Periodic trends: atomic radius and electronegativity

Early October 2026: in Year 3 at the science high school (liceo scientifico) the class has just finished (or is finishing) atomic models and electron configuration, and the periodic table is the next chapter. The test on atoms, the table and periodic properties falls between November and December, so five to nine weeks from now. In the meantime bonding and nomenclature arrive, which all rest on this topic: if you have a hole here, you meet it again in every chapter after.

Periodic properties are the part of Year 3 chemistry where the difference between learning and understanding shows best. Radius, ionisation energy, electron affinity and electronegativity are four trends, and whoever studies them as four arrows to remember by heart mixes them up at the first “why” question. Whoever instead has a single explanation in hand, the effective nuclear charge, rebuilds them in a few seconds, even when the teacher changes the elements in the exercise.

This post starts from the explanation and from there arrives at a method for comparing any two elements. I don’t repeat how to write electron configurations, which you’ll find in Electron configuration and quantum numbers, nor the introductory version (for the human sciences track) in Chemistry in Year 3: atoms and bonds: here the centre is the why of the trends.

Why they matter: bonds and reactivity

An atom reacts with another according to how strongly it holds its own outer electrons and how strongly it attracts those of others. Sodium gives up an electron easily, chlorine accepts it gladly: that is why $\mathrm{NaCl}$ is an ionic compound. Fluorine attracts electrons more than almost anything else, which is why its bonds are very polar. All the prediction of the type of bond, which you will see shortly, is a reading of these properties. The geometry of molecules (VSEPR) comes later still.

The single explanation: effective nuclear charge and shielding

Take an electron in the outermost level. It is subject to two opposite effects:

• the attraction of the nucleus, which has $Z$ protons;
• the repulsion of the other electrons, particularly those of the inner levels, which put themselves “in between” it and the nucleus. This effect is called shielding.

The net result is the effective nuclear charge, $Z_{\text{eff}}$: the positive charge that the outer electron really feels. As a first approximation, for main-group elements:

In practice $Z_{\text{eff}}$ is roughly equal to the number of valence electrons (1 for sodium, 7 for chlorine). It is a rough estimate, but enough to predict all the trends.

Two rules follow from this:

1. Along a period (left to right) the number of shells doesn’t change, but $Z$ goes up by one at each step and the electrons being added sit in the same level, so they shield each other little. $Z_{\text{eff}}$ grows: the outer electrons are pulled harder.
2. Down a group (top to bottom) $Z_{\text{eff}}$ stays more or less the same, but a whole shell is added: the outer electron sits further from the nucleus.

All the properties that follow are consequences of these two lines.

Atomic radius

The atomic radius is, roughly, the distance between the nucleus and the outer electrons.

• It decreases along the period. $Z_{\text{eff}}$ grows and the electron cloud is pulled towards the nucleus.
• It increases down the group. Each extra row is an extra shell.

Ionic radii

When an atom becomes an ion it changes size:

• A cation is always smaller than the neutral atom: it loses electrons, often the whole outer shell, and the electrons that remain feel less mutual repulsion and more attraction per electron.
• An anion is always larger than the neutral atom: the same protons have to hold more electrons, the repulsion between electrons increases and the cloud swells.

In short: $\text{cation}<\text{atom}<\text{anion}$.

Ionisation energy

It is the energy needed to pull an electron off an isolated atom (in the gaseous state). Unless you specify otherwise, the first ionisation energy is meant, that is that of the first electron.

• It increases along the period: higher $Z_{\text{eff}}$, electron closer and more tightly held.
• It decreases down the group: the outer electron is further away and more shielded, so it comes off more easily.

Successive energies are always greater than the first: after removing one electron, the total positive charge weighs on fewer electrons.

The two anomalies

If you take the second period the increase is not perfectly regular. There are two inversions that teachers love:

Beryllium and boron: $\mathrm{Be}>\mathrm{B}$. Beryllium has configuration $1s^{2}2s^2$ and loses a $2s$ electron. Boron is $1s^{2}2s^{2}2p^1$ and loses the $2p$ electron. A $p$ orbital has slightly higher energy than an $s$ of the same level and is more shielded by the $2s$ electrons: the boron electron is less tightly bound, even though boron has one more proton.

Nitrogen and oxygen: $\mathrm{N}>\mathrm{O}$. Nitrogen is $2p^3$: three electrons in three different $p$ orbitals, each on its own. Oxygen is $2p^4$: one of the three orbitals holds two paired electrons. These two repel each other because they occupy the same region of space, so removing one of the two costs less. The rest of the exercise is done with Hund’s rule (unpaired electrons as long as possible), which you’ll find in the post on configuration.

Electron affinity

It is the energy involved when an isolated atom gains an electron, forming an anion. Watch the direction: ionisation energy is about losing an electron, affinity is about gaining one. They are two different processes, and whoever swaps them loses marks even with the right trend.

The general trend is that affinity, understood as the tendency to accept an electron, increases along the period (halogens have it very high) and, broadly, falls down the group. Unlike the other properties it has more exceptions: noble gases don’t have it to any appreciable degree (the shell is closed, an extra electron would go into a new level), and in the halogen group fluorine, which is tiny, has a slightly lower affinity than chlorine because the repulsion between electrons is strong in such a small space. In a Year 3 test the general trend is usually asked, and these exceptions are mentioned only if the teacher has covered them in class.

