Why Cations Are Usually Smaller Than Their Neutral Atoms

A sodium atom has 11 protons and 11 electrons. When it loses one electron to form \(\mathrm{Na}^{+}\), its nucleus still contains 11 protons, but the ion is noticeably smaller than the original atom.

Why should removing a negatively charged particle make the whole atom smaller?

The answer comes down to three interacting changes: the number of electrons decreases, electron-electron repulsion is reduced, and the unchanged nucleus attracts the remaining electron cloud more effectively. In many cases, the atom also loses its entire outer electron shell.

Understanding these changes explains why cations are generally smaller than their neutral atoms, why some ions contract more than others, and why simply counting the positive charge is not enough to predict ionic size.

01Why losing electrons makes an atom smaller

Start with sodium.

A neutral sodium atom has the electron configuration:

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

Its shell arrangement is \(2,8,1\). The single electron in the third shell is the outermost electron.

When sodium forms a positive ion, it loses that electron:

\[ \ce{Na -> Na+ + e-} \]

The resulting sodium ion has the electron configuration:

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

Its shell arrangement is now \(2,8\).

Notice what has changed:

The sodium nucleus has not changed. It still contains 11 protons, so its nuclear charge remains \(+11\).

However, the third electron shell has disappeared. The remaining electrons occupy a smaller region around the nucleus.

There are three connected reasons for this contraction.

1. The outer electron shell can disappear

This is often the biggest change when a main-group metal forms a cation.

Think about the sodium atom. Its outermost electron occupies the third principal energy level, which extends further from the nucleus than the first and second levels.

When that electron is removed, the third occupied shell is gone. The outer boundary of the electron cloud is now associated with the second shell.

The same pattern occurs when magnesium forms a magnesium ion:

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

Magnesium has 12 protons and the electron arrangement \(2,8,2\). Its \(2+\) ion has 12 protons and the arrangement \(2,8\).

Both electrons in the third shell are removed, leaving the ion with only two occupied shells.

This explains why losing an electron can produce a much larger change in size than you might initially expect. The electron cloud does not simply lose one tiny particle at its edge. Its entire outermost occupied shell may disappear.

2. Electron-electron repulsion decreases

Electrons are all negatively charged. They repel one another.

In a neutral atom, every electron contributes to the overall pattern of repulsion within the electron cloud. When an atom loses electrons, there are fewer electrons repelling the remaining ones.

This allows the remaining electron cloud to contract towards the nucleus.

A useful way to picture this is a crowded group of people trying to occupy a space. Remove some of the crowd, and the remaining people have more room to move closer together.

The analogy has limits: electrons are not people with fixed positions, and their arrangement is governed by quantum mechanics. However, it captures the idea that removing electrons reduces mutual repulsion.

3. The remaining electrons experience a stronger effective nuclear attraction

The nucleus contains positively charged protons, which attract negatively charged electrons.

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

This effect is called electron shielding.

For example, sodium’s outermost \(3s\) electron experiences attraction from the \(+11\) nucleus, but the other ten electrons partially shield it.

When sodium loses that outer electron, the nuclear charge remains unchanged while the electron arrangement becomes smaller and the remaining electrons experience less opposition from electron-electron interactions.

The net attraction experienced by the remaining electron cloud becomes stronger, causing it to contract.

This net attraction is described using effective nuclear charge, often written as \(Z_{\mathrm{eff}}\).

You do not need to calculate an exact effective nuclear charge to explain this HSC concept. The important reasoning is:

• Nuclear charge stays the same because no protons are removed.
• The number of electrons decreases.
• Electron-electron repulsion and shielding contributions change.
• The remaining electron cloud is pulled closer to the nucleus.

02Why the contraction is not always equally large

So far, sodium and magnesium have given us a particularly clear example: losing their outer electrons removes an entire occupied shell.

But does every cation lose its outermost shell?

No. That is where the explanation needs another layer of precision.

The amount of contraction depends partly on which electrons are removed and how the remaining electrons are arranged.

When an entire shell is removed

For many main-group metals, forming a stable cation involves removing all the electrons from the outermost occupied shell.

Consider these examples:

In each case, the ion has one fewer occupied electron shell than its neutral atom.

The reduction in size is therefore substantial.

When electrons are removed without eliminating another shell

Transition metals provide a useful contrast.

