Why Shielding Changes Periodic Trends Down a Group

Sodium has 11 protons. Lithium has only 3. So which atom should hold its outer electron more tightly?

If you focus only on nuclear charge, sodium seems like the obvious answer. Its nucleus has considerably more positive charge, so surely it should attract its electrons more strongly.

But sodium’s first ionisation energy is approximately 496 kJ mol⁻¹, compared with 520 kJ mol⁻¹ for lithium. Sodium actually loses its outer electron more easily.

The reason is electron shielding. Moving down a group adds new occupied electron shells. These inner electrons shield the outer electrons from some of the nucleus’s attraction. The outer electrons also move further away from the nucleus.

Although nuclear charge increases down a group, increased shielding and electron distance weaken the attraction experienced by the valence electrons. This is why first ionisation energy generally decreases down a group, while atomic radius increases.

Let’s look at how the electron arrangements create this effect, and how experimental measurements provide evidence for the explanation.

01Why shielding increases down a group

Start with lithium, sodium, and potassium

Consider three Group 1 elements:

• Lithium (Li)
• Sodium (Na)
• Potassium (K)

All three have one valence electron. However, that electron occupies a different principal energy level in each atom.

Their electron configurations are:

Notice that each atom has one electron in its outermost shell. This is why they belong to the same group.

However, the number of inner-shell electrons increases:

• Lithium has 2 inner-shell electrons.
• Sodium has 10 inner-shell electrons.
• Potassium has 18 inner-shell electrons.

These inner electrons sit between the nucleus and the outermost electron.

As you move from lithium to sodium to potassium, two important things happen:

1. The outer electron occupies a higher principal energy level and is generally further from the nucleus.
2. More inner-shell electrons are present, increasing the shielding experienced by the outer electron.

Both changes reduce the strength of the attraction holding the outer electron.

What does electron shielding actually mean?

The nucleus contains positively charged protons, while electrons carry negative charge.

Opposite charges attract, so the nucleus attracts the electrons. However, electrons also repel one another because they have the same negative charge.

This electron-electron repulsion is important.

Imagine standing near a loudspeaker with a crowd of people between you and the speaker. The crowd reduces how much sound reaches you.

Inner-shell electrons have a similar screening effect on an outer electron. Their negative charge repels the outer electron, reducing the net positive attraction it experiences from the nucleus.

This effect is called electron shielding, also known as the screening effect.

The analogy has a limitation: electrons are not stationary objects forming a physical barrier. They occupy regions of electron probability, and shielding arises from electrostatic interactions between their charges.

The important point is that inner electrons do not remove the positive charge of the nucleus. They reduce the net attraction experienced by an outer electron.

Effective nuclear charge: how much attraction does the electron experience?

Chemists use the term effective nuclear charge to describe the net positive attraction experienced by an electron after accounting for shielding.

It is represented by:

\[ Z_{\mathrm{eff}}\approx Z-S \]

Where:

• \(Z_{\mathrm{eff}}\) is the effective nuclear charge.
• \(Z\) is the atomic number, representing the number of protons.
• \(S\) represents the shielding effect of other electrons.

This is a simplified qualitative relationship, not an exact calculation.

For example, sodium has 11 protons and 10 inner-shell electrons. It would be tempting to calculate \(11-10=1\) and conclude that its outer electron experiences a nuclear charge of exactly +1.

That is not correct.

Electrons do not perfectly cancel individual protons. Different electrons also provide different amounts of shielding depending on their positions and orbitals.

The relationship is useful because it reminds us that the outer electron does not experience the full nuclear charge.

Why the extra protons don’t win down a group

Here is the important comparison.

You might initially predict that potassium should hold its outer electron more tightly than lithium because potassium has 19 protons compared with lithium’s 3.

But that ignores the electron arrangement.

Potassium’s outer electron occupies the fourth principal energy level, while lithium’s occupies the second. Potassium also has substantially more inner-shell electron density shielding its outer electron.

The increased distance and shielding outweigh the increased nuclear charge.

This is why the outer electron becomes easier to remove down Group 1.

The same reasoning applies to other groups, although the precise trends can vary with electron configuration.

