How to Rank Isoelectronic Ions by Size: Worked Examples

You are given five ions: \(\mathrm{S}^{2-}\), \(\mathrm{Cl}^{-}\), \(\mathrm{K}^{+}\), \(\mathrm{Ca}^{2+}\), and \(\mathrm{Sc}^{3+}\). Which is largest, and which is smallest? A student might try to rank them by ionic charge, but that doesn't give a reliable answer.

All five ions have 18 electrons. Their proton numbers are different, though, and that is what determines their relative sizes in this isoelectronic series.

From largest to smallest:

\[
\boxed{
\mathrm{S}^{2-}>
\mathrm{Cl}^{-}>
\mathrm{K}^{+}>
\mathrm{Ca}^{2+}>
\mathrm{Sc}^{3+}
}
\]

The rule is straightforward: **among isoelectronic ions, the ion with more protons is smaller.** Its nucleus attracts the same electron arrangement more strongly, pulling the electron cloud closer.

The important part is recognising when this rule applies, understanding why it works, and knowing when you should not use it.

## What makes ions isoelectronic?

Isoelectronic species have the same number of electrons. They can be atoms, positive ions, or negative ions.

For example, consider the five ions in our opening problem.

The atomic number tells you how many protons an element has. A positive ion has lost electrons, while a negative ion has gained them.

Use:

\[
N_e=Z-q
\]

where:
- \(N_e\) is the number of electrons,
- \(Z\) is the atomic number, and
- \(q\) is the signed ionic charge.

For a negative ion, subtracting a negative charge is equivalent to adding electrons.

### Worked example: Check whether five ions are isoelectronic

Show that \(\mathrm{S}^{2-}\), \(\mathrm{Cl}^{-}\), \(\mathrm{K}^{+}\), \(\mathrm{Ca}^{2+}\), and \(\mathrm{Sc}^{3+}\) are isoelectronic.

**Step 1**: Find the atomic number of each element.

| Ion | Atomic number, \(Z\) | Ionic charge, \(q\) |
|---|---:|---:|
| \(\mathrm{S}^{2-}\) | 16 | -2 |
| \(\mathrm{Cl}^{-}\) | 17 | -1 |
| \(\mathrm{K}^{+}\) | 19 | +1 |
| \(\mathrm{Ca}^{2+}\) | 20 | +2 |
| \(\mathrm{Sc}^{3+}\) | 21 | +3 |

**Step 2**: Calculate the electron count.

\[
\begin{aligned}
\mathrm{S}^{2-}:&\quad 16-(-2)=18\\
\mathrm{Cl}^{-}:&\quad 17-(-1)=18\\
\mathrm{K}^{+}:&\quad 19-(+1)=18\\
\mathrm{Ca}^{2+}:&\quad 20-(+2)=18\\
\mathrm{Sc}^{3+}:&\quad 21-(+3)=18
\end{aligned}
\]

**Step 3**: Confirm their electron arrangements.

All five ions have the configuration:

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

Their shell arrangement is \(2,8,8\). Each has three occupied electron shells.

Therefore, all five ions are isoelectronic. We can now compare their nuclear charges to rank their sizes.

[IMAGE: Five side-by-side schematic electron-cloud illustrations for S2-, Cl-, K+, Ca2+ and Sc3+. Show the same 18-electron arrangement in three occupied shells for every ion, with nuclei labelled 16p+, 17p+, 19p+, 20p+ and 21p+ respectively. Illustrate progressively contracting electron clouds from left to right. Include a clear largest-to-smallest direction. Electron positions should be schematic probability-cloud representations, not fixed circular orbits.]

## Why more protons make an isoelectronic ion smaller

The key is to look at what stays the same and what changes.

In our five-ion series, every ion has:
- 18 electrons,
- three occupied electron shells,
- the same electron configuration, \([Ar]\),
- the same general arrangement of inner and outer electrons.

What changes is the number of protons in the nucleus.

### Nuclear charge and electron shielding

The nucleus contains positively charged protons. Electrons are negatively charged, so the nucleus attracts them.

However, an outer electron does not experience the full attraction of the nucleus. Other electrons partially shield it from the positive nuclear charge.

