At GCSE you learned the periodic table is arranged by atomic number, in groups and periods, with electronic structure explaining the pattern. A-Level sharpens that into blocks (s, p, d, f) and asks you to explain the trends across a period — atomic radius, first ionisation energy and melting point across Period 3 — in terms of nuclear charge, shielding, and each element’s structure and bonding. Every explanation on this page reuses the toolkit from Atomic Structure (3.1.1) — electron configuration, sub-shells and ionisation energy — and Bonding (3.1.3) for the melting points. The recurring exam skill is language: name the particles, and say “ion” when you mean an ion.
s, p, d and f blocks
The periodic table divides into four blocks, named after the sub-shell that holds an element’s highest-energy electron. Because that sub-shell follows from the electron configuration — which follows from the proton number — an element’s position fixes its block.
- s block — Groups 1 and 2 (and helium): highest-energy electron in an s sub-shell.
- p block — Groups 13–18 (except helium): highest-energy electron in a p sub-shell.
- d block — the transition series: highest-energy electron in a d sub-shell.
- f block — the lanthanides and actinides: highest-energy electron in an f sub-shell.
Interactive — name the block
- Block = the sub-shell of the highest-energy electron — read it from the electron configuration, not just the group.
- Helium is s block (its electrons are in 1s), despite being placed above the Group 18 noble gases.
🧪 Exam-style questions
Which element is classified as a d block element? Tick (✓) one box.
Which block in the periodic table contains the element samarium (Sm)? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Atomic radius
Across Period 3, the atomic radius decreases from sodium to chlorine. To explain the trend you need to name the particles involved: the nuclear charge, the shielding, and the shell the outer electrons are in. Argon is left out of the comparison — as a noble gas it forms no bonds, so it has no covalent radius to measure, only a much larger van der Waals one.
Across the period the number of protons increases, so the nuclear charge increases. The electrons are added to the same shell, so the shielding stays roughly the same. The stronger nuclear attraction therefore pulls the outer electrons closer, and the atom gets smaller.
The same logic sizes ions. For an isoelectronic set — ions with the same electron configuration — the one with the most protons is the smallest, because the same electrons are pulled in harder.
Put O2−, F−, Na+ and Mg2+ in order of increasing ionic radius.
All four have 10 electrons (the same configuration). The radius is set by the proton count: O (8) < F (9) < Na (11) < Mg (12), so more protons means a smaller ion:
Mg2+ < Na+ < F− < O2− (increasing radius)
Say “ion” throughout — these are ionic radii, not atomic radii.
Interactive — slide across Period 3 and watch the atom shrink
Across the period the nuclear charge rises while the electrons stay in the same shell, so shielding is about the same — the outer electrons are pulled in and the radius falls.
- If it is an ion, say “ion”. Examiner reports withhold marks for “fluoride atom” or “fluoride has an atomic radius” — write “the fluoride ion has a smaller ionic radius”.
- Name the particles. Explain with protons / nuclear charge, shielding and the outer electrons — never “stronger force” on its own.
- Same shell → similar shielding. Across a period the extra electrons go into the same principal shell, so shielding barely changes — that is why nuclear charge wins.
🧪 Exam-style questions
Which represents the correct order of increasing radius of the ions? Tick (✓) one box.
Which ion has the largest radius? Tick (✓) one box.
Explain why the atomic radius decreases across Period 3, from sodium to chlorine.
Show answer
The number of protons (the nuclear charge) increases. 1 mark
The shielding is similar / the electrons are added to the same shell, so the outer electrons are pulled in more strongly. 1 mark
Source: AQA A-Level Chemistry past papers.
First ionisation energy
The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
That definition, the three factors that control it, and the successive-ionisation evidence for shells are covered in full in Atomic Structure (3.1.1). This page uses it to read the trend across Period 3.
Across Period 3 the first ionisation energy generally increases: the same rising nuclear charge and shrinking radius that pulled the atoms in also hold the outer electron more tightly, so more energy is needed to remove it. But the trend has two dips, and you need to be able to explain both.
- Al < Mg. Aluminium’s outer electron is in a 3p sub-shell, slightly higher in energy and more shielded than magnesium’s 3s electron — so it is easier to remove.
- S < P. In sulfur the 3p electrons start to pair up (3p4); the two paired electrons repel, making one easier to remove. Phosphorus has a stable half-filled 3p3.
“Give one element in Period 3 that deviates from the general increase in first ionisation energy, and explain why.”
Element: sulfur. Explanation: sulfur’s outermost 3p electrons begin to pair (3p4); the paired electrons repel each other, so one is easier to remove than expected — giving a lower first ionisation energy than phosphorus.
Aluminium works too: its outer electron is in 3p, higher in energy and more shielded than the 3s of magnesium. Name the sub-shell either way.
The ionisation processes are written per electron, with state symbols. The third ionisation energy of sodium, for example, is:
Na2+(g) → Na3+(g) + e−
Interactive — first ionisation energy, element by element
Pick a Period 3 element to see its outer configuration and why it sits on, above or below the general trend.
- The two dips have different causes. Al < Mg is a 3p-above-3s effect; S < P is 3p4 pairing and repulsion. Do not swap them.
