4.2 Bonding, structure, and the properties of matter — notes on one page

Every AQA GCSE Chemistry 8462 specification point in 4.2, read straight through: 18 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 8462 section 4.2. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

4.2.1.1 · Chemical bonds

The three types of strong chemical bond are ionic, covalent and metallic. Ionic bonding is the electrostatic attraction between oppositely charged ions, usually formed when electrons transfer from metal atoms to non-metal atoms.
A covalent bond is a shared pair of electrons between non-metal atoms.
Metallic bonding is the strong electrostatic attraction between positive metal ions and delocalised electrons throughout a giant structure.
Use the elements and particles shown to identify the bonding: metal plus non-metal suggests ionic; non-metals joined in molecules or networks are covalent; and a metal element or alloy is metallic.
Examiners expect both the correct particles and the attraction between them, not only a bond name.

Particle models for ionic, covalent and metallic bonding.

Worked example

Identify the strong bonding in sodium chloride, chlorine and copper.

1.Sodium is a metal and chlorine is a non-metal, so sodium chloride contains oppositely charged ions and has ionic bonding.
2.Chlorine atoms are non-metals joined by shared electron pairs, so chlorine has covalent bonding.
3.Copper is a metal with positive ions and delocalised electrons, so it has metallic bonding.

Answer: Sodium chloride is ionic, chlorine is covalent and copper is metallic.

Common mistakes

Don't say ionic bonding is electron transfer, although transfer forms the ions and the bond is the attraction between opposite charges.
Don't describe metallic bonding as positive ions attracting each other instead of positive ions attracting delocalised electrons.
Don't call the weak forces between molecules covalent bonds.

Exam tip. For an ‘explain the bonding’ question, name the charged particles or shared pair and state the electrostatic attraction.

4.2.1.2 · Ionic bonding

Ionic compounds form when electrons are transferred so that atoms gain the electronic structures of noble gases. Metal atoms lose outer-shell electrons and become positive ions; non-metal atoms gain electrons and become negative ions.
Group 1 and Group 2 metals form $1+$ and $2+$ ions, while Group 6 and Group 7 non-metals form $2-$ and $1-$ ions.
The compound must be electrically neutral, so the total positive and negative charges balance.
In a dot-and-cross diagram, show outer-shell electrons only, distinguish their origins, enclose each ion in square brackets and write its charge.
The ionic bond is then the electrostatic attraction between the resulting oppositely charged ions.

Electron transfer forms oppositely charged ions that attract.

Worked example

Describe the ions formed when one magnesium atom reacts with chlorine atoms.

1.A magnesium atom loses two outer-shell electrons to form Mg $^{2+}$ .
2.Two chlorine atoms each gain one electron to form two Cl $^-$ ions.
3.The charges balance: one $2+$ ion is matched by two $1-$ ions.

Answer: One Mg $^{2+}$ ion and two Cl $^-$ ions form, giving MgCl $_2$ .

Common mistakes

Don't draw the transferred electron outside the recipient ion’s full outer shell.
Don't omit square brackets or charges from a dot-and-cross diagram of ions.
Don't use the number of transferred electrons as the formula without first balancing total charge.

Exam tip. In a dot-and-cross question, check electron origin, complete outer shells, brackets, charges and overall charge balance.

4.2.1.3 · Ionic compounds

An ionic compound has a regular giant lattice containing positive and negative ions.
Strong electrostatic forces of attraction act in all directions between oppositely charged ions; it is not a collection of separate molecules.
Ionic structures must be recognised from dot-and-cross, ball-and-stick and two- or three-dimensional diagrams.
To deduce an empirical formula from a model, count each ion type and simplify the ratio to the smallest whole numbers, checking that the total charge balances.
Models are useful representations but not literal pictures: they may show incorrect relative sizes and separations, suggest directional sticks, omit charges or display only a tiny section of the repeating giant lattice.

Worked example

A lattice model contains $12$ X $^{2+}$ ions and $24$ Y $^-$ ions. Deduce the empirical formula.

1.Count the ions to obtain the ratio X:Y $=12:24$ .
2.Divide both numbers by $12$ to obtain the simplest ratio $1:2$ .
3.Check charge balance: $1(2+)+2(1-)=0$ .

Answer: The empirical formula is XY $_2$ .

Common mistakes

Don't write the unsimplified ion count as the formula instead of the smallest whole-number ratio.
Don't describe an ionic lattice as molecules joined by sticks.
Don't claim a two-dimensional model shows the full three-dimensional lattice.

