Did you know that sodium’s second ionisation energy is far greater than magnesium’s, even though magnesium has the greater nuclear charge? The key is not simply how many protons are present. It is which electron must be removed next. After losing its first electron, Na⁺ has the stable [Ne] electron configuration. Removing another electron means breaking into a filled inner shell. After losing one electron, Mg⁺ still has an electron in its outer 3s subshell, so the second ionisation removes another valence electron rather than a core electron.
For students searching for an H2 Chemistry tutor in Hougang, JC1 ionisation energy trends, successive ionisation energies, sodium versus magnesium, electron configuration worked questions or Singapore A-Level Chemistry 9476 atomic structure, this comparison is a particularly good test. A student may correctly memorise that first ionisation energy generally rises across Period 3 yet fail to realise that second ionisation energy compares the removal of electrons from singly charged ions, not from neutral atoms.
My approach is to pause before calculating and write the actual species. Are we ionising Na or Na⁺? Mg or Mg⁺? Which occupied orbital contains the next electron? If the student can answer those questions, the apparently surprising data become chemically sensible.
At eduKate, our wider Chemistry teaching approach combines concept rebuilding with changed problem practice and small-group feedback. Families considering our three-student tutorial format should check the actual JC Chemistry timetable, venue and commuting arrangements. A Hougang-targeted guide is educational support for learners in that area, not a claim that there is a physical branch in Hougang.
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First ionisation is not the same experiment as second ionisation
First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms in their ground states, forming one mole of gaseous singly charged cations.
For sodium:
Na(g) → Na⁺(g) + e⁻.
For magnesium:
Mg(g) → Mg⁺(g) + e⁻.
The important word is gaseous. An ionisation energy concerns isolated gaseous atoms or ions, not the energy of dissolving a salt or extracting sodium from a metal lattice.
Second ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous singly charged positive ions, forming gaseous doubly charged cations.
For sodium:
Na⁺(g) → Na²⁺(g) + e⁻.
For magnesium:
Mg⁺(g) → Mg²⁺(g) + e⁻.
The second equations begin with ions. That single observation changes the entire explanation.
Do not write that the second ionisation energy removes two electrons simultaneously from a neutral atom. The process is successive: the second electron is removed from a cation already produced by the first ionisation.
Write the neutral atom electron configurations
Sodium has proton number 11 and ground-state electronic configuration:
Na: 1s² 2s² 2p⁶ 3s¹ = [Ne]3s¹.
Magnesium has proton number 12 and ground-state electronic configuration:
Mg: 1s² 2s² 2p⁶ 3s² = [Ne]3s².
Both atoms have valence electrons in the third principal electron shell. The sodium atom has one 3s electron; magnesium has two.
In first ionisation, the outer 3s electron is removed from each neutral atom. Magnesium generally has the higher first ionisation energy because its greater nuclear charge acts on a broadly similar shell and shielding arrangement, attracting its outer electron more strongly.
The following step, however, is dramatically different because the singly charged ions no longer have matching outer electron arrangements.
Track what Na becomes after its first electron leaves
Remove the one sodium valence electron:
Na: [Ne]3s¹ → Na⁺: [Ne].
Na⁺ has 10 electrons and the same closed-shell electron configuration as neon:
Na⁺: 1s² 2s² 2p⁶.
There is now no occupied 3s valence orbital. The next electron to remove must come from the n = 2 shell, commonly described using its filled 2p subshell.
This electron is much closer to the nucleus on average and experiences a very different effective nuclear attraction from the previously removed 3s electron.
Therefore the second ionisation energy of sodium makes a huge jump. Sodium is not suddenly acquiring more protons; the electron being removed has changed from an outer-shell electron to a tightly bound inner-shell electron.
That is the central physical explanation.
Track what Mg becomes after its first electron leaves
Now remove one outer electron from magnesium:
Mg: [Ne]3s² → Mg⁺: [Ne]3s¹.
Unlike Na⁺, the singly charged magnesium ion still has one electron in the 3s subshell.
Second ionisation is:
Mg⁺: [Ne]3s¹ → Mg²⁺: [Ne].
We remove the remaining outer 3s electron, not a 2p core electron. The energy is greater than the first ionisation because the ion is positively charged and its remaining electrons are more strongly attracted, but it is not the massive core-shell jump encountered for the second ionisation of sodium.
