OCR GCSE Chemistry A: Properties of materials (J248)

This study guide teaches sub-topic C2.3 Properties of materials from Topic C2 of OCR GCSE (9-1) Chemistry A (Gateway Science). Following OCR’s Version 4.0 J248 specification (August 2026), it works through statements C2.3a to C2.3j, the mathematical learning outcomes CM2.3i to CM2.3vi and the practical suggestion for this sub-topic. None of C2.3 is Higher Tier only: every statement can be tested on both tiers. You meet this content on Paper 1 if you sit Foundation Tier, or Paper 3 if you sit Higher Tier.

C2.3g to C2.3j and CM2.3ii to CM2.3vi carry the specification’s separate-chemistry symbol: they are in this course but not in combined science.

Links: OCR GCSE Chemistry hub, printable checklist, free 10-minute diagnostics, atomic structure guide and bonding guide.

Coverage map

Ref	You should be able to
C2.3a	Recall that carbon forms four covalent bonds
C2.3b	Explain why carbon forms so many organic compounds (families, chains, rings)
C2.3c	Explain properties of diamond, graphite, fullerenes and graphene
C2.3d	Explain the temperatures of changes of state using energy and bond or force strength
C2.3e	Use data to predict states under given conditions
C2.3f	Link bulk properties of five structure types to their bonding
C2.3g	Compare nano dimensions with atoms and molecules
C2.3h	Surface area to volume and its effect on properties (dissolving tablets, PAG C8)
C2.3i	Link properties of nanoparticulate materials to uses
C2.3j	Explain possible risks of nanoparticulate materials
CM2.3i to vi	2D and 3D representations, scale, standard form, ratios, cube area and volume

Carbon: four bonds, endless compounds (C2.3a, C2.3b)

A carbon atom has four outer electrons and shares all four, so carbon forms four covalent bonds, including strong bonds to other carbon atoms. Carbon atoms therefore link into:

• chains, which can be short or thousands of atoms long, straight or branched
• rings, such as the six-carbon ring found in many medicines and dyes
• families of similar compounds, where each member differs from the next by one CH₂ unit. Methane, ethane and propane (CH₄, C₂H₆, C₃H₈) are the first three of one such family.

Chain length, branching, rings and attached groups can all vary, so the number of possible compounds is enormous: natural ones such as sugars, fats and proteins, and synthetic ones such as plastics and drugs.

Carbon structures (C2.3c, CM2.3i)

The same atoms give very different materials depending on bonding and arrangement.

Form	Structure	Key properties	Why
Diamond	Giant covalent; each C bonded to four others in a rigid 3D tetrahedral network	Very hard, very high melting point, does not conduct	Many strong covalent bonds must break; no free electrons
Graphite	Stacked hexagonal layers, giant covalent; every C joined to three neighbours	Soft and slippery, high melting point, conducts	Weak forces between layers let them slide; one delocalised electron per C moves along the layers
Graphene	One sheet of graphite, just a single atom deep	Strong yet light, flexible, conducts	Strong covalent bonds across the sheet; delocalised electrons
Fullerenes	Hollow molecules: spheres such as C₆₀, or tubes (nanotubes)	C₆₀ is soft and changes state at a far lower temperature than diamond; nanotubes are strong and conduct	C₆₀ molecules are held together only by weak intermolecular forces; nanotubes have strong bonds along their length

Representing 3D structures in 2D (CM2.3i). A diagram is a flat projection of a 3D object. In diamond, a carbon with four lines to it stands for a tetrahedron, with bonds pointing out of the page. A graphite drawing shows stacked hexagonal sheets; the gaps stand for weak forces between layers. Given a model, count the bonds on one atom: four means diamond, three means graphite, graphene or a fullerene.

Worked example 1. Explain why graphite is used in pencils and as an electrode, but diamond is used on drill tips.