Exam tip. For ‘deduce the empirical formula’, show the counted ratio, simplify it and confirm that the charges balance.

4.2.1.4 · Covalent bonding

A covalent bond is a shared pair of electrons between atoms, and bonds within molecules or giant structures are strong.
Covalent substances may consist of small molecules, very large polymer molecules or giant covalent structures such as diamond and silicon dioxide.
Required skills include recognising these structures from formulae and bonding diagrams, drawing dot-and-cross diagrams for the specified simple molecules, and using one line to represent each single covalent bond.
Counting atoms in a complete molecular model gives its molecular formula.
Diagrams are simplified: lines are not physical sticks, atom sizes and bond angles may be unrealistic, and a small section may represent a much larger polymer or giant structure.

Covalent substances can be small molecules, polymers or giant structures.

Worked example

Describe the covalent bonding in a methane molecule, CH $_4$ .

1.Place one carbon atom at the centre with four hydrogen atoms around it.
2.Carbon shares one pair of electrons with each hydrogen atom.
3.Represent the four shared pairs as four single C–H bonds.

Answer: Methane contains four strong covalent bonds, each formed from one shared pair of electrons.

Common mistakes

Don't call a covalent bond weak because a simple molecular substance has a low boiling point.
Don't count lines instead of atoms when deducing a molecular formula.
Don't treat the displayed section of a polymer or giant structure as the whole particle.

Exam tip. Before interpreting a covalent diagram, decide whether it shows a complete molecule, a repeating polymer unit or part of a giant structure.

4.2.1.5 · Metallic bonding

Metals consist of giant structures of atoms arranged in a regular pattern. Their outer-shell electrons become delocalised, so they are free to move through the whole structure rather than belonging to one atom or one pair of atoms.
The remaining positive metal ions form a regular lattice. Strong metallic bonding arises from the electrostatic attraction between these positive ions and the negatively charged delocalised electrons.
A diagram should be interpreted by identifying both components and the repeated giant arrangement.
Metals must not be described as separate molecules or as positive ions attracting each other.
Examiners reuse this model in explanations of conductivity and the strength of metals.

A regular lattice of positive metal ions surrounded by delocalised electrons.

Worked example

Explain why the bonding in a piece of magnesium metal is strong.

1.Magnesium forms a giant regular structure of positive metal ions.
2.Its outer-shell electrons are delocalised through the whole structure.
3.There is strong electrostatic attraction between the positive ions and negative delocalised electrons.

Answer: Strong metallic bonds hold the giant magnesium structure together.

Common mistakes

Don't say metallic bonding is attraction between positive metal ions.
Don't draw electrons attached to particular neighbouring ion pairs rather than delocalised through the structure.
Don't describe a piece of metal as many separate molecules.

Exam tip. For ‘explain why metallic bonding is strong’, state the strong electrostatic attraction between positive ions and delocalised electrons.

4.2.2.1 · The three states of matter

Definition. The three states of matter are solid, liquid and gas. Melting and freezing both take place at the melting point; boiling and condensing both take place at the boiling point.

The simple particle model represents particles as small solid spheres and explains melting, boiling, freezing and condensing. In a solid the particles are packed closely in a regular arrangement and vibrate about fixed positions; in a liquid they can move past one another; in a gas they are far apart and move randomly and quickly.
Changing state does not change the particles themselves — only their energy, and so how far apart and how freely they move.
The energy needed to change state depends on the strength of the forces between particles, set by bonding and structure: the stronger those forces, the higher the melting and boiling points.
Predict a state from a data table: solid below the melting point, liquid between the two, gas above the boiling point.
Higher tier: AQA restricts "explain the limitations of the particle theory in relation to changes of state" to Higher tier — solid, inelastic spheres with no forces between them.

The simple particle model for a solid, liquid and gas.

Worked example

A substance melts at $-20\,^{\circ}\mathrm{C}$ and boils at $70\,^{\circ}\mathrm{C}$ . State its physical state at $25\,^{\circ}\mathrm{C}$ .

1. $25\,^{\circ}\mathrm{C}$ is above the melting point, so the substance is not solid.
2. $25\,^{\circ}\mathrm{C}$ is below the boiling point, so the substance is not a gas.
3.The temperature lies between the two points.

Answer: The substance is liquid at $25\,^{\circ}\mathrm{C}$ .