02How scientists investigate shielding experimentally

So how can scientists tell whether an outer electron is more or less tightly held?

One useful experimental technique is photoelectron spectroscopy (PES).

It provides evidence about electron binding energies and the arrangement of electrons within atoms.

Importantly, PES does not directly measure a shielding constant. Instead, scientists measure electron energies and use those measurements, together with electron configurations, to understand the effects of shielding.

The problem: measuring how tightly electrons are held

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

The process is represented by:

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

The energy required is usually expressed in kJ mol⁻¹.

A useful experimental question is:

If an electron is held less tightly by its nucleus, how much energy does it need to escape?

We can investigate this by supplying energy to the atom and measuring the energy of the escaping electron.

How photoelectron spectroscopy works

A photoelectron spectrometer uses photons to eject electrons from gaseous atoms.

The apparatus operates in a controlled, low-pressure environment.

The process follows a clear sequence:

Step 1: A sample is converted into gaseous atoms and introduced into an evacuated chamber.

Step 2: A source directs ultraviolet or X-ray photons towards the gaseous atoms.

Step 3: The atoms absorb photons. If a photon supplies enough energy, an electron can escape from its atom.

Step 4: The ejected electrons travel into an energy analyser, which measures their kinetic energies.

Step 5: A detector records the electrons and produces a spectrum showing the number of detected electrons at different energies.

The key measurement is the kinetic energy of the ejected electrons.

If an electron is strongly attracted to the nucleus, more energy is required to remove it. With the same photon energy, the escaping electron therefore has less kinetic energy.

If an electron is weakly attracted to the nucleus, less energy is required to remove it. The escaping electron has more kinetic energy.

This relationship follows from conservation of energy:

\[ E_{\mathrm{bind}}=E_{\gamma}-E_k \]

Where:

• \(E_{\mathrm{bind}}\) is the binding energy of the electron.
• \(E_{\gamma}\) is the energy of the incoming photon.
• \(E_k\) is the kinetic energy of the ejected electron.

All three quantities can be expressed in electronvolts (eV) per electron.

What should the experiment show?

Let’s compare lithium, sodium, and potassium.

If the outer electron becomes less tightly held down Group 1, its binding energy should decrease.

For the same incoming photon energy, its kinetic energy should increase.

The following table uses rounded gas-phase ionisation-energy data expressed in eV per atom. The corresponding values are approximately 520, 496, and 419 kJ mol⁻¹ respectively. The reference values are available from the National Institute of Standards and Technology (NIST).

The binding energy decreases down the group.

This means the outer electron in potassium requires less energy to remove than the outer electron in sodium or lithium.

The experimental evidence supports the explanation:

• The outer electron in lithium is more tightly bound.
• The outer electron in sodium is less tightly bound.
• The outer electron in potassium is less tightly bound again.

This is consistent with the increase in shielding and distance down the group.

The spectrum can also provide information about occupied electron subshells. Different groups of electrons produce peaks at different binding energies.

However, remember the distinction: the experiment measures binding energies. Shielding is part of the chemical model used to explain those measurements.

Why the experimental design matters

There is a problem with a tempting first approach.

Suppose scientists simply shine ultraviolet light onto solid pieces of lithium, sodium, and potassium and compare the electrons escaping from their surfaces.

The results would not directly represent the first ionisation energies of isolated atoms.

The metals contain atoms joined in a metallic structure, and the energy needed for electrons to escape a solid surface is influenced by the material’s electronic structure and work function.

A better design uses isolated gaseous atoms.

This removes the complications associated with metallic bonding and surface effects, allowing scientists to investigate atomic electron binding more directly.

Using the same photon energy and a calibrated analyser also makes the comparison meaningful.

For HSC Chemistry, the important lesson is not that you must memorise every component of the instrument. It is understanding how measured electron energies support the model of electron arrangement and shielding.

03Worked examples: applying shielding to periodic trends

04Are there exceptions to the trend?

The general trend down a group is that first ionisation energy decreases.

However, this pattern is not perfectly consistent across every group.

The reason is that electron shielding depends on more than simply counting the number of electrons.

