Why Anions Are Usually Larger Than Their Neutral Atoms
The reason is that the added electrons increase electron-electron repulsion and reduce the effective nuclear attraction experienced by the electrons in the outer shell. The nucleus still has the same positive charge, but it now has more electrons to attract.
An oxygen atom and an oxide ion both contain eight protons. However, the oxide ion, \(\mathrm{O}^{2-}\), is significantly larger than the neutral oxygen atom. How can adding just two electrons make an atom noticeably bigger when its nucleus hasn’t changed?
The reason is that the added electrons increase electron-electron repulsion and reduce the effective nuclear attraction experienced by the electrons in the outer shell. The nucleus still has the same positive charge, but it now has more electrons to attract. The electron cloud spreads out, increasing the ionic radius.
This is why anions are usually larger than their neutral atoms. The key is not that the nucleus gets bigger. It is that the balance between nuclear attraction and electron-electron repulsion changes when electrons are added.
01Why adding electrons increases atomic size
Imagine adding extra people to a small room. The walls haven’t moved, but everyone has less space, and people are more likely to push against one another.
Electrons behave differently from people, but the analogy captures one useful idea: adding electrons to an existing electron shell increases electron-electron repulsion.
There are two forces to keep track of:
- Nuclear attraction: the positively charged nucleus attracts negatively charged electrons.
- Electron-electron repulsion: negatively charged electrons repel one another.
When a neutral atom gains electrons, its number of protons stays constant, but its number of electrons increases. The extra electrons occupy the existing outer shell, increasing repulsion within the electron cloud.
The outer electrons are therefore held less tightly overall, and the electron cloud expands.
The role of nuclear charge
The nuclear charge is determined by the number of protons in the nucleus.
For oxygen:
| Species | Protons | Electrons | Net charge |
|---|---|---|---|
| Neutral oxygen, \(\mathrm{O}\) | 8 | 8 | 0 |
| Oxide ion, \(\mathrm{O}^{2-}\) | 8 | 10 | −2 |
Notice that the nuclear charge is identical in both species. Each nucleus contains eight protons, giving it a charge of \(+8\) in units of the elementary charge.
However, the oxide ion has two additional electrons. The same nuclear charge must now attract ten electrons instead of eight.
This changes the balance of attraction and repulsion. The electrons experience greater competition from other electrons, making the outer electron cloud more diffuse.
A useful distinction is that the nuclear charge has not decreased. Rather, the effective nuclear attraction experienced by an outer electron is reduced by shielding and electron-electron interactions.
Electron shielding: why outer electrons feel less attraction
Electrons in an atom do not all experience the full nuclear charge equally.
Electrons closer to the nucleus partially shield outer electrons from its positive charge. This is called the shielding effect.
Think of an outer electron as trying to respond to the nucleus while other electrons occupy the space between them. Those inner electrons partially reduce the net attraction experienced by the outer electron.
Adding electrons to an existing outer shell increases electron-electron repulsion. It also changes the shielding experienced by electrons in that shell, although electrons in the same shell do not shield one another as effectively as electrons in inner shells.
The important point is that the added electrons do not come with additional protons. The nuclear charge stays fixed while the electron interactions change.
Electron configurations: checking whether a new shell is added
Electron configurations make the explanation much easier to see.
Consider sulfur and the sulfide ion.
| Species | Electron arrangement | Number of protons | Number of electrons |
|---|---|---|---|
| Sulfur, \(\mathrm{S}\) | 2, 8, 6 | 16 | 16 |
| Sulfide, \(\mathrm{S}^{2-}\) | 2, 8, 8 | 16 | 18 |
Sulfur gains two electrons to form \(\mathrm{S}^{2-}\).
Notice something important: the added electrons enter the existing outer shell. Sulfur does not gain an entirely new electron shell.
The outer shell now contains eight electrons instead of six. This increases electron-electron repulsion, causing the electron cloud to expand.
This is a useful distinction for HSC questions. If an atom gains electrons, don’t automatically assume that another shell has been added. Check where those electrons actually go.
How nuclear charge and shielding work together
Let’s bring the two effects together.
1. Nuclear charge
Nuclear charge depends on the number of protons.
When an atom becomes an anion, its proton number stays constant. The nucleus therefore has the same positive charge before and after electron gain.
2. Electron shielding
Inner-shell electrons partially shield outer-shell electrons from the nucleus.
The additional electrons in an anion occupy the outer shell. They do not shield one another as effectively as electrons in an inner shell would, but they do contribute to electron-electron repulsion.
3. Effective nuclear attraction
The effective nuclear attraction is the net attraction an electron experiences from the nucleus after accounting for shielding and interactions with other electrons.
