Why Isoelectronic Ions Have Different Radii | HSC Chemistry

Place \(\mathrm{O}^{2-}\), \(\mathrm{F}^{-}\), \(\mathrm{Na}^{+}\), \(\mathrm{Mg}^{2+}\), and \(\mathrm{Al}^{3+}\) in a row. Every ion has exactly 10 electrons. If they have the same number of electrons, should they have the same radius?

Predict the order before reading on.

They don’t. Their effective ionic radii decrease from approximately 140 pm for \(\mathrm{O}^{2-}\) to just 53.5 pm for \(\mathrm{Al}^{3+}\). The reason is that their nuclei contain different numbers of protons. As nuclear charge increases, the nucleus attracts the same electron arrangement more strongly, pulling its electron cloud closer.

Isoelectronic ions have the same number of electrons, but they do not necessarily have the same nuclear charge or electron cloud size.

01Why the same electron configuration gives different radii

Two atoms or ions are called isoelectronic when they have the same number of electrons.

For example, all five ions in our opening comparison have the electron configuration:

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

This is also the ground-state electron configuration of neon.

However, their nuclei are different:

• \(\mathrm{O}^{2-}\) has 8 protons.
• \(\mathrm{F}^{-}\) has 9 protons.
• \(\mathrm{Na}^{+}\) has 11 protons.
• \(\mathrm{Mg}^{2+}\) has 12 protons.
• \(\mathrm{Al}^{3+}\) has 13 protons.

Each nucleus attracts the 10 electrons surrounding it. A nucleus containing 13 protons produces a stronger electrostatic attraction than one containing only 8 protons.

The electrons are therefore held more tightly in \(\mathrm{Al}^{3+}\), producing a smaller electron cloud.

The important point is that an electron configuration tells you which orbitals contain electrons. It does not tell you that those orbitals have identical spatial distributions in every atom or ion.

The \(2p\) orbitals in \(\mathrm{O}^{2-}\) and \(\mathrm{Al}^{3+}\) have the same electron occupancy, but the electrons in \(\mathrm{Al}^{3+}\) are attracted more strongly towards the nucleus.

Comparing their actual radii

The following values are representative Shannon effective ionic radii for six-coordinate ions. All measurements are in picometres (pm), where \(1\text{ pm}=10^{-12}\text{ m}\).

The coordination number is six for every ion, meaning each is surrounded by six nearest neighbouring ions of opposite charge in the crystal structure.

Notice the consistent pattern:

\[
\boxed{
\mathrm{O}^{2-}>
\mathrm{F}^{-}>
\mathrm{Na}^{+}>
\mathrm{Mg}^{2+}>
\mathrm{Al}^{3+}
}
\]

The ionic radius decreases as the number of protons increases.

These are effective ionic radii, not universal measurements of a hard physical boundary. We’ll examine how scientists obtain these values shortly.

02First, make sure the ions have the correct electron configurations

Before comparing ionic radii, you need to construct each ion’s electron configuration correctly.

This is where a surprisingly common mistake occurs: students sometimes use the neutral atom’s electron configuration without accounting for the electrons gained or lost.

Remember two rules:

• When an atom gains electrons, add them to the next available orbitals according to the normal filling order.
• When an atom loses electrons, remove them from the highest occupied energy level first.

Let’s apply these rules to our five ions.

There is an important detail in the aluminium example.

Neutral aluminium has three electrons in its outermost shell:

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

To form \(\mathrm{Al}^{3+}\), it loses the \(3p\) electron first, followed by both \(3s\) electrons.

The resulting configuration is:

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

It does not retain the \(3s^2\) electrons.

If you need to revise why \(1s\), \(2s\), and \(2p\) represent different electron arrangements, see How to Distinguish Shells, Subshells and Orbitals in Chemistry.

Don’t confuse nuclear charge with ionic charge

This distinction is essential.

The nuclear charge depends on the number of protons in the nucleus. It does not change when an atom forms an ion.

For example:

• An aluminium nucleus always contains 13 protons.
• Neutral aluminium has 13 electrons.
• \(\mathrm{Al}^{3+}\) has lost three electrons and now has 10 electrons.

The \(+3\) charge is the ion’s net charge. It is not the total positive charge of its nucleus.