Different subshells provide different amounts of shielding. In particular, d and f electrons do not shield outer electrons as effectively as a simple shell-counting model might suggest.

Extension: The aluminium and gallium comparison

Consider aluminium and gallium, which are both in Group 13.

Their electron configurations are:

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

\[ \mathrm{Ga}:[Ar]\,3d^{10}\,4s^2\,4p^1 \]

You might expect gallium to have a lower first ionisation energy because its outer electron occupies a higher principal energy level.

However, the approximate experimental values are:

Gallium actually has a slightly higher first ionisation energy.

Why?

Gallium has ten additional 3d electrons. These electrons provide relatively poor shielding of the outer electrons.

As a result, the outer electron experiences a greater effective nuclear attraction than a simple shell-counting model would predict.

This effect slightly increases the energy required to remove gallium’s outer electron.

The comparison is a useful example of why shielding is not a perfect one-electron-for-one-proton cancellation.

The aluminium and gallium values, and the explanation involving imperfect shielding by 3d electrons, are also examined in the American Chemical Society’s 2023 US National Chemistry Olympiad examination.

For a standard HSC explanation, concentrate on the general trend and the roles of nuclear charge, shielding, and distance. Treat the gallium example as an extension that demonstrates the limits of the simplified model.

05Questions and solutions

Question 1

Lithium, sodium, and potassium belong to Group 1.

Their electron arrangements are:

• Lithium: 2, 1
• Sodium: 2, 8, 1
• Potassium: 2, 8, 8, 1

Rank these elements from highest to lowest first ionisation energy.

Explain your answer using nuclear charge, shielding, and electron arrangement.

Solution 1

The correct order is:

\[ \boxed{\mathrm{Li>Na>K}} \]

Lithium has the highest first ionisation energy, while potassium has the lowest.

All three elements have one valence electron, but that electron occupies a different principal energy level.

Moving from lithium to potassium:

• The nuclear charge increases.
• The number of inner-shell electrons increases.
• Shielding becomes greater.
• The outer electron occupies a higher principal energy level and is further from the nucleus.

Although the increased nuclear charge strengthens attraction, the increased shielding and distance have a greater overall effect.

Consequently, the outer electron becomes easier to remove down the group.

A common mistake is to rank potassium highest because it has the most protons. This ignores the electron’s position and the shielding caused by the additional inner shells.

Question 2

A photoelectron spectrometer uses photons with an energy of 12.00 eV to investigate gaseous lithium, sodium, and potassium atoms.

The measured kinetic energies of the ejected valence electrons are:

(a) Calculate the binding energy of the valence electron for each element.

(b) Convert each binding energy into kJ mol⁻¹ using:

\[ 1\ \mathrm{eV\ per\ atom} =96.485\ \mathrm{kJ\,mol^{-1}} \]

(c) Explain how the results support the general trend in first ionisation energy down Group 1.

Solution 2

(a) The binding energies are 5.39 eV for lithium, 5.14 eV for sodium, and 4.34 eV for potassium.

Use the relationship:

\[ E_{\mathrm{bind}}=E_{\gamma}-E_k \]

For lithium:

\[ E_{\mathrm{bind}}=12.00-6.61 =5.39\ \mathrm{eV} \]

For sodium:

\[ E_{\mathrm{bind}}=12.00-6.86 =5.14\ \mathrm{eV} \]

For potassium:

\[ E_{\mathrm{bind}}=12.00-7.66 =4.34\ \mathrm{eV} \]

(b) The corresponding first ionisation energies are approximately 520, 496, and 419 kJ mol⁻¹.

Convert each result using the supplied conversion factor.

Lithium:

\[ 5.39\times96.485 =520\ \mathrm{kJ\,mol^{-1}} \]

Sodium:

\[ 5.14\times96.485 =496\ \mathrm{kJ\,mol^{-1}} \]

Potassium:

\[ 4.34\times96.485 =419\ \mathrm{kJ\,mol^{-1}} \]

(c) The results demonstrate that first ionisation energy decreases down Group 1.

The binding energy decreases from lithium to potassium.

This means less energy is required to remove the outer electron from potassium than from lithium.

The reason is that potassium’s outer electron occupies a higher principal energy level and experiences greater shielding from inner-shell electrons.