This effect is called *electron shielding*.

Think of the inner electrons as a screen between the nucleus and the outer electrons. They reduce the attraction that the outer electrons experience. The analogy is useful for picturing the effect, although electrons are not stationary barriers, and shielding is an electrostatic effect rather than a physical wall.

For our isoelectronic series, the inner-shell arrangement stays the same. There are still 10 inner-shell electrons and 8 electrons in the third shell.

Now compare the nuclei:

| Ion | Protons | Electrons | Inner-shell electrons |
|---|---:|---:|---:|
| \(\mathrm{S}^{2-}\) | 16 | 18 | 10 |
| \(\mathrm{Cl}^{-}\) | 17 | 18 | 10 |
| \(\mathrm{K}^{+}\) | 19 | 18 | 10 |
| \(\mathrm{Ca}^{2+}\) | 20 | 18 | 10 |
| \(\mathrm{Sc}^{3+}\) | 21 | 18 | 10 |

The number of protons increases, but the number and arrangement of electrons remain the same.

The shielding arrangement is therefore broadly similar. It does not increase enough to cancel the additional nuclear charge.

The outer electrons experience a stronger net attraction as we move from \(\mathrm{S}^{2-}\) towards \(\mathrm{Sc}^{3+}\).

This net attraction is described using the term *effective nuclear charge*. It is the positive nuclear attraction experienced by an electron after accounting for shielding by other electrons.

A greater effective nuclear charge pulls the electron cloud closer to the nucleus.

[IMAGE: An annotated comparison of S2- and Sc3+ showing their identical 18-electron arrangements, with 10 inner-shell electrons and 8 outer-shell electrons. Show the 16-proton nucleus of S2- beside the 21-proton nucleus of Sc3+. Illustrate how the stronger nuclear attraction in Sc3+ contracts the electron cloud, while keeping the electron count and shell arrangement identical. Clearly distinguish the nucleus, inner electrons and outer electron cloud.]

### The ranking method

Once you've confirmed that the ions are isoelectronic, you can use a very reliable three-step method.

**Step 1**: Count the electrons in every ion.

Use atomic number and ionic charge. Don't assume that ions with similar charges have the same electron count.

**Step 2**: Confirm that the electron configurations match.

For the common main-group ions used in HSC questions, check that they have the same occupied shells and electron arrangement.

**Step 3**: Compare the number of protons.

- Fewer protons means weaker nuclear attraction and a larger radius.
- More protons means stronger nuclear attraction and a smaller radius.

To rank from largest to smallest, arrange the proton numbers from lowest to highest.

### Worked example: Rank a five-ion isoelectronic series

Rank \(\mathrm{S}^{2-}\), \(\mathrm{Cl}^{-}\), \(\mathrm{K}^{+}\), \(\mathrm{Ca}^{2+}\), and \(\mathrm{Sc}^{3+}\) from largest to smallest ionic radius. Explain your reasoning using nuclear charge and electron shielding.

**Step 1**: Calculate the electron count of each ion.

\[
\begin{aligned}
\mathrm{S}^{2-}:&\quad 16+2=18\\
\mathrm{Cl}^{-}:&\quad 17+1=18\\
\mathrm{K}^{+}:&\quad 19-1=18\\
\mathrm{Ca}^{2+}:&\quad 20-2=18\\
\mathrm{Sc}^{3+}:&\quad 21-3=18
\end{aligned}
\]

All five ions contain 18 electrons.

**Step 2**: Identify their electron arrangement.

Each ion has the configuration:

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

Their electron arrangements are identical, with three occupied shells.

**Step 3**: Compare their nuclear charges.

The proton numbers are:

\[
16,\ 17,\ 19,\ 20,\ 21
\]

As proton number increases, the effective nuclear attraction on the shared electron arrangement increases.

The electron cloud contracts, so ionic radius decreases.

**Step 4**: Write the final ranking.

\[
\boxed{
\mathrm{S}^{2-}>
\mathrm{Cl}^{-}>
\mathrm{K}^{+}>
\mathrm{Ca}^{2+}>
\mathrm{Sc}^{3+}
}
\]

This is the order from largest to smallest.

