Why Successive Ionisation Energies Show a Large Jump

Magnesium’s first two ionisation energies are approximately 738 and 1,451 kJ mol⁻¹. Its third ionisation energy is about 7,733 kJ mol⁻¹.

That’s more than five times the energy needed to remove the second electron. Why does removing just one more electron suddenly become so difficult?

The key is that the third electron is coming from a different electron shell. After magnesium loses its two outer-shell electrons, the next electron is much closer to the nucleus and experiences a much stronger attraction.

Before looking at the numbers, predict what has changed between the second and third ionisations. Has the nucleus gained more protons, or has the position of the electron being removed changed?

The nucleus still contains 12 protons. What changes is the electron arrangement, and therefore how strongly the remaining electrons are held.

01Why successive ionisation energies increase

First, remember what ionisation energy measures.

Ionisation energy is the energy required to remove an electron from a gaseous atom or ion. Successive ionisation energies describe removing electrons one at a time, with each removal starting from the positively charged species produced by the previous step.

For magnesium, the first three ionisations are:

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

\[
\ce{Mg+(g) -> Mg^2+(g) + e-}
\]

\[
\ce{Mg^2+(g) -> Mg^3+(g) + e-}
\]

The first ionisation energy is written as \(IE_1\), the second as \(IE_2\), and the third as \(IE_3\).

All ionisation energies are positive because energy must be supplied to remove an electron. They are commonly measured in \(\mathrm{kJ\,mol^{-1}}\).

There are two related effects to understand:

1. Each successive removal makes the remaining electrons harder to remove.
2. Removing an electron from a new, inner shell produces a particularly large increase.

Let’s separate these effects.

1. Why each successive ionisation energy increases

Imagine a neutral magnesium atom.

It has:

• 12 protons in its nucleus.
• 12 electrons surrounding the nucleus.
• An electron arrangement of \(2,8,2\).

The positively charged nucleus attracts the negatively charged electrons. At the same time, electrons repel one another.

When the first electron is removed, magnesium becomes \(\mathrm{Mg}^{+}\).

The nucleus still contains 12 protons, but there are now only 11 electrons.

This changes the balance of electrostatic forces.

Three things happen:

• The nucleus has the same positive charge, but there are fewer electrons sharing its attraction.
• Electron-electron repulsion decreases.
• The remaining electron cloud contracts because the electrons experience a stronger effective attraction to the nucleus.

The remaining electrons are therefore held more tightly.

Removing the second electron requires more energy than removing the first.

The same reasoning applies to subsequent ionisations. For ordinary successive ionisation of an isolated ground-state atom:

\[
IE_1 < IE_2 < IE_3 < IE_4
\]

However, these increases are not necessarily equal.

2. Why there is a particularly large jump

Now consider what happens after both outer-shell electrons have been removed from magnesium.

Its electron arrangement changes as follows:

The first two electrons removed occupy the third principal energy level.

The third electron must come from the second principal energy level.

This is the crucial change.

An electron in the second shell is generally:

• Closer to the nucleus.
• Less shielded from the nuclear charge.
• More strongly attracted to the positively charged nucleus.

The third electron is therefore much more difficult to remove.

This explains the large jump between \(IE_2\) and \(IE_3\).

Nuclear charge, shielding, and distance

To explain this properly, you need to understand three connected ideas.

Nuclear charge

Nuclear charge is the positive charge of the nucleus, determined by the number of protons.

Magnesium has 12 protons. This number does not change when electrons are removed.

A greater nuclear charge generally produces a stronger attraction between the nucleus and an electron, provided the other relevant factors are comparable.

Electron shielding

Electrons repel one another.

Inner-shell electrons reduce the effective attraction that outer-shell electrons experience from the nucleus. This effect is called electron shielding.

For magnesium, the two outer 3s electrons are shielded by the ten electrons in the inner shells.

The inner 2p electrons, however, are much closer to the nucleus. They are not shielded by the outer 3s electrons in the same way that the outer electrons are shielded by the inner electrons.