Mark-scheme wording to use

“hydrogen is not shown as H₂ / molecules … particles are shown as spheres … particles are shown as solid … does not show the (weak) forces (between particles) … does not show the movement / speed (of particles) … is only two-dimensional” — AQA 8462, November 2020 Paper 1H, Q07.4
“the intermolecular forces are weak” — AQA 8462, November 2021 Paper 1H, Q01.5
“-35 (°C) … any value between -35 °C and -100 °C” — AQA 8462, November 2020 Paper 1F, Q01.2

What the examiner wants. For a state prediction, the temperature compared against both the melting and the boiling point, not just one. For a limitations question, features of the model itself — never features of the real substance.

Common mistakes

Don't decide the state using only the boiling point and ignore the melting point.
Don't say particles themselves melt or boil rather than the substance changing state.
Don't claim the simple model shows forces between particles even though it does not (Higher tier).

Exam tip. When predicting state, write the temperature comparison with both the melting point and boiling point.

4.2.2.2 · State symbols

State symbols show the physical state of each substance in a chemical equation: (s) means solid, (l) means liquid, (g) means gas and (aq) means dissolved in water.
Place each symbol immediately after its formula and use the conditions or information in the question rather than guessing from the substance name.
A solution is aqueous, while a precipitate is solid and a gas released during a reaction is gaseous.
Aqueous does not mean the same as liquid: NaCl(aq) is sodium chloride dissolved in water, whereas NaCl(l) is molten sodium chloride.
Examiners may require suitable state symbols as part of a complete balanced equation.

Worked example

Add state symbols to the reaction of solid magnesium with hydrochloric acid solution, producing magnesium chloride solution and hydrogen gas.

1.Magnesium is stated to be solid, so write Mg(s).
2.Both solutions are aqueous, so write HCl(aq) and MgCl $_2$ (aq).
3.Hydrogen is released as a gas, so write H $_2$ (g), then balance the equation.

Answer: Mg(s) + 2HCl(aq) → MgCl $_2$ (aq) + H $_2$ (g).

Common mistakes

Don't use (l) for a substance dissolved in water instead of (aq).
Don't place a state symbol after a coefficient rather than immediately after the formula.
Don't assume every product formed in solution is aqueous even when it is described as a precipitate.

Exam tip. For a symbol-equation question, balance the formulae and then check that every substance has the correct state symbol.

4.2.2.3 · Properties of ionic compounds

Ionic compounds have regular giant lattices with strong electrostatic forces of attraction in all directions between oppositely charged ions. Their melting and boiling points are high because large amounts of energy are needed to overcome the many strong ionic bonds.
Electrical conductivity depends on whether charged particles can move.
In a solid ionic compound, the ions are held in fixed lattice positions, so charge cannot flow.
When the compound is molten or dissolved in water, its ions are free to move and carry charge, so it conducts.
Examiners expect a full structure–bonding–property chain and the correct charge carrier: ions move, not electrons.

Ions are fixed in a solid ionic lattice but mobile when molten.

Worked example

Explain why sodium chloride conducts when molten but not when solid.

1.In solid sodium chloride, oppositely charged ions are held in fixed lattice positions.
2.The fixed ions cannot move, so charge cannot flow through the solid.
3.Melting frees the ions to move and carry charge through the liquid.

Answer: Molten sodium chloride conducts because its ions are mobile; solid sodium chloride does not because its ions are fixed.

Common mistakes

Don't say solid sodium chloride conducts because it contains charged ions, ignoring that the ions cannot move.
Don't name electrons as the charge carriers in a molten ionic compound.
Don't explain a high melting point using weak attractions or only one ionic bond.

Exam tip. For a property explanation, link giant lattice → many strong attractions → energy, or ion mobility → charge flow.

4.2.2.4 · Properties of small molecules

Definition. Substances made of small molecules are usually gases or liquids with relatively low melting and boiling points. Melting or boiling them overcomes the weak intermolecular forces between molecules, not the covalent bonds inside them.

This distinction is tested repeatedly across 4.2. The covalent bonds within a molecule are strong; the intermolecular forces between separate molecules are weak.
When a molecular substance melts or boils, only the intermolecular forces are overcome, so melting and boiling points are low even though the covalent bonds are strong. Intermolecular forces increase with molecule size, so larger molecules have higher melting and boiling points — which is why methane is a gas, longer-chain hydrocarbons are liquids and poly(ethene) is a solid.
A full-mark explanation is a chain of linked steps: name the molecule size, name the intermolecular forces, say how much energy is needed to overcome them, then state the melting or boiling point.
Substances made of small molecules do not conduct electricity, because the molecules have no overall charge — no ions and no delocalised electrons to carry it.
Boiling point is a bulk property and cannot belong to a single molecule.

Weak intermolecular forces act between molecules; strong covalent bonds act within them.