Iron has the neutral electron configuration:

\[ \mathrm{Fe}:\ [Ar]\,3d^6\,4s^2 \]

When it forms an iron(II) ion, the two \(4s\) electrons are removed:

\[ \mathrm{Fe}^{2+}:\ [Ar]\,3d^6 \]

Now consider the formation of iron(III):

\[ \mathrm{Fe}^{3+}:\ [Ar]\,3d^5 \]

The additional electron is removed from the \(3d\) subshell. No further principal electron shell disappears.

Yet \(\mathrm{Fe}^{3+}\) is smaller than \(\mathrm{Fe}^{2+}\).

Why?

Both ions have the same 26-proton nucleus, but \(\mathrm{Fe}^{3+}\) has one fewer electron. Its remaining electrons experience a stronger effective nuclear attraction, and the electron cloud contracts further.

This is an important qualification: losing an entire shell is a strong reason for contraction, but it is not a requirement. A cation can become smaller even when electron removal occurs within a shell that remains occupied.

03How do scientists actually measure ionic size?

There is a measurement problem hiding in this topic.

An atom does not have a solid outer surface. Its electrons occupy regions of probability, and the electron cloud gradually becomes less dense as you move away from the nucleus.

So where exactly does an atom end?

Chemists cannot simply place a ruler against an atom and measure its radius. Instead, they use experimental evidence from the distances between atoms and ions in crystals.

X-ray diffraction: measuring distances inside a crystal

The development of X-ray crystallography gave scientists a way to investigate the arrangement and spacing of particles inside crystalline solids.

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

Its ions form a regular three-dimensional lattice. Each \(\mathrm{Na}^{+}\) ion has six nearest-neighbour \(\mathrm{Cl}^{-}\) ions, and each \(\mathrm{Cl}^{-}\) ion has six nearest-neighbour \(\mathrm{Na}^{+}\) ions.

The experiment works as follows:

1. An X-ray source produces a beam of X-rays with a known wavelength.
2. The beam is directed towards a single crystal mounted in the apparatus.
3. The regularly arranged ions scatter the incoming X-rays.
4. The scattered waves reinforce one another at particular angles, producing diffraction peaks.
5. A detector records the diffraction pattern. Scientists use the positions of these peaks to determine the dimensions of the crystal’s repeating unit cell.

What should this experiment show?

If the ions are arranged in a regular repeating structure, the diffraction pattern should contain peaks corresponding to that regular spacing. The measured pattern allows scientists to calculate the dimensions of the unit cell and the distances between neighbouring ions.

It gives a measurable distance. It does not automatically give an individual ionic radius.

From crystal spacing to ionic radius

Look at the arrangement of ions along an edge of the cubic sodium chloride unit cell.

The ions alternate:

\[ \mathrm{Na}^{+}\;-\;\mathrm{Cl}^{-}\;-\;\mathrm{Na}^{+} \]

The distance from the first sodium ion to the chloride ion is the same as the distance from that chloride ion to the next sodium ion.

Therefore, if the unit cell edge has length \(a\), the nearest-neighbour distance \(d\) is:

\[ d=\frac{a}{2} \]

Why one measurement is not enough

This is an important limitation of the original measurement approach.

Suppose the distance between a cation and an anion is \(280\ \mathrm{pm}\).

Several combinations of ionic radii could produce that distance:

The diffraction measurement alone cannot distinguish between these possibilities.

Scientists therefore developed consistent ionic-radius scales by comparing distances across many crystal structures and using reference values and structural assumptions.

One important consideration is the coordination number. This is the number of nearest-neighbour ions surrounding a particular ion.

An ion surrounded by six neighbouring ions may have a different assigned ionic radius from the same ion surrounded by eight.

Ionic radius is therefore an effective value, not a perfectly defined physical boundary.

This is the modern interpretation: X-ray diffraction measures crystal spacing, while ionic-radius models use those measurements to assign useful, consistent sizes to ions.

04How to predict whether a cation is smaller

When you encounter a question comparing a neutral atom with its cation, use the following reasoning process.

There is one distinction worth remembering.

The positive charge of an ion is not the same thing as its nuclear charge.

For example, \(\mathrm{Mg}^{2+}\) has a net charge of \(+2\), but its nucleus still has a charge of \(+12\), because magnesium has 12 protons.

The nucleus does not become more positively charged when magnesium loses electrons. The number of protons simply stays the same while the number of electrons decreases.

