Why Atomic Radius Decreases Across a Period in HSC Chemistry

Imagine comparing a lithium atom with a fluorine atom. Both have electrons in the second energy level, but fluorine has a much smaller atomic radius.

Why? Fluorine has more electrons, so you might expect its electron cloud to take up more space. Instead, it is smaller.

Atomic radius generally decreases from left to right across a period because nuclear charge increases, while shielding by inner-shell electrons stays approximately constant. The additional protons attract the electrons in the outer energy level more strongly, pulling the electron cloud closer to the nucleus.

The important part is understanding why the extra electrons do not cancel out the extra protons. Once you understand that, the trend becomes much easier to predict and explain in an HSC Chemistry response.

01Why does atomic radius decrease across a period?

Start with the atoms themselves

Consider lithium (Li), carbon (C), and fluorine (F). All three are in Period 2 of the periodic table.

Their electron configurations are:

Notice something important: all three atoms have two occupied electron shells. As you move from lithium to fluorine, the additional electrons enter the second energy level rather than a completely new shell.

However, the number of protons in the nucleus increases from 3 to 9.

Before reading on, predict what happens to the attraction between the nucleus and the outer electrons as you move from lithium to fluorine. Does it stay constant because both the number of protons and electrons increase, or does one effect become more important?

The attraction increases. The additional electrons do create more electron-electron repulsion, but they do not shield one another from the nucleus as effectively as electrons in an inner shell do.

1. Nuclear charge increases

Nuclear charge is the total positive charge of an atom’s nucleus. It depends on the number of protons.

Each proton has a positive charge, so adding protons increases the electrostatic attraction between the nucleus and negatively charged electrons.

Across Period 2, the nuclear charge changes as follows:

The positive charge increases by one at every step.

You can picture the nucleus as a stronger magnet attracting the electrons. The analogy is useful for visualising the attraction, although electrons and nuclei interact through electrostatic forces, not magnetism.

If nuclear charge were the only factor, the electron cloud would contract as more protons were added. But nuclear charge is not the only factor. We also need to consider shielding.

2. Electron shielding increases only slightly

Electrons repel one another because they all carry negative charge.

This repulsion reduces the net attraction that an outer electron experiences from the positively charged nucleus. This effect is called electron shielding.

Think of an inner-shell electron as providing a partial screen between the nucleus and an outer electron. The outer electron experiences less of the nucleus’s full positive charge because other electrons are also present.

However, not all electrons provide the same amount of shielding.

• Inner-shell electrons shield outer electrons relatively effectively.
• Electrons in the same principal energy level provide less effective shielding.
• Electrons do not perfectly cancel the positive charge of individual protons.

This last point is the key to understanding the periodic trend.

3. Effective nuclear charge increases

The combined effect of nuclear charge and shielding is described by effective nuclear charge.

Effective nuclear charge is the net positive attraction experienced by an electron after accounting for shielding by the other electrons.

It is commonly represented by \(Z_{\mathrm{eff}}\).

A simplified relationship is:

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

where:

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

This is a qualitative model. It is not an exact calculation in which you simply subtract the total number of other electrons from the number of protons. Different electrons shield one another to different extents.

Across a period, the number of protons increases, but the number of inner-shell electrons remains approximately constant. The extra electrons enter the same principal energy level and do not provide enough additional shielding to cancel the increased nuclear charge.

Therefore, effective nuclear charge generally increases.

4. The electron cloud contracts

Now connect the three ideas:

1. Nuclear charge increases because the number of protons increases.
2. Inner-shell shielding remains approximately constant across the period.
3. Additional electrons in the same shell do not shield one another effectively enough to cancel the increased nuclear charge.
4. Effective nuclear charge increases.
5. The stronger electrostatic attraction pulls the electron cloud closer to the nucleus.
6. Atomic radius decreases.

This is the complete explanation for the general decrease in atomic radius across a period.

02How is atomic radius actually measured?

Before using atomic-radius values, there’s an important complication: an atom doesn’t have a solid outer surface.

Its electrons occupy regions of probability around the nucleus. The electron cloud gradually becomes less dense as you move further from the nucleus, rather than stopping at a precise boundary.