This means the inner electrons experience a stronger effective nuclear attraction.

Distance from the nucleus

Electrostatic attraction becomes weaker as the separation between charged particles increases.

Electrons in a higher principal energy level are generally further from the nucleus.

Consequently, an electron in the third shell is generally easier to remove than an electron in the second shell, assuming comparable nuclear attraction and shielding.

The important point is that these factors work together.

When the outer shell has been emptied, the next electron comes from a region closer to the nucleus, where it experiences a much stronger effective attraction.

It takes substantially more energy to remove.

A useful way to picture the process

Think about removing layers from an onion.

You can peel away the outer layers relatively easily. Eventually, however, you reach the more tightly packed layers closer to the centre.

Removing those requires a different level of effort.

The analogy helps you picture why removing outer electrons and inner electrons can require very different amounts of energy.

However, electrons do not occupy rigid physical layers. Electron shells and orbitals describe regions of probability, not solid surfaces.

02How experiments reveal the large jump

The large jump is not simply predicted from electron arrangements. Ionisation energies can also be determined experimentally.

One approach uses threshold photoionisation.

The central idea is straightforward: gradually increase the energy supplied to gaseous atoms or ions and detect when an electron is first removed.

How the apparatus works

Picture a low-pressure chamber containing gaseous atoms or ions.

A beam of photons is directed through the chamber. The photon energy can be controlled by changing the wavelength of the light.

The energy of a photon is:

\[
E=\frac{hc}{\lambda}
\]

where:

• \(E\) is photon energy, measured in joules.
• \(h\) is Planck’s constant.
• \(c\) is the speed of light.
• \(\lambda\) is the wavelength of the photon.

The photon beam interacts with the gaseous particles.

If a photon supplies enough energy to remove an electron, a positive ion is produced.

The resulting ions are directed into a mass analyser, which separates ions according to their mass-to-charge ratio. A detector then records the number of ions produced.

The experiment is repeated at different photon energies.

What should the experiment show?

Suppose the starting species is gaseous \(\mathrm{Mg}^{+}\).

The process being investigated is:

\[
\ce{Mg+(g) -> Mg^2+(g) + e-}
\]

Below the required energy threshold, this ionisation process cannot occur from the ground state.

As the photon energy reaches the threshold, \(\mathrm{Mg}^{2+}\) ions begin to appear.

The threshold energy corresponds to the second ionisation energy of magnesium.

The experiment can then be repeated using \(\mathrm{Mg}^{2+}\) as the starting species to determine the third ionisation energy.

Notice something important: the starting species must be identified correctly.

If you begin with neutral magnesium, you measure its first ionisation energy. If you begin with \(\mathrm{Mg}^{+}\), you measure its second.

Why was a more precise energy source useful?

Electron-impact experiments can also determine ionisation thresholds.

In these experiments, an electron beam is accelerated towards the gaseous sample. Its energy is varied, and the formation of positive ions is monitored.

One difficulty is that the incoming electrons can have a spread of energies. This makes the exact threshold harder to identify.

An important refinement was to use photons with more precisely defined energies. Monochromators allowed researchers to select a narrower range of photon energies, improving the determination of ionisation thresholds.

NIST describes the development of these threshold measurement approaches, including the use of photoionisation. Modern experiments can also use laser and synchrotron light sources. citeturn426280search6

There are still experimental limitations. Finite energy resolution, excited-state populations, and detector sensitivity can affect how clearly the onset is observed.

The threshold is inferred from the measured signal; it is not simply the energy at which every particle suddenly ionises.

03What the magnesium data actually show

The National Institute of Standards and Technology (NIST) provides evaluated ionisation-energy data for atoms and atomic ions.

For magnesium, the first three values are approximately:

These values are rounded conversions of NIST data originally reported in electronvolts. citeturn426280search0

Look at the difference between the second and third ionisation energies.

\[
7733-1451=6282\ \mathrm{kJ\,mol^{-1}}
\]

The third ionisation requires approximately 6,282 kJ mol⁻¹ more energy than the second.

