Magnesium has a first ionisation energy of approximately 738 kJ mol⁻¹, while aluminium has a lower value of about 578 kJ mol⁻¹—even though aluminium has one more proton. This apparent exception to the usual rise across Period 3 has a clear chemical explanation: magnesium loses a 3s electron, whereas aluminium loses a more energetic 3p electron. Once you understand the relative energies of these subshells, the fall is no longer mysterious.
This H1 Chemistry Tutor Hougang guide helps JC1 and JC2 students explain the first ionisation energy pattern from sodium to chlorine, including the decrease from magnesium to aluminium and the smaller decrease from phosphorus to sulfur. We’ll use electron configurations, shielding, nuclear attraction, orbital occupancy and numerical data. The revised Singapore–Cambridge H1 Chemistry syllabus 8873 expects students to interpret periodic trends and variations from electronic structure, and this is an ideal way to learn the reasoning rather than memorising a jagged graph.
- Main trend: First ionisation energy generally increases from sodium to chlorine as effective nuclear attraction rises across Period 3.
- Mg to Al: Aluminium’s electron removed is in 3p, a higher-energy subshell than magnesium’s filled 3s subshell.
- P to S: Pairing of electrons in a sulfur 3p orbital introduces repulsion, making an electron easier to remove than from phosphorus’s half-filled 3p arrangement.
- Exam method: Define the gaseous-ion process, write both configurations, identify the electron being removed and explain the competing effects before comparing data.
First ionisation energy: what the measurement means
The first ionisation energy of an element is the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous singly positive ions. The atoms are in the ground state, and the process is endothermic in the usual definition.
The general equation is X(g) → X⁺(g) + e⁻. The positive enthalpy value is expressed in kJ mol⁻¹. It describes a precisely defined process, not the heat released when a metal reacts with water, or the energy of every electron in the atom combined.
This definition matters because a good H1 answer starts with the correct process. If you leave out the gaseous state or describe removing a proton from the nucleus, you have changed the phenomenon.
Why an electron does not leave for free
The electron is negatively charged and attracted to the positively charged atomic nucleus. Removing it far enough to form an isolated gaseous ion requires overcoming that attraction.
How tightly an electron is held depends on more than the number of protons. The electron’s average distance from the nucleus, its orbital or subshell, shielding by other electrons and electron–electron repulsions all matter.
This gives us a systematic explanation of trends and exceptions. The periodic table is not a promise that a single variable controls everything; it organises atoms whose electronic structures change in related but sometimes competing ways.
The general Period 3 rise
Across the third period from sodium to chlorine, atoms gain protons in the nucleus while additional electrons mainly enter the same principal shell. Their inner-shell shielding does not increase as dramatically as it would when starting a new period.
Consequently the outer electrons generally experience greater effective nuclear attraction, atomic radii tend to decrease and first ionisation energy generally rises.
The word “generally” matters. It describes the overall tendency without erasing the aluminium and sulfur exceptions, both of which arise from details of electronic structure within that same third shell.
The real Period 3 data
Representative first ionisation energies, rounded to whole kJ mol⁻¹, are Na 496, Mg 738, Al 578, Si 789, P 1012, S 1000, Cl 1251 and Ar 1521. The exact values used in your examination should follow the supplied Data Booklet or table.
These are measured quantities. If you plot them against proton number, the graph has an overall upward trend but dips at Al and S. A straight rising line drawn through all points would conceal the two important exceptions.
A useful independent numerical reference is Creative Chemistry’s Period 3 first-ionisation-energy table. It gives the dataset and its electronic explanations.
Why sodium starts at a low value
Sodium’s ground-state electron configuration is 1s²2s²2p⁶3s¹, commonly abbreviated [Ne]3s¹. Its outer electron occupies a 3s orbital beyond the filled neon core.
That electron is relatively shielded from the full nuclear charge and can be removed to form the familiar Na⁺ ion. This supports sodium’s relatively low first ionisation energy compared with elements further across Period 3.
Do not claim sodium has a small nuclear charge in any absolute sense or that its electron feels no attraction. The comparison is about the balance of nuclear charge, distance and shielding among the relevant atoms.
Why magnesium is higher than sodium
Magnesium has configuration [Ne]3s². Compared with sodium, magnesium’s nucleus contains one more proton while the added electron enters the existing 3s subshell.