Worked example

Explain why a small molecular substance can boil at a low temperature even though it contains strong covalent bonds.

1.Boiling separates whole molecules from one another.
2.Only the weak intermolecular forces between molecules are overcome.
3.The strong covalent bonds within each molecule remain intact.

Answer: Little energy is needed to overcome the weak intermolecular forces, so the boiling point is low.

Mark-scheme wording to use

“small molecules … weak intermolecular forces … little energy is needed to overcome the intermolecular forces” — AQA 8462, June 2019 Paper 1H, Q05.3
“Do not accept reference to weak(er) covalent bonds; do not accept reference to breaking covalent bonds” — AQA 8462, June 2023 Paper 1H, Q04.5
“molecules have no overall electric charge; allow no ions or no delocalised electrons” — AQA 8462, June 2025 Paper 1H, Q02.4

What the examiner wants. The words 'intermolecular forces', never 'weak covalent bonds'. The scheme carries an explicit do-not-accept for breaking covalent bonds, so one careless phrase can cost the whole four marks.

Common mistakes

Don't say covalent bonds break when a molecular substance boils.
Don't call the intermolecular forces strong when explaining a low boiling point.
Don't claim neutral molecules conduct by carrying charge through the substance.

Exam tip. Use the exact phrase ‘intermolecular forces are overcome’ for melting or boiling a simple molecular substance.

4.2.2.5 · Polymers

Polymers have very large molecules. Within each polymer molecule, atoms are linked by strong covalent bonds in a long repeating chain.
Separate chains remain separate molecules and are held near one another by intermolecular forces. These forces are relatively strong compared with those between small molecules, so polymer substances are solids at room temperature.
A polymer diagram can be recognised by tracing a repeated bonding pattern along a long chain.
It must not be confused with a giant covalent structure: a polymer contains separate very large molecules, whereas covalent bonds extend throughout a giant network.
Melting a polymer overcomes intermolecular forces between chains, not the covalent bonds along each chain.

Separate long polymer molecules compared with a continuous giant covalent network.

Worked example

Explain why a substance made from long polymer chains is solid at room temperature.

1.The polymer contains very large molecules formed from long chains.
2.Intermolecular forces between these large molecules are relatively strong.
3.Room-temperature energy is insufficient to overcome enough of these forces for the chains to move past one another.

Answer: The relatively strong intermolecular forces keep the polymer solid at room temperature.

Common mistakes

Don't call every polymer a giant covalent structure even though polymer chains are separate molecules.
Don't say weak covalent bonds are overcome when a polymer melts.
Don't identify one displayed repeating unit as the complete polymer molecule.

Exam tip. For ‘recognise the structure’, trace whether covalent bonding ends at separate chains or continues through one giant network.

4.2.2.6 · Giant covalent structures

A giant covalent structure is a continuous network in which all atoms are linked to other atoms by strong covalent bonds.
Diamond, graphite and silicon dioxide are examples.
A diagram shows only part of the structure, so the key feature is that the bonding pattern continues rather than ending at the boundary of a separate molecule.
Giant covalent substances are solids with very high melting points because many strong covalent bonds must be overcome to melt or boil them, requiring a large energy transfer.
Intermolecular forces must not be used in this explanation because a giant covalent substance does not consist of small, separate molecules.

Worked example

Silicon dioxide remains solid at $1500\,^{\circ}\mathrm{C}$ . Explain this observation.

1.Silicon dioxide has a giant covalent structure.
2.Strong covalent bonds link atoms throughout the continuous network.
3.A large amount of energy is needed to overcome many of these bonds.

Answer: Silicon dioxide has a very high melting point, so it remains solid at $1500\,^{\circ}\mathrm{C}$ .

Common mistakes

Don't describe a giant covalent substance as many molecules joined by intermolecular forces.
Don't say only one strong bond needs to be overcome during melting.
Don't assume a drawn fragment is the whole structure rather than part of a repeating network.

Exam tip. A high-melting-point explanation needs the full chain: giant structure → many strong covalent bonds → large energy transfer.

4.2.2.7 · Properties of metals and alloys

Definition. An alloy is a mixture of a metal with one or more other elements. Alloys are harder than pure metals because the different-sized atoms distort the regular layers, so the layers cannot slide over each other easily.