- Name the sub-shell (3s, 3p) and, for sulfur, say the electrons pair and repel — not just “p4 vs p3”.
- Ionisation equations need (g) state symbols and remove one electron at a time.
🧪 Exam-style questions
Which element has a first ionisation energy lower than that of sulfur? Tick (✓) one box.
There is a general increase in first ionisation energy across Period 3. Give one element that deviates from this trend, and explain why the deviation occurs.
Show answer
Element: aluminium (or sulfur). 1 mark
Aluminium: its outer electron is in a 3p sub-shell, 1 mark which is higher in energy / slightly more shielded than the 3s, so it is easier to remove than magnesium’s. 1 mark
Sulfur scores the same way: its 3p electrons begin to pair (M2), and the paired electrons repel, so one is easier to remove (M3).
Give an equation, including state symbols, to represent the process that occurs when the third ionisation energy of sodium is measured.
Show answer
Na2+(g) → Na3+(g) + e− 1 mark
State symbols are essential, and only one electron is removed.
Source: AQA A-Level Chemistry past papers.
Melting point
Melting point across Period 3 is not one smooth trend — it follows the structure and bonding of each element, so it splits into three regions. The three attractions that break on melting — metallic bonding, covalent bonds and van der Waals forces — are covered in Bonding (3.1.3).
| Elements | Structure & bonding | Melting point | What breaks on melting |
|---|---|---|---|
| Na, Mg, Al | giant metallic lattice | rises Na→Al | metallic bonding — stronger with more delocalised electrons & higher charge |
| Si | giant covalent (macromolecular) | highest | many strong covalent bonds |
| P4, S8, Cl2, Ar | simple molecular (Ar monatomic) | low (Ar lowest) | weak van der Waals forces between molecules |
Explain why the melting point increases from sodium to aluminium.
All three are giant metallic lattices, so melting them means overcoming metallic bonding — the attraction between the positive metal ions and the delocalised electrons. From Na to Mg to Al, each atom releases more electrons into that sea (1, then 2, then 3), and the ion left behind is smaller and more highly charged (Na+, Mg2+, Al3+).
So the attraction between the ions and the delocalised electrons is stronger, and more energy is needed to break the lattice — a higher melting point.
Name the structure (giant metallic) and the particles (metal ions, delocalised electrons). “Stronger metallic bonding” on its own does not score.
Interactive — melting point by structure
Pick a Period 3 element to see its structure and bonding, and what has to be overcome to melt it.
- Name the structure and bonding for each region — giant metallic, giant covalent, simple molecular. A melting point “because of stronger bonds” without the structure scores nothing.
- Silicon is highest (giant covalent) — the classic trap is forgetting it or calling it molecular.
- For P4, S8, Cl2 it is the van der Waals forces between molecules that break, not the covalent bonds inside them (S8 > P4 > Cl2 as the molecules get more electrons).
🧪 Exam-style questions
Which element in Period 3 has the highest melting point? Tick (✓) one box.
Which statement is not correct about the Period 3 elements sodium to chlorine? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Capstone: reading Period 3 across
Line the three trends up along the same row and the story is one of nuclear charge rising while the electrons stay in the same shell — until, for melting point, the structure takes over.
| Property, Na → Ar | The pattern | Because… |
|---|---|---|
| Atomic radius | decreases | nuclear charge ↑, same shell (shielding ≈ same) → outer electrons pulled in |
| First ionisation energy | increases (dips at Al, S) | outer electron held more tightly; 3p-above-3s (Al), 3p4 pairing (S) |
| Melting point | rises to Si, then crashes | metallic (Na–Al) → giant covalent Si (highest) → molecular (van der Waals) |
Build every trend answer from these particles, not vague forces: nuclear charge (number of protons), shielding, the outer electron and its sub-shell, ionic radius and charge density (for ions), delocalised electrons (metals), and van der Waals forces (molecular elements). And when the species is an ion, call it an ion.
Interactive — larger or higher, and the reason that scores
now pick the reason a mark scheme would credit
🧪 Exam-style questions
Which row correctly shows the general trends across Period 3? Tick (✓) one box.
Consider the Period 3 elements from sodium to chlorine. Which statement is correct? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
- Blocks (by the highest-energy electron’s sub-shell): s (Groups 1–2 + He), p (13–18), d (transition), f (lanthanides/actinides).
- Atomic radius decreases across Period 3: more protons (nuclear charge ↑), electrons added to the same shell (shielding ≈ same), so the outer electrons are pulled in closer.
- Ionic radius: for an isoelectronic set, more protons → smaller ion (O2− > F− > Na+ > Mg2+). Say “ion” when it is an ion.
- First ionisation energy increases across Period 3, with two dips: Al < Mg (outer electron in 3p, higher/more shielded than 3s) and S < P (3p4 electrons pair and repel; P has a stable 3p3).
- Melting point by structure: Na–Al giant metallic (rising with more delocalised electrons); Si giant covalent — highest; P4, S8, Cl2, Ar molecular — low (weak van der Waals; Ar lowest).
- Language wins marks: name the particles (nuclear charge, shielding, outer electron), and never “stronger force” unqualified.