Why Aluminium Has a Lower First Ionisation Energy Than Magnesium

Magnesium has 12 protons. Aluminium has 13. If aluminium has a greater nuclear charge, you'd expect its outer electrons to be held more tightly. So why does it actually take *less* energy to remove an electron from aluminium?

The first ionisation energy of magnesium is approximately 738 kJ mol⁻¹, while aluminium's is only 578 kJ mol⁻¹. This is a genuine dip in the general increase in first ionisation energy across Period 3. The explanation is that the electron being removed changes from a 3s electron in magnesium to a 3p electron in aluminium. That change in electron arrangement outweighs the effect of the extra proton.

01The measurements: where the trend bends

Before looking at the electron configurations, consider the prediction.

Moving from magnesium to aluminium increases the atomic number from 12 to 13. Both elements are in Period 3, so you might expect aluminium’s first ionisation energy to be higher.

The measured values tell a different story.

Data: NIST periodic table. Values converted from eV to kJ mol⁻¹ and rounded to the nearest whole number. NIST reference data.

Notice three important changes:

• Sodium to magnesium: ionisation energy increases by 242 kJ mol⁻¹.
• Magnesium to aluminium: ionisation energy decreases by 160 kJ mol⁻¹.
• Aluminium to silicon: ionisation energy increases by 209 kJ mol⁻¹.

The middle comparison is the exception we need to explain.

There is a useful clue in the electron configurations. Sodium and magnesium both lose a 3s electron. Aluminium, however, loses a 3p electron.

That change of subshell is the key.

02Why a 3p electron is easier to remove than a 3s electron

First, let’s establish what actually happens during first ionisation.

First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms.

For magnesium:

\[ \ce{Mg(g) -> Mg+(g) + e-} \]

For aluminium:

\[ \ce{Al(g) -> Al+(g) + e-} \]

Both processes require energy. The question is how strongly each atom holds the electron being removed.

Step 1: Identify the electron being removed

Magnesium has 12 electrons, with the configuration:

\[ \mathrm{Mg}: [Ne]\\,3s^2 \]

Its two outer electrons occupy the 3s subshell. Removing the first electron means removing one of these 3s electrons.

Aluminium has 13 electrons:

\[ \mathrm{Al}: [Ne]\\,3s^2\\,3p^1 \]

Its outermost electron occupies the 3p subshell. This is the electron removed during first ionisation.

Both electrons belong to the third principal energy level, \(n=3\). However, they occupy different subshells.

This distinction matters because electrons in different subshells do not necessarily have the same energy or experience the same nuclear attraction.

Step 2: Understand orbital penetration

A 3s orbital and a 3p orbital have different shapes and electron probability distributions.

• A 3s orbital is spherical.
• A 3p orbital has two main regions of electron probability on opposite sides of the nucleus, separated by a nodal plane.

More importantly, a 3s electron has greater penetration than a 3p electron.

Penetration describes how much an electron’s orbital probability distribution extends into regions close to the nucleus, including regions occupied by inner-shell electrons.

A 3s electron has a greater probability of being found close to the nucleus than a 3p electron in the same principal energy level.

You can picture this as having access to a more central region, where the attractive force from the nucleus is stronger.

The analogy has a limit: electrons do not travel along fixed tracks. Orbitals describe probability distributions, not physical paths.

Step 3: Connect penetration to shielding

Remember that electrons repel one another because they have the same negative charge.

The inner electrons therefore partially shield outer electrons from the positive nucleus.

This is called electron shielding.

A 3s electron penetrates further into the inner-electron region. It therefore experiences less shielding on average and a stronger effective nuclear attraction.

A 3p electron penetrates less. It experiences more shielding and a weaker effective nuclear attraction.

The result is that the 3p subshell has a higher energy than the 3s subshell in these multi-electron atoms.

An electron in the higher-energy 3p subshell is less tightly bound. Less energy is needed to remove it.

Step 4: Decide which effect wins

Now return to magnesium and aluminium.

Aluminium does have the greater nuclear charge. This increases its attraction to electrons.

However, the electron removed from aluminium occupies a higher-energy, less penetrating 3p subshell. It experiences greater shielding and is less tightly bound than magnesium’s 3s electron.