The important reasoning is that all five ions have the same electron arrangement. The increasing nuclear charge is therefore the main factor controlling their relative sizes.

## A contrasting case: When proton number gives the wrong shortcut

Here's where students sometimes take the rule too far.

Suppose you compare \(\mathrm{K}^{+}\) and \(\mathrm{Mg}^{2+}\).

Potassium has 19 protons, while magnesium has only 12. If more protons always meant a smaller ion, you might predict that \(\mathrm{K}^{+}\) must be smaller.

That prediction is wrong.

The electron configurations are:

\[
\mathrm{K}^{+}:[Ar]
\]

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

Their shell arrangements are:

| Ion | Protons | Electrons | Shell arrangement |
|---|---:|---:|---|
| \(\mathrm{K}^{+}\) | 19 | 18 | 2, 8, 8 |
| \(\mathrm{Mg}^{2+}\) | 12 | 10 | 2, 8 |

Potassium has three occupied shells, while magnesium has only two.

The outermost occupied shell of \(\mathrm{K}^{+}\) is the third shell. For \(\mathrm{Mg}^{2+}\), it is the second shell.

Consequently, \(\mathrm{K}^{+}\) is larger than \(\mathrm{Mg}^{2+}\).

[IMAGE: Side-by-side electron-shell comparison of K+ and Mg2+. Show K+ with 19 protons, 18 electrons and shell arrangement 2,8,8. Show Mg2+ with 12 protons, 10 electrons and shell arrangement 2,8. Make the additional occupied third shell in K+ visually clear. Do not suggest the ions are isoelectronic.]

The lesson is important:

**More protons means a smaller radius only when the electron arrangement is comparable, as it is in an isoelectronic series.**

When the number of electrons differs, you must consider the occupied shells and electron arrangement before comparing nuclear charge.

For a broader explanation of how losing and gaining electrons changes ionic size, see [Why Ions Change Size: How to Rank Ionic Radii](https://tutorgum.com/learn/resources/why-ions-change-size-comparing-cations-anions-and-their-atoms/).

## How ionic radii are actually estimated

There is another interesting detail behind the numbers used in chemistry textbooks.

An ion does not have a solid outer surface. Its electron cloud gradually becomes less dense as you move away from the nucleus.

So how can chemists assign an ionic radius to something without a clearly defined edge?

One important approach uses X-ray diffraction to investigate the distances between ions in crystalline solids.

### The experimental problem

Imagine a crystal made from positively and negatively charged ions arranged in a repeating lattice.

If we model the ions as approximately spherical particles, the distance between the centres of two neighbouring ions can be represented as:

\[
d=r_+ + r_-
\]

where:
- \(d\) is the centre-to-centre distance,
- \(r_+\) is the effective radius of the cation,
- \(r_-\) is the effective radius of the anion.

The difficulty is that measuring the distance between two ion centres does not automatically tell us the individual radius of each ion.

We need a consistent way to divide that distance between them.

### How X-ray diffraction helps

The experimental arrangement follows a simple sequence.

1. An X-ray source directs a beam towards a carefully prepared crystal.
2. The crystal's regularly arranged particles scatter the X-rays.
3. A detector records the resulting diffraction pattern.
4. Scientists analyse the pattern to determine the crystal structure and the distances between neighbouring ions.

[IMAGE: Scientific illustration of an X-ray diffraction experiment. Show an X-ray source on the left directing a beam towards a mounted ionic crystal in the centre, with diffracted beams travelling towards a detector on the right. Include an enlarged inset of the crystal lattice showing alternating positive and negative ions, their centres, and the measured nearest-neighbour distance d. Show the relationship d = r+ + r- as an approximate spherical-ion model.]

What should the measurements show if the spherical-ion model is useful?

The measured distances should be approximately consistent with the sums of the assigned ionic radii.

Early crystallographic work helped establish this approach. William Lawrence Bragg's work on crystal structures contributed to the development of ionic-radius estimates, and later compilations refined the values.

A particularly influential development was the work of Shannon and Prewitt, followed by R. D. Shannon's revised effective ionic radii in 1976.

These compilations account for factors such as oxidation state and coordination number. Modern crystallographic research continues to refine how ion sizes are assigned to different chemical environments.