What about comparisons between different elements?

Be careful here.

The rule that cations are smaller than their neutral atoms compares an ion with the neutral atom of the same element.

It does not mean that every cation is smaller than every neutral atom.

For example, \(\mathrm{K}^{+}\) can be larger than a neutral hydrogen atom, even though \(\mathrm{K}^{+}\) is smaller than neutral potassium.

If a question asks you to compare ions from different elements, you need to investigate their electron configurations and nuclear charges as well.

A particularly useful case occurs when the species have the same number of electrons. These are called isoelectronic species.

When the electron arrangements are the same, the species with more protons generally has the smaller radius because its nucleus attracts the same electron cloud more strongly.

This is a separate comparison from the neutral-atom-to-cation contraction discussed here. You can practise that method in How to Rank Isoelectronic Ions by Size: Worked Examples.

05Common misconceptions

Anions provide a useful contrast. When an atom gains electrons, electron-electron repulsion and shielding generally increase, causing its electron cloud to expand. This is why anions are usually larger than their corresponding neutral atoms.

06Questions and solutions

The following questions move from straightforward electron configurations to interpreting experimental measurements and evaluating competing explanations.

Question 1

A neutral calcium atom has 20 protons and the electron arrangement \(2,8,8,2\).

It forms a calcium ion by losing two electrons.

a) Write the electron arrangement of the calcium ion.

b) State whether the calcium ion is larger or smaller than the neutral calcium atom.

c) Explain your answer using nuclear charge, electron shielding, and occupied electron shells.

Solution 1

The calcium ion, \(\mathrm{Ca}^{2+}\), has the electron arrangement \(2,8,8\) and is smaller than the neutral calcium atom.

a) Calcium loses two electrons:

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

Its electron arrangement changes from:

\[ 2,8,8,2 \longrightarrow 2,8,8 \]

The ion has 18 electrons and 20 protons.

b) The calcium ion is smaller.

c) The neutral calcium atom has four occupied electron shells, whereas \(\mathrm{Ca}^{2+}\) has only three.

The two outermost electrons have been removed, eliminating the fourth occupied shell.

The nucleus still contains 20 protons, so its nuclear charge is unchanged. However, the remaining electrons experience less electron-electron repulsion and a stronger effective nuclear attraction.

The electron cloud contracts towards the nucleus.

The key point is that removing the outermost shell produces a substantial reduction in ionic size.

Question 2

Iron forms two common cations, \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\).

Their electron configurations are:

\[ \begin{aligned} \mathrm{Fe}&:\ [Ar]\,3d^6\,4s^2\\ \mathrm{Fe}^{2+}&:\ [Ar]\,3d^6\\ \mathrm{Fe}^{3+}&:\ [Ar]\,3d^5 \end{aligned} \]

A student argues that \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) should have approximately the same size because both have already lost their outer \(4s\) electrons.

Evaluate this argument and rank the three species from largest to smallest.

Solution 2

The student’s argument is incorrect. The correct order is:

\[ \boxed{\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}} \]

The key is that losing an additional electron can cause further contraction even when no additional principal electron shell disappears.

Step 1: Compare neutral iron with iron(II).

Neutral iron has two electrons in its \(4s\) subshell. These are removed when it forms \(\mathrm{Fe}^{2+}\).

The resulting ion has fewer electrons, reduced electron-electron repulsion, and a smaller electron cloud.

Therefore, \(\mathrm{Fe}^{2+}\) is smaller than neutral iron.

Step 2: Compare iron(II) with iron(III).

Iron(III) forms when another electron is removed from the \(3d\) subshell:

\[ \ce{Fe^2+ -> Fe^3+ + e-} \]

Both ions have 26 protons, but \(\mathrm{Fe}^{3+}\) has one fewer electron.

The reduction in electron-electron repulsion and shielding allows the remaining electron cloud to contract further.

Therefore:

\[ r(\mathrm{Fe}^{3+})<r(\mathrm{Fe}^{2+}) \]

Step 3: Evaluate the misconception.

The student assumes that losing the outermost principal shell is the only mechanism that reduces atomic size.

That is the missing piece of reasoning.

Removing an entire shell can cause a particularly large contraction, but further electron removal can still reduce the radius within the remaining shells.