The large shell jump for magnesium comes at its third ionisation, when Mg²⁺ = [Ne] would have to lose a core electron.
That is why sodium and magnesium have very different successive-ionisation patterns.
Compare typical ionisation energy values
Representative gas-phase experimental values in common data sets are approximately:
| Step | Sodium / kJ mol⁻¹ | Magnesium / kJ mol⁻¹ |
|---|---|---|
| First IE | 496 | 738 |
| Second IE | 4562 | 1451 |
| Third IE | 6910 | 7733 |
Use the actual values supplied in an examination question or data booklet; rounded numbers can differ slightly among sources.
Notice the striking comparison:
Second IE of Na ≈ 4562 kJ mol⁻¹.
Second IE of Mg ≈ 1451 kJ mol⁻¹.
Despite magnesium having 12 protons to sodium’s 11, the electron being removed during sodium’s second ionisation is a core 2p electron, whereas magnesium’s second ionisation removes a valence 3s electron.
This difference in shell is much more important for the comparison than simply counting the extra proton in magnesium.
Why the third ionisation energies matter too
For sodium, the first ionisation is comparatively modest, then the second is very large. Its third ionisation is larger again because Na²⁺ is an even more positively charged species with electrons left in an inner shell.
For magnesium, the first two removals concern 3s valence electrons. The third removal would need to break into the [Ne] core and therefore causes a dramatic increase.
This gives a useful pattern:
- Group 1 atom: large jump immediately after removal of one outer electron.
- Group 2 atom: large jump immediately after removal of two outer electrons.
- Element with three outer electrons: large jump may occur after three valence-electron removals, subject to the actual electron configuration.
The first large discontinuity can therefore help deduce the number of outer electrons in an unknown main-group atom.
Do not simply identify a group from which individual ionisation energy number is largest. Compare successive values and look for a jump much greater than the preceding increases.
Worked example 1: identify an unknown element’s likely group
An unknown Period 3 element has the following illustrative successive ionisation energies:
| Ionisation step | Energy / arbitrary consistent units |
|---|---|
| IE₁ | 500 |
| IE₂ | 4600 |
| IE₃ | 6900 |
| IE₄ | 9500 |
The first huge jump occurs between IE₁ and IE₂. That means the first electron removed was the only readily removable valence electron, after which a core electron would be next.
The pattern is consistent with a Group 1 main-group atom. If the question also tells us the unknown is in Period 3, sodium is the likely identity.
The exact numerical values are only illustrative. The reasoning is the grouping of energies and the fact that a large jump signals entry into an inner electron shell.
Worked example 2: magnesium-like successive energies
Suppose a second Period 3 unknown has illustrative values:
IE₁ = 740; IE₂ = 1450; IE₃ = 7700 kJ mol⁻¹.
Both first and second removals are much less demanding than the third. The largest discontinuity is between IE₂ and IE₃.
That suggests the original neutral atom had two outer electrons. For a Period 3 main-group element, magnesium is the expected candidate.
The learner should not say merely “it is magnesium because 1450 looks like a familiar number”. A robust answer is: two valence electrons are removed before reaching the stable inner-shell configuration; the third ionisation is a core-electron removal.
That explanation transfers to a changed dataset even if the exact values differ.
Ratios can help make jumps visible
Students sometimes see three or four large numbers and struggle to decide where the real discontinuity lies. Calculating approximate ratios can help.
For sodium:
IE₂/IE₁ ≈ 4562/496 ≈ 9.20.
For magnesium:
IE₂/IE₁ ≈ 1451/738 ≈ 1.97, while IE₃/IE₂ ≈ 7733/1451 ≈ 5.33.
The ratios reinforce where the sharp increases occur.
However, there is no universal rule that “a ratio over exactly five proves a new shell.” The number of protons, ionic charge, electron configuration and dataset all matter. Ratios are a diagnostic aid, not a replacement for the electronic structure.
Why successive ionisation energies generally increase
Even without a shell transition, successive ionisation energy values tend to rise as electrons are removed. After an atom loses an electron, the same nucleus attracts a smaller number of remaining electrons, and the positive ion often holds them more tightly.
That explains why magnesium’s second ionisation is larger than its first even though both are removals from the 3s subshell.