Similarly, \(\mathrm{O}^{2-}\) has a nucleus containing 8 protons, but it has gained two electrons, giving it a net charge of \(-2\).

This explains why the five ions can have different ionic charges while sharing the same electron configuration.

03The role of nuclear attraction and effective nuclear charge

Now we can explain the actual physical reason for the size difference.

A positively charged nucleus attracts negatively charged electrons. Increasing the number of protons increases this electrostatic attraction.

However, electrons also repel one another.

This means an electron does not experience the full nuclear attraction without interference. Other electrons partially shield it from the nucleus.

Chemists describe the net attraction experienced by an electron using the term effective nuclear charge.

A simplified relationship is:

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

where:

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

You don’t need to calculate \(S\) for this comparison. The equation helps explain the underlying idea.

In an isoelectronic series, the total number of electrons remains constant. Therefore, the number of electrons available to shield one another does not increase as you move through the series.

But the number of protons does increase.

The additional nuclear attraction is not cancelled by adding more electrons, so the electron cloud contracts.

This gives us a reliable rule:

Within an isoelectronic series, the species with the greater nuclear charge has the smaller radius.

Be careful with the wording. The electron-electron repulsion does not disappear. It remains part of the system, and the electron distribution changes as the cloud contracts. The observed trend is the result of the stronger net attraction from the nucleus.

A useful decision process

When an exam question gives you several ions, use this sequence:

1. Calculate the number of electrons in each ion.
2. Identify which ions are isoelectronic.
3. Compare their atomic numbers (proton numbers).
4. Within an isoelectronic series, place the ion with the most protons at the smallest radius.
5. If the species are not isoelectronic, consider their occupied shells and electron-electron repulsion as well.

The fourth step is only reliable when you’re comparing species with the same electron count.

04A contrasting case: why the same ionic charge does not mean the same radius

Consider these two ions:

\[
\mathrm{O}^{2-}\qquad \mathrm{S}^{2-}
\]

Both have a charge of \(-2\).

Does that mean they have similar radii?

No. In fact, the sulfide ion is considerably larger than the oxide ion.

The reason is that they are not isoelectronic.

Oxygen gains two electrons to form an ion with the configuration:

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

Sulfur gains two electrons to form:

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

The sulfide ion has an additional occupied electron shell. Its outer electrons occupy the third shell rather than the second.

Consequently, its electron cloud extends further from the nucleus.

This comparison reveals an important limitation of the isoelectronic rule.

You cannot simply say that more protons always produce a smaller ion when comparing any two ions. The number of electrons, occupied shells, and electron-electron repulsion must also be considered.

For the isoelectronic comparison, the electron count is held constant. That’s what makes the nuclear charge trend so useful.

05How scientists determined ionic radii using X-ray diffraction

There is another interesting problem with ionic radii.

An electron cloud does not have a sharp outer boundary. Electrons are described by probability distributions, so there isn’t a precise point where an ion suddenly ends.

How, then, can scientists assign an ionic radius in picometres?

One important method involves X-ray crystallography.

The measurement problem

Imagine a crystal containing a regular arrangement of positive and negative ions.

If you could determine the distance between neighbouring ion centres, you would have a way to estimate their relative sizes.

The difficulty is that you cannot simply measure the boundary of each ion. Instead, you need to determine the positions of the ions within the crystal structure.

X-ray diffraction provides a way to do this.

How the apparatus works

The experiment follows a specific path:

1. An X-ray source produces a beam with a known wavelength.
2. The beam travels towards a crystal mounted on a rotating stage.
3. Electrons within the crystal scatter the incoming X-rays.
4. The scattered waves interfere with one another.
5. A detector records the intensity of the diffracted beams at different angles.
6. Scientists analyse the diffraction pattern to determine the crystal’s repeating structure and lattice dimensions.

What should the experiment show?

If the crystal has a regular repeating structure, the scattered X-rays should produce distinct diffraction peaks at particular angles.

These peaks occur because waves scattered from different crystal planes interfere constructively at certain angles.

The relationship is described by Bragg’s law:

\[
n\lambda=2d\sin\theta
\]

where:

• \(n\) is the diffraction order (a positive integer).
• \(\lambda\) is the X-ray wavelength, measured in metres or picometres.
• \(d\) is the spacing between crystal planes.
• \(\theta\) is the angle between the incident X-ray beam and the crystal planes.