1. Graphite has layers held together by weak forces, so layers slide off onto paper.
2. Graphite’s carbons bond to three neighbours each, so one outer electron per atom is delocalised; it moves and carries charge, making graphite a good electrode.
3. Diamond’s carbons each take part in four strong covalent bonds, locked into a 3D network, so it is extremely hard, which suits a cutting tip.

Changes of state and bond strength (C2.3d, C2.3e)

Heating transfers energy to the particles. At the melting point that energy overcomes the forces holding particles in place, so the temperature stays constant until melting is complete; the same happens at the boiling point. Energy is transferred out when a gas condenses or a liquid freezes.

The temperature of a change of state depends on what has to be overcome:

• Simple molecules (such as water or methane): heating only has to overcome the weak intermolecular forces; each molecule’s own covalent bonds stay intact. Little energy is needed, so melting and boiling points are low.
• Giant structures (ionic lattices, giant covalent networks, metals): strong bonds throughout the structure must be broken or loosened. Much more energy is needed, so melting points are high.

Larger simple molecules usually have stronger intermolecular forces and higher boiling points.

Predicting states (C2.3e). Compare the temperature you are given with the melting point (mp) and boiling point (bp):

• below the mp: solid
• between the mp and the bp: liquid
• above the bp: gas

Worked example 2. Use these values to give the state of each substance at 25 °C and at −200 °C.

Substance	mp / °C	bp / °C
A	−117	79
B	63	360
C	−218	−183
D	1538	2862

At 25 °C: A is between −117 and 79, so liquid. B is below 63, so solid. C is above −183, so gas. D is below 1538, so solid.

At −200 °C: A, B and D are below their melting points, so all solid. C is between −218 and −183, so liquid. C is liquid over a range of only 35 °C. D has by far the highest melting point, which points to a giant structure; C, with very low values, is a simple molecular substance with weak intermolecular forces.

Bulk properties and bonding (C2.3f)

Bulk properties (hardness, melting point, conductivity) belong to a large sample. They depend on bond type, bond strength compared with intermolecular forces, and arrangement.

Structure	Bonding and arrangement	Typical bulk properties
Ionic compounds	Giant lattice of oppositely charged ions held by strong electrostatic attraction	High mp; conduct only when molten or dissolved, as ions are then free to move; brittle
Simple molecules	Strong covalent bonds inside molecules; weak intermolecular forces between them	Low mp and bp; often gases or liquids at room temperature; do not conduct
Giant covalent	Every atom held by strong covalent bonds in a continuous network	Very high mp; hard (diamond); usually do not conduct (graphite and graphene are exceptions)
Polymers	Very long chains; strong covalent bonds within chains; intermolecular forces between chains	Usually solid at room temperature; longer chains give stronger forces between chains; do not conduct
Metals	Giant lattice of metal cations surrounded by delocalised electrons	High mp; conduct heat and electricity; malleable, as layers of ions slide

Atoms do not have bulk properties. A single copper atom is not shiny, malleable or a conductor. Those properties come from billions of atoms bonded together with delocalised electrons. Diamond and graphite prove the point: both contain only carbon atoms, yet one is the hardest natural material and the other is soft. The difference is the bonding and arrangement, not the atoms.

Nanoparticles: size and scale (C2.3g, CM2.3ii to CM2.3iv)

One nanometre (1 nm) is 1 × 10⁻⁹ m. Nanoparticles are usually taken as particles between 1 nm and 100 nm across, so each contains anything from a few tens to millions of atoms. An atom has a radius of about 0.1 nm, so an atom is roughly 0.2 nm across; small molecules are a little larger, and a C₆₀ molecule is about 1 nm across. A nanoparticle is therefore bigger than an atom or a small molecule, but far smaller than anything you can see.

Worked example 3. A nanoparticle is 32 nm across. Take the diameter of one atom as 0.2 nm.