So how can chemists assign an atom a radius?

They infer it from distances between atomic nuclei. The particular definition depends on how the atoms are interacting.

Three useful definitions are:

For comparing main-group elements across a period, covalent radii provide a useful consistent set of values.

How an experiment determines the distance

One experimental technique is gas electron diffraction.

Imagine a chamber containing gaseous diatomic molecules, such as \(\ce{Cl2}\).

The apparatus works in this order:

1. The gaseous molecules enter a low-pressure chamber.
2. A beam of electrons passes through the gas sample.
3. The electrons are scattered by the molecules.
4. A detector records the intensity of scattered electrons at different angles.
5. Scientists analyse the diffraction pattern to determine the average distance between the two nuclei.

The prediction is straightforward: molecules with different internuclear separations produce different diffraction patterns.

The observation is a pattern of scattered electrons, not a photograph of the atoms’ boundaries. Scientists use a physical model to extract the internuclear distance.

For two identical atoms joined by a covalent bond, the covalent radius is half their internuclear separation:

\[ r_{\mathrm{cov}}=\frac{d}{2} \]

Here:

• \(r_{\mathrm{cov}}\) is the covalent radius, measured in picometres (pm).
• \(d\) is the distance between the two nuclei, also measured in pm.

One picometre is \(10^{-12}\) metres.

03What do the atomic-radius values show?

Now that we know how radii are estimated, we can examine the actual pattern.

The following table gives representative covalent radii for selected elements in Periods 2 and 3.

Representative rounded covalent radii in pm. Values are from the SQA Higher and Advanced Higher Chemistry Data Booklet. Other compilations may differ slightly because atomic radius depends on bonding environment and measurement convention.

Two patterns stand out.

Pattern 1: Radius decreases across each period.

• Period 2: lithium decreases from \(130\ \mathrm{pm}\) to fluorine at \(60\ \mathrm{pm}\).
• Period 3: sodium decreases from \(160\ \mathrm{pm}\) to chlorine at \(100\ \mathrm{pm}\).

Both periods show the same general trend. This supports the explanation based on increasing effective nuclear charge.

Pattern 2: Radius increases when moving down to a new period.

Compare lithium and sodium.

Sodium has a greater nuclear charge, but its atomic radius is larger. Why?

Sodium has an additional occupied principal energy level.

Lithium has electrons in two occupied shells, while sodium has electrons in three. Sodium’s outer electrons occupy the third energy level and are further from the nucleus on average.

The additional inner shell also increases shielding.

This is why the periodic trend changes direction when you move down to the next period.

04Are there exceptions to the trend?

The general decrease in atomic radius is reliable, but there are some important qualifications.

1. Noble gases and different radius definitions

You might find a periodic table that shows the radius increasing sharply from fluorine to neon.

That doesn’t necessarily mean the underlying periodic trend has reversed.

The problem is that noble gases generally do not form the ordinary covalent bonds used to determine covalent radii.

Instead, many reference tables list their van der Waals radii.

Remember:

• Covalent radius comes from the distance between bonded atoms.
• van der Waals radius comes from the closest approach of non-bonded atoms.

Non-bonded atoms generally remain further apart than atoms joined by a covalent bond.

Consequently, van der Waals radii are usually larger than covalent radii.

If a table switches from covalent radius to van der Waals radius at the noble gases, the values are no longer directly comparable.

When interpreting a periodic trend, always check which type of atomic radius is being reported.

2. Transition metals show a less pronounced trend

The simple explanation works particularly well for the main-group elements.

Across a transition-metal series, additional electrons enter a \(d\) subshell. These electrons affect shielding and the distribution of electron density differently from the straightforward filling of an outer \(s\) or \(p\) subshell.

As nuclear charge increases, transition-metal radii generally change less dramatically, and some irregularities occur.

The main idea still matters: increasing nuclear charge tends to contract the electron cloud, but the competing effects of electron arrangement and shielding make the trend less straightforward.

3. Do not confuse atomic radius exceptions with ionisation energy exceptions

This is a particularly useful distinction.

You may have encountered two famous decreases in first ionisation energy across Period 2:

• Beryllium to boron.
• Nitrogen to oxygen.

