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

This study guide teaches sections 3.2.3 (Structure and bonding of carbon) and 3.2.4 (Nanoparticles) of the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. There are no tiers in this qualification: everyone sits the same papers and learns all of this content. No required practical belongs to these two sections.

Once you have read it, use the revision notes to condense it, and check your understanding with the practice questions. The full course is on the OxfordAQA International GCSE Chemistry hub, and you can tick off each statement on the printable checklist. A free 10-minute diagnostic will show you which ideas need the most time.

What this unit covers

Statement	You should be able to
3.2.3 a	State that carbon can form four covalent bonds
3.2.3 b	Describe the giant covalent structure of diamond and explain why it is very hard
3.2.3 c	Describe the layers in graphite and explain why graphite is soft and slippery, using weak forces between the layers
3.2.3 d	Explain why graphite conducts heat and electricity, and compare it with metals
3.2.3 e	State that fullerenes have different numbers of carbon atoms, are based on hexagonal rings, and list their uses
3.2.4	Define nanoscience and nanoparticles, explain why they behave differently from bulk material, and relate given information about a nanomaterial to its use

These sections build on 3.2.2, where giant covalent structures are introduced. If the outer-electron ideas below feel shaky, revisit the atomic structure and periodic table guide first.

3.2.3 a: Why carbon forms four bonds

A carbon atom has six electrons, arranged 2,4. It has four electrons in its outer shell. By sharing each of these four electrons with another atom, carbon makes four covalent bonds and gains a full outer shell.

The same element can join its atoms in different arrangements. Diamond, graphite and the fullerenes are all pure carbon. Their properties differ because the atoms are bonded in different patterns, not because the atoms themselves are different.

3.2.3 b: Diamond

In diamond, each carbon atom forms four covalent bonds with four other carbon atoms. The pattern repeats in all three directions, giving a giant covalent structure with no separate molecules.

Why diamond is very hard:

1. Every atom is held in place by four strong covalent bonds.
2. The bonds point in every direction, so the network is rigid.
3. To scratch or dent diamond you must break many strong covalent bonds, which needs a great deal of energy.

Two linked points follow from the same structure. Diamond has a very high melting point, because melting means breaking many strong covalent bonds (see 3.2.2 f). Diamond does not conduct electricity, because all four outer electrons of every carbon atom are used in bonds, so no electrons are free to move.

3.2.3 c: Graphite and its layers

In graphite, each carbon atom bonds covalently to three others. These bonds join the atoms into flat layers made of hexagonal rings.

There are no covalent bonds between the layers. Only weak forces hold one layer to the next. The layers can therefore slide over each other easily, which makes graphite soft and slippery.

Keep two kinds of force separate in your answers:

• Within a layer: strong covalent bonds. This is why graphite still has a very high melting point.
• Between layers: weak forces. This is why graphite is soft and slippery.

A frequent slip is to blame weak covalent bonds for the softness. That cannot be right, since graphite melts only at a very high temperature. The bonds inside each layer are strong; it is the attraction between layers that is weak.

3.2.3 d: Delocalised electrons in graphite

Carbon has four outer electrons, but in graphite each atom uses only three of them in covalent bonds. The fourth electron from each atom is delocalised. It is not tied to one atom or one bond and can move through the layer.

These moving electrons let graphite conduct electricity and heat. In this way graphite is similar to a metal: both contain delocalised electrons that can move through the structure.

Worked example 1: counting delocalised electrons

A pencil lead contains 0.60 g of graphite. Assume it is pure carbon (Aᵣ of C = 12) and that one electron per carbon atom is delocalised. Calculate the number of delocalised electrons. Avogadro’s constant is 6.02 × 10²³ per mole.

Amount of carbon = mass ÷ Aᵣ = 0.60 ÷ 12 = 0.050 mol
Number of atoms  = 0.050 × 6.02 × 10²³ = 3.01 × 10²² atoms
One delocalised electron per atom, so
Delocalised electrons = 3.01 × 10²²

The mole step comes from section 3.6. If you need practice on it, use the quantitative chemistry study guide.

Comparing the two giant forms

Property	Diamond	Graphite
Bonds per carbon atom	4	3
Arrangement	Rigid network in three dimensions	Flat layers of hexagonal rings
Hardness	Very hard	Soft and slippery
Reason	Many strong bonds in all directions	No covalent bonds between layers; weak forces let layers slide
Conducts electricity?	No	Yes
Reason	No free electrons	One delocalised electron per atom

Worked example 2: explaining a use from structure

A technician uses graphite blocks as sliding contacts that carry current in an electric motor. Explain why graphite suits this job.