When electrons are added without increasing the proton number, the balance shifts towards greater electron-electron repulsion. The outer electron cloud becomes more spread out.
You can summarise the relationship like this:
| Factor | When an atom gains electrons |
|---|---|
| Number of protons | Stays constant |
| Nuclear charge | Stays constant |
| Number of electrons | Increases |
| Electron-electron repulsion | Increases |
| Shielding within the outer shell | Changes slightly |
| Effective nuclear attraction experienced by outer electrons | Generally decreases |
| Electron cloud size | Increases |
Be careful with one particular misconception: the nucleus does not become less positively charged when an anion forms. Its charge is unchanged. It is the balance of attraction and repulsion experienced by the electrons that changes.
02A useful contrast: what happens when nuclear charge increases?
So far, we’ve kept the number of protons constant and added electrons.
But what happens if we keep the number of electrons constant and increase the number of protons?
This gives us a particularly useful comparison called an isoelectronic series.
Isoelectronic species have the same number of electrons, even though they may have different numbers of protons.
Consider these four ions:
| Ion | Protons | Electrons | Electron arrangement |
|---|---|---|---|
| \(\mathrm{O}^{2-}\) | 8 | 10 | 2, 8 |
| \(\mathrm{F}^{-}\) | 9 | 10 | 2, 8 |
| \(\mathrm{Na}^{+}\) | 11 | 10 | 2, 8 |
| \(\mathrm{Mg}^{2+}\) | 12 | 10 | 2, 8 |
All four species have exactly ten electrons.
However, their nuclei have different positive charges.
Which species has the largest radius?
It is \(\mathrm{O}^{2-}\).
All four species have the same electron arrangement, but the oxygen ion has the fewest protons. Its nucleus exerts the weakest attraction on the ten electrons.
As nuclear charge increases, the electrons are pulled more strongly towards the nucleus, and the radius decreases.
The predicted order from largest to smallest is:
\[
\mathrm{O}^{2-} >
\mathrm{F}^{-} >
\mathrm{Na}^{+} >
\mathrm{Mg}^{2+}
\]
This gives you an important comparison:
- Adding electrons to the same atom generally increases its radius.
- Increasing nuclear charge while keeping the electron count constant generally decreases the radius.
These are different situations. Don’t mix up the two trends.
Worked example: comparing an isoelectronic series
Arrange \(\mathrm{O}^{2-}\), \(\mathrm{F}^{-}\), and \(\mathrm{Mg}^{2+}\) in order of decreasing ionic radius.
Step 1
- \(\mathrm{O}^{2-}\): 8 protons and 10 electrons.
- \(\mathrm{F}^{-}\): 9 protons and 10 electrons.
- \(\mathrm{Mg}^{2+}\): 12 protons and 10 electrons.
Each species has ten electrons.
Step 2
The number of protons increases in this order:
\[
\mathrm{O}^{2-} \rightarrow
\mathrm{F}^{-} \rightarrow
\mathrm{Mg}^{2+}
\]
Step 3
Because the electron count is constant, the species with more protons attracts its electrons more strongly.
Its electron cloud is therefore more contracted.
Step 4
\[
\boxed{\mathrm{O}^{2-} >
\mathrm{F}^{-} >
\mathrm{Mg}^{2+}}
\]
The oxide ion is largest, while the magnesium ion is smallest.
The key is that all three have the same electron arrangement. The changing nuclear charge determines the size trend.
03How do chemists actually determine ionic radius?
There is a measurement problem hiding behind the word radius.
An atom isn’t a tiny solid ball with a clearly defined surface. Its electrons occupy a probability distribution, which gradually becomes less dense further from the nucleus.
So how can scientists assign an atom or ion a radius?
One important experimental method is X-ray diffraction.
Using X-ray diffraction to investigate crystal structure
Imagine a crystal made from regularly arranged positive and negative ions.
An X-ray beam is directed towards the crystal. The X-rays interact with the electrons in the crystal, producing a diffraction pattern that is recorded by a detector.
The angles and intensities in this pattern provide information about the arrangement and spacing of the ions.
What should this measurement tell us?
If we approximate the ions as touching spheres, the distance between neighbouring ion centres should equal the sum of their radii.
For a simple cubic crystal structure of the rock-salt type, the nearest-neighbour distance between a positive and negative ion is half the unit-cell edge length.
The diffraction experiment can therefore provide the centre-to-centre separation.
However, there is a limitation.
The experiment measures the repeating crystal geometry. It does not directly identify the precise boundary of each individual ion.
