OxfordAQA International GCSE Chemistry: Structure and bonding of carbon, and nanoparticles (9202) – Revision Notes

These notes condense sections 3.2.3 and 3.2.4 of the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. Everyone takes the same untiered papers, and no required practical sits in these sections. Explanations from first principles, with worked examples, are in the study guide.

Other links: course hub, printable checklist, practice questions and the free diagnostics.

Key definitions

Covalent bond – the bond formed when two atoms share a pair of electrons.

Giant covalent structure – a very large network of atoms, all joined by strong covalent bonds, with no separate molecules.

Delocalised electron – an electron not fixed to one atom or one bond, free to move through the structure.

Fullerene – a form of carbon whose structure is based on hexagonal rings of carbon atoms; different fullerenes contain different numbers of carbon atoms.

Nanoscience – the study of structures 1–100 nm in size, of the order of a few hundred atoms.

Nanoparticle – a particle in the 1–100 nm range. Its properties differ from the same material in bulk.

3.2.3 a: Carbon’s bonding

• Electron arrangement 2,4, so four outer electrons.
• Carbon can form four covalent bonds.
• Diamond, graphite and fullerenes are all pure carbon. Different bonding patterns give different properties.

3.2.3 b–d: Diamond and graphite side by side

	Diamond	Graphite
Bonds per C atom	4	3
Shape	Rigid 3-D giant covalent network	Layers of hexagonal rings
Between layers	Not layered	No covalent bonds; weak forces only
Hard or soft?	Very hard	Soft and slippery
Delocalised electrons	None	One per C atom
Conducts heat and electricity?	Electricity: no	Yes, both
Melting point	Very high	Very high

Method: explain a property in three steps

1. Name the structure (giant covalent; layers; delocalised electrons).
2. Name the bonds or forces involved (strong covalent bonds; weak forces between layers).
3. Link to the property (energy needed to break bonds; layers slide; electrons move and carry charge).

Small reminders

• Diamond is hard: every carbon atom is locked in place by four strong covalent bonds pointing in all directions, so scratching it means breaking many strong bonds.
• Graphite is slippery: no covalent bonds between layers, only weak forces, so the layers slide.
• Graphite conducts: one electron from each carbon atom is delocalised and moves through the structure.
• Graphite is like a metal in one way: both have delocalised electrons.
• Graphite still melts at a very high temperature: melting breaks the strong covalent bonds inside the layers.

Worked reminder: delocalised electrons from a mass

Mass of graphite 0.024 g, Aᵣ C = 12, Avogadro’s constant 6.02 × 10²³ per mol.

n = 0.024 ÷ 12 = 0.0020 mol
atoms = 0.0020 × 6.02 × 10²³ = 1.204 × 10²¹
delocalised electrons = 1.20 × 10²¹ (one per atom)

3.2.3 e: Fullerenes

• Different numbers of carbon atoms.
• Structure based on hexagonal rings of carbon atoms. You need no more structural detail than that.
• Four uses to learn word for word:

Use	Example of linking a use to given information
Drug delivery into the body	A cage shape can carry a molecule inside
Lubricants	Molecules can roll or slide past each other
Catalysts	Large surface for reacting particles to meet
Nanotubes for reinforcing materials, eg tennis rackets	Strong for their mass, so they add strength without much weight

The right-hand column is the style of reasoning to use when a question supplies information. It is not extra content to memorise.

3.2.4: Nanoparticles

Scale

Length	In metres	In nanometres
1 nm	1 × 10⁻⁹ m	1 nm
100 nm	1 × 10⁻⁷ m	100 nm
1 µm	1 × 10⁻⁶ m	1000 nm
1 mm	1 × 10⁻³ m	1 000 000 nm

To convert metres to nm, divide by 1 × 10⁻⁹. To convert µm to nm, multiply by 1000.

Surface area to volume ratio

For a cube of side L:

surface area = 6 × L²
volume       = L³
ratio        = 6 ÷ L

Smaller particles have a higher ratio. Example: a 50 nm cube has a ratio of 6 ÷ 50 = 0.12 nm⁻¹.

Why it matters

• More of the material is at the surface, so it can react or interact more.
• Properties differ from the bulk material.
• The ratio is the reason to quote when explaining new catalysts, sensors or coatings.