• The contact must carry current: graphite has delocalised electrons, one from each carbon atom, which can move through the structure, so it conducts electricity.
• The contact must slide without wearing the motor: graphite’s layers have no covalent bonds between them, only weak forces, so layers slide over each other and the surface is slippery.

Notice that each property is linked to a separate feature of the structure. Two properties need two explanations.

3.2.3 e: Fullerenes

Carbon can also form fullerenes. Fullerenes contain different numbers of carbon atoms, and their structures are based on hexagonal rings of carbon atoms. That is the only structural detail the specification requires.

Uses you must know:

• drug delivery into the body
• lubricants
• catalysts
• nanotubes for reinforcing materials, for example in tennis rackets

In an exam you may be given extra information about a fullerene and asked to link it to a use. For example, if you are told that a fullerene forms hollow cages, you could suggest it can carry a drug molecule inside. If you are told nanotubes are very strong for their mass, you could explain why they reinforce a racket frame without making it heavy. Use only the information you are given; you do not need to memorise extra structural details.

3.2.4: Nanoscience and nanoparticles

Nanoscience is the study of structures that are 1–100 nm in size. A particle in this range contains of the order of a few hundred atoms. Such particles are called nanoparticles.

The nanometre is tiny:

1 nm   = 1 × 10⁻⁹ m
100 nm = 1 × 10⁻⁷ m = 0.1 µm
1000 nm = 1 µm
1 000 000 nm = 1 mm

Worked example 3: is it nanoscale?

Particle A has a diameter of 4.0 × 10⁻⁸ m. Particle B has a diameter of 0.35 µm. Which is a nanoparticle?

A: 4.0 × 10⁻⁸ m ÷ 1 × 10⁻⁹ m per nm = 40 nm   → inside 1–100 nm, nanoparticle
B: 0.35 µm × 1000 nm per µm        = 350 nm  → larger than 100 nm, not a nanoparticle

Why nanoparticles behave differently

Nanoparticles show properties different from those of the same material in bulk. One key reason is that they have a high surface area to volume ratio. A much larger fraction of their atoms sit at the surface, where they can react or interact with other substances.

For a cube of side L, surface area = 6L² and volume = L³, so surface area to volume ratio = 6 ÷ L. Halving the side doubles the ratio.

Worked example 4: surface area to volume ratio

Compare a nanoparticle modelled as a cube of side 5 nm with a grain of the same material modelled as a cube of side 5 µm.

Nanoparticle (L = 5 nm)
  Surface area = 6 × 5²  = 150 nm²
  Volume       = 5³      = 125 nm³
  Ratio        = 150 ÷ 125 = 1.2 nm⁻¹

Grain (L = 5 µm = 5000 nm)
  Ratio        = 6 ÷ 5000 = 0.0012 nm⁻¹

1.2 ÷ 0.0012 = 1000

The nanoparticle’s surface area to volume ratio is 1000 times larger. Both sides must be in the same unit before you divide; mixing nm and µm gives a wrong factor.

Applications

The specification lists possible developments from nanoparticles:

• 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 to learn specific nanoparticles or their properties. Instead, expect to be given information about a material and asked to relate its use to its structure.

Worked example 5: using given information

You are told: “Nanoparticles of metal oxide M speed up a reaction that cleans exhaust gases. The same mass of M as a coarse powder has much less effect.”

Explain this. The nanoparticles have a much higher surface area to volume ratio than the coarse powder. So, for the same mass, much more of oxide M is exposed at the surface, where the gas molecules can make contact with it. This is why nanoparticles are useful as new catalysts.

Common errors

• Putting graphite’s softness down to weak covalent bonds. The weakness is in the attraction between layers.
• Writing that graphite has “free ions” or “free atoms”. It is electrons that are delocalised.
• Forgetting that diamond’s hardness comes from four bonds per atom in a rigid network, not just “strong bonds”.
• Giving only three fullerene uses when asked for four, or inventing uses outside the specification list.
• Quoting the nanoscale as “under 100 nm” without the lower limit. Give 1–100 nm.
• Dividing surface area by volume with the two lengths in different units.
• Saying nanoparticles are “smaller so more reactive” without linking it to surface area to volume ratio.

Next steps

Read the revision notes for a one-page summary, then work through the practice questions. The exam preparation guide covers wider exam technique, and the diagnostics page helps you decide 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.