But an ordinary increase within a shell is different from a giant jump into an inner shell. The large jump is especially informative about the valence-shell boundary.
Students should learn to distinguish the normal rise due to increasing positive charge from a structural discontinuity due to the next electron occupying a different shell.
What nuclear charge and shielding explain—and what they do not
Nuclear charge is the attraction source from positively charged protons. Shielding arises because other electrons partially reduce the net attraction experienced by a particular electron.
Across Period 3, nuclear charge generally increases while shielding from filled inner shells changes relatively little. This helps explain the overall rise in first ionisation energy, with known exceptions due to subshell energies and electron pairing.
In the Na second IE versus Mg second IE comparison, the central difference is that we are not removing two electrons from the same kind of outer shell in the two species.
Na⁺ has no n = 3 electron left. Mg⁺ has one. So the relevant electron’s principal shell and distance from the nucleus must be mentioned before a generic nuclear-charge comparison can make sense.
If the exam asks for the relative ionisation energies, “magnesium has more protons” is incomplete and in this instance leads to the wrong prediction.
Link to the stability of Na⁺ and Mg²⁺ in compounds
Sodium typically forms Na⁺ in familiar ionic compounds. Magnesium commonly forms Mg²⁺. Their stable ionic charges align with removal of one and two valence electrons respectively before the very large next ionisation-energy step.
That does not mean ionisation energy alone determines whether a compound is thermodynamically stable. Lattice enthalpy, hydration, bonding and other energy contributions also matter.
For example, magnesium oxide can form a strong ionic solid even though removing two electrons from magnesium requires substantial energy. The formation of a compound involves electron transfer together with other energetic steps.
Students sometimes reason “a large positive ionisation energy means ionic compounds cannot form.” That is false because the relevant overall process is an energy balance, not the isolated ionisation step alone.
This concept prepares students for Born–Haber cycles and energetic comparisons.
Common mistakes to correct
| Student statement | Problem | Better question |
|---|---|---|
| Second IE removes two electrons at once | Confused successive steps | Which ion is the reactant in the second IE equation? |
| Magnesium’s second IE must be bigger because Mg has more protons | Ignored the orbital being ionised | Is Na⁺ losing a 2p or 3s electron? |
| Na⁺ still has a 3s electron | Failed to remove the first Na electron | What is the Na⁺ electron configuration? |
| Any rise means a new shell is reached | Ignored charge effect on remaining electrons | How exceptional is the increase compared with neighbouring steps? |
| Group follows the largest numeric IE directly | Did not identify the jump position | How many electrons were removed before the big discontinuity? |
| High IE prohibits ionic bonding | Confused one energetic step with total lattice formation | What other energy terms contribute when a compound forms? |
Practice question 1: compare two ions
Explain why IE₂(Na) > IE₂(Mg).
A model answer is:
Na⁺ has the stable [Ne] configuration, so its second ionisation removes a tightly bound inner-shell 2p electron. Mg⁺ still has a valence 3s electron, so its second ionisation removes that less tightly bound outer electron. Consequently the second ionisation of sodium requires much more energy despite magnesium’s greater nuclear charge.
This includes the starting species, the relevant orbital and the physical reason for the energy difference.
Practice question 2: infer outer electron count
An unknown main-group element has successive energies:
IE₁ 600, IE₂ 1200, IE₃ 2100, IE₄ 10,000 in the same arbitrary units.
Where is the dominant jump? Between IE₃ and IE₄.
What does it imply? Three outer electrons could be removed before entering a much more tightly bound inner shell.
For a Period 3 candidate, aluminium is a plausible identity. But if the period is not supplied, more information may be needed to identify a unique element.
A strong student distinguishes group evidence from proof of exact identity.
Practice question 3: find the inconsistency
A question says the second ionisation of sodium removes a 3s electron.
Why is this impossible in its ground-state sequential ionisation?
Because sodium had only one 3s electron to begin with. It was removed during the first ionisation. Therefore the sodium ion entering the second ionisation has the [Ne] configuration and cannot lose another 3s electron from an unoccupied orbital.
This is an excellent ten-second diagnostic of electron-configuration understanding.
Practice question 4: why doesn’t every element make a +2 ion?
Suppose a student says sodium should form Na²⁺ because it only needs to lose two electrons.