This is not simply another gradual increase caused by removing an electron from an increasingly positive ion.

The third ionisation removes an electron from a different principal energy level.

04An important exception: Not every increase represents a shell change

Here is where students can make an incorrect prediction.

You might assume that whenever successive ionisation energies increase substantially, the next electron must be coming from an inner shell.

That isn’t necessarily true.

Electrons within the same principal energy level can occupy different subshells. These subshells do not all have identical energies.

Aluminium provides a useful example.

Its electron configuration is:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^1
\]

Its outer-shell arrangement is \(2,8,3\).

The first electron removed is a 3p electron. The next two electrons come from the 3s subshell.

A 3p electron is generally higher in energy and less strongly held than a 3s electron in the same principal energy level.

This means the second ionisation energy can be substantially greater than the first, even though both electrons are removed from the third shell.

Aluminium’s successive ionisation energies

The following are rounded values from NIST data:

The first increase is already substantial:

\[
\frac{1817}{578}\approx3.15
\]

However, this is not the major change from the outer shell to an inner shell.

The fourth ionisation energy is much higher because all three electrons in the third shell have now been removed.

The next electron must come from the second shell.

\[
\frac{11578}{2745}\approx4.22
\]

The important distinction is between a change in subshell and a change in principal energy level. Both can affect ionisation energy, but they are not the same thing. citeturn996910search2

Other factors that can affect individual values

There are two further details worth knowing.

Electron pairing

Electrons occupying the same orbital repel one another. This additional repulsion can make a paired electron easier to remove than an electron in a comparable singly occupied orbital.

As a result, electron pairing can affect the size of individual increases.

Transition metals

Transition-metal electron arrangements can be more complicated because orbital energies can change as electrons are removed.

For example, 4s electrons are generally removed before 3d electrons, even though the 4s subshell fills before the 3d subshell.

This is an extension of the simple main-group examples. It is a reminder that you should track the actual electron configuration of the ion rather than blindly applying a shell-counting rule.

05A reliable way to explain a large jump

When you encounter successive ionisation-energy data, use this reasoning sequence.

1. Write down the electron configuration of the neutral atom.
2. Remove electrons one at a time, updating the electron configuration after each removal.
3. Identify the shell and subshell from which each electron is removed.
4. Look for the transition from removing outer-shell electrons to removing an inner-shell electron.
5. Explain the increase using distance from the nucleus, shielding, and effective nuclear charge.

A useful summary is:

The nuclear charge itself does not increase during successive ionisation. The number of protons stays constant. It is the attraction experienced by the remaining electrons that changes.

For a refresher on how these same factors explain comparisons between different neutral atoms, see First Ionisation Energy Trends for HSC Chemistry.

06Questions and solutions

Question 1

A neutral atom has the electron arrangement \(2,8,2\).

Between which two successive ionisation energies would you expect a large jump? Explain your answer using the electron arrangement.

Solution 1

The large jump should occur between the second and third ionisation energies.

The atom has two electrons in its outermost shell. The first two ionisations remove these electrons.

After the second ionisation, the electron arrangement becomes \(2,8\).

The third ionisation must therefore remove an electron from the second shell. This electron is closer to the nucleus and experiences a stronger effective nuclear attraction.

Consequently, substantially more energy is required to remove it.

Question 2

An unknown gaseous atom Y has the electron arrangement \(2,8,4\).

Its fourth and fifth successive ionisation energies are:

• \(IE_4=4356\ \mathrm{kJ\,mol^{-1}}\)
• \(IE_5=16091\ \mathrm{kJ\,mol^{-1}}\)

Calculate:

(a) The increase in ionisation energy between the fourth and fifth ionisations.

(b) The ratio of the fifth ionisation energy to the fourth.

(c) Explain why the fifth ionisation requires substantially more energy.