These occur because of changes in subshell energy and electron pairing.

However, they do not mean atomic radius must increase at those same points.

Atomic radius and first ionisation energy measure different properties:

• Atomic radius describes the spatial extent of an atom, using a specified radius convention.
• First ionisation energy measures the energy needed to remove an electron from a gaseous atom.

Both are influenced by effective nuclear charge, but their detailed patterns are not identical.

For the related energy trend, see Why First Ionisation Energy Generally Increases Across a Period.

05A reliable way to explain the trend in an HSC response

If an exam asks why atomic radius decreases across a period, don’t stop at saying that nuclear charge increases.

You need to explain what happens to the shielding and the attraction experienced by the electrons.

Use this sequence:

A strong written response could be:

Across a period, nuclear charge increases because the number of protons increases. Additional electrons enter the same principal energy level, while the number of inner-shell electrons remains approximately constant. Although electron shielding increases slightly, electrons in the same energy level do not shield one another as effectively as inner-shell electrons. Therefore, effective nuclear charge increases, producing a stronger electrostatic attraction between the nucleus and the electron cloud. The electron cloud contracts, causing atomic radius to decrease.

That explanation identifies the cause, accounts for shielding, and connects both to the observed change in radius.

06Questions and solutions

Question 1

Sodium and chlorine are both in Period 3.

Their representative covalent radii are:

• Sodium: \(160\ \mathrm{pm}\)
• Chlorine: \(100\ \mathrm{pm}\)

Which element has the smaller atomic radius? Explain your answer using nuclear charge, shielding, and electron arrangement.

Solution 1

Chlorine has the smaller atomic radius.

Sodium has 11 protons, while chlorine has 17 protons.

Both elements have three occupied principal energy levels. The additional electrons in chlorine enter the third energy level rather than a new shell.

The number of inner-shell electrons remains eight:

• Sodium: \(2,8,1\)
• Chlorine: \(2,8,7\)

Although chlorine has more electrons, these additional electrons do not shield one another effectively enough to cancel the increased nuclear charge.

Therefore, chlorine experiences a greater effective nuclear charge. The stronger attraction pulls its electron cloud closer to the nucleus, resulting in a smaller atomic radius.

The radius difference is:

\[ 160-100=60\ \mathrm{pm} \]

This is consistent with the general decrease in atomic radius across a period.

Question 2

Magnesium and sulfur are both in Period 3.

Their electron arrangements and representative covalent radii are:

(a) Calculate the absolute decrease in atomic radius from magnesium to sulfur.

(b) Calculate the percentage decrease in atomic radius.

(c) A student argues that sulfur should have a larger radius because it contains four more electrons than magnesium. Explain why this prediction is incorrect.

Solution 2

Sulfur has a radius \(36\ \mathrm{pm}\) smaller than magnesium, representing a decrease of approximately \(25.7\%\). Its greater nuclear charge outweighs the additional shielding from its outer-shell electrons.

(a) Calculate the absolute decrease.

Use:

\[ \Delta r=r_{\mathrm{Mg}}-r_{\mathrm{S}} \]

Substitute:

\[ \Delta r=140-104 \]

\[ \boxed{\Delta r=36\ \mathrm{pm}} \]

(b) Calculate the percentage decrease.

Use:

\[ \%\text{ decrease} = \frac{r_{\mathrm{Mg}}-r_{\mathrm{S}}} {r_{\mathrm{Mg}}}\times100 \]

Substitute:

\[ \begin{aligned} \%\text{ decrease} &=\frac{140-104}{140}\times100\\ &=\frac{36}{140}\times100\\ &=25.7\% \end{aligned} \]

Therefore, the percentage decrease is approximately \(25.7\%\).

(c) Explain the effect of the additional electrons.

The student’s prediction is incorrect because the number of electrons alone does not determine atomic radius.

Magnesium and sulfur both have three occupied principal energy levels. Their eight inner-shell electrons provide approximately the same shielding.

Sulfur has four additional protons, increasing its nuclear charge from \(+12\) to \(+16\). Its additional outer-shell electrons provide some extra shielding, but not enough to cancel the increased nuclear attraction.