The outer electrons experience stronger effective nuclear attraction in the usual comparison. The first ionisation energy therefore rises from approximately 496 for sodium to 738 kJ mol⁻¹ for magnesium.
At this point, a student may conclude that the rise must continue smoothly for every new proton. Aluminium is the best test of why that explanation needs to include subshell energy as well.
The apparent magnesium–aluminium puzzle
Aluminium has one more proton than magnesium, yet its first ionisation energy is lower. The values are approximately Mg 738 and Al 578 kJ mol⁻¹.
If nuclear charge were the only relevant factor, this might seem impossible. But the electron removed is not in the same kind of orbital. Magnesium loses a 3s electron; aluminium’s outermost electron is in 3p.
The aluminium 3p electron is higher in energy and less penetrating than a 3s electron. That difference outweighs the expected increase from nuclear attraction for this particular step across the period.
Write the configurations before explaining
Magnesium: [Ne]3s². Aluminium: [Ne]3s²3p¹.
In a first-ionisation event, Mg(g) → Mg⁺(g) + e⁻ removes a 3s electron. For aluminium, Al(g) → Al⁺(g) + e⁻ removes its 3p electron first.
The electron that leaves is the one most readily removed under the ground-state configuration. A student who compares magnesium’s 3s electron with aluminium’s 3s electron has chosen the wrong electron for the aluminium first-ionisation process.
Why 3p is higher in energy than 3s
In multi-electron atoms, 3s and 3p orbitals do not have identical energies. A 3s electron penetrates closer to the nucleus more effectively and can experience a stronger attraction than a 3p electron within the same principal shell.
The 3p electron is therefore generally higher in energy and more readily removed in the familiar magnesium–aluminium comparison.
You do not need a full quantum mechanical calculation to explain this H1 observation. Name the subshells and the relative electron energy or attraction, then connect the difference to the measured drop in ionisation energy.
Do not say aluminium gained a new electron shell
Both magnesium and aluminium have their outer electrons in the third principal shell, n = 3. The transition is from a filled 3s subshell to the first electron in 3p, not from the third shell to the fourth shell.
Writing “aluminium has more shielding because it has one extra shell” is therefore incorrect. The additional proton and electron are accompanied by a change of subshell within the same principal shell.
A precise explanation refers to 3p being higher in energy, with a different penetration and shielding environment, rather than inventing a new principal shell.
An answer fit for a structured question
A useful H1 explanation is: “Mg has electron configuration [Ne]3s², while Al is [Ne]3s²3p¹. The electron removed from Al is in a higher-energy, less penetrating 3p orbital than the 3s electron removed from Mg. It is therefore easier to remove, so Al has a lower first ionisation energy despite its larger nuclear charge.”
That is more informative than “aluminium is an exception” because it identifies the actual physical mechanism.
It is also more accurate than “aluminium is a larger atom” as a complete reason. The relevant comparison requires the subshell and electron energy.
Why silicon rises again
Silicon has configuration [Ne]3s²3p². Both silicon and aluminium lose a 3p electron during first ionisation in the simple comparison.
Moving from Al to Si increases nuclear charge while remaining in the same principal shell and subshell family. The increased effective nuclear attraction raises the first ionisation energy again, from roughly 578 to 789 kJ mol⁻¹.
This reinforces the general trend. The aluminium dip was a local subshell transition, not evidence that the overall Period 3 nuclear-charge effect had disappeared.
Why phosphorus rises further
Phosphorus has configuration [Ne]3s²3p³. Its three 3p electrons occupy three separate p orbitals singly in the usual ground-state orbital-box representation, following Hund’s rule.
The first ionisation energy is about 1012 kJ mol⁻¹, higher than silicon’s, consistent with greater nuclear attraction and the relatively stable half-filled p-subshell arrangement.
This sets up the second exception. The next element sulfur has more protons, yet its first ionisation energy falls slightly.
The phosphorus–sulfur puzzle
Sulfur has configuration [Ne]3s²3p⁴. Four electrons occupy the three 3p orbitals, so one orbital contains a pair of electrons.
The two electrons sharing that orbital repel one another. This repulsion makes it easier to remove an electron from sulfur’s paired 3p orbital than from phosphorus’s singly occupied 3p orbitals in the usual comparison.
