Why Ions Change Size: How to Rank Ionic Radii

You look at the periodic table, see fluorine sitting above and to the right of sodium, and predict that \(\mathrm{F}^{-}\) must be smaller than \(\mathrm{Na}^{+}\). That sounds reasonable if you remember that atomic radii generally decrease towards the top right of the periodic table. But this is exactly the kind of shortcut that can cost you marks in an HSC Chemistry response.

Before you commit to that prediction, consider what the nuclei are actually pulling on. Are you comparing neutral atoms, or have the electrons changed?

Fluorine has 9 protons, and \(\mathrm{F}^{-}\) has gained one electron, giving it 10 electrons. Sodium has 11 protons, and \(\mathrm{Na}^{+}\) has lost one electron, also leaving it with 10 electrons. Both ions have the electron configuration \(1s^2\,2s^2\,2p^6\).

The sodium nucleus has more protons pulling on the same number of electrons. Its electron cloud is therefore held more tightly, making \(\mathrm{Na}^{+}\) smaller than \(\mathrm{F}^{-}\).

The problem wasn’t that you forgot a periodic trend. It was that you applied a trend for neutral atoms without checking how ion formation changed the electron arrangement.

Here’s how to fix that mistake reliably.

01Why losing or gaining electrons changes an ion’s size

When an atom forms an ordinary ion, its number of protons stays the same. It gains or loses electrons.

That changes two things you need to track:

• The number of occupied electron shells.
• The balance between attraction from the nucleus and repulsion between electrons.

These effects explain why ions can have very different sizes from their parent atoms.

When an atom loses electrons: cations

Consider a neutral sodium atom, \(\ce{Na}\).

It has 11 protons and 11 electrons, with the electron configuration

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

The single electron in the third shell is its outermost electron.

When sodium loses that electron, it forms \(\mathrm{Na}^{+}\). Its configuration becomes

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

The third shell is now empty. The outermost occupied shell changes from \(n=3\) to \(n=2\), so the electron cloud contracts substantially.

There is another effect. With fewer electrons, electron-electron repulsion is reduced. The remaining electrons also experience a stronger effective attraction to the nucleus because there is less electron shielding and repulsion.

The result is a smaller ion.

For a given element, losing electrons to form a cation generally decreases its radius. If the highest occupied shell becomes empty, the contraction is especially significant.

When an atom gains electrons: anions

Now consider chlorine.

A neutral chlorine atom has 17 protons and 17 electrons:

\[ 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \]

When it gains an electron, it forms \(\mathrm{Cl}^{-}\), with 18 electrons:

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

Notice something important: the additional electron enters the third shell. It does not create a fourth shell.

The nucleus still has 17 protons, but there is now an additional electron sharing the outer shell. Electron-electron repulsion increases, and the electron cloud expands.

Therefore, \(\mathrm{Cl}^{-}\) is larger than neutral \(\ce{Cl}\).

Think of electrons in the same shell as people sharing a crowded lift. Adding another passenger increases the crowding; removing one eases it. This helps picture the effect of electron-electron repulsion, although electrons are quantum distributions, not tiny people following fixed paths.

The key distinction is that gaining an electron generally expands an occupied shell, while losing the only electron in the outermost shell can remove that shell from the configuration altogether.

02The neutral-atom trend is still useful

Your original shortcut wasn’t completely unreasonable. It came from a real trend, but that trend applies to a particular comparison.

Across Period 3, neutral atoms generally become smaller from sodium to magnesium to aluminium.

Why?

All three atoms have electrons occupying the third shell. As you move across the period, the number of protons increases, while the additional electrons enter the same principal shell. The increase in nuclear attraction is not fully cancelled by the additional shielding.

The electron cloud is pulled closer to the nucleus.

This is why a neutral-atom trend can be a useful starting point. The mistake is treating it as a universal rule that automatically ranks ions, too.

03A dependable method for ranking ionic radii

When an exam gives you several ions, don’t start by looking for a trend arrow. Start by counting electrons.