Sulfur therefore experiences a greater effective nuclear charge, causing its electron cloud to contract.

This is why sulfur is smaller despite containing more electrons.

Question 3

A chemistry data sheet gives the following values:

A student plots these values on a graph and concludes that atomic radius increases sharply from chlorine to argon.

(a) Explain why this conclusion is not justified.

(b) Describe how the two radius measurements differ.

(c) Explain what information you would need before making a valid comparison between chlorine and argon.

Solution 3

The student’s conclusion is not justified because the two values use different definitions of atomic radius. The apparent increase could result from the change in measurement convention rather than a genuine reversal of the periodic trend.

(a) Explain why the conclusion is not justified.

The chlorine value is a covalent radius, while the argon value is a van der Waals radius.

These values describe different types of interatomic distances. They cannot be treated as directly equivalent measurements.

The larger argon value does not, by itself, prove that its electron cloud is larger than chlorine’s.

(b) Describe the measurement difference.

Chlorine’s covalent radius is derived from half the internuclear distance in a \(\ce{Cl2}\) molecule.

Argon’s van der Waals radius is derived from the closest approach of non-bonded argon atoms.

Because non-bonded atoms generally remain further apart than covalently bonded atoms, van der Waals radii are generally larger.

(c) Identify the information required.

We would need values determined using a consistent radius convention.

For example, if comparing van der Waals radii, both chlorine and argon should have van der Waals radii.

This would provide a more meaningful comparison, although differences in measurement conditions and atomic interactions would still need to be considered.

The important lesson is that a periodic trend is only meaningful when the quantities being compared have been defined consistently.

Question 4

Beryllium and boron are neighbouring elements in Period 2.

Their electron configurations are:

\[ \mathrm{Be}:1s^2\,2s^2 \]

\[ \mathrm{B}:1s^2\,2s^2\,2p^1 \]

Their representative covalent radii are:

• Beryllium: \(99\ \mathrm{pm}\)
• Boron: \(84\ \mathrm{pm}\)

However, boron has a lower first ionisation energy than beryllium.

A student argues:

If boron’s outer electron is easier to remove, boron must have a larger atomic radius.

Evaluate this argument. Explain why the first ionisation energy decreases from beryllium to boron while the atomic radius also decreases.

Solution 4

The student’s argument is incorrect. Boron has a smaller covalent radius than beryllium, even though its first ionisation energy is lower.

The mistake is assuming that atomic radius and first ionisation energy must change together in exactly the same way.

They measure different properties.

First, consider the electron configurations.

Beryllium has two electrons in its \(2s\) subshell. Boron has one additional proton and one additional electron, which enters the \(2p\) subshell.

The \(2p\) electron in boron is higher in energy and less penetrating than a \(2s\) electron. It is therefore less strongly attracted to the nucleus and easier to remove.

This explains why boron has a lower first ionisation energy.

However, the increased nuclear charge still produces an overall increase in effective nuclear charge across the period.

The additional electron does not provide enough shielding to cancel the extra proton.

Consequently, the electron cloud contracts, and the covalent radius decreases from \(99\ \mathrm{pm}\) to \(84\ \mathrm{pm}\).

The two observations are consistent:

• Boron’s outermost \(2p\) electron is easier to remove.
• Boron’s overall atomic radius is smaller.

A local exception in ionisation energy does not automatically produce an exception in atomic radius.

Question 5

A student is investigating atomic radius and finds the following three values in different reference tables:

The student plots all three values on the same graph and concludes that atomic radius increases dramatically from fluorine to neon before decreasing again for sodium ions.

A second student argues that the following four species must all have exactly the same radius because they each contain 10 electrons:

\[ \mathrm{O}^{2-},\quad \mathrm{F}^{-},\quad \mathrm{Na}^{+},\quad \mathrm{Mg}^{2+} \]

Assume that the ionic radii of these four species are determined using the same radius convention and coordination number.

(a) Explain why the first student’s graph cannot establish a valid periodic trend.

(b) All four ions in the second student’s comparison are isoelectronic. Write their electron configurations, identify their proton numbers, and rank them from largest to smallest ionic radius.