Question 3

Aluminium forms three successive cations with the following electron arrangements:

A student suggests that neutral aluminium, \(\mathrm{Al}^{+}\), and \(\mathrm{Al}^{2+}\) should have approximately the same radius because all three have three occupied electron shells.

a) Explain why this prediction is incorrect.

b) Rank all four species from largest to smallest.

c) Identify the stage at which you would expect a particularly substantial contraction and explain why.

Solution 3

The student’s prediction is incorrect. The expected order from largest to smallest is:

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

Having the same number of occupied shells does not mean having the same ionic radius.

a) Consider the first two ionisation steps.

When aluminium loses its first electron, its electron arrangement changes from \(2,8,3\) to \(2,8,2\).

It still has three occupied shells, but it now has fewer electrons.

This reduces electron-electron repulsion and changes the shielding experienced by the remaining electrons. The electron cloud contracts.

The same reasoning applies when aluminium forms \(\mathrm{Al}^{2+}\).

Therefore, the radii do not remain constant just because the number of occupied shells is unchanged.

b) Each successive ion is smaller because the proton number stays at 13 while the electron count decreases.

The expected order is:

\[ \mathrm{Al}> \mathrm{Al}^{+}> \mathrm{Al}^{2+}> \mathrm{Al}^{3+} \]

c) The particularly substantial contraction is expected when \(\mathrm{Al}^{2+}\) becomes \(\mathrm{Al}^{3+}\).

The electron arrangement changes from:

\[ 2,8,1 \longrightarrow 2,8 \]

The final electron in the third shell is removed, eliminating that entire occupied shell.

This is a more substantial structural change than simply removing an electron while leaving the outer shell occupied.

The important distinction is that electron removal can cause contraction at every stage, but removing the last electron from an occupied outer shell produces an especially significant change.

Question 4

An X-ray diffraction experiment investigates an illustrative ionic crystal with a rock-salt structure.

The measured cubic unit cell edge is:

\[ a=572\pm2\ \mathrm{pm} \]

Along each unit cell edge, the ions alternate between positive and negative ions.

a) Calculate the nearest-neighbour distance between oppositely charged ions.

b) Calculate the uncertainty in this distance.

c) A student claims that each ion must have a radius of half the nearest-neighbour distance. Explain why this conclusion is not justified.

Solution 4

The nearest-neighbour distance is \(286\pm1\ \mathrm{pm}\). The individual ionic radii cannot be determined from this measurement alone.

a) In the rock-salt structure, two equal nearest-neighbour distances fit along one unit cell edge.

Therefore:

\[ d=\frac{a}{2} \]

Substitute:

\[ d=\frac{572}{2} \]

\[ \boxed{d=286\ \mathrm{pm}} \]

b) The uncertainty in the lattice edge is \(2\ \mathrm{pm}\).

Since the distance is calculated by dividing the lattice edge by two, the absolute uncertainty is also divided by two:

\[ \Delta d=\frac{2}{2}=1\ \mathrm{pm} \]

Therefore:

\[ \boxed{d=286\pm1\ \mathrm{pm}} \]

c) The student’s conclusion is not justified because the measurement gives the distance between the centres of two different ions.

Under the simplified touching-sphere model:

\[ r_+ + r_-=286\ \mathrm{pm} \]

It does not establish that the two ions have equal radii.

For example, one ion could have a radius of \(100\ \mathrm{pm}\), while the other has a radius of \(186\ \mathrm{pm}\).

Both would give the same combined distance.

The missing information is the individual contribution of each ion to the measured separation.

This is why ionic-radius scales require additional structural evidence and consistent assumptions.

Question 5

Two hypothetical ionic crystals, \(\mathrm{NaX}\) and \(\mathrm{KX}\), are investigated using X-ray diffraction.

Both crystals:

• Have the rock-salt structure.
• Have six-coordinate cations and anions.
• Contain the same monovalent anion, \(\mathrm{X}^{-}\).
• Are measured at the same temperature.
• Are analysed using the same additive ionic-radius model.

The following illustrative measurements are obtained:

A separate reference table gives the following illustrative radius values:

A student makes three claims:

Claim A: \(\mathrm{K}^{+}\) is 30 pm larger than \(\mathrm{Na}^{+}\).

Claim B: X-ray diffraction proves that \(\mathrm{Na}^{+}\) is exactly 80 pm smaller than neutral sodium.