The detector measures the angle between the incident and diffracted beams, which is \(2\theta\).

By analysing the peak positions, scientists can calculate crystal plane spacings and refine the crystal structure.

This gives information about the positions of the ions within the repeating lattice.

From crystal spacing to ionic radius

Consider sodium chloride, \(\ce{NaCl}\).

It has a crystal structure known as the rock-salt structure. Each sodium ion is surrounded by six chloride ions, and each chloride ion is surrounded by six sodium ions.

In its conventional cubic unit cell, the nearest-neighbour separation between a sodium ion and a chloride ion is half the unit-cell edge length.

A rounded unit-cell edge length for NaCl is approximately 563 pm.

Therefore:

\[
\begin{aligned}
d_{\mathrm{Na-Cl}}&=\frac{a}{2}\\
&=\frac{563\text{ pm}}{2}\\
&=281.5\text{ pm}
\end{aligned}
\]

The nearest sodium-chloride ion centres are approximately 281.5 pm apart.

Now compare this with their six-coordinate Shannon effective ionic radii:

\[
r_{\mathrm{Na}^{+}}=102\text{ pm}
\]

\[
r_{\mathrm{Cl}^{-}}=181\text{ pm}
\]

If we add the two radii:

\[
\begin{aligned}
r_{\mathrm{Na}^{+}}+r_{\mathrm{Cl}^{-}}
&=102+181\\
&=283\text{ pm}
\end{aligned}
\]

This is close to the separation obtained from the crystal structure.

The small difference is reasonable because ionic radii are effective values, not perfectly defined boundaries of rigid spheres. The tabulated values are also rounded.

What the experiment could not determine by itself

There is an important limitation here.

X-ray diffraction can help determine the distance between neighbouring ion centres. However, a single sodium-chloride separation does not independently tell you the radius of sodium and the radius of chloride.

It gives the combined distance:

\[
d\approx r_+ + r_-
\]

You need an additional reference or a consistent system of radius assignments to separate the two contributions.

This was an important limitation of early hard-sphere descriptions of ionic crystals. Treating ions as rigid spheres with universal radii also failed to account adequately for differences in coordination and crystal distortion.

Scientists therefore developed more systematic radius scales using structural information from many compounds.

In 1976, R. D. Shannon published a major revision of effective ionic radii based on crystallographic data. The work accounted for coordination number and other structural influences.

The resulting values are widely used when comparing ionic sizes in crystals.

You can consult Shannon’s original 1976 publication and the compiled Shannon ionic radii table.

The important modern qualification is that an ionic radius depends partly on its chemical environment. A six-coordinate radius should not automatically be treated as the exact radius of that ion in a four-coordinate crystal.

For our isoelectronic comparison, using the same coordination number makes the trend much more meaningful.

08Questions and solutions

Question 1

The ions \(\mathrm{Na}^{+}\) and \(\mathrm{Mg}^{2+}\) are isoelectronic.

Which ion has the smaller ionic radius? Explain your answer using their proton numbers.

Solution 1

\(\mathrm{Mg}^{2+}\) has the smaller ionic radius.

Both ions contain 10 electrons, but their nuclei contain different numbers of protons.

• \(\mathrm{Na}^{+}\): 11 protons.
• \(\mathrm{Mg}^{2+}\): 12 protons.

Because \(\mathrm{Mg}^{2+}\) has one additional proton attracting the same number of electrons, it has a stronger nuclear attraction.

Its electron cloud is therefore more contracted.

Representative six-coordinate radii are:

\[
r_{\mathrm{Na}^{+}}=102\text{ pm}
\]

\[
r_{\mathrm{Mg}^{2+}}=72\text{ pm}
\]

Therefore, \(\mathrm{Mg}^{2+}\) is smaller.

Question 2

A student is asked to write the electron configuration of \(\mathrm{Al}^{3+}\).

They start with neutral aluminium:

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

They remove the \(3p\) electron and write the resulting configuration as:

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

Explain the error, determine the correct electron configuration, and compare the radius of \(\mathrm{Al}^{3+}\) with \(\mathrm{O}^{2-}\).