(c) Explain why adding an electron to a neutral fluorine atom to form \(\mathrm{F}^{-}\) generally increases its radius, while changing \(\mathrm{O}^{2-}\) into \(\mathrm{F}^{-}\) decreases its radius.

(d) Can you use the reported covalent radius of neutral fluorine to calculate the ionic radius of \(\mathrm{F}^{-}\)? Explain your reasoning.

Solution 5

The first student’s graph is invalid because it combines three different radius definitions. The second student is also incorrect: the four isoelectronic ions have different radii. Their order from largest to smallest is:

\[ \boxed{ \ce{O^2->F^->Na^+>Mg^2+} } \]

This problem highlights why electron count, nuclear charge, and the measurement convention must all be considered.

(a) Evaluate the first student’s graph.

The three reported values describe different quantities:

• Fluorine’s covalent radius is derived from a bonded pair of fluorine atoms.
• Neon’s van der Waals radius is derived from non-bonded atomic contacts.
• Sodium’s ionic radius describes a positively charged ion in an ionic environment.

These measurements are not directly interchangeable.

In particular, neon’s larger van der Waals radius does not demonstrate that the normal decrease in atomic radius across a period has reversed.

The sodium value introduces another complication: \(\mathrm{Na}^{+}\) is an ion, not a neutral atom.

To make a meaningful comparison, the student would need to use a consistent radius definition and compare equivalent types of species.

(b) Rank the four isoelectronic ions.

First, determine their electron configurations.

All four ions have the same electron configuration.

However, their nuclei have different positive charges.

The proton number increases in this order:

\[ 8<9<11<12 \]

Because all four ions have the same number of electrons, the comparison isolates the effect of increasing nuclear charge.

As proton number increases, the nucleus attracts the same electron distribution more strongly. The electron cloud contracts.

Therefore, ionic radius decreases in this order:

\[ \boxed{ \ce{O^2->F^->Na^+>Mg^2+} } \]

Oxygen ions are largest because their nuclei have the fewest protons attracting the ten electrons.

Magnesium ions are smallest because their nuclei have the greatest proton number.

The important point is that having the same number of electrons does not mean having the same radius.

(c) Explain the two different changes.

This part is where the comparison becomes interesting.

First, consider fluorine becoming a fluoride ion:

\[ \ce{F + e- -> F^-} \]

The nucleus still contains nine protons, but the electron count increases from nine to ten.

The additional electron increases electron-electron repulsion and shielding. The electron cloud expands, so the fluoride ion is generally larger than the neutral fluorine atom.

Now consider the comparison between oxide and fluoride ions:

\[ \mathrm{O}^{2-} \quad \text{and} \quad \mathrm{F}^{-} \]

Both contain ten electrons.

However, the fluoride ion has nine protons, compared with eight in the oxide ion.

The additional proton increases the attraction experienced by the same electron population. The fluoride ion therefore has a smaller radius than the oxide ion.

These comparisons demonstrate two different effects:

• Adding an electron while keeping nuclear charge constant generally expands the electron cloud.
• Adding a proton while keeping electron number constant contracts the electron cloud.

There is no contradiction. The outcome depends on which quantity changes.

(d) Can the covalent radius of fluorine be used to calculate the ionic radius of fluoride?

No. The covalent radius of neutral fluorine cannot be directly converted into the ionic radius of fluoride using a simple numerical relationship.

The two values describe different species in different chemical environments.

The covalent radius is inferred from the distance between bonded neutral fluorine atoms. The ionic radius is inferred from interatomic distances in ionic structures and depends on the coordination environment.

We can predict the qualitative relationship that \(\mathrm{F}^{-}\) is larger than neutral F because the added electron increases repulsion within the electron cloud.

However, this does not provide an exact numerical ionic radius.

This is why a correct comparison needs both the electronic structure and the radius definition.

The next useful periodic trend to connect to atomic radius is first ionisation energy. As atomic radius generally decreases across a period, the outer electrons experience stronger nuclear attraction, which helps explain why more energy is generally required to remove one. The important next step is learning why changes in subshell occupancy sometimes interrupt that energy trend without producing the same pattern in atomic radius.