Although potassium has a greater nuclear charge, these effects reduce the net attraction holding its outer electron.

An important observation is that potassium’s ejected electron has the greatest kinetic energy. This does not mean the incoming photon was stronger. All three atoms received photons with the same energy, so the difference comes from the energy required to remove the electron.

Question 3

A student makes the following claim:

Lithium has 3 protons and 2 inner-shell electrons. Sodium has 11 protons and 10 inner-shell electrons. Potassium has 19 protons and 18 inner-shell electrons.

Therefore, each atom has exactly one unit of effective nuclear charge acting on its outer electron. Their outer electrons should experience the same attraction, so their first ionisation energies should be identical.

Evaluate this explanation.

Identify two problems with the student’s reasoning and explain the actual trend.

Solution 3

The student’s explanation is incorrect because inner electrons do not shield the nucleus perfectly, and the outer electrons occupy different principal energy levels.

There are two important problems.

Problem 1: Shielding is not perfect cancellation.

The student assumes that each inner electron cancels the effect of exactly one proton.

This is not how electron shielding works.

Electrons repel one another, reducing the net attraction experienced by an outer electron. However, the extent of shielding depends on electron distribution and orbital characteristics.

The number of inner electrons cannot simply be subtracted from the proton number to obtain an exact effective nuclear charge.

Problem 2: The distance of the outer electron is ignored.

Even if two atoms had similar effective nuclear charges, their outer electrons could occupy different principal energy levels.

For lithium, sodium, and potassium, the outer electrons occupy the second, third, and fourth principal energy levels respectively.

The outer electron in potassium is generally further from the nucleus than the outer electron in lithium.

This reduces the electrostatic attraction between the nucleus and the outer electron.

The correct explanation is that first ionisation energy generally decreases down Group 1.

Increasing nuclear charge tends to increase attraction, but increasing shielding and electron distance weaken the attraction experienced by the outer electron.

The student’s mistake is treating shielding as an exact numerical cancellation and ignoring the spatial distribution of electrons.

Question 4

Magnesium and calcium form the ions \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\).

Their electron arrangements are:

\[ \mathrm{Mg}^{2+}:2,8 \]

\[ \mathrm{Ca}^{2+}:2,8,8 \]

Both ions have an overall charge of +2.

(a) Predict which ion has the larger ionic radius.

(b) Explain your prediction using electron arrangement, nuclear charge, and shielding.

(c) A student argues that calcium must form the smaller ion because its nucleus has 20 protons compared with magnesium’s 12 protons. Explain why this argument is incomplete.

Solution 4

(a) The \(\mathrm{Ca}^{2+}\) ion has the larger ionic radius.

(b) Calcium has an additional occupied electron shell, increasing both the distance of its outer electrons from the nucleus and the shielding they experience.

Magnesium’s ion has two occupied electron shells, while calcium’s ion has three.

The outermost electrons in \(\mathrm{Ca}^{2+}\) occupy the third principal energy level. Those in \(\mathrm{Mg}^{2+}\) occupy the second.

Calcium therefore has a larger electron cloud.

Although calcium has a greater nuclear charge, the additional shell and shielding allow its electron cloud to extend further from the nucleus.

Consequently:

\[ r(\mathrm{Ca}^{2+})>r(\mathrm{Mg}^{2+}) \]

(c) The student’s argument ignores electron arrangement and shielding.

Calcium does have more protons, which increases the attractive force exerted by its nucleus.

However, its ion also has more electrons and an additional occupied shell.

The extra inner-shell electrons increase shielding, while the outermost electrons occupy a higher principal energy level.

The nuclear charge alone cannot determine ionic radius.

It is also important not to confuse these ions with isoelectronic ions. Magnesium and calcium ions do not have the same number of electrons, so their electron arrangements cannot be compared by considering nuclear charge alone.

Question 5

Consider the following data for three gaseous species:

For calcium, the energy listed for \(\mathrm{Ca^+}\) is its second ionisation energy.

Two students make different claims.

Student A:

Potassium has a lower first ionisation energy than sodium because its extra shell increases shielding. Therefore, shielding always outweighs an increase in nuclear charge.