The result is a small fall from approximately 1012 kJ mol⁻¹ for phosphorus to 1000 kJ mol⁻¹ for sulfur, despite sulfur having one more proton.
Draw the orbital boxes
Phosphorus 3p³ can be represented as three 3p boxes, each with one electron of parallel spin before any pairing: ↑ | ↑ | ↑.
Sulfur 3p⁴ has four electrons in those three boxes, so a typical arrangement is ↑↓ | ↑ | ↑.
The paired electrons in one sulfur 3p orbital experience additional repulsion. That helps explain the lower first ionisation energy. The boxes make the mechanism visible rather than simply declaring that sulfur breaks a rule.
Why “half-filled stability” is not the whole explanation
Saying “phosphorus is half-filled and therefore stable” identifies an observation about orbital occupancy, but students should connect it to the absence of paired-electron repulsion in the 3p orbital arrangement.
The stronger answer explains that sulfur’s electron pair creates additional electron–electron repulsion, making an electron easier to remove. This produces the measured local decrease.
A word such as “stable” is useful when it summarises physical reasoning, but not when it replaces the reasoning.
Why the sulfur dip is smaller
The fall from Mg to Al is relatively pronounced because the ionised electron changes from a 3s to a 3p subshell. The fall from P to S is smaller and is associated with pairing within the same 3p subshell.
These are different mechanisms and should not be treated as one generic “pairing” explanation.
If an exam asks you to account for both dips, write two separate explanations. Magnesium–aluminium requires subshell energy; phosphorus–sulfur requires electron pairing and repulsion.
Chlorine resumes the overall increase
Chlorine has configuration [Ne]3s²3p⁵. Its nucleus has a greater positive charge than sulfur’s, and its first ionisation energy rises to around 1251 kJ mol⁻¹.
Although chlorine contains paired 3p electrons as well, the increase in effective nuclear attraction through the period now dominates the comparison.
This is a reminder that electron repulsion does not turn off nuclear attraction. Several physical influences act together; the observed property reflects their balance.
First ionisation energy is not electron affinity
First ionisation energy refers to removing an electron from a gaseous atom. Electron affinity, in its usual definition, concerns an electron being added to a gaseous atom and the associated energy change.
Those are not the same process. A student cannot use first ionisation energy values directly as a complete ranking of how strongly atoms attract an incoming electron.
Keep the process equation beside the quantity. That often prevents errors before any numerical work begins.
Electronegativity is different too
Electronegativity describes an atom’s tendency to attract electron density in a covalent bond. It is a relative property used in bonding and polarity arguments.
First ionisation energy is an enthalpy change for a specified isolated gaseous-atom ionisation process. These often show related periodic tendencies, but they measure different things.
If a question asks for bond polarity, you need electronegativity and molecular geometry. If it asks for the energy to remove an electron from gaseous atoms, use ionisation energy.
Why electron shielding matters
Inner-shell electrons repel outer electrons and screen them from the full positive charge of the nucleus. This is the idea of shielding.
Across Period 3, electrons added mainly occupy the third principal shell while the number of deeply inner-shell electrons remains largely unchanged. Effective nuclear attraction on the outer electrons generally increases.
Shielding is not a single switch that is either present or absent. It depends on electron distributions and orbital penetration, which is why 3s and 3p removal energies differ even within the same principal shell.
Atomic radius and the same evidence
Atomic radii generally decrease across Period 3 because the larger nuclear charge attracts the outer electron distribution more strongly within the same principal shell.
The first ionisation energy generally increases for related reasons. But the two observables are not identical, so small ionisation-energy anomalies can arise through subshell occupancy and pairing even as the broad radius trend remains useful.
A well-written answer keeps the trend and its exception in their proper roles rather than claiming the radius graph must have precisely the same local shape as the energy graph.
What about ionic radii?
Ion formation changes electron count and sometimes the occupied shell pattern. Na⁺, Mg²⁺ and Al³⁺ are isoelectronic with neon in the simple electron-count comparison, each containing ten electrons.
With similar electron counts but increasing nuclear charge, their ionic radii generally decrease in the sequence Na⁺, Mg²⁺, Al³⁺.
This is not the same comparison as the first ionisation energies of neutral Mg and Al. One concerns ionic sizes after electron removal; the other concerns the energy of the first electron-removal step.
