice contribution is so large
A basic ionic electrostatic model says that the attraction between charged ions grows with the magnitudes of their charges and generally becomes stronger at smaller separations.
Compare a simple 1+/1− ionic lattice with a 2+/2− lattice.
Magnesium oxide contains Mg²⁺ and O²⁻. The product of charge magnitudes is larger than in an otherwise similar 1+/1− system.
Its ions also have their own characteristic sizes and packing geometry.
That means its lattice formation can be very exothermic.
However, a precise lattice energy cannot be calculated simply by multiplying charges and ignoring crystal structure, polarisation and repulsion.
For H2, the required skill is to describe the qualitative effects of ionic charge and radius, then use supplied lattice-energy or Born–Haber data in calculations.
Why an isolated O²⁻ ion is not the same as O²⁻ inside MgO
An isolated gaseous dianion is a highly unfavourable species in the simple electron-addition picture. The second electron experiences repulsion from the already negative ion.
In a magnesium oxide crystal, the oxide ion is surrounded by positively charged magnesium ions. These strong favourable electrostatic interactions stabilise the ionic arrangement.
Therefore the O²⁻ formal oxidation state can be part of a stable solid even though creating isolated O²⁻(g) through electron attachment is endothermic.
This is the difference between a free particle in the gas phase and an ion embedded in a solid lattice.
A student who says “O²⁻ can’t exist in a stable compound because the second electron affinity is positive” has forgotten the environment in which the ion exists.
Many thermochemical puzzles disappear when the physical state and neighbouring particles are made explicit.
Why oxygen atomisation requires half an O₂ molecule
The direct standard formation reaction for one mole of MgO consumes:
Mg(s) + ½O₂(g) → MgO(s).
One MgO formula unit contains one oxygen atom, while one O₂ molecule contains two oxygen atoms.
Therefore the Born–Haber path needs:
½O₂(g) → O(g).
If a data booklet provides the enthalpy for:
O₂(g) → 2O(g),
you must use half that value when constructing one mole of MgO.
The factor of ½ comes from stoichiometry, not an arbitrary energy discount.
This is one of the most common reasons students get a Born–Haber numerical answer wrong while using all the right named terms.
The useful repair is to balance the overall formation equation before building the energy cycle.
What if the electron-affinity data are reported using the opposite sign convention?
Some chemistry resources use the term “electron affinity” for an energy released, reporting an exothermic electron attachment as a positive magnitude.
Other school data lists express electron-affinity enthalpy changes, where an exothermic step has a negative signed ΔH.
These conventions can confuse students when they mix sources.
The safest method is to write the actual electron addition equation and then attach the correct enthalpy sign for that process.
For the oxygen-first step as an enthalpy change, use a negative value such as −141 kJ mol⁻¹ in our stated teaching set.
For oxygen-second attachment, use a positive value such as +744 kJ mol⁻¹ in that same set.
If the reference instead defines affinity as energy released and lists unsigned magnitudes, translate carefully before adding the values to a Hess cycle.
Never mix conventions silently.
A worked consistency check with the overall chemical equation
Our conceptual steps begin with:
Mg(s) + ½O₂(g).
After atomisation, ionisation and two electron additions, we have:
Mg²⁺(g) + O²⁻(g).
The electrons released by magnesium’s two ionisation steps are consumed by oxygen’s first and second attachment steps.
The lattice step combines the ions into:
MgO(s).
When everything is added and the gaseous intermediates cancel, the net equation is:
Mg(s) + ½O₂(g) → MgO(s).
This matches the definition of standard enthalpy of formation for one mole of MgO from its elements in their standard states.
If your own cycle leaves an extra electron or an O atom on one side, you probably forgot a coefficient or wrote a step in the wrong direction.
Does the lattice step show actual electrons flying through the crystal?
Not necessarily.
A Born–Haber cycle is a thermochemical construction used to connect well-defined states with Hess’s Law.
It is not necessarily the actual microscopic pathway for the formation of MgO when magnesium burns in oxygen.
There is no requirement that real magnesium oxide formation proceed through a flask full of separately isolated Mg²⁺(g) and O²⁻(g) ions and then a neat sudden lattice-collapse event.
The cycle is useful because enthalpy is a state function: the total enthalpy difference between specified initial and final states can be found from an alternative route.
Students should distinguish an energy accounting path from a reaction mechanism.
That distinction also applies to enthalpy-of-formation cycles, combustion enthalpy cycles and other applications of Hess’s Law.
Common mistakes and the first repair
| Student statement | First misconception | Repair question |
|---|---|---|
| Oxygen’s second electron affinity is exothermic because oxygen wants a full shell | Ignored electron repulsion from O⁻ | Is the starting particle neutral or negative? |
| Positive second EA means MgO cannot be stable | Ignored lattice energy and total Hess cycle | What happens when gaseous ions form a solid? |
| Ionisation energy and electron affinity are identical | Confused removing with adding electrons | Which side of each equation contains e⁻? |
| Lattice formation energy must be positive | Confused formation and dissociation conventions | Which direction is the lattice arrow pointing? |
| Use full O₂ dissociation enthalpy for one MgO | Ignored ½O₂ in formation equation | How many oxygen atoms are required? |
| The two Mg ionisations are exothermic | Ignored the energy required to remove electrons | Does electron removal require energy? |
| Born–Haber steps are the real stepwise burning mechanism | Confused a Hess-law cycle with kinetics | What property makes an alternative path valid? |
| Every textbook must use exactly +744 for EA₂(O) | Ignored data-set variation and convention | What values are supplied for this question? |
Independent practice questions
- Write the first and second electron-affinity equations for oxygen, including gaseous state symbols.