Solution 2

The correct electron configuration of \(\mathrm{Al}^{3+}\) is:

\[
\boxed{1s^2\,2s^2\,2p^6}
\]

The student has removed only one electron, even though the aluminium ion has a charge of \(+3\).

Neutral aluminium contains three electrons in its outermost shell:

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

To form \(\mathrm{Al}^{3+}\), it must lose all three:

1. Remove the \(3p\) electron.
2. Remove the first \(3s\) electron.
3. Remove the second \(3s\) electron.

This leaves 10 electrons.

Now consider \(\mathrm{O}^{2-}\).

Oxygen has 8 protons and gains two electrons, also producing 10 electrons.

Both ions therefore have the configuration:

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

However, aluminium has 13 protons while oxygen has only 8.

The stronger nuclear attraction in \(\mathrm{Al}^{3+}\) produces a smaller electron cloud.

Therefore:

\[
\boxed{r_{\mathrm{Al}^{3+}}<r_{\mathrm{O}^{2-}}}
\]

The mistake in the original answer was forgetting that forming a \(3+\) ion requires removing three electrons, not just the single \(3p\) electron.

Question 3

The ions \(\mathrm{O}^{2-}\) and \(\mathrm{S}^{2-}\) both have a charge of \(-2\).

A student argues that their radii should be similar because they have the same ionic charge.

Explain why this reasoning is incorrect. Include their electron configurations and identify which ion has the larger radius.

Solution 3

\(\mathrm{S}^{2-}\) has the larger radius because it has an additional occupied electron shell.

The electron configurations are:

For oxide:

\[
\mathrm{O}^{2-}:1s^2\,2s^2\,2p^6
\]

For sulfide:

\[
\mathrm{S}^{2-}:1s^2\,2s^2\,2p^6\,3s^2\,3p^6
\]

The oxide ion contains 10 electrons, whereas the sulfide ion contains 18.

Their outermost occupied shells are different:

• \(\mathrm{O}^{2-}\) has its outer electrons in the second shell.
• \(\mathrm{S}^{2-}\) has its outer electrons in the third shell.

The third shell extends further from the nucleus, giving the sulfide ion a larger electron cloud.

Representative six-coordinate radii are:

\[
r_{\mathrm{O}^{2-}}=140\text{ pm}
\]

\[
r_{\mathrm{S}^{2-}}=184\text{ pm}
\]

Therefore:

\[
\boxed{r_{\mathrm{S}^{2-}}>r_{\mathrm{O}^{2-}}}
\]

The student’s mistake is assuming that the same ionic charge means the same electron configuration.

It doesn’t.

The isoelectronic rule requires the same number of electrons, not merely the same ionic charge.

Question 4

A crystal is examined using monochromatic X-ray diffraction.

The X-ray wavelength is 154 pm, and a first-order diffraction peak occurs at an angle of \(15.0^\circ\). The angle is measured between the incident beam and the crystal planes.

Calculate the spacing between the crystal planes.

A student then claims that half this distance must be the radius of an ion. Explain why the second conclusion is incorrect.

Solution 4

The spacing between the crystal planes is approximately 297.5 pm, but half this distance is not automatically an ionic radius.

Step 1: Write Bragg’s law.

\[
n\lambda=2d\sin\theta
\]

For a first-order diffraction peak:

\[
n=1
\]

Step 2: Rearrange to make \(d\) the subject.

\[
d=\frac{n\lambda}{2\sin\theta}
\]

Step 3: Substitute the given values.

\[
d=\frac{1\times154}{2\sin15^\circ}
\]

Using:

\[
\sin15^\circ\approx0.2588
\]

We obtain:

\[
d=\frac{154}{0.5176}
\]

\[
\boxed{d\approx297.5\text{ pm}}
\]

Step 4: Explain the limitation.

The calculated value is the spacing between crystal planes.

It is not necessarily the distance between neighbouring ion centres.

The crystal’s geometry must first be determined to convert plane spacings into actual interionic distances.

Even when the distance between two neighbouring ion centres is known, it represents the combined effective radii of the two ions:

\[
d\approx r_+ + r_-
\]

It does not independently determine either radius.