For further reading, see [Shannon's 1976 paper](https://doi.org/10.1107/S0567739476001551) and the more recent [Hawthorne and Gagné study of ion radii (2024)](https://doi.org/10.1107/S2052520624005080).

### Worked example: Estimate ionic radii from diffraction measurements

Two hypothetical, isostructural crystals contain cations \(M^+\) and \(N^{2+}\). Both cations are isoelectronic, each containing 18 electrons.

The crystals have the same anion, \(X^-\), and the cations have the same coordination number.

For this illustrative calculation, assume the effective radius of \(X^-\) is 181 pm.

X-ray diffraction gives the following average distances:

| Measurement | Crystal containing \(M^+\) | Crystal containing \(N^{2+}\) |
|---|---:|---:|
| Cation-anion distance | 319 pm | 281 pm |
| Anion radius | 181 pm | 181 pm |

Estimate both cation radii and determine which cation is smaller.

*All numerical data are illustrative, not measurements from actual crystals.*

**Step 1**: Start with the approximate radius-addition relationship.

\[
d=r_+ + r_-
\]

Rearrange to find the cation radius:

\[
r_+=d-r_-
\]

**Step 2**: Calculate the radius of \(M^+\).

\[
\begin{aligned}
r_{M^+}&=319-181\\
&=138\text{ pm}
\end{aligned}
\]

**Step 3**: Calculate the radius of \(N^{2+}\).

\[
\begin{aligned}
r_{N^{2+}}&=281-181\\
&=100\text{ pm}
\end{aligned}
\]

**Step 4**: Compare the results.

\[
138-100=38\text{ pm}
\]

The effective radius of \(N^{2+}\) is 38 pm smaller in this illustrative comparison.

This agrees with the isoelectronic trend. Since both ions have 18 electrons, the ion with more protons should have the smaller radius.

In this example, \(M^+\) has 19 protons, while \(N^{2+}\) has 20.

The calculation also shows why the experimental conditions matter. We used the same anion, coordination number, and radius convention, allowing a meaningful comparison within the simplified model.

## When the isoelectronic rule needs care

The ranking rule is powerful, but there are two important limitations.

### 1. Ionic radii are not fixed hard-sphere measurements

The effective radius assigned to an ion can depend on its surroundings.

For example, the coordination number describes how many neighbouring ions surround a particular ion in a crystal.

An ion surrounded by four neighbours may have a different effective radius from the same ion surrounded by six neighbours.

Other factors, including electronic spin state, covalent character, and crystal structure, can also affect measured interatomic distances.

This does not invalidate the isoelectronic trend. It means that experimental ionic-radius values should be compared using consistent conditions.

For typical HSC ranking questions, use the qualitative electron-cloud model unless the question supplies experimental radius data or additional environmental information.

### 2. Isoelectronic means the same electron count, not necessarily identical chemistry

For the common main-group ions used in HSC questions, an isoelectronic series generally has the same ground-state electron configuration.

However, the term *isoelectronic* strictly means that the species have the same number of electrons. It does not, by itself, guarantee identical detailed orbital occupations in every possible atomic or ionic species.

More advanced transition-metal examples can require checking the full electron configuration, including subshell occupation.

Don't assume that counting electrons is the end of the problem if the question gives you unusual transition-metal species.

## A reliable exam decision process

When you see a question asking you to rank isoelectronic ions, work through this sequence.

| Stage | What to do | What it tells you |
|---|---|---|
| 1. Electron count | Calculate \(N_e=Z-q\) | Whether the species are isoelectronic |
| 2. Electron arrangement | Check the configurations and occupied shells | Whether the same arrangement is being compared |
| 3. Nuclear charge | Compare proton numbers | Which nucleus attracts the electrons more strongly |
| 4. Shielding | Consider whether the electron arrangements are comparable | Whether the extra nuclear attraction is offset |
| 5. Ranking | Put fewer protons first for largest-to-smallest order | The final ionic-radius sequence |

A strong HSC explanation should connect all three ideas:

- The ions have the same electron arrangement.
- Their nuclear charges increase.
- The stronger effective nuclear attraction contracts the electron cloud, decreasing ionic radius.