To assign separate ionic radii, chemists use a consistent radius convention. The assigned values depend on factors such as coordination number, which describes how many neighbouring ions surround a particular ion.
This is why ionic radius values can differ between tables.
A more reliable comparison keeps the coordination environment consistent and uses the same radius convention.
Worked example: estimating an ionic radius from crystal measurements
A hypothetical rock-salt-type crystal has a cubic unit-cell edge length of \(564\ \mathrm{pm}\).
The nearest positive and negative ions are separated by half this distance. If the assigned radius of the positive ion is \(102\ \mathrm{pm}\), estimate the radius of the negative ion.
Assume the ions are represented as touching spheres.
Step 1
The separation is half the unit-cell edge length:
\[
d = \frac{564}{2}
\]
\[
d = 282\ \mathrm{pm}
\]
Step 2
For two touching spherical ions:
\[
d = r_+ + r_-
\]
where \(r_+\) is the positive ion’s radius and \(r_-\) is the negative ion’s radius.
Step 3
\[
r_- = d-r_+
\]
Substitute the supplied values:
\[
r_- = 282-102
\]
\[
\boxed{r_- = 180\ \mathrm{pm}}
\]
The estimated negative ionic radius is \(180\ \mathrm{pm}\).
This is an illustrative calculation, not a reported experimental measurement. The value depends on the spherical-contact model and the assigned radius of the positive ion.
The important experimental distinction is that diffraction provides the separation between ion centres. The individual radii are inferred using a model.
04When does the anion-size rule need qualification?
The statement that anions are larger than their neutral atoms is a useful general rule, but you need to understand exactly what it compares.
It applies to the same element when a stable monatomic anion forms.
There are three important qualifications.
1. Not every atom forms a stable monatomic anion
Consider neon.
A neutral neon atom has the electron arrangement:
\[
2,8
\]
Its outer shell is already full.
If it gained another electron, that electron would need to occupy a higher-energy shell rather than simply fill the existing outer shell.
Neon does not normally form a stable monatomic anion under ordinary conditions.
This means you cannot automatically apply the usual anion-size explanation to every element. You first need to establish that the anion can exist as a stable species.
2. Not every anion is larger than every neutral atom
This is a particularly important distinction.
The rule compares an anion with the neutral atom of the same element. It does not say that every anion is larger than every neutral atom.
Consider fluoride, \(\mathrm{F}^{-}\), and neutral sodium, \(\mathrm{Na}\).
Their electron arrangements are:
| Species | Electron arrangement | Occupied shells |
|---|---|---|
| \(\mathrm{F}^{-}\) | 2, 8 | 2 |
| \(\mathrm{Na}\) | 2, 8, 1 | 3 |
Fluoride has ten electrons, while sodium has eleven.
More importantly, sodium has an electron in the third shell. Fluoride’s outermost electrons are in the second shell.
The sodium atom can therefore be larger than the fluoride ion.
This does not contradict the rule. We are comparing two different elements, with different electron arrangements.
3. Ionic radius depends on the chemical environment
An ionic radius is not a completely fixed measurement independent of its surroundings.
The arrangement of neighbouring ions affects the electron distribution, and the assigned ionic radius depends partly on coordination number.
For example, an ion surrounded by six neighbouring ions may have a different assigned radius from the same ion in a four-coordinate environment.
Neutral atomic radii also have different definitions, including covalent and van der Waals radii.
Therefore, be cautious when comparing numerical values taken from tables that use different definitions.
The general chemical trend remains useful, but the precise numerical difference depends on how the radii are defined and measured.
05A decision method for HSC Chemistry questions
When an exam asks you to compare atomic and ionic radii, use the following reasoning.
| Situation | What to investigate | Expected trend |
|---|---|---|
| An atom gains electrons | Same nuclear charge, greater electron repulsion | Radius increases |
| An atom loses electrons | Fewer electrons, reduced repulsion and often a contracted outer shell | Radius decreases |
| Isoelectronic species have different proton numbers | Greater nuclear charge attracts the same number of electrons more strongly | Radius decreases as proton number increases |
| Neutral atoms are compared across a period | Increasing nuclear charge with relatively similar shielding | Atomic radius generally decreases |
| Neutral atoms are compared down a group | Additional occupied shells and greater shielding | Atomic radius generally increases |
One final exam tip: always distinguish between the number of protons and the number of electrons.
The proton number determines the nuclear charge. The electron number determines how many electrons are interacting with that nucleus.
Both matter, but they affect atomic and ionic size in different ways.
06Questions and solutions
Question 1
A neutral oxygen atom gains two electrons to form an oxide ion, \(\mathrm{O}^{2-}\).