Step 1: Count the electrons in each species

The atomic number, \(Z\), is the number of protons in the nucleus. For a neutral atom, the number of electrons equals \(Z\).

For an ion, use

\[ N_e=Z-q \]

where \(N_e\) is the number of electrons and \(q\) is the signed ionic charge, treated as a number such as \(+1\), \(+2\), or \(-1\).

For example:

• \(\mathrm{Na}^{+}\): \(N_e=11-(+1)=10\).
• \(\mathrm{F}^{-}\): \(N_e=9-(-1)=10\).
• \(\mathrm{Mg}^{2+}\): \(N_e=12-(+2)=10\).
• \(\mathrm{O}^{2-}\): \(N_e=8-(-2)=10\).

These calculations reveal a useful pattern: all four species have 10 electrons.

Step 2: Check the occupied shells

Write the electron configuration if you’re unsure.

For example, a species with configuration \(1s^2\,2s^2\,2p^6\) has two occupied shells. A species with configuration \(1s^2\,2s^2\,2p^6\,3s^1\) has three.

A species with an additional occupied outer shell will generally be larger than a comparable species whose outermost occupied shell is lower.

Don’t confuse the number of electrons with the number of occupied shells. Gaining an electron does not automatically create a new shell.

Step 3: Check whether the species are isoelectronic

Isoelectronic species have the same number of electrons.

For these species, the electron arrangements are the same in the usual ground-state shell model. The key difference is the number of protons.

Among isoelectronic species, more protons means a smaller radius.

The extra positive charge in the nucleus attracts the same electron arrangement more strongly, contracting the cloud.

This is one of the most dependable rules for HSC ionic-radius questions.

Step 4: If the electron counts differ, compare the shell structures

You cannot automatically use proton number alone when the electron arrangements differ.

First check the highest occupied shell. Then consider how the number of electrons and the nuclear attraction affect the cloud.

For example, removing an electron may leave the outer shell occupied, or it may empty that shell completely. Those are different structural changes.

Step 5: Explain the physical reason, not just the trend

A strong exam response identifies the relevant electron change and connects it to size.

05Shortcuts that fail in exams

“More protons always means a smaller ion.”

This is reliable for an isoelectronic comparison, where the electron arrangement is the same. It is not a universal rule for comparing species with different occupied shells.

A higher nuclear charge can pull electrons in more strongly, but an additional occupied outer shell can make another species larger despite its lower proton count.

“Every cation is smaller than every anion.”

The useful rule is narrower: a cation is generally smaller than its own neutral atom, and an anion is generally larger than its own neutral atom.

That does not mean every cation is smaller than every anion. For example, \(\mathrm{K}^{+}\) has electrons occupying three shells, whereas \(\mathrm{F}^{-}\) has electrons occupying only two. The shell difference matters when comparing these different elements.

“A more positive charge always means a smaller ion.”

A more highly charged cation of the same element is generally smaller because electrons have been removed and repulsion has decreased. Removing the last electron from the outer shell can produce a particularly large contraction.

Across different elements, however, charge alone is insufficient. Check the electron configuration before making the comparison.

“Gaining an electron creates another shell.”

Not necessarily. The added electron can enter an orbital in a shell that is already occupied, as happens when \(\ce{Cl}\) becomes \(\mathrm{Cl}^{-}\).

If the highest occupied shell remains the same, the change in size comes mainly from the balance between nuclear attraction and electron-electron repulsion, rather than from adding a whole new shell.

06Before you submit your answer

Use this quick self-check whenever a question asks you to rank ionic radii.

• [ ] Have I counted the electrons correctly using the ion’s charge?
• [ ] Have I identified the highest occupied shell?
• [ ] If the species are isoelectronic, have I compared their proton counts?
• [ ] If their electron arrangements differ, have I considered shell structure before using proton number?
• [ ] Have I explained the attraction or repulsion responsible for the size difference, rather than just stating a trend?

If you can justify every comparison in the ranking, you’re much less likely to fall for a memorised shortcut.