Simply writing that the ion with more protons is smaller may earn credit for the trend, but explaining why makes your reasoning much stronger.

## Questions and solutions

### Question 1

Rank the following ions from largest to smallest ionic radius:

\[
\mathrm{S}^{2-},\quad
\mathrm{Cl}^{-},\quad
\mathrm{K}^{+},\quad
\mathrm{Ca}^{2+}
\]

Show that the ions are isoelectronic and explain your ranking.

### Solution 1

The correct order from largest to smallest is:

\[
\boxed{
\mathrm{S}^{2-}>
\mathrm{Cl}^{-}>
\mathrm{K}^{+}>
\mathrm{Ca}^{2+}
}
\]

First, calculate the electron counts:

- \(\mathrm{S}^{2-}\): \(16+2=18\)
- \(\mathrm{Cl}^{-}\): \(17+1=18\)
- \(\mathrm{K}^{+}\): \(19-1=18\)
- \(\mathrm{Ca}^{2+}\): \(20-2=18\)

All four ions have 18 electrons and the same electron configuration, \([Ar]\).

Their proton numbers are 16, 17, 19, and 20 respectively.

As proton number increases, nuclear attraction increases. Since the electron arrangement remains the same, shielding is broadly similar, and the additional nuclear charge pulls the electron cloud closer.

Therefore, \(\mathrm{S}^{2-}\) is the largest, while \(\mathrm{Ca}^{2+}\) is the smallest.

### Question 2

A student is comparing the following 18-electron ions:

\[
\mathrm{S}^{2-},\quad
\mathrm{Cl}^{-},\quad
\mathrm{Ca}^{2+},\quad
\mathrm{Sc}^{3+}
\]

The student makes two statements:

- \(\mathrm{S}^{2-}\) is the largest because it has the greatest negative charge.
- \(\mathrm{Sc}^{3+}\) is the smallest because it has the greatest nuclear charge.

Evaluate both statements. Explain the role of nuclear charge, shielding, and electron arrangement in determining the actual trend.

### Solution 2

Both statements identify the correct largest and smallest ions, but the reasoning needs to be more precise.

The correct order is:

\[
\boxed{
\mathrm{S}^{2-}>
\mathrm{Cl}^{-}>
\mathrm{Ca}^{2+}>
\mathrm{Sc}^{3+}
}
\]

All four ions contain 18 electrons, so they have the same ground-state electron configuration:

\[
[Ar]
\]

Their proton numbers are:

| Ion | Protons |
|---|---:|
| \(\mathrm{S}^{2-}\) | 16 |
| \(\mathrm{Cl}^{-}\) | 17 |
| \(\mathrm{Ca}^{2+}\) | 20 |
| \(\mathrm{Sc}^{3+}\) | 21 |

The explanation for the trend is increasing nuclear charge.

The inner-shell electrons shield the outer electrons from some of the nuclear attraction. However, the number and arrangement of electrons remain the same across this series.

Therefore, the increasing proton number produces a greater effective nuclear attraction, contracting the electron cloud.

The negative charge on \(\mathrm{S}^{2-}\) is not an independent reason for its larger radius. It reflects the fact that sulfur has gained electrons while retaining a relatively low proton number.

Similarly, the positive charge on \(\mathrm{Sc}^{3+}\) is not the fundamental explanation. Its nucleus has the greatest number of protons attracting the same 18-electron arrangement.

This distinction matters: charge helps you calculate the electron count, but nuclear charge explains the size trend once the species are confirmed to be isoelectronic.

### Question 3

Consider the following three ions:

\[
\mathrm{K}^{+},\quad
\mathrm{Ca}^{2+},\quad
\mathrm{Mg}^{2+}
\]

A student argues that \(\mathrm{Ca}^{2+}\) must be the smallest because it has a greater positive charge than \(\mathrm{K}^{+}\).

Another student argues that \(\mathrm{K}^{+}\) must be the smallest because potassium has the greatest proton number.

Determine the correct order from largest to smallest. Explain why both students have used incomplete reasoning.

