Lesson 1.6 · Unit 1 · CED topic 1.7

Periodic trends

Dmitri Mendeleev arranged the elements by mass and noticed that properties recur at regular intervals. We now know why: the table is organized by electron configuration, and properties such as size and the grip on outer electrons repeat whenever a new shell begins. This lesson gives you the working tools of periodicity, effective nuclear charge and shielding, and applies them to four properties: ionization energy, atomic and ionic radius, electron affinity and electronegativity. Everything is an application of Coulomb’s law.

Learning objective · 1.7.A

Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity.

Required content · EK 1.7.A.1 to 1.7.A.3

The organization of the periodic table is based on patterns of recurring properties of the elements, which are explained by patterns of ground-state electron configurations and the presence of completely or partially filled shells (and subshells) of electrons in atoms.
Trends in atomic properties within the periodic table (periodicity) can be predicted by the position of the element on the periodic table and qualitatively understood using Coulomb’s law, the shell model, and the concepts of shielding and effective nuclear charge. These properties include ionization energy, atomic and ionic radii, electron affinity and electronegativity.
The periodicity is useful to predict or estimate values of properties in the absence of data.
Exclusion: writing the configuration of elements that are exceptions to the Aufbau principle is not assessed.

Shielding and effective nuclear charge

A valence electron is attracted by the nucleus but repelled by the other electrons. The inner (core) electrons shield it: they sit between the nucleus and the valence electron and cancel part of the nuclear charge. The net pull the valence electron feels is the effective nuclear charge, \(Z_{\text{eff}}\), roughly the nuclear charge minus the number of core electrons. For sodium that is about 11 − 10 = +1; for chlorine about 17 − 10 = +7.

Figure 1.6c Effective nuclear charge. Core electrons shield valence electrons from part of the nuclear charge, so Z_eff ≈ Z − (core electrons). Both atoms have 10 core electrons, but chlorine’s larger Z gives a stronger pull on its valence shell.

Two facts organize all of the trends:

Across a period (left to right) protons are added but the new electrons go into the same shell, which shields almost not at all. \(Z_{\text{eff}}\) increases, and the valence electrons are pulled in and held more tightly.
Down a group a whole new shell is added. The valence electrons are farther from the nucleus (larger \(r\)) and shielded by more core electrons, so they are held more loosely even though the nuclear charge is larger.

Ionization energy

The first ionization energy is the energy needed to remove one electron from a gaseous atom: \(\text{X}(g) \rightarrow \text{X}^+(g) + e^-\). A high value means the electron is held tightly. By the two facts above, ionization energy increases across a period and decreases down a group.

Figure 1.6a First ionization energies across period 2 (kJ/mol). The general rise comes from increasing effective nuclear charge; the dips at B (a higher-energy 2p subshell) and O (a paired 2p electron that is repelled) are the exceptions you are expected to explain.

The graph for period 2 rises overall from lithium (520 kJ/mol) to neon (2081 kJ/mol), with two dips. Beryllium (900) to boron (800): the boron electron being removed is in a 2p subshell, which is higher in energy and slightly farther out than the 2s electrons of beryllium, and partly shielded by them. Nitrogen (1402) to oxygen (1314): in oxygen the electron removed comes from a 2p orbital that holds two electrons; the repulsion between the pair makes that electron easier to remove than the unpaired electron in nitrogen.

Successive ionization energies

Removing a second, third and later electron costs more each time, because the remaining electrons feel a larger net pull from the same nuclear charge. The key pattern is a huge jump when the first core electron is removed. Sodium needs 496 kJ/mol for the first electron and 4562 kJ/mol for the second, a jump of about nine times. That tells you sodium has one valence electron, because the second comes from the inner shell, which is much closer to the nucleus.

Figure 1.6e Successive ionization energies (kJ/mol). A large jump shows that the next electron comes from an inner shell: the jump after IE₁ for Na, after IE₂ for Mg and after IE₃ for Al shows the number of valence electrons (1, 2, 3).

Worked example · the group from successive ionization energies

An element has ionization energies (kJ/mol) of 578, 1817, 2745 and 11,577 for the first four electrons. How many valence electrons does it have, and what element in period 3 is it?

The big jump is between the third and fourth, so three electrons are valence electrons (group 13). In period 3, group 13 is aluminum.

Atomic radius

Atomic radius is half the distance between nuclei of two bonded atoms. It decreases across a period (rising \(Z_{\text{eff}}\) pulls the shell in) and increases down a group (added shells). Sodium (186 pm) to chlorine (99 pm), and lithium (152 pm) to cesium (265 pm) show the two directions.

Figure 1.6b Atomic radius increases down a group (more shells) and decreases across a period (greater effective nuclear charge pulls the same shell closer). Values are approximate.