07Questions and solutions

Question 1

A chlorine atom gains one electron to form \(\mathrm{Cl}^{-}\). Which species has the larger radius, \(\ce{Cl}\) or \(\mathrm{Cl}^{-}\)? State their electron counts and explain whether the additional electron occupies a new shell.

Solution 1

\(\mathrm{Cl}^{-}\) is larger than neutral \(\ce{Cl}\). Chlorine has 17 protons and 17 electrons, with the configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^5\). The ion has 18 electrons and the configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\).

The added electron enters the third shell, so no fourth shell is created. The nuclear charge remains unchanged, while electron-electron repulsion increases. This causes the electron cloud to expand, making the anion larger than its parent atom.

Question 2

Rank the following species from largest to smallest radius:

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

Use the standard qualitative radius model for these main-group species. Explain why the neutral aluminium atom is still larger than the sodium ion.

Solution 2

The order from largest to smallest is

\[ \boxed{ \ce{Na}>\ce{Mg}>\ce{Al}> \mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} } \]

The neutral atoms all have occupied electrons in the third shell. Moving from Na to Mg to Al increases the proton count, and the additional electrons enter the same principal shell. The increasing nuclear attraction contracts the electron cloud, so the neutral-atom radii decrease across the sequence.

Now consider the ions. Each loses its outer-shell electrons and ends with 10 electrons:

• \(\mathrm{Na}^{+}\): 11 protons, 10 electrons.
• \(\mathrm{Mg}^{2+}\): 12 protons, 10 electrons.
• \(\mathrm{Al}^{3+}\): 13 protons, 10 electrons.

These ions are isoelectronic. Their radii decrease as the number of protons increases, giving \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\).

The neutral atoms still have an occupied third shell, while the cations have only two occupied shells. In this qualitative comparison, that places even neutral \(\ce{Al}\) above \(\mathrm{Na}^{+}\) in size.

The trap is to apply only one trend to all six species. The correct ranking combines the neutral-atom trend, the isoelectronic trend, and the change in occupied shells during ion formation.

Question 3

Magnesium undergoes successive ionisation to form \(\mathrm{Mg}^{+}\) and then \(\mathrm{Mg}^{2+}\).

The electron configurations are:

• \(\ce{Mg}\): \(1s^2\,2s^2\,2p^6\,3s^2\)
• \(\mathrm{Mg}^{+}\): \(1s^2\,2s^2\,2p^6\,3s^1\)
• \(\mathrm{Mg}^{2+}\): \(1s^2\,2s^2\,2p^6\)

Rank the three species from largest to smallest. Explain why the two successive removals should not be assumed to produce equal decreases in radius, and identify which removal empties an occupied shell.

Solution 3

The order is

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

Both ionisation steps remove one electron, but their structural effects are different.

When \(\ce{Mg}\) becomes \(\mathrm{Mg}^{+}\), one electron is removed from the third shell. That shell remains occupied because one \(3s\) electron is left. The reduction in electron-electron repulsion allows the remaining electrons to be held more tightly, so the radius decreases.

When \(\mathrm{Mg}^{+}\) becomes \(\mathrm{Mg}^{2+}\), the last \(3s\) electron is removed. The third shell becomes empty, and the highest occupied shell changes from \(n=3\) to \(n=2\). This gives an additional structural reason for a pronounced contraction.

The two decreases in radius should not be treated as equal. Removing one electron does not guarantee the same change in size each time. The shell configuration matters, and the configurations alone do not provide exact numerical radius changes.

Question 4

Rank \(\mathrm{K}^{+}\), \(\mathrm{F}^{-}\), and \(\mathrm{Na}^{+}\) from largest to smallest radius.

A student argues that \(\mathrm{K}^{+}\) must be the smallest because it is a cation, while \(\mathrm{F}^{-}\) must be the largest because it is an anion. Assess both claims using electron counts and occupied shells.

Solution 4

The order from largest to smallest is

\[ \boxed{\mathrm{K}^{+}>\mathrm{F}^{-}>\mathrm{Na}^{+}} \]

The student’s first claim is incorrect because a cation is not necessarily smaller than an anion from a different element.