Therefore, dividing the plane spacing by two and calling the result an ionic radius is not justified.

Question 5

A student examines the six-coordinate ionic radii of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\).

They correctly identify that both ions have 10 electrons and that \(\mathrm{Al}^{3+}\) should be smaller because it has a greater nuclear charge.

However, they then propose a mathematical model:

\[
r\propto\frac{1}{Z}
\]

where \(r\) is ionic radius and \(Z\) is atomic number.

Using the radius of \(\mathrm{Mg}^{2+}\), which is 72 pm, they predict:

\[
r_{\mathrm{Al}^{3+}}
=72\times\frac{12}{13}
=66.5\text{ pm}
\]

A consistent six-coordinate data set gives an effective radius of 53.5 pm for \(\mathrm{Al}^{3+}\).

The student concludes that the isoelectronic trend must be incorrect because the measured value does not match their prediction.

Evaluate the student’s reasoning.

Your response should:

• Identify which part of their reasoning is correct.
• Explain why the inverse relationship \(r\propto1/Z\) is not a valid general model for these ions.
• Explain how electron shielding, electron-electron repulsion, and the nature of experimentally assigned ionic radii affect the comparison.

Solution 5

The student’s qualitative prediction is correct, but their mathematical model is not.

\(\mathrm{Al}^{3+}\) is smaller than \(\mathrm{Mg}^{2+}\), even though its radius does not follow the proposed inverse relationship.

Step 1: Confirm that the ions are isoelectronic.

Magnesium has 12 protons and loses two electrons:

\[
12-2=10\text{ electrons}
\]

Aluminium has 13 protons and loses three electrons:

\[
13-3=10\text{ electrons}
\]

Both ions have the configuration:

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

Therefore, the comparison is valid for an isoelectronic series.

Step 2: Identify the correct prediction.

The aluminium nucleus has 13 protons, compared with 12 in magnesium.

Its stronger nuclear attraction pulls the same general electron arrangement closer to the nucleus.

Therefore:

\[
r_{\mathrm{Al}^{3+}}<r_{\mathrm{Mg}^{2+}}
\]

The measured values support this:

\[
53.5\text{ pm}<72\text{ pm}
\]

The qualitative isoelectronic trend is correct.

Step 3: Identify the problem with the inverse relationship.

The student’s mistake is treating the ions as though they were one-electron, hydrogen-like species.

For a hydrogen-like ion containing only one electron, the orbital radius at a fixed principal energy level has an inverse dependence on nuclear charge.

However, \(\mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) each contain 10 electrons.

They are multi-electron systems.

Their electrons experience:

• Attraction from the positively charged nucleus.
• Repulsion from the other electrons.
• Shielding from other electrons, which reduces the effective nuclear attraction.

As the nuclear charge increases, the electron cloud contracts, but the effective nuclear attraction does not follow a simple inverse relationship with the atomic number.

The electron distribution must be considered as a whole.

Step 4: Consider how the radii were obtained.

The values of 72 pm and 53.5 pm are effective ionic radii derived from crystal structures.

They are not exact measurements of the outermost position of an isolated electron.

Although both values refer to six-coordinate ions, the assigned radii still reflect the conventions used to analyse crystal structures.

This provides another reason not to assume a simple mathematical relationship between ionic radius and atomic number.

Step 5: State the correct conclusion.

The isoelectronic trend predicts the direction of the size change:

\[
\boxed{
\text{Increasing nuclear charge}
\Rightarrow
\text{Decreasing ionic radius}
}
\]

It does not establish that ionic radius is exactly inversely proportional to atomic number.

The student’s predicted value of 66.5 pm is therefore not a reliable result.

The actual effective radius of \(\mathrm{Al}^{3+}\) in the stated six-coordinate data set is 53.5 pm.

The important distinction is between a qualitative trend and an exact mathematical law. A trend can be correct without following a simple proportional relationship.

Once you can identify an isoelectronic series, comparing proton numbers gives you a reliable way to predict the relative ionic sizes.

The next step is to recognise when that shortcut no longer applies. When electron counts differ, return to electron configurations, occupied shells, and electron-electron repulsion before making a prediction.

For the broader comparison of neutral atoms, cations, and anions, see Why Ions Change Size: How to Rank Ionic Radii.