This effect outweighs the increase in nuclear charge.

Therefore, aluminium has a lower first ionisation energy than magnesium.

One important detail: aluminium has not gained an entirely new electron shell. Both atoms have their outer electrons in the third principal energy level. The difference is between the 3s and 3p subshells.

03How could an experiment test this explanation?

We have an explanation based on electron arrangements. But how could we investigate whether the electrons really have different binding energies?

One useful experimental technique is photoelectron spectroscopy (PES).

The basic idea is straightforward: provide atoms with photons of known energy, measure the kinetic energy of the electrons they release, and calculate how much energy was needed to remove those electrons.

The apparatus

Imagine a chamber containing a dilute beam of gaseous magnesium atoms.

A source of ultraviolet radiation sends photons of a known energy into the chamber. The photons strike the atoms and can transfer enough energy to remove electrons.

The emitted electrons travel into an energy analyser. An electric field inside the analyser directs electrons according to their kinetic energy.

A detector records the electrons, allowing scientists to produce a spectrum showing the number of detected electrons at different kinetic energies.

For a controlled comparison, the same photon energy and calibrated analyser are used for magnesium and aluminium.

What should the experiment show?

Let’s make two competing predictions.

Prediction 1: Nuclear charge is the only important factor.

Aluminium has one more proton, so its outer electron should be harder to remove. With identical photon energies, its emitted electron should have lower kinetic energy.

Prediction 2: The subshell difference matters.

Aluminium’s 3p electron is less tightly bound than magnesium’s 3s electron. It should therefore require less energy to remove and leave with greater kinetic energy when the photon energy is identical.

The second prediction agrees with the measured ionisation energies.

The relevant first ionisation energies are:

• Magnesium: 7.646 eV per atom.
• Aluminium: 5.986 eV per atom.

These correspond to approximately 738 and 578 kJ mol⁻¹, respectively.

Since the aluminium electron requires less energy to remove, it retains more of the incoming photon energy as kinetic energy.

Why must the atoms be gaseous?

This is an important experimental detail.

First ionisation energy is defined for isolated gaseous atoms. If we instead shine light onto a solid piece of magnesium or aluminium, we are investigating electrons in a metallic solid.

The electrons in a metal participate in its electronic structure and experience interactions with neighbouring atoms. Surface effects and the metal’s electronic bands also influence the measured energies.

A solid-metal photoemission experiment therefore does not directly measure the first ionisation energy of an isolated gaseous atom.

Using a controlled source of gaseous atoms removes this complication and makes the experiment relevant to the quantity we want to measure.

The distinction is between measuring an electron’s binding energy in a solid and measuring the energy needed to ionise an isolated atom.

04Worked examples

05Common mistakes to avoid

A strong HSC explanation should identify the subshells, discuss penetration and shielding, and then explain why the change outweighs the increased nuclear charge.

For a broader explanation of the overall pattern, see First Ionisation Energy Trends for HSC Chemistry.

06Questions and solutions

Question 1

Magnesium has the electron configuration \([Ne]3s^2\), while aluminium has the configuration \([Ne]3s^23p^1\).

Explain why aluminium has a lower first ionisation energy than magnesium, despite having a greater nuclear charge.

Solution 1

Aluminium has the lower first ionisation energy because its outermost electron occupies a higher-energy 3p subshell rather than the 3s subshell occupied by magnesium’s outermost electrons.

The 3p electron has lower penetration and experiences greater shielding than a 3s electron in the same principal energy level.

It is therefore less strongly attracted to the nucleus and easier to remove.

Although aluminium has one additional proton, the change from removing a 3s electron to removing a 3p electron has the greater effect.

Question 2

A student makes the following statement:

“Magnesium and aluminium both have their outermost electrons in the third shell. Therefore, their electrons must experience approximately the same nuclear attraction. Aluminium has one extra proton, so its first ionisation energy should be higher. The measured result must be due entirely to magnesium having a stable, full 3s subshell.”

Evaluate this explanation. Identify what is correct, what is incorrect, and provide a more scientifically accurate explanation.