Metals have giant structures of atoms held by strong metallic bonding, which is why most have high melting and boiling points: much energy is needed to overcome the strong electrostatic attraction between the positive metal ions and the sea of delocalised electrons.
Those delocalised electrons also make metals good conductors of electricity and of thermal energy, because they are free to move through the whole structure and carry charge or energy.
In a pure metal the atoms are the same size and arranged in layers, so the layers slide over one another — that is why pure metals are malleable, and also why they are too soft for many uses.
Introducing atoms of a different element disrupts that regularity: the different-sized atoms distort the layers, the layers can no longer slide easily, and the alloy is harder than the pure metal.
The same distortion restricts the delocalised electrons, so alloys are usually poorer conductors.

Regular layers in a pure metal and distorted layers in an alloy.

Worked example

Explain why an alloy containing atoms of two different sizes is harder than the pure metal.

1.The pure metal has regular layers of equal-sized atoms.
2.Different-sized atoms in the alloy distort these layers.
3.The distorted layers cannot slide over one another as easily when a force is applied.

Answer: More force is needed to change the alloy’s shape, so the alloy is harder.

Mark-scheme wording to use

“different-sized atoms. 1 mark: layers are distorted. 1 mark: layers/atoms cannot easily slide over each other” — AQA 8462, June 2024 Paper 1H, Q07.2
“sizes/diameters; do not accept shapes” — AQA 8462, June 2023 Paper 1F, Q04.2
“in alloys different sized atoms distort the layers/structure. 1 mark: (so) the movement of (delocalised) electrons is restricted” — AQA 8462, alloy-conductivity two-marker, June 2023 Paper 1H, Q06.3

What the examiner wants. The three-step chain in order: different-sized atoms, distorted layers, layers cannot slide. Saying the atoms are a different shape is explicitly not accepted, and 'stronger bonds' scores nothing.

Common mistakes

Don't say an alloy is harder because it contains more atoms or more bonds.
Don't claim layers of atoms cannot move at all in a pure metal.
Don't use increased particle mass rather than layer distortion to explain hardness.

Exam tip. For ‘explain why an alloy is harder’, include different atom sizes, distorted layers and reduced sliding.

4.2.2.8 · Metals as conductors

Metallic structures contain delocalised electrons that are free to move through the whole giant structure. These electrons explain both electrical and thermal conduction.
During electrical conduction, mobile delocalised electrons carry electrical charge through the metal. During thermal conduction, the delocalised electrons transfer energy through the structure.
The positive metal ions remain in fixed lattice positions in a solid wire and are not the moving charge carriers.
An exam answer must identify the delocalised electrons, state that they can move, and link their movement to the property being explained.
The word ‘electrons’ alone does not establish why charge or energy can pass through the material.

Worked example

Explain why copper is suitable for both electrical wiring and a saucepan base.

1.Copper contains delocalised electrons that can move through its metallic structure.
2.The electrons carry electrical charge through a wire.
3.The same mobile electrons transfer thermal energy rapidly through a saucepan base.

Answer: Mobile delocalised electrons make copper a good conductor of both electricity and thermal energy.

Common mistakes

Don't name positive metal ions as the particles moving along a solid wire.
Don't state that metals contain electrons without saying the electrons are delocalised and mobile.
Don't explain electrical conduction using energy transfer but never mentions charge.

Exam tip. Match the final phrase to the property: delocalised electrons carry charge electrically and transfer energy thermally.

4.2.3.1 · Diamond

Diamond is a giant covalent structure in which every carbon atom forms four strong covalent bonds with other carbon atoms. This produces a rigid network extending throughout the structure.
Many strong covalent bonds must be overcome to melt diamond, so it has a very high melting point.
The rigid arrangement also makes diamond very hard.
Diamond does not conduct electricity because all four outer electrons from each carbon atom are used in covalent bonds, leaving no delocalised electrons free to move and carry charge.
Property explanations must select the relevant structural feature: bond strength and network rigidity explain melting point and hardness, while absence of mobile electrons explains conductivity.

Part of diamond’s rigid giant covalent network.

Worked example

Explain why diamond is hard and does not conduct electricity.

1.Each carbon atom forms four strong covalent bonds in a rigid giant structure.
2.The rigid network is difficult to deform, so diamond is hard.
3.All four outer electrons are used in bonds, leaving no delocalised electrons to carry charge.

Answer: Diamond is hard because of its rigid covalent network and does not conduct because it has no mobile delocalised electrons.

Common mistakes

Don't say diamond has strong intermolecular forces instead of a giant network of covalent bonds.
Don't assume every form of carbon conducts electricity.
Don't explain hardness only by saying carbon atoms are close together.

Exam tip. For a multi-property question, give a separate structure-to-property link for hardness, melting point and conductivity.

4.2.3.2 · Graphite

In graphite, each carbon atom forms three strong covalent bonds with