Ionic radius

A cation is smaller than its atom: it has lost its outermost shell (or, for a few cases, the electron count falls so repulsion drops) and the same nuclear charge now holds fewer electrons. Na (186 pm) becomes Na⁺ (102 pm). An anion is larger than its atom: electrons are added to the same shell with the same nuclear charge, so repulsion rises and the cloud expands. Cl (99 pm) becomes Cl⁻ (181 pm). For isoelectronic species (the same number of electrons), the one with the larger nuclear charge is smaller: O²⁻ (8 protons, 140 pm) is bigger than F⁻, which is bigger than Na⁺ (11 protons, 102 pm), which is bigger than Mg²⁺ (12 protons, 72 pm).

Figure 1.6d Atoms and their ions (radii in pm, approximate). Losing electrons makes a cation smaller (fewer electrons, and often the outer shell is gone); gaining electrons makes an anion larger (more electron-electron repulsion for the same nuclear charge).

Electron affinity and electronegativity

Electron affinity is the energy change when a gaseous atom gains an electron. Elements near the top right (the halogens especially) release the most energy because the added electron meets a high \(Z_{\text{eff}}\) (chlorine releases 349 kJ/mol). Noble gases and some elements with filled subshells do not add an electron favorably. Electronegativity is a different idea: how strongly an atom in a bond attracts the shared electrons. It follows the same pattern as ionization energy: it increases across a period and decreases down a group, so fluorine is the most electronegative element and the elements at the lower left (cesium, francium) the least. The AP course uses electronegativity as a qualitative guide to bond polarity in Unit 2.

Property	Across a period (→)	Down a group (↓)	Reason
Atomic radius	Decreases	Increases	Z_eff up across; new shells down
First ionization energy	Increases (with small dips)	Decreases	Tighter grip across; farther and shielded down
Electronegativity	Increases	Decreases	Same as ionization energy
Electron affinity	Generally more negative (exceptions)	Generally less negative	Z_eff and distance

Worked example · a full free-response explanation

Explain why the first ionization energy of sodium (496 kJ/mol) is lower than that of lithium (520 kJ/mol).

Both atoms have one valence electron. In sodium it is in the third shell and in lithium in the second, so the sodium valence electron is farther from the nucleus. It is also shielded by ten core electrons rather than two. Although sodium has a larger nuclear charge (+11 against +3), the greater distance and shielding mean that by Coulomb’s law the attraction to the valence electron is weaker, so less energy is needed to remove it.

Exam focus

Every answer needs three parts: the comparison (nuclear charge, distance, shielding), the Coulomb’s law consequence (weaker or stronger attraction), and the result (easier or harder to remove).
Do not say “it wants a full shell.” Atoms do not want; explain with forces and distances.
Cations shrink, anions grow. Compare isoelectronic ions by nuclear charge.
Use the jump. A very large increase in successive ionization energy marks the end of the valence electrons.
Know the two dips in period 2 ionization energy and the reason for each (subshell change; paired electrons).

Practice

Rank Li, Na and K from lowest to highest first ionization energy and justify the ranking.
Show answer
K < Na < Li. Down the group the valence electron is in a higher shell, farther from the nucleus and shielded by more core electrons, so it is held less tightly and removed more easily.
Rank Na, Mg, Al and Cl by atomic radius, largest first, and explain.
Show answer
Na > Mg > Al > Cl. All are in period 3, so the valence electrons are in the same shell, but the nuclear charge increases from 11 to 17 with almost no extra shielding, so the electron cloud is pulled in.
The successive ionization energies of an element (kJ/mol) are 738, 1451, 7733 and 10,540. Identify its group and a likely ion charge.
Show answer
The jump is between the second and third ionization energy, so there are two valence electrons (group 2). It will form an ion with charge 2+ (and is magnesium, if in period 3).
Explain why the Cl⁻ ion (181 pm) is larger than the Cl atom (99 pm).
Show answer
Cl⁻ has one more electron than the atom but the same nuclear charge (+17). The added electron increases electron-electron repulsion in the third shell and each electron feels a smaller share of the nuclear attraction, so the electron cloud expands.
Arrange K⁺, Ca²⁺ and Cl⁻ from smallest to largest and explain.
Show answer
Ca²⁺ < K⁺ < Cl⁻. All have 18 electrons (isoelectronic), so the ion with the most protons pulls the electrons in the most: Ca (20) > K (19) > Cl (17).
The first ionization energy of oxygen (1314 kJ/mol) is lower than that of nitrogen (1402 kJ/mol), even though oxygen is farther right. Explain.
Show answer
The electron removed from oxygen comes from a 2p orbital that already holds two electrons. The repulsion between the paired electrons raises the energy of that electron, so it is easier to remove than the unpaired 2p electron of nitrogen, which is in an orbital by itself.