Explain why the oxide ion has a larger radius than the neutral oxygen atom. Refer to nuclear charge, electron arrangement, and electron-electron repulsion.
Solution 1
The oxide ion is larger than the neutral oxygen atom because the additional electrons increase electron-electron repulsion without changing the nuclear charge.
Oxygen has eight protons and eight electrons. Its electron arrangement is 2, 6.
After gaining two electrons, the oxide ion has eight protons and ten electrons, giving it an arrangement of 2, 8.
The added electrons enter the existing outer shell. They increase electron-electron repulsion and partially shield other outer-shell electrons from the nucleus.
Because the number of protons remains constant, the nucleus does not gain any additional positive charge to compensate.
The effective nuclear attraction experienced by the outer electrons decreases, allowing the electron cloud to expand.
Therefore:
\[
r(\mathrm{O}^{2-}) > r(\mathrm{O})
\]
The important point is that oxygen gains electrons, not protons. Its nuclear charge stays unchanged.
Question 2
Three ions have the following properties:
| Ion | Number of protons | Number of electrons |
|---|---|---|
| \(\mathrm{S}^{2-}\) | 16 | 18 |
| \(\mathrm{Cl}^{-}\) | 17 | 18 |
| \(\mathrm{K}^{+}\) | 19 | 18 |
Assume the ions are being compared using the same coordination environment.
(a) Explain why these ions are isoelectronic.
(b) Arrange them in order of decreasing ionic radius.
(c) Explain why the ion with the greatest negative charge is not the only factor determining the order.
Solution 2
(a) All three ions are isoelectronic because they contain 18 electrons each.
Their electron arrangements are identical:
\[
2,8,8
\]
Although their nuclei have different numbers of protons, their electron counts are the same.
(b) The correct order is:
\[
\boxed{\mathrm{S}^{2-} >
\mathrm{Cl}^{-} >
\mathrm{K}^{+}}
\]
(c) The order is determined by nuclear charge.
Sulfur has 16 protons, chlorine has 17, and potassium has 19.
Because the electron count is constant, increasing the number of protons increases the attraction between the nucleus and the electron cloud.
The potassium ion therefore has the strongest nuclear attraction and the smallest radius.
A tempting explanation is that the sulfide ion is largest simply because it has a \(2-\) charge. However, ionic charge alone does not explain the complete trend.
The more useful comparison is between proton number and electron number. In an isoelectronic series, increasing nuclear charge produces a smaller radius.
Question 3
A hypothetical ionic crystal has a rock-salt structure and a cubic unit-cell edge length of \(572\ \mathrm{pm}\).
The nearest positive and negative ions are separated by half the unit-cell edge length.
A consistent radius convention assigns the positive ion a radius of \(104\ \mathrm{pm}\).
(a) Calculate the nearest-neighbour separation.
(b) Calculate the assigned radius of the negative ion.
(c) Explain why the X-ray diffraction measurement does not directly establish the physical boundary of the negative ion.
Solution 3
(a) The nearest-neighbour separation is \(286\ \mathrm{pm}\).
For a rock-salt structure, the nearest-neighbour distance is half the unit-cell edge length.
\[
d = \frac{a}{2}
\]
Substitute:
\[
d = \frac{572}{2}
\]
\[
\boxed{d=286\ \mathrm{pm}}
\]
(b) The assigned radius of the negative ion is \(182\ \mathrm{pm}\).
Assuming the ions are touching spheres:
\[
d=r_+ + r_-
\]
Rearrange:
\[
r_-=d-r_+
\]
Substitute:
\[
r_-=286-104
\]
\[
\boxed{r_-=182\ \mathrm{pm}}
\]
(c) X-ray diffraction measures the repeating crystal structure, not the exact edge of an individual ion.
The diffraction pattern provides information about the positions and separations of ions within the crystal.
The individual ionic radii are then assigned using a model in which the ions are treated approximately as touching spheres.
The exact radius depends on the convention used, including the coordination environment.
Therefore, \(182\ \mathrm{pm}\) is an assigned ionic radius based on the supplied assumptions, not a direct measurement of a sharp physical boundary.
Question 4
A student claims:
“Every anion must be larger than every neutral atom because anions contain additional electrons, which increase electron-electron repulsion.”
Use fluoride ions, \(\mathrm{F}^{-}\), and neutral sodium atoms, \(\mathrm{Na}\), to evaluate this statement.
Your answer should refer to electron arrangements, occupied shells, and the difference between comparing the same element and comparing different elements.
Solution 4
The student’s statement is incorrect because anions are generally larger than their own neutral atoms, not necessarily larger than neutral atoms of other elements.
