OCR A Level Chemistry A: Periodicity (H432) – Revision Notes

These notes condense section 3.1.1, Periodicity, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover learning outcomes 3.1.1(a) to (g) in Module 3, Periodic table and energy. H432 is a full A level; the specification names components 01 and 03 as the papers that examine Module 3. The Periodicity study guide holds the longer explanations and every worked example.

Test yourself afterwards with the Periodicity practice questions, check other topics on the OCR chemistry hub, and mark progress on the printable checklist. The free diagnostics show which topics need the most time.

Key terms

Term	Meaning
Atomic (proton) number	Number of protons; the order of the modern table, 3.1.1(a)(i)
Period	Row; properties change in a pattern that repeats in the next row (periodicity), (a)(ii)
Group	Column; same outer electron arrangement, similar chemical properties, (a)(iii)
Block	s, p or d: the sub-shell holding the highest-energy electron, (b)(ii)
First ionisation energy	Energy for 1 mol of gaseous atoms to lose 1 mol of electrons, giving 1 mol of gaseous 1+ ions, (c)
Metallic bonding	Strong electrostatic attraction between cations and delocalised electrons, (d)(i)
Giant covalent lattice	Network of atoms joined by strong covalent bonds, (e)

3.1.1(a): The periodic table and its history

• Order: increasing proton number. Rows show repeating trends; columns share chemical properties.
• HSW: Mendeleev (1869) grouped elements by properties, left gaps and predicted missing elements. The discovery of germanium (1886), matching his forecast, helped the Periodic Law gain acceptance.
• HSW: new elements join only after independent checking. IUPAC approved names for elements 113, 115, 117 and 118 in November 2016, completing the seventh row.

3.1.1(b): Configurations and blocks

• Period 2 fills 2s then 2p, from Li (2s¹) to Ne (2s² 2p⁶). Period 3 repeats the pattern with 3s and 3p, from Na (3s¹) to Ar (3s² 3p⁶).
• Same outer arrangement, one shell further out, means the same group.

Method in steps: classifying an element

1. Write the configuration in filling order (4s before 3d, 5s before 4d).
2. Find the sub-shell the last electron enters: s, p or d gives the block.
3. Highest shell number = period.
4. s-block: group = outer s electrons. p-block: group = 10 + outer s and p electrons.

3.1.1(c): Ionisation energy

Equations. First: X(g) → X⁺(g) + e⁻. Second: X⁺(g) → X²⁺(g) + e⁻. You must be able to write any successive one; only the first needs a written definition.

Three factors. Nuclear charge (up = harder to remove), atomic radius (up = easier), shielding by inner shells (up = easier).

Trend	Direction	Reason in one line
Across Periods 2 and 3	General increase	More protons, same shell, similar shielding, smaller radius
Group 2 → 13 (Be→B, Mg→Al)	Small fall	Electron removed from p sub-shell, higher in energy than s
Group 15 → 16 (N→O, P→S)	Small fall	Electron removed from a paired p orbital; pair repulsion
Down a group	Decrease	Extra shell: larger radius and more shielding outweigh more protons

Method in steps: successive ionisation energies

1. Work out the ratio (or the rise) between each value and the next.
2. Find the largest jump. The electrons before it are the outer shell.
3. Number of outer electrons gives the group (s- and p-block).
4. For a full set, each large jump marks a new, inner shell; count electrons between jumps.
5. On a log-scale graph, look for steps rather than reading exact values.

Worked reminder

Argon (kJ mol⁻¹): 1521, 2666, 3931, 5771, 7238, 8781, 11995, 13842, 40760. Ratios: 1.75, 1.47, 1.47, 1.25, 1.21, 1.37, 1.15, 2.94. The largest jump comes after the 8th value, so eight outer electrons: Group 18. The ratio 1.37 between the 6th and 7th values is small and reflects the change from the 3p electrons to the 3s pair in the same shell, not a new shell.

HSW link. Jumps between groups of ionisation energies were evidence for shells, supporting the Bohr model. The same first and second ionisation energies explain reactivity in Group 2.

3.1.1(d) to (f): Giant lattices

	Giant metallic	Diamond	Graphite	Graphene	Silicon
Particles	Cations + delocalised electrons	C atoms	C atoms	C atoms	Si atoms
Bonds per atom	Not applicable	4	3	3	4
Arrangement	Regular lattice of ions	Tetrahedral network	Hexagonal layers, London forces between layers	One hexagonal layer	Tetrahedral network
Melting point	High (varies)	Very high	Very high	Very high	Very high
Conducts?	Yes, solid and liquid	No	Yes, along layers	Yes	Weakly (semiconductor)
Soluble?	No	No	No	No	No

Graphene (HSW). Strong because every atom is held by strong covalent bonds across the sheet; conducts because delocalised electrons move along it. Suggested uses: flexible electronics, sensors, light and strong composites.