Solution 2

The explanation is incomplete and its prediction is incorrect. Magnesium and aluminium have outer electrons in the same principal energy level, but those electrons occupy different subshells.

The student is correct that aluminium has one additional proton. This increases its nuclear charge and would tend to increase its first ionisation energy if the electron being removed experienced otherwise similar conditions.

However, the student’s assumption about electron attraction is incorrect.

Magnesium loses a 3s electron, while aluminium loses a 3p electron. The 3s orbital has greater penetration, meaning its electrons experience less shielding and stronger effective nuclear attraction.

The 3p electron in aluminium is higher in energy and more shielded, so it is easier to remove.

The full 3s subshell in magnesium is a feature of its electron arrangement, but simply describing it as stable does not adequately explain the observed difference.

A complete explanation must account for the different subshells, orbital penetration, shielding, and nuclear charge.

Question 3

Consider the following first ionisation energies:

A student predicts that ionisation energy should increase at every step from sodium to silicon because each successive element has an additional proton.

Calculate the change in first ionisation energy for each consecutive pair and explain why the student’s prediction is incorrect.

Solution 3

The first ionisation energy increases from sodium to magnesium, decreases from magnesium to aluminium, and increases again from aluminium to silicon.

The numerical changes are:

Sodium to magnesium:

\[ \begin{aligned} \Delta IE_1&=738-496\\\ &=+242\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Both atoms lose a 3s electron. Magnesium has one additional proton, and the shielding changes relatively little. Its outer electron is therefore more strongly attracted to the nucleus.

Magnesium to aluminium:

\[ \begin{aligned} \Delta IE_1&=578-738\\\ &=-160\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

The electron being removed changes from 3s to 3p. Aluminium’s 3p electron has greater energy, lower penetration, and greater shielding than magnesium’s 3s electron.

This makes aluminium easier to ionise despite its greater nuclear charge.

Aluminium to silicon:

\[ \begin{aligned} \Delta IE_1&=787-578\\\ &=+209\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Both elements lose a 3p electron. Silicon has an additional proton, and the increase in effective nuclear attraction outweighs the additional shielding from its electron.

Therefore, the student’s prediction fails because it considers nuclear charge without accounting for changes in electron arrangement.

Question 4

A researcher wants to compare the first ionisation energies of magnesium and aluminium.

They place a piece of solid magnesium and a piece of solid aluminium inside a vacuum chamber. Both samples are illuminated using photons with an energy of 10.00 eV. The researcher measures the kinetic energies of the emitted electrons and claims that these measurements directly represent the first ionisation energies of isolated atoms.

Explain the problem with this experimental design and propose a more appropriate method.

A revised experiment using isolated gaseous atoms produces the following illustrative measurements:

Calculate the first ionisation energy of each element in kJ mol⁻¹.

Assume the atoms begin in their ground states, the resulting ions are in their ground states, and recoil energy is negligible.

Solution 4

The original experiment does not directly measure atomic first ionisation energies because the samples are solids rather than isolated gaseous atoms.

In a solid metal, electrons participate in the metallic electronic structure. Their measured energies are influenced by the solid and its surface, so they cannot simply be treated as the ionisation energies of individual gaseous atoms.

A more appropriate experiment uses isolated gaseous atoms in a controlled vacuum chamber.

Step 1: Establish the energy relationship.

\[ IE_1=E_{\gamma}-KE \]

The incoming photon energy is divided between removing the electron and giving it kinetic energy.

Step 2: Calculate magnesium’s first ionisation energy.

\[ \begin{aligned} IE_1&=10.00-2.35\\\ &=7.65\ \mathrm{eV\\,atom^{-1}} \end{aligned} \]

Convert to kJ mol⁻¹:

\[ \begin{aligned} IE_1&=7.65\times96.485\\\ &=738.1\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Step 3: Calculate aluminium’s first ionisation energy.

\[ \begin{aligned} IE_1&=10.00-4.01\\\ &=5.99\ \mathrm{eV\\,atom^{-1}} \end{aligned} \]

Convert to kJ mol⁻¹:

\[ \begin{aligned} IE_1&=5.99\times96.485\\\ &=578.0\ \mathrm{kJ\\,mol^{-1}} \end{aligned} \]

Final answer:

• Magnesium: \(738.1\ \mathrm{kJ\\,mol^{-1}}\).
• Aluminium: \(578.0\ \mathrm{kJ\\,mol^{-1}}\).