Electronegativity

Electronegativity describes how strongly an atom, inside a bond, attracts the shared electrons to itself. It is not a quantity measurable directly like ionisation energy: it is a scale constructed by Pauling, with no unit of measurement. Values you will often meet:

It follows the same trend as $Z_{\text{eff}}$: it increases along the period and falls down the group. Fluorine, at the top right (excluding the noble gases), is the most electronegative element; the alkali metals, at the left and bottom, are among the least electronegative.

The difference in electronegativity between two atoms indicates whether the bond is nonpolar, polar or ionic. For the rest of the reasoning (which thresholds to use, how to move on to the polarity of the molecule) I refer you to the posts on bonds and geometry.

Summary table

Once it is clear that the radius and the other three properties go in opposite directions, the table is learned in ten seconds: a small atom means electrons held tightly.

The method: comparing two elements in three steps

1. Position. Find the two elements in the table: same period, same group, or neither?
2. $Z_{\text{eff}}$ and shells. Same period: same shells, the one with the higher $Z_{\text{eff}}$ wins (further right). Same group: about the same $Z_{\text{eff}}$, the one with more shells wins (further down). If they share neither, bring one of the two into the same period or group as the other through an intermediate step.
3. Reasoned conclusion. Write the answer with the why in one sentence: ”$X$ has the smaller radius because it has a higher $Z_{\text{eff}}$ with the same number of shells”.

Typical mistakes

• “The radius increases along the period because there are more electrons.” It is the most frequent mistake, and it has a wrong explanation and a wrong conclusion. Electrons do increase, yes, but they sit in the same shell and the protons increase along with them: what decides is the nuclear charge, which pulls everything closer.
• Confusing ionisation energy and electron affinity. The first is losing, the second is gaining. If in the test you read “remove” and answer with affinity, you get it wrong even with the right trend.
• Forgetting the anomalies. Whoever writes “ionisation energy always increases along the period” falls on Be/B and on N/O. A comparison exercise between adjacent elements is built precisely to bring them out.
• Applying the rules to a cation or anion as if they were neutral atoms. For ions the comparison changes: what matters is how many electrons there are compared with protons.

Three worked exercises

1. Arrange in order of increasing atomic radius: Na, Mg, Al, Cl

The four elements are all in the third period: same shells (three). Only $Z$ changes, and so $Z_{\text{eff}}$: sodium group 1, magnesium group 2, aluminium group 13, chlorine group 17. Going right $Z_{\text{eff}}$ grows and the radius falls. The increasing order is therefore the reverse of right to left:

Chlorine is the smallest because its outer electrons feel the highest charge; sodium is the largest because it has a single valence electron and the lowest $Z_{\text{eff}}$.

2. Compare Na with $\mathrm{N}{\mathrm{a}}^{+}$ and Cl with $\mathrm{C}{\mathrm{l}}^{-}$

Sodium. Na has 11 electrons: $1s^{2}2s^{2}2p^{6}3s^1$, three shells. $\mathrm{N}{\mathrm{a}}^{+}$ has lost the $3s$ electron: $1s^{2}2s^{2}2p^6$, only two shells. It has lost a whole level and the remaining 10 electrons are held by 11 protons. So $\mathrm{N}{\mathrm{a}}^{+}$ is much smaller than Na.

Chlorine. Cl has 17 electrons: $\dots 3s^{2}3p^5$. $\mathrm{C}{\mathrm{l}}^{-}$ has 18: $\dots 3s^{2}3p^6$, same shell, one extra electron, but still 17 protons. With the same nuclear charge and more repulsion between electrons, the cloud expands: $\mathrm{C}{\mathrm{l}}^{-}$ is larger than Cl.

3. Why is the first ionisation of O lower than that of N?

I write the valence configurations: nitrogen $2s^{2}2p^3$, oxygen $2s^{2}2p^4$. If I look only at $Z_{\text{eff}}$ I would expect oxygen, which is further right, to have the higher ionisation energy. It doesn’t, because the occupation of the three $2p$ orbitals changes.

In nitrogen the three $2p$ electrons are in three different orbitals, all unpaired. In oxygen the fourth electron must pair up with one of the other three: two electrons in the same orbital repel each other more than two in different orbitals. When oxygen loses one, it loses precisely one of these two paired electrons and the internal repulsion eases: it costs less energy. Conclusion: $E_{\text{ion}}(\mathrm{O})<E_{\text{ion}}(\mathrm{N})$, even though $Z_{\text{eff}}$ is higher in oxygen. The same logic explains Be and B, with the role of the $p$ orbitals in place of pairing.

Test the method

On the site there is the stoichiometry and mole simulator: it is the site’s chemistry simulator, but honestly it doesn’t cover periodic properties. On this topic there is no dedicated practice test. I mention it because moles and stoichiometry are the topic that follows right after in the syllabus, and the work on method (steps, units, checking the result) is the same. For periodic properties, the best training is to redo the exercises above with different elements, changing period and group each round.

When you move on to the mole, look at Moles and stoichiometry in Year 3 science high school: that is where in the syllabus the most marks are lost in the whole of Year 3 chemistry. For nomenclature, which depends on oxidation numbers and so again on position in the table, there is the guide to IUPAC and traditional nomenclature.

How we work in lessons

I always start from $Z_{\text{eff}}$ and the number of shells, and I ask the student to predict the trend before giving it to them. Then we do pairs of elements, chosen to trigger the anomalies, and every answer must have its “why” in one sentence. If last year’s test or an already marked paper shows where you lose marks, we start from that. Other materials and paths on science are on the sciences page.

If you want to see where you stand, write to me from the contact page and we’ll arrange a free introductory call. For costs and lesson formulas there is the pricing page; payment is made offline, after the lessons.