Student B:

Potassium and calcium ions have the same electron configuration, so their outer electrons should have the same binding energy. Their ionisation energies should therefore be equal.

Evaluate both claims.

In your response:

• Explain why potassium has a lower first ionisation energy than sodium.
• Explain why \(\mathrm{Ca^+}\) requires considerably more energy to ionise than neutral potassium.
• Identify which factors change in each comparison.
• Explain why the calcium value is its second ionisation energy rather than its first.

Solution 5

Both students make incorrect generalisations. Potassium is easier to ionise than sodium because shielding and distance outweigh the increased nuclear charge, but \(\mathrm{Ca^+}\) is much harder to ionise than potassium because it has the same electron configuration but a greater nuclear charge.

The key is that the two comparisons involve different changes to the atoms.

Comparison 1: Sodium and potassium

Sodium has 11 protons and an outer electron in the third principal energy level.

Potassium has 19 protons and an outer electron in the fourth principal energy level.

Moving from sodium to potassium increases:

• Nuclear charge.
• The number of inner-shell electrons.
• Electron shielding.
• The distance of the outer electron from the nucleus.

The extra proton attraction would tend to increase the first ionisation energy.

However, potassium’s additional occupied shell increases shielding and places its outer electron further from the nucleus.

These effects outweigh the increased nuclear charge.

Therefore:

\[ IE_1(\mathrm{K})<IE_1(\mathrm{Na}) \]

Student A identifies the correct trend but makes an incorrect generalisation.

Shielding does not always outweigh nuclear charge. It depends on which other factors change at the same time.

Comparison 2: Potassium and calcium ion

Now compare neutral potassium with \(\mathrm{Ca^+}\).

Both have 19 electrons and the same electron configuration:

\[ \mathrm{K}:[Ar]4s^1 \]

\[ \mathrm{Ca^+}:[Ar]4s^1 \]

However, their nuclei are different.

Potassium has 19 protons, while \(\mathrm{Ca^+}\) has 20.

Because the electron configurations are the same, the two species have broadly similar shell arrangements and shielding patterns.

However, \(\mathrm{Ca^+}\) has an additional proton.

This increases the effective nuclear attraction experienced by its remaining 4s electron. Its electron cloud is also contracted by the stronger attraction.

Consequently, its outer electron is held much more tightly.

The measured energies demonstrate this:

\[ \begin{aligned} IE_1(\mathrm{K})&=419\ \mathrm{kJ\,mol^{-1}}\\ IE_2(\mathrm{Ca})&=1145\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Student B is therefore incorrect. Having the same electron configuration does not mean two species have the same electron binding energies.

The number of protons still matters.

Why is the calcium value its second ionisation energy?

The first ionisation of calcium removes one electron from neutral calcium:

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

Neutral calcium initially has the configuration:

\[ \mathrm{Ca}:[Ar]4s^2 \]

After losing one electron, it becomes:

\[ \mathrm{Ca^+}:[Ar]4s^1 \]

The next electron removed comes from this already positively charged ion:

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

This is the second ionisation process.

The electron is now being removed from a positively charged species, and it experiences a stronger attraction to the nucleus.

That is why the second ionisation energy is much greater than potassium’s first ionisation energy.

The deeper lesson: shielding does not automatically cancel extra protons. When comparing sodium and potassium, the new shell and increased shielding dominate. When comparing potassium and calcium ions, the electron arrangement is held similar while nuclear charge increases, so nuclear attraction becomes stronger.

The outcome depends on the interaction between nuclear charge, shielding, and electron distance. You cannot predict it by counting protons or electrons alone.

06Where this leads next

You now have a useful way to approach periodic trends: identify the electron being considered before deciding which atom holds it more tightly.

The next useful comparison is between isoelectronic ions. These species have the same number of electrons but different numbers of protons, making them particularly useful for investigating the effect of nuclear charge without introducing a different electron configuration.

Practise this reasoning with How to Rank Isoelectronic Ions by Size: Worked Examples.

For a broader explanation of first ionisation energy, including trends across a period and subshell-related exceptions, see First Ionisation Energy Trends for HSC Chemistry.