Fluoride has the electron arrangement:
\[
\mathrm{F}^{-}: 2,8
\]
It has two occupied electron shells.
Neutral sodium has the arrangement:
\[
\mathrm{Na}: 2,8,1
\]
It has three occupied electron shells.
Sodium’s outermost electron occupies the third shell, which is further from the nucleus than the outermost electrons in fluoride.
Consequently, neutral sodium can have a larger atomic radius than the fluoride ion.
The student’s mistake is applying a rule for comparing two species of the same element to a comparison involving different elements.
When fluorine gains an electron, its proton number stays at nine, while its electron number increases from nine to ten. This increases electron-electron repulsion and expands its electron cloud.
However, sodium and fluoride have different nuclear charges and different electron arrangements.
The correct conclusion is that the anion-size rule must be applied to the neutral atom and its corresponding anion, not indiscriminately to all atoms and ions.
Question 5
An examiner presents three proposed electron-gain comparisons.
| Comparison | Initial species | Proposed final species |
|---|---|---|
| A | Chlorine: 17 protons, 17 electrons | Chloride: 17 protons, 18 electrons |
| B | Neon: 10 protons, 10 electrons | Hypothetical \(\mathrm{Ne}^{-}\): 10 protons, 11 electrons |
| C | Sulfide: 16 protons, 18 electrons | Chloride: 17 protons, 18 electrons |
Three students make the following claims:
- Student 1: Chloride must be larger than chlorine because it has gained an electron.
- Student 2: Neon must form a stable, larger anion for exactly the same reason.
- Student 3: Sulfide and chloride must have identical radii because they contain the same number of electrons.
Evaluate all three claims.
For each comparison, discuss electron arrangement, nuclear charge, electron-electron repulsion, and whether a stable monatomic anion exists. For comparison C, assume the ions have the same coordination number in comparable ionic environments.
Solution 5
Student 1 is correct, while Students 2 and 3 are incorrect. The three comparisons involve different electron arrangements and different conditions for forming stable ions.
Comparison A: chlorine and chloride
Student 1 is correct.
Neutral chlorine has the electron arrangement:
\[
2,8,7
\]
When chlorine gains one electron, it forms chloride:
\[
2,8,8
\]
Both species have 17 protons, so their nuclear charges are identical.
The additional electron enters the existing third shell. This increases electron-electron repulsion and changes the shielding experienced by the outer electrons.
The electron cloud expands, so:
\[
\boxed{r(\mathrm{Cl}^{-})>r(\mathrm{Cl})}
\]
This is a valid application of the usual anion-size rule.
Comparison B: neon and hypothetical neon anion
Student 2 is incorrect.
Neutral neon already has a complete outer shell:
\[
2,8
\]
If another electron were added, it would need to occupy a higher-energy shell. A simple electron-arrangement representation would place it in the third shell.
However, this does not mean that a stable neon anion forms.
Neon has a particularly stable closed-shell electron arrangement, and an additional electron is not bound to the atom under ordinary conditions.
Therefore, there is no stable monatomic \(\mathrm{Ne}^{-}\) ion to which we can assign an ordinary equilibrium ionic radius.
The important trap is assuming that every neutral atom can accept an electron simply because electron gain increases repulsion.
Before comparing radii, you must establish that the proposed ion is stable.
Comparison C: sulfide and chloride
Student 3 is incorrect.
Both ions contain 18 electrons, but their nuclear charges are different.
Sulfide has 16 protons, while chloride has 17.
Their electron arrangements are identical:
\[
2,8,8
\]
However, the chloride ion has one additional proton. Its nucleus attracts the same number of electrons more strongly.
Consequently, the chloride ion has the smaller radius:
\[
\boxed{r(\mathrm{S}^{2-})>r(\mathrm{Cl}^{-})}
\]
This conclusion assumes comparable coordination environments, as specified in the question.
The fact that both ions contain 18 electrons does not make their radii identical. Nuclear charge still matters.
The deeper lesson
These comparisons reveal three different questions that must not be confused:
- Can the atom form a stable anion?
- If it does, how does adding electrons change its size when nuclear charge stays constant?
- If two species have the same electron count, how does a difference in nuclear charge affect their relative sizes?
A strong HSC explanation considers all three.
07Where to go next
The next useful step is connecting the electron structure of anions to their chemical behaviour in aqueous solutions.
For example, chloride, bromide, and iodide ions have different chemical properties despite all carrying a single negative charge.
See Qualitative Tests for Aqueous Anions: HSC Chemistry Guide to explore how chemists identify these ions through their reactions and observable evidence.