The supplied values are illustrative, rounded data.

Solution 2

(a) The increase is \(11,735\ \mathrm{kJ\,mol^{-1}}\).

Calculate the difference:

\[
\Delta IE=IE_5-IE_4
\]

\[
=16091-4356
\]

\[
\boxed{\Delta IE=11735\ \mathrm{kJ\,mol^{-1}}}
\]

(b) The fifth ionisation energy is approximately 3.69 times the fourth.

\[
\text{Ratio}=\frac{IE_5}{IE_4}
\]

\[
=\frac{16091}{4356}
\]

\[
\boxed{\text{Ratio}\approx3.69}
\]

(c) The fifth electron comes from an inner shell.

The neutral atom has four electrons in its outermost shell.

The first four ionisations remove those four electrons. Its electron arrangement changes from \(2,8,4\) to \(2,8\).

The fifth ionisation must remove an electron from the second shell.

This electron is closer to the nucleus and experiences a stronger effective nuclear attraction than the electrons removed previously.

The large increase therefore reflects a change in the principal energy level of the electron being removed, not simply the fact that the ion has become more positively charged.

Question 3

Aluminium has the electron configuration:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^1
\]

Its successive ionisation energies are:

A student argues that the large increase between the first and second ionisation energies must indicate that aluminium has started losing an inner-shell electron.

Evaluate this claim. Identify the subshell from which each of the first four electrons is removed.

Solution 3

The student’s claim is incorrect. The first four ionisations remove three electrons from the third shell before reaching the second shell.

Track the electron configuration after each removal:

The first ionisation removes the 3p electron.

The second removes a 3s electron. Although both electrons belong to the third principal energy level, they occupy different subshells.

The 3p electron is higher in energy and easier to remove than a 3s electron. This explains the substantial increase between the first and second ionisation energies.

The third ionisation removes the remaining 3s electron.

The fourth ionisation is different. Aluminium has now lost all three third-shell electrons, so the next electron comes from the second shell.

This produces the large shell-related jump between \(IE_3\) and \(IE_4\).

The trap is assuming that a substantial increase must always indicate a change of principal energy level. Subshell differences can also affect the values.

Question 4

A researcher wants to determine the third ionisation energy of magnesium.

The researcher initially uses an electron-impact apparatus. However, the electron beam contains a range of energies, and the detector records the formation of \(\mathrm{Mg}^{3+}\) over a relatively broad range of accelerating voltages.

The researcher then replaces the electron beam with a source of photons whose energies can be selected more precisely.

Explain:

(a) Why the original electron beam makes the threshold difficult to determine accurately.

(b) Why the photon-based approach improves the measurement.

(c) What the appearance of \(\mathrm{Mg}^{3+}\) tells the researcher, and why it does not mean that every magnesium ion has been ionised.

Solution 4

(a) The spread of electron energies makes the threshold less precise.

The original electron beam does not deliver exactly the same energy to every collision.

Some electrons may have enough energy to cause ionisation while others do not. Consequently, the onset of ion production becomes spread across a range of accelerating voltages.

This makes it harder to identify the minimum energy required for ionisation.

(b) The photon-based approach provides better control over the energy supplied.

Photons can be selected within a narrow energy range.

By gradually increasing the photon energy and monitoring the production of \(\mathrm{Mg}^{3+}\), the researcher can identify the threshold more precisely.

The relevant process is:

\[
\ce{Mg^2+(g) -> Mg^3+(g) + e-}
\]

The threshold corresponds to the third ionisation energy of magnesium, assuming the starting ions are in their ground state.

(c) The appearance of \(\mathrm{Mg}^{3+}\) indicates that the ionisation process has become energetically accessible.

It does not mean that every \(\mathrm{Mg}^{2+}\) ion has been ionised.

The number of product ions detected also depends on factors such as the number of starting ions, interaction probability, detector efficiency, and experimental conditions.