Aluminium’s emitted electron has greater kinetic energy because less energy was required to remove it.

The results support the explanation based on the difference between the 3s and 3p subshells.

Question 5

The following table gives the first three successive ionisation energies of gaseous magnesium and aluminium atoms.

All values are in kJ mol⁻¹ and are rounded.

A student notices that aluminium has a lower first ionisation energy than magnesium. They conclude that aluminium should also have lower second and third ionisation energies.

However, the table shows that the relative order changes.

Determine which element has the higher ionisation energy at each of the three stages. Explain all three comparisons using the electron configurations of the atoms and the ions formed after each removal.

Assume each ion is in its ground electronic state before the next electron is removed.

Solution 5

The relative order changes at each successive ionisation:

• First ionisation: magnesium has the higher ionisation energy.
• Second ionisation: aluminium has the higher ionisation energy.
• Third ionisation: magnesium has the much higher ionisation energy.

The key is to track which electron is removed at every stage. The electron arrangement changes after each ionisation, so the first-ionisation trend cannot simply be carried over.

First ionisation:

Magnesium starts with:

\[ \mathrm{Mg}:[Ne]\\,3s^2 \]

Aluminium starts with:

\[ \mathrm{Al}:[Ne]\\,3s^2\\,3p^1 \]

Magnesium loses a 3s electron, whereas aluminium loses a 3p electron.

The 3p electron is higher in energy, less penetrating, and more shielded. It is easier to remove.

Therefore:

\[ IE_1(\mathrm{Mg})>IE_1(\mathrm{Al}) \]

This is the original magnesium-aluminium exception.

Second ionisation:

After their first ionisation, the configurations become:

\[ \mathrm{Mg^+}:[Ne]\\,3s^1 \]

\[ \mathrm{Al^+}:[Ne]\\,3s^2 \]

Notice what has changed.

The second electron removed from both ions now comes from the 3s subshell.

The difference is that aluminium has 13 protons, while magnesium has 12. Their inner \([Ne]\) cores are the same.

Aluminium’s greater nuclear charge results in stronger attraction on its remaining 3s electrons. Although electron-electron repulsion also exists, the increased nuclear attraction is the dominant effect in this comparison.

Therefore:

\[ IE_2(\mathrm{Al})>IE_2(\mathrm{Mg}) \]

This reverses the first-ionisation ordering.

Third ionisation:

Now consider the ions after their second ionisation.

For magnesium:

\[ \mathrm{Mg^{2+}}:[Ne] \]

For aluminium:

\[ \mathrm{Al^{2+}}:[Ne]\\,3s^1 \]

This is the crucial difference.

Magnesium has now lost both of its 3s electrons. Its remaining electrons form a neon-like configuration.

Removing a third electron from magnesium means removing an electron from the inner \(n=2\) shell.

Aluminium, however, still has one 3s electron remaining. Its third ionisation removes this outer electron rather than an inner-shell electron.

Removing an inner-shell electron requires substantially more energy because it is much closer to the nucleus and experiences less shielding.

Therefore:

\[ IE_3(\mathrm{Mg})\gg IE_3(\mathrm{Al}) \]

This explains the dramatic difference between 7733 and 2745 kJ mol⁻¹.

The deeper lesson: Ionisation energy depends on the particular electron being removed and the electronic structure of the ion at that stage. A comparison that is correct for neutral atoms does not necessarily predict what happens after one or more electrons have already been removed.

For more practice tracking these changes, see How to Find Valence Electrons from Successive Ionisation Energies.

07Where to take this next

The magnesium-aluminium exception shows why periodic trends must be explained using actual electron configurations rather than memorised arrows.

When comparing first ionisation energies, identify the subshell supplying the electron before considering nuclear charge and shielding. When comparing successive ionisation energies, go one step further: write the configuration of the ion after each electron is removed.

That second skill lets you recognise when an electron has to be removed from an inner shell, which explains the enormous jumps in successive ionisation energy.