Why successive ionisation energy graphs look different
First ionisation energy removes one electron from the neutral atom. The second removes another from the singly charged gaseous ion, and so on.
For a Group 1 element such as sodium, there is a large jump after the first electron has been removed because the next electron belongs to an inner filled shell. Such jumps help deduce the original valence-electron count.
That is a related H1 atomic-structure skill, but do not confuse a successive-ionisation plot for one element with a first-ionisation plot across several elements.
A data table question
Suppose an original exam question provides the following values: Mg 738, Al 578, Si 789 and P 1012 kJ mol⁻¹. It asks which neighbouring pair contradicts a simple uninterrupted increase across the period.
The answer is Mg and Al. The aluminium value is lower despite a higher proton number.
A strong explanation uses [Ne]3s² and [Ne]3s²3p¹ to show that the aluminium electron is removed from a higher-energy subshell. No conclusion about an extra electron shell should be drawn.
A second data question
Another table gives P 1012, S 1000 and Cl 1251 kJ mol⁻¹. The local decrease is from P to S.
Phosphorus has three singly occupied 3p orbitals; sulfur has a paired electron in one 3p orbital. The paired electrons’ repulsion helps make removal easier.
The chlorine value then continues the overall upward trend. The correct answer combines a local explanation with the broader nuclear-charge pattern.
An error that costs marks: confusing proton number with electron count
Neutral aluminium has 13 protons and 13 electrons. After first ionisation, Al⁺ has the same 13 protons but only 12 electrons.
Ionisation has not removed a proton or changed aluminium into magnesium. The element’s identity is determined by proton number, which stays unchanged in ordinary chemical electron removal.
This basic check is valuable because a student can do complex orbital work yet write an impossible element change in a final equation.
Another error: describing the wrong subshell
If you describe aluminium losing a 3s electron first, the explanation for its lower energy than magnesium will be wrong. Its outermost electron in the ground-state configuration is 3p¹.
Similarly, sulfur’s first electron removal is from 3p, but the relevant comparison with phosphorus involves paired-versus-unpaired 3p occupancy.
The first task in both cases is to write the correct configuration before explaining what the measurement means.
An unseen element challenge
Suppose a hypothetical element within Period 3 has configuration [Ne]3s²3p². Identify its position relative to aluminium and predict whether its first ionisation energy is likely higher than aluminium’s.
It corresponds to silicon. Its outermost electron is also in 3p, but the nucleus has one additional proton relative to aluminium. The first ionisation energy is generally higher, as seen in the supplied periodic data.
This is the sort of transfer question that measures understanding beyond memorising four values.
A why-not question
A student claims, “Aluminium has a lower first ionisation energy because it has fewer protons than magnesium.” That is false: aluminium has 13 protons, while magnesium has 12.
A good correction does not simply replace “fewer” with “more” and stop. It identifies the higher-energy 3p electron in aluminium and why it is easier to remove than magnesium’s 3s electron.
I encourage students to challenge every explanation by checking whether its claimed atomic numbers and electron configurations are correct.
Connection to real materials
Ionisation-energy trends contribute to our understanding of metallic behaviour, electron transfer and the broader chemical properties of elements. Magnesium and aluminium both have industrial uses but differ in oxidation behaviour, compounds and material characteristics.
That does not mean a first-ionisation-energy value alone predicts every aspect of corrosion resistance or extraction energy. Real materials depend on oxide films, bonding, thermodynamics and kinetics.
This is the 21st-century scientific habit worth teaching: use one measured property for what it establishes, not as a magical answer to every different question.
A seven-day H1 revision plan
Day one: write the first ionisation energy definition and gaseous equation accurately. Day two: build Period 3 electron configurations from Na to Cl. Day three: explain why nuclear attraction generally increases and atomic radius decreases.
Day four: reconstruct the magnesium–aluminium 3s-versus-3p exception. Day five: draw phosphorus and sulfur 3p orbital boxes and explain pairing repulsion.
Day six: interpret a fresh numerical first-ionisation graph without labels. Day seven: mix first and successive ionisation questions, checking each student’s first incorrect reasoning step.
What a tutor should diagnose
If a student writes the wrong Al configuration, repair subshell filling before teaching the trend. If the configurations are correct but the answer discusses an extra principal shell, repair orbital versus shell vocabulary.