Claim C: These measurements prove that every cation must be smaller than every neutral atom.

Evaluate all three claims.

In your response, explain which comparisons are valid, which are not, and how the electron arrangement of sodium supports the expected size change when it forms \(\mathrm{Na}^{+}\).

Solution 5

Claim A is supported by the supplied model and assumptions. Claims B and C are not justified by the evidence.

This question tests an important distinction: measuring distances accurately does not automatically mean that every interpretation of those distances is valid.

Claim A: \(\mathrm{K}^{+}\) is 30 pm larger than \(\mathrm{Na}^{+}\).

This conclusion is supported under the stated assumptions.

Step 1: Calculate the nearest-neighbour distance in each crystal.

For the rock-salt structure:

\[ d=\frac{a}{2} \]

For \(\mathrm{NaX}\):

\[ d_{\mathrm{NaX}}=\frac{560}{2} =280\ \mathrm{pm} \]

For \(\mathrm{KX}\):

\[ d_{\mathrm{KX}}=\frac{620}{2} =310\ \mathrm{pm} \]

Step 2: Apply the additive-radius model.

For \(\mathrm{NaX}\):

\[ r_{\mathrm{Na}^{+}}+r_{\mathrm{X}^{-}} =280\ \mathrm{pm} \]

For \(\mathrm{KX}\):

\[ r_{\mathrm{K}^{+}}+r_{\mathrm{X}^{-}} =310\ \mathrm{pm} \]

Step 3: Compare the equations.

The anion is the same in both crystals, and the coordination number is unchanged.

Subtracting the first equation from the second eliminates the common anion radius:

\[ r_{\mathrm{K}^{+}}-r_{\mathrm{Na}^{+}} =310-280 \]

\[ \boxed{r_{\mathrm{K}^{+}}-r_{\mathrm{Na}^{+}} =30\ \mathrm{pm}} \]

Therefore, Claim A is supported within the specified radius model.

Notice the important constraint: the same anion and coordination environment allow the cation-radius difference to be isolated.

If the anions or coordination numbers were different, this simple subtraction would not necessarily be valid.

Claim B: X-ray diffraction proves that \(\mathrm{Na}^{+}\) is exactly 80 pm smaller than neutral sodium.

This conclusion is incorrect.

The student calculates:

\[ 180-100=80\ \mathrm{pm} \]

The arithmetic is correct, but the comparison is not a direct measurement of the same type of radius.

The neutral sodium value is a metallic radius, while the \(\mathrm{Na}^{+}\) value is an ionic radius assigned using a six-coordinate crystal model.

These values come from different definitions and physical environments.

Although both are expressed in picometres, that does not make them directly interchangeable.

The values may support the expected qualitative trend, but they do not establish an exact 80 pm contraction.

Claim C: These measurements prove that every cation must be smaller than every neutral atom.

This conclusion is also incorrect.

The measurements only investigate particular ionic crystals. They do not establish a universal size ranking across all elements and species.

A large cation can still be larger than a small neutral atom.

The correct comparison is between a cation and the neutral atom of the same element.

For sodium, the electron configurations provide the relevant explanation:

\[ \mathrm{Na}:\ 2,8,1 \]

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

Sodium loses its outermost electron, removing the third occupied shell. Its nucleus retains all 11 protons, while the remaining electron cloud experiences reduced electron-electron repulsion and a stronger effective nuclear attraction.

Therefore, the expected conclusion is that \(\mathrm{Na}^{+}\) is smaller than neutral sodium.

The deeper lesson is that three different things must not be confused:

• Electron configuration explains why the electron cloud contracts.
• X-ray diffraction measures distances between ions in a crystal.
• Ionic-radius conventions turn those distances into useful effective radius values.

A strong chemistry explanation connects all three without claiming that one measurement proves more than it actually does.

07Where to go next

You now have the reasoning behind the general rule: losing electrons usually contracts an atom because the remaining electron cloud experiences less electron-electron repulsion and a stronger effective nuclear attraction.

The next useful step is comparing ions from different elements.

In particular, when two ions have the same number of electrons, you can hold their electron arrangement constant and investigate how nuclear charge determines their relative sizes.

That is the key idea behind How to Rank Isoelectronic Ions by Size: Worked Examples.

You can also connect electron removal to the energy required to remove an electron by studying Why First Ionisation Energy Generally Increases Across a Period.