Method in steps: any properties explanation

1. Name the structure (giant metallic, giant covalent, simple molecular).
2. Name the particles.
3. Name what must be overcome (metallic bonds, covalent bonds or London forces) and say strong or weak.
4. For conductivity, name the mobile charged particle, or say there is none.

3.1.1(g): Melting points across Periods 2 and 3

Shape: rise across the metals, peak at Group 14, sharp drop to simple molecules, lowest at the noble gas.

Structure	Period 2	Period 3	What melting overcomes
Giant metallic	Li, Be	Na, Mg, Al	Metallic bonding; stronger as ion charge and delocalised electrons rise
Giant covalent	B, C	Si	Many strong covalent bonds
Simple molecular	N₂, O₂, F₂	P₄, S₈, Cl₂	Weak London forces only
Single atoms	Ne	Ar	Very weak London forces

Period 3 molecular order: S₈ (128 electrons) > P₄ (60) > Cl₂ (34) > Ar (18).

Worked reminder

Why is silicon’s melting point far above phosphorus’s? Silicon: giant covalent, so melting breaks many strong Si–Si bonds. Phosphorus: simple molecular P₄, so melting overcomes only weak London forces between molecules; the P–P bonds stay intact.

Must-know distinctions

• Fall from Group 2 to 13 (sub-shell energy) versus fall from Group 15 to 16 (pair repulsion). Do not swap the reasons.
• Shielding stays similar across a period but increases down a group.
• Ionisation energy equation: gaseous atom loses an electron. Not a solid, not a gained electron.
• Graphite conducts along layers; diamond does not conduct.
• Melting a simple molecular solid breaks London forces, not covalent bonds.
• Metallic bonding is between cations and delocalised electrons, not between atoms.

Quick self-test

1. What property of an atom fixes its position in the modern periodic table?
2. Give the block, group and period of the element with Z = 37.
3. Give the block, group and period of the element with Z = 49.
4. Write an equation for the first ionisation energy of sulfur.
5. Write an equation for the second ionisation energy of aluminium.
6. Explain why aluminium’s first ionisation energy is lower than magnesium’s.
7. An element has successive ionisation energies of 590, 1145, 4912 and 6491 kJ mol⁻¹. Deduce its group.
8. Beryllium’s four ionisation energies are 900, 1757, 14849 and 21007 kJ mol⁻¹. How many electrons are in each shell?
9. Why is argon’s first ionisation energy lower than neon’s?
10. Which of diamond, graphite, graphene and silicon are good electrical conductors?
11. Why does sulfur melt at a higher temperature than phosphorus?
12. Why does aluminium melt at a higher temperature than sodium?

Answers

1. Its atomic (proton) number.
2. [Kr] 5s¹: s-block, Group 1, Period 5 (rubidium).
3. [Kr] 4d¹⁰ 5s² 5p¹: p-block, Group 13, Period 5 (indium).
4. S(g) → S⁺(g) + e⁻
5. Al⁺(g) → Al²⁺(g) + e⁻
6. Aluminium’s outer electron is in 3p, which is higher in energy than magnesium’s 3s, so less energy is needed to remove it.
7. Largest jump after the 2nd value (ratio 4.29), so two outer electrons: Group 2.
8. Jump after the 2nd value (ratio 8.45): 2 in the inner shell, 2 in the outer shell.
9. Argon’s outer electron is in a shell further out, with more shielding; this outweighs its larger nuclear charge.
10. Graphite and graphene (silicon is only a semiconductor; diamond does not conduct).
11. Both are simple molecular. S₈ has 128 electrons and P₄ has 60, so London forces between S₈ molecules are stronger.
12. Al³⁺ has a higher charge and is smaller than Na⁺, and aluminium gives three delocalised electrons per atom against sodium’s one, so the metallic bonding is stronger.

Where marks are usually lost

• Leaving (g) off the atom or ion in an ionisation equation.
• Defining first ionisation energy for “an atom” without the moles, or for a solid.
• Giving “half-filled sub-shell stability” as the reason for the fall from N to O instead of repulsion between paired p electrons.
• Saying shielding increases across a period.
• Reading the group from the electrons after the largest jump instead of before it.
• Treating the small rise between an s and a p electron in the same shell as a new shell.
• Saying silicon or diamond conducts because “electrons are free”.
• Stating that melting chlorine or phosphorus breaks covalent bonds.
• Using “atoms” in a metallic bonding answer instead of cations and delocalised electrons.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.1 Periodicity, learning outcomes (a) to (g). Published by OCR.