The trouble is the energetic cost. The second ionisation of sodium removes a core electron from Na⁺ and is enormously more costly than removal of its outer 3s electron. Ordinary sodium compounds strongly favour the familiar +1 oxidation state for energetic reasons considered across the full compound-forming system.
This is a qualitative chemical explanation. It is not a blanket claim that a formal Na²⁺ species could never exist under any exotic physical condition.
For normal H2 examples, the large second IE helps explain why sodium most commonly forms Na⁺.
An independent five-question test
- Write the gaseous equations defining first and second ionisation energy for sodium.
- Give the electron configurations of Na⁺ and Mg⁺.
- State which subshell supplies the electron in IE₂(Na) and IE₂(Mg).
- Why does magnesium experience its largest early shell-boundary jump at IE₃ rather than IE₂?
- A main-group unknown shows its first giant jump between IE₂ and IE₃. How many valence electrons did it most likely have?
Answers
1. Na(g) → Na⁺(g) + e⁻; Na⁺(g) → Na²⁺(g) + e⁻. 2. Na⁺ = [Ne]; Mg⁺ = [Ne]3s¹. 3. Na⁺ loses a core 2p electron; Mg⁺ loses its remaining 3s electron. 4. The first two ionisations remove the two valence 3s electrons; the third accesses the [Ne] core. 5. Two outer electrons.
If the learner can complete these without notes, change the comparison from sodium and magnesium to two unfamiliar adjacent main-group elements. The method should still work.
How a week of JC study can strengthen this topic
- Day 1: write configurations for Period 3 atoms from Na to Cl.
- Day 2: write the cation configurations after successive electron removals.
- Day 3: compare IE₁ trends across a period and explain known subshell or electron-pair exceptions.
- Day 4: compare IE₂(Na) and IE₂(Mg), naming the shell of the next electron.
- Day 5: infer outer electron counts from two anonymised successive-IE datasets.
- Day 6: connect ion formation to ionic compound stability without confusing one step with the entire energy cycle.
- Day 7: answer a mixed data question with no element labels and defend the group assignment.
There is little benefit in repeating a long explanation if the student cannot do the altered question. The objective is to shift from memorised numbers to confident electronic-structure reasoning.
What parents looking for H2 Chemistry tuition around Hougang should check
Ask the tutor to show how a wrong successive-ionisation answer is diagnosed. Did the student start from the wrong gaseous ion? Did they retain an electron that was already removed? Did they ignore a shell boundary? Or were they simply unfamiliar with the first- versus second-ionisation notation?
These are different misconceptions. A useful three-pax class or private lesson gives each student a fresh set of ionisation data to interpret independently before comparing explanations. Confirm current syllabus familiarity, lesson timing, actual venue and travel needs rather than inferring the existence of a physical Hougang branch from an article title.
Why this matters after JC
The stability of ions, electron configurations, bonding and energy changes appear throughout chemistry, materials science and related engineering courses. Understanding why a core electron is harder to remove is a foundation for thinking about oxidation states and what chemical transformations are practical.
The long-term skill is not knowing sodium’s numerical ionisation energies forever. It is recognising how the starting species and occupied electron shell determine the physical calculation.
Frequently asked questions
Why is the second ionisation energy of sodium so high?
The first ionisation removes its only 3s electron. The second must remove a tightly held inner-shell electron from Na⁺ = [Ne].
Why is magnesium’s second ionisation energy lower?
Mg⁺ still contains one 3s valence electron, which can be removed before a core-shell boundary is crossed.
Which successive ionisation is the big jump for magnesium?
The third, because Mg²⁺ already has the [Ne] configuration after its first two valence electrons are removed.
Does second ionisation remove two electrons simultaneously?
No. The first has already been removed; the second process removes one more electron from the singly charged gaseous cation.
How do I find an unknown element’s group from successive IEs?
Look for the first very large discontinuity. The number of relatively accessible electrons removed before that jump suggests the original number of valence electrons for a main-group element.
Continue learning
JC1 electron configuration, orbitals and ionisation energy · Why first ionisation energy dips from Mg to Al and P to S · Hess’s law and Born–Haber cycles · Hougang Chemistry learning library · eduKate Singapore Chemistry enquiries.
All ionisation-energy values are rounded illustrative reference values in kJ mol⁻¹. Use the data and examination-year conventions specified in a graded question.