### Solution 3

The correct order from largest to smallest is:

\[
\boxed{
\mathrm{K}^{+}>
\mathrm{Ca}^{2+}>
\mathrm{Mg}^{2+}
}
\]

Start by calculating the electron counts:

\[
\begin{aligned}
\mathrm{K}^{+}:&\quad 19-1=18\\
\mathrm{Ca}^{2+}:&\quad 20-2=18\\
\mathrm{Mg}^{2+}:&\quad 12-2=10
\end{aligned}
\]

The first two ions are isoelectronic.

Both have the configuration \([Ar]\), but calcium has 20 protons compared with potassium's 19.

Therefore, \(\mathrm{Ca}^{2+}\) has a stronger effective nuclear attraction and a smaller radius than \(\mathrm{K}^{+}\).

Magnesium is different.

Its electron configuration is \([Ne]\), meaning it has only two occupied shells, compared with the three occupied shells in the other two ions.

The outermost occupied shell is closer to the nucleus, producing a much smaller ion.

Therefore, \(\mathrm{Mg}^{2+}\) is the smallest of the three.

The first student has incorrectly used ionic charge as a universal size rule. The second has incorrectly used proton number without checking electron arrangement.

The correct approach is to compare proton numbers only within the isoelectronic pair, then use shell structure to compare magnesium with the other ions.

### Question 4

Two hypothetical ionic crystals contain the cations \(M^+\) and \(N^{2+}\).

Both cations have 18 electrons and occupy equivalent six-coordinate sites in their respective crystal structures. The same anion, \(X^-\), is present in both crystals.

X-ray diffraction produces the following illustrative measurements:

| Measurement | Crystal M | Crystal N |
|---|---:|---:|
| Mean cation-anion distance | 305 pm | 292 pm |
| Effective anion radius | 165 pm | 165 pm |
| Cation coordination number | 6 | 6 |

Calculate the effective radius of each cation and explain whether the results agree with the isoelectronic trend.

[IMAGE: Two side-by-side simplified ionic crystal lattice sections, each showing a central cation surrounded by six neighbouring anions in an octahedral arrangement. Label the first central ion M+ with an M-X centre-to-centre distance of 305 pm, and the second N2+ with an N-X distance of 292 pm. Show identical anion types, equivalent coordination arrangements and the supplied anion radius of 165 pm. Do not include the calculated cation radii.]

### Solution 4

The effective radius of \(N^{2+}\) is smaller than that of \(M^+\), and the measurements agree with the isoelectronic trend.

Use the approximate radius-addition relationship:

\[
d=r_+ + r_-
\]

Rearranging:

\[
r_+=d-r_-
\]

**Step 1**: Calculate the effective radius of \(M^+\).

\[
\begin{aligned}
r_{M^+}&=305-165\\
&=140\text{ pm}
\end{aligned}
\]

**Step 2**: Calculate the effective radius of \(N^{2+}\).

\[
\begin{aligned}
r_{N^{2+}}&=292-165\\
&=127\text{ pm}
\end{aligned}
\]

**Step 3**: Compare the results.

\[
140-127=13\text{ pm}
\]

The effective radius of \(N^{2+}\) is 13 pm smaller.

Both cations have 18 electrons, so they are isoelectronic.

Since \(M^+\) has 19 protons and \(N^{2+}\) has 20, the nucleus of \(N^{2+}\) attracts the shared electron arrangement more strongly.

The smaller calculated radius is therefore consistent with the increase in nuclear charge.

The comparison is meaningful within the assumptions because the same anion and coordination number have been used.

### Question 5

Two researchers investigate the relative sizes of two isoelectronic cations, \(A^+\) and \(B^{2+}\).

Both ions have 18 electrons.

Their proton numbers are:

- \(A^+\): 19 protons
- \(B^{2+}\): 20 protons

The researchers use X-ray diffraction to examine two different crystal structures.

Their illustrative results are:

| Measurement | Crystal A | Crystal B |
|---|---:|---:|
| Cation | \(A^+\) | \(B^{2+}\) |
| Mean cation-anion distance | 276 pm | 282 pm |
| Cation coordination number | 6 | 8 |
| Anion | \(X^-\) | \(Y^-\) |
| Supplied effective anion radius | 130 pm | 150 pm |

All radii and distances are illustrative and use the same general effective-radius convention, with the supplied anion values applying to their respective local environments.