Applications listed in the specification

New computers; new catalysts; new coatings; highly selective sensors; stronger and lighter construction materials; new cosmetics such as suntan creams and deodorants.

You do not need specific nanoparticle examples. Questions may give you information about a material and ask you to relate its use to its structure.

Must-know distinctions

• Covalent bonds within graphite layers vs forces between layers: strong vs weak. Hardness is about the second; melting point is about the first.
• Diamond vs graphite conduction: no free electrons vs one delocalised electron per atom.
• Graphite vs metal conduction: both use delocalised electrons. Graphite is not a metal; it is a giant covalent structure.
• Nanoparticle vs bulk: same substance, different size, different properties, because of the far higher surface area to volume ratio.
• Nanoscale range: 1–100 nm, not “anything small”.

Answer frames to adapt

“Compare the hardness of diamond and graphite.” Diamond: four covalent bonds join every carbon atom to its neighbours, making a rigid giant structure, so many strong bonds must break to deform it. Graphite: each carbon atom bonds to three others in layers; there are no covalent bonds between the layers, so the layers slide over each other.

“Suggest why material X is used as a coating.” (information given) Pick out the property in the data (for example, very thin layer, high surface area to volume ratio, does not let water through). Say what the property allows the coating to do. Then name the use. One sentence per link keeps the reasoning clear.

“Is a particle of 6.5 × 10⁻⁸ m a nanoparticle?” 6.5 × 10⁻⁸ ÷ 1 × 10⁻⁹ = 65 nm. This lies within 1–100 nm, so yes. Always show the converted value and then compare it with the range.

Quick self-test

1. How many covalent bonds can a carbon atom form?
2. How many carbon atoms is each carbon atom bonded to in diamond?
3. Why is diamond very hard?
4. Why is graphite soft and slippery?
5. Why does graphite conduct electricity?
6. In what way is graphite similar to a metal?
7. State the shape of the rings that fullerene structures are based on.
8. Give the four uses of fullerenes named in the specification.
9. State the size range for nanoscience.
10. Convert 1 mm to nanometres.
11. A nanoparticle is a cube with sides of 2 nm. Work out its surface area to volume ratio.
12. Give three possible applications of nanoparticles.

Answers

1. Four.
2. Four.
3. Each carbon atom forms four strong covalent bonds in a rigid giant structure, so a lot of energy is needed to break the bonds.
4. There are no covalent bonds between the layers, only weak forces, so the layers slide over each other.
5. One electron from each carbon atom is delocalised and can move through the structure, carrying charge.
6. Both have delocalised electrons that move through the structure.
7. Hexagonal.
8. Drug delivery into the body; lubricants; catalysts; nanotubes for reinforcing materials, eg tennis rackets.
9. 1–100 nm.
10. 1 mm = 1 × 10⁻³ m ÷ 1 × 10⁻⁹ m = 1 000 000 nm (1 × 10⁶ nm).
11. Surface area = 6 × 2² = 24 nm²; volume = 2³ = 8 nm³; ratio = 24 ÷ 8 = 3 nm⁻¹.
12. Any three of: new computers, new catalysts, new coatings, highly selective sensors, stronger and lighter construction materials, new cosmetics such as suntan creams and deodorants.

Where marks are usually lost

• Blaming graphite’s softness on weak covalent bonds. Write “weak forces between layers” or “no covalent bonds between layers” instead.
• Writing “free electrons” with no link to charge moving. Say the delocalised electrons move through the structure and carry the current.
• Saying diamond is hard because it has “strong bonds” without “four bonds per atom” or “giant structure”. A full explanation needs both ideas.
• Writing that ions or atoms move in graphite. Only electrons move.
• Leaving the lower limit off the nanoscale range.
• Treating “smaller particle” and “higher surface area to volume ratio” as the same point. Say the ratio is higher, then say what that allows.
• Converting units wrongly: mixing µm and nm in a ratio, or dividing when you should multiply.
• In information-based questions, ignoring the data given and writing a memorised answer. Quote the information and link it to the use.
• Listing fullerene uses vaguely (“medicine”, “sport”). Use the specification wording: drug delivery, lubricants, catalysts, reinforcing nanotubes.

Next steps

Try the practice questions, then refresh mole calculations with the quantitative chemistry revision notes. Use the diagnostics to choose what to revise next.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.2.3 and 3.2.4.