Potassium has 19 protons. The ion \(\mathrm{K}^{+}\) has 18 electrons, with the configuration

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

Its highest occupied shell is the third shell.

By contrast, \(\mathrm{F}^{-}\) has 10 electrons and the configuration \(1s^2\,2s^2\,2p^6\). It has only two occupied shells. The additional occupied shell in \(\mathrm{K}^{+}\) makes it larger than \(\mathrm{F}^{-}\) in the standard qualitative shell model.

The second comparison is different. Both \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\) have 10 electrons and the same electron configuration. Fluoride has 9 protons, while sodium has 11. The greater nuclear charge in \(\mathrm{Na}^{+}\) pulls the shared electron arrangement in more strongly, making \(\mathrm{Na}^{+}\) smaller.

The key is to use shell structure when the electron arrangements differ, and proton number when the species are isoelectronic.

Question 5

Consider four isolated species: \(\ce{Mg}\), \(\mathrm{Mg}^{+}\), \(\mathrm{Na}^{+}\), and \(\mathrm{Mg}^{2+}\).

A student applies the rule “more protons means a smaller radius” to every comparison. They predict that \(\mathrm{Mg}^{+}\) must be smaller than \(\mathrm{Na}^{+}\), while also predicting that \(\mathrm{Mg}^{2+}\) is smaller than \(\mathrm{Na}^{+}\).

Rank all four species from largest to smallest. Explain why one prediction fails while the other works. Use electron configurations to identify the shell effect that changes the result.

Solution 5

The correct order from largest to smallest is

\[ \boxed{ \ce{Mg}>\mathrm{Mg}^{+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+} } \]

The rule about proton number works for \(\mathrm{Na}^{+}\) and \(\mathrm{Mg}^{2+}\), but it fails for \(\mathrm{Mg}^{+}\) and \(\mathrm{Na}^{+}\). The reason is that the first pair is isoelectronic, while the second pair has different electron arrangements.

Start with magnesium and its ions:

• \(\ce{Mg}\) has 12 protons and 12 electrons, with configuration \([Ne]3s^2\).
• \(\mathrm{Mg}^{+}\) has 12 protons and 11 electrons, with configuration \([Ne]3s^1\).
• \(\mathrm{Mg}^{2+}\) has 12 protons and 10 electrons, with configuration \([Ne]\).

Here, \([Ne]\) represents the electron configuration \(1s^2\,2s^2\,2p^6\).

Removing the first electron contracts the electron cloud because electron-electron repulsion decreases and the remaining electrons are held more tightly. Removing the second electron empties the third shell altogether, producing a further contraction. Therefore, \(\ce{Mg}>\mathrm{Mg}^{+}>\mathrm{Mg}^{2+}\).

Now compare \(\mathrm{Mg}^{+}\) with \(\mathrm{Na}^{+}\).

The tempting argument is that magnesium has 12 protons while sodium has only 11, so \(\mathrm{Mg}^{+}\) should be smaller. But these species do not have the same electron arrangement.

The \(\mathrm{Mg}^{+}\) ion still has an electron in the third shell, \(3s^1\). The \(\mathrm{Na}^{+}\) ion has the configuration \([Ne]\), so its highest occupied shell is the second. The additional occupied shell makes \(\mathrm{Mg}^{+}\) larger despite its greater nuclear charge.

Finally, compare \(\mathrm{Na}^{+}\) and \(\mathrm{Mg}^{2+}\). Both have 10 electrons and the same configuration, \([Ne]\). Magnesium has 12 protons compared with sodium’s 11, so the stronger nuclear attraction makes \(\mathrm{Mg}^{2+}\) smaller.

The exact constraint that changes the result is the electron arrangement. Proton count is a dependable comparison for isoelectronic species, but it cannot be used on its own when one species retains an occupied shell that the other has lost.

The next useful step is to connect ionic size to electrostatic attraction in ionic compounds. For a given pair of charges, bringing the ion centres closer generally strengthens their attraction. Radius is one part of that explanation; ionic charge and the arrangement of ions in the crystal also matter.