- Explain why the second process has a positive enthalpy change.
- Give the sign of magnesium’s first and second ionisation enthalpies.
- For the standard formation of one mole MgO, explain why oxygen atomisation begins with ½O₂(g).
- State the sign of MgO lattice formation enthalpy.
- Using the illustrative values, show that the total non-lattice enthalpy is +3189 kJ mol⁻¹ and deduce lattice formation enthalpy if ΔH°f = −602 kJ mol⁻¹.
- Why does a positive second electron affinity not determine the sign of the complete MgO formation reaction?
Answers
1. O(g) + e⁻ → O⁻(g); O⁻(g) + e⁻ → O²⁻(g). 2. The incoming electron experiences strong repulsion from the already negatively charged O⁻ ion. 3. Both positive. 4. One MgO contains one oxygen atom, so it uses half an O₂ molecule. 5. Negative for gaseous ions forming the solid lattice. 6. L = −602 − 3189 = −3791 kJ mol⁻¹. 7. The full Hess cycle contains other enthalpy contributions, especially the large exothermic lattice formation.
The crucial transfer is to solve Question 6 after the numerical values have been rearranged on the page, not only in the order presented in this article.
A seven-session JC1–JC2 revision route
- Session 1: write electron configuration and ionisation equations for magnesium.
- Session 2: distinguish first and second oxygen electron affinities, including their signs.
- Session 3: define lattice formation and lattice dissociation enthalpies by writing the actual equations.
- Session 4: balance the standard MgO formation equation and build a labelled Born–Haber cycle.
- Session 5: calculate lattice formation enthalpy from a complete internally consistent dataset.
- Session 6: solve the reverse problem for an unknown second electron affinity, checking the sign.
- Session 7: change the ionic solid to a different oxide or chloride and identify which equations and coefficients change.
The last step tests whether the learner understands the cycle rather than memorising seven numbers for MgO.
What parents seeking H2 Chemistry tuition near Hougang can ask
Bring a marked Born–Haber question and ask the tutor where the student’s explanation first failed: did they reverse the lattice sign, confuse electron removal with addition, forget oxygen’s second-electron repulsion or use the full O₂ atomisation energy instead of half?
A three-student tutorial works when every learner must label a different energy cycle independently before sharing answers. One-to-one teaching should also finish with a genuinely changed dataset.
Check the tutor’s current 9476 experience, class venue, actual lesson timetable and commuting requirements directly. This Hougang-focused learning guide does not imply a physical classroom is situated in Hougang.
Why the concept matters beyond JC
Born–Haber reasoning connects atomic electron structure with the physical stability of ionic materials. It is useful for thinking about crystal energetics, material formation, ionic bonding and the balance of competing energy contributions.
The broad scientific habit is to distinguish isolated species from species stabilised by their surroundings, and a single energy contribution from the sum that determines a complete reaction.
That habit transfers to Chemistry, materials science and chemical engineering, irrespective of the student’s eventual admissions pathway.
Frequently asked questions
Why is oxygen’s second electron affinity endothermic?
Because the incoming negative electron is added to an already negative O⁻ ion and experiences strong electron–electron repulsion.
Why does magnesium oxide still form exothermically?
Formation of the stable Mg²⁺/O²⁻ ionic lattice releases enough energy to outweigh the positive steps in the overall Born–Haber cycle.
What is the sign of lattice formation enthalpy?
Negative when gaseous ions combine to form the solid ionic lattice.
Is the second electron affinity of oxygen the same as magnesium’s second ionisation energy?
No. Electron affinity adds an electron to oxygen; ionisation removes an electron from a magnesium cation.
Why use half an O₂ molecule in the cycle?
One MgO formula unit contains one oxygen atom while an O₂ molecule contains two.
Are the numerical values universal?
No. The calculations here use a rounded internally consistent dataset. Use values and sign conventions provided in the examination question.
Continue learning and references
Born–Haber cycles and lattice-energy sign errors · Why sodium’s second ionisation energy is so high · Hess’s Law and Born–Haber cycles · Why ammonium nitrate dissolves endothermically · Royal Society of Chemistry: Born–Haber cycles teaching toolkit · Hougang Chemistry Tutor Library · eduKate Singapore JC Chemistry enquiries.
Data note: The numerical cycle is explicitly illustrative. Different reference sets use different oxygen second-electron-affinity and lattice-energy values; signs must be consistent throughout each calculation. The gaseous-ion pathway is an energy cycle, not a laboratory preparation procedure.