If the student explains Mg–Al using the sulfur pairing argument, practise distinguishing the two exception mechanisms. If the student writes the correct conceptual argument but the gaseous ionisation equation is wrong, address chemical symbols and charge balance.
A useful tutor adapts the next question to the exact weak link rather than treating every error as lack of effort or lack of memory.
What parents can ask at home
Ask why aluminium has a lower first ionisation energy than magnesium although it has more protons. The answer should mention the higher-energy 3p electron in aluminium.
Then ask why sulfur is slightly lower than phosphorus. The answer should mention electron pairing and repulsion in the 3p orbitals.
These two short explanations reveal a great deal about whether the student can use electronic structure. A learner who says only “because the periodic trend has an exception” has not yet explained the chemistry.
A 90-minute tutorial example
The session begins with a retrieval exercise on definitions and electron configurations. Next, students examine a numerical table and predict the broad trend before seeing the measured graph.
The tutor asks each student to account independently for one local decrease, then provides a parallel unseen comparison. Later practice connects the explanation to atomic radii and successive ionisation data.
The final task is to write a complete comparison without referring to notes. For families travelling from Hougang, verify the current class location, teaching arrangements and availability rather than assuming the search phrase indicates a local branch.
What the official syllabus asks
The revised 2025 pre-university H1 Chemistry syllabus 8873, applicable to the 2027 A-Level examination, requires qualitative explanations of atomic radius, ionic radius, first ionisation energy and electronegativity across Period 3 and down relevant groups.
It also requires explaining factors influencing ionisation energies and using successive-ionisation data to deduce electronic configurations and periodic position. The magnesium–aluminium and phosphorus–sulfur comparisons are valuable applications of those foundations.
Use the official MOE revised H1/H2/H3 Chemistry syllabus for exact learning outcomes and the SEAB A-Level syllabus directory for examination-year documents.
University and science pathways
Understanding electron configurations supports materials science, spectroscopy, energy technologies and broader chemical analysis. But one ionisation-energy trend does not determine whether a student qualifies for a university degree.
Check programme-specific subject requirements, including any H2 Chemistry or Mathematics prerequisites, for the actual admissions intake.
The immediate goal is much more concrete: when a new dataset shows a small dip inside a broad periodic trend, the student should examine the electron being removed before assuming the data are wrong.
Frequently asked questions
Why is aluminium’s first ionisation energy lower than magnesium’s?
Aluminium loses a 3p electron, which is higher in energy and easier to remove than magnesium’s 3s electron, despite aluminium’s larger nuclear charge.
What are the electronic configurations of Mg and Al?
Mg is [Ne]3s²; Al is [Ne]3s²3p¹.
What are their approximate first ionisation energies?
About 738 kJ mol⁻¹ for Mg and 578 kJ mol⁻¹ for Al, using common rounded reference data.
Why is sulfur lower than phosphorus?
Sulfur has a paired 3p orbital with additional electron–electron repulsion. Phosphorus has three singly occupied 3p orbitals.
Does first ionisation energy generally increase across Period 3?
Yes, because effective nuclear attraction generally increases, but there are local decreases at aluminium and sulfur.
Does aluminium have an extra electron shell compared with magnesium?
No. Both have outer electrons in the third principal shell; the key difference is the 3s versus 3p subshell.
Is ionisation energy always negative because electrons are released?
No. Removing an electron from a gaseous atom requires energy; the conventional first ionisation enthalpy is positive.
Is ionisation energy the same as electronegativity?
No. Ionisation energy concerns electron removal from gaseous atoms, whereas electronegativity concerns attraction for shared electron density in bonding.
Must I memorise every ionisation-energy value?
Learn the explanation and use the supplied Data Booklet or numerical table as directed. The ability to interpret values is more transferable than memorising a long dataset.
Continue learning through the Hougang Chemistry library
For the broader electron structure, use H1 atomic orbitals, 4s and electronic configurations. For group deduction from ionisation jumps, read JC H1 successive ionisation energies and group position. For general Period 3 trends and physical properties, continue with JC1 H1 periodic trends and ionisation energy.
For H1 Chemistry lesson enquiries, see eduKate’s parent–student consultation information. The question to test at the end is simple: if I hide the names Mg, Al, P and S, can you explain the two drops from electron configurations alone?