• In standard form: 32 nm = 32 × 10⁻⁹ m = 3.2 × 10⁻⁸ m.
• Atoms across the particle: 32 ÷ 0.2 = 160 atoms.
• Compare with a 1 mm grain of salt: 1 × 10⁻³ ÷ 3.2 × 10⁻⁸ = 31 250. The grain is about 31 000 times wider.

Put both numbers in the same unit before you divide.

Surface area to volume (C2.3h, CM2.3v, CM2.3vi)

For a cube of side x:

• surface area = 6x²
• volume = x³
• surface area to volume ratio = 6x² ÷ x³ = 6 ÷ x

So as particles get smaller, the surface area to volume ratio gets larger. If the side is made 10 times smaller, the ratio becomes 10 times larger.

Worked example 4. Compare a cube of side 1 µm (1000 nm) with a cube of side 20 nm.

• 1000 nm cube: surface area = 6 × 1000² = 6 000 000 nm²; volume = 1000³ = 1 000 000 000 nm³; ratio = 0.006 nm⁻¹.
• 20 nm cube: surface area = 6 × 20² = 2400 nm²; volume = 20³ = 8000 nm³; ratio = 0.3 nm⁻¹.
• 0.3 ÷ 0.006 = 50, so the smaller cube has 50 times as much surface for each unit of volume.

How this affects properties. A much larger fraction of a nanoparticle’s atoms sit at the surface, where reactions and dissolving happen, so nanoparticles react faster and make good catalysts.

Practical suggestion: dissolving tablets (PAG C8). Drop identical soluble tablets into the same volume of water at the same temperature: one whole, one broken into halves, one crushed to powder. Time how long each takes to dissolve. Repeat each test and calculate a mean.

Worked example 5. Times to dissolve in seconds: whole tablet 74, 70, 72; halves 51, 55, 53; powder 24, 22, 26. Means: whole = 216 ÷ 3 = 72 s, halves = 159 ÷ 3 = 53 s, powder = 72 ÷ 3 = 24 s. The smaller the pieces, the larger the total surface exposed to water, so the faster the tablet dissolves. The independent variable is particle size; the mass of tablet, volume and temperature of water must be kept the same.

Uses and risks of nanoparticles (C2.3i, C2.3j)

Uses come from properties:

• High surface area to volume ratio: catalysts, where less material gives the same effect.
• Silver nanoparticles are antibacterial: wound dressings and socks.
• Titanium dioxide and zinc oxide nanoparticles absorb or block ultraviolet light but are too small to leave a white layer: sunscreens.
• Carbon nanotubes are strong, light and conduct: reinforcing sports equipment and in electronic components.
• Fullerene cages can carry other molecules: drug delivery is being developed.

Possible risks:

• Nanoparticles are small enough to be breathed in, and some may pass through skin or into cells.
• Their large surface area can make them more reactive than the bulk material, possibly toxic.
• Long-term effects on health are not fully known, because these materials are new.
• Nanoparticles washed out of products could harm useful bacteria and other organisms in the environment.

Link each risk to a property (size, surface area, reactivity), weigh it against the benefit, and note that testing and peer-reviewed research reduce uncertainty.

Common errors

• Claiming that boiling a simple molecular substance breaks covalent bonds; only the forces between molecules are overcome.
• Saying graphite conducts because of “free ions”. It is delocalised electrons.
• Calling C₆₀ a giant structure. It is molecular.
• Giving an atom a bulk property, such as “a copper atom conducts”.
• Mixing nm and m in one calculation.
• Saying a smaller particle has a larger surface area. A single smaller particle has a smaller area; it is the ratio, or the total area of many small pieces, that is larger.

When you finish, check recall using the revision notes, attempt the practice questions, and read OCR GCSE Chemistry exam preparation for paper structure and timing.

Official syllabus

OCR specification J248, Chemistry A (Gateway Science), Cambridge OCR Level 1/Level 2 GCSE (9-1): Version 4.0, August 2026; first assessment 2018. This page draws on sub-topic C2.3 Properties of materials.