The experiment identifies an energy threshold, not the proportion of particles that must ionise at that energy.

Question 5

Consider four gaseous species:

\[
\ce{Ne},\quad \mathrm{Na}^{+},\quad \mathrm{Mg}^{2+},\quad \mathrm{Al}^{3+}
\]

Each species contains exactly ten electrons and has the electron arrangement \(2,8\).

Their nuclei contain different numbers of protons:

The approximate energies required to remove one electron from each species are:

All four ionisations remove an electron from the second shell.

A student makes the following argument:

“All four species have the same electron arrangement, so they should require approximately the same energy to remove an electron. Their electron shielding and electron positions are identical.”

Evaluate this argument.

In your response:

(a) Explain why the four ionisation energies are not equal, despite the identical electron arrangements.

(b) Explain how these four measurements relate to successive ionisation energies of the neutral elements.

(c) Explain why the large jump in successive ionisation energies and the increase across this four-species comparison are not contradictory.

Solution 5

The ionisation energies increase in the order Ne < Na⁺ < Mg²⁺ < Al³⁺, even though all four species have the same electron arrangement.

The student’s argument is incomplete because identical electron arrangements do not mean identical nuclear charges.

(a) Greater nuclear charge produces stronger attraction.

All four species contain ten electrons.

Their electron arrangements are identical:

\[
2,8
\]

However, their nuclei contain different numbers of protons.

Ne has 10 protons, whereas \(\mathrm{Al}^{3+}\) has 13.

The electrons in these species experience different effective nuclear attractions.

Because the electron populations are the same, the differences in nuclear charge become particularly important.

As nuclear charge increases, the attraction between the nucleus and the electrons becomes stronger.

The electrons are therefore more difficult to remove.

This explains the increasing ionisation energies:

\[
IE(\ce{Ne})<
IE(\mathrm{Na}^{+})<
IE(\mathrm{Mg}^{2+})<
IE(\mathrm{Al}^{3+})
\]

The fact that all four species contain ten electrons does not mean their electrons are held with equal strength.

(b) Each measurement corresponds to a different successive ionisation energy.

Consider the neutral elements before ionisation.

Sodium has the electron arrangement \(2,8,1\).

Its first ionisation removes the outer 3s electron, leaving \(\mathrm{Na}^{+}\) with an arrangement of \(2,8\).

The next removal is therefore sodium’s second ionisation energy.

The same reasoning applies to magnesium and aluminium:

• \(\mathrm{Na}^{+}\) corresponds to the starting species for \(IE_2\) of sodium.
• \(\mathrm{Mg}^{2+}\) corresponds to the starting species for \(IE_3\) of magnesium.
• \(\mathrm{Al}^{3+}\) corresponds to the starting species for \(IE_4\) of aluminium.

Ne is slightly different. Its first ionisation removes an electron from its outermost occupied shell, which is already the second shell.

(c) The two observations describe different comparisons.

This is the central reasoning challenge.

When comparing successive ionisation energies of one element, the nuclear charge stays constant.

For magnesium, for example, the first two electrons come from the third shell. The third comes from the second shell.

The large increase occurs because the electron being removed is now closer to the nucleus and experiences a stronger effective nuclear attraction.

However, when comparing the four species in this question, the electron arrangement stays constant while the nuclear charge increases.

All four species are removing an electron from the same principal shell and the same general subshell.

The increasing ionisation energies are therefore explained primarily by the increasing nuclear charge.

There is no contradiction.

The large jump within an element is explained by changing the shell from which an electron is removed. The increase across the isoelectronic species is explained by increasing nuclear charge while the electron arrangement remains the same.

This is a much more complete explanation than simply saying that fewer electrons make an atom harder to ionise.

Understanding why the jump occurs is the foundation for interpreting unfamiliar ionisation-energy data. The next useful skill is using the position of the first major jump to determine the number of valence electrons in an unknown element.

Practise that method in How to Find Valence Electrons from Successive Ionisation Energies.