A student observes that the cation-anion distance is greater in Crystal B and concludes:

\[
r_{B^{2+}}>r_{A^+}
\]

The student then argues that this disproves the rule that more protons produce smaller isoelectronic ions.

Evaluate the student's conclusion.

Calculate the effective cation radii using the supplied data, and explain why the measured distances cannot be interpreted as direct measurements of isolated-ion size. State what experimental conditions would make the comparison more reliable.

[IMAGE: Two contrasting ionic crystal structures side by side. Crystal A should show a central A+ ion surrounded by six X- neighbours, with a centre-to-centre distance labelled 276 pm. Crystal B should show a central B2+ ion surrounded by eight Y- neighbours, with a centre-to-centre distance labelled 282 pm. Label the supplied effective anion radii as 130 pm for X- and 150 pm for Y-. Clearly show the different coordination environments without drawing or labelling inferred cation radii.]

### Solution 5

The student's conclusion is incorrect. The greater measured cation-anion distance in Crystal B does not demonstrate that \(B^{2+}\) is larger.

In fact, the supplied radius model gives a smaller effective radius for \(B^{2+}\), although the different coordination environments mean that the two calculated values are not a strict like-for-like comparison.

There are several steps to resolving this.

**Step 1**: Establish the expected isolated-ion trend.

Both ions contain 18 electrons.

However:

- \(A^+\) has 19 protons.
- \(B^{2+}\) has 20 protons.

Their electron arrangements are the same in the simple ground-state model.

Therefore, \(B^{2+}\) experiences stronger effective nuclear attraction and is expected to have the smaller isolated-ion radius.

**Step 2**: Calculate the effective radius of \(A^+\).

Use:

\[
r_+=d-r_-
\]

Substitute the Crystal A measurements:

\[
\begin{aligned}
r_{A^+}&=276-130\\
&=146\text{ pm}
\end{aligned}
\]

**Step 3**: Calculate the effective radius of \(B^{2+}\).

\[
\begin{aligned}
r_{B^{2+}}&=282-150\\
&=132\text{ pm}
\end{aligned}
\]

The supplied radius model therefore gives:

\[
146-132=14\text{ pm}
\]

The effective radius calculated for \(B^{2+}\) is 14 pm smaller under its stated local-environment convention.

**Step 4**: Explain why Crystal B has the longer measured distance.

The centre-to-centre distance depends on both ions:

\[
d=r_+ + r_-
\]

Crystal B has a larger supplied anion radius:

\[
150-130=20\text{ pm}
\]

Its larger anion contributes to the longer measured distance.

The coordination environments also differ. Crystal A has six-coordinate cations, whereas Crystal B has eight-coordinate cations.

Coordination number can affect the effective radii assigned to ions in a crystal. The resulting values describe the local structural environment rather than an absolute, universal boundary around an isolated ion.

**Step 5**: Evaluate the experimental comparison.

The subtraction gives useful effective-radius estimates within the supplied model, but the two coordination environments are not equivalent.

The measured distances alone cannot establish a universal ranking of isolated-ion radii.

For a stronger experimental comparison, researchers would seek:
- the same anion,
- equivalent coordination numbers,
- comparable crystal structures,
- consistent methods for assigning effective ionic radii.

The isoelectronic electron-cloud model still predicts that \(B^{2+}\) is smaller because it has one additional proton attracting the same number of electrons.

The central mistake was treating a measured cation-anion distance as though it were the radius of the cation alone.

This is the deeper distinction: **an isoelectronic trend describes how nuclear charge affects an electron cloud, while an experimentally assigned crystal radius also reflects the ion's chemical environment.**

## Where to take this next

Once you can rank isoelectronic ions, you can use the same reasoning to understand why ions of different charges and sizes form different crystal arrangements.

The next useful connection is lattice energy: ionic charge and the distance between neighbouring ions both influence the strength of electrostatic attraction in an ionic solid.

That takes you from predicting which ion is smaller to explaining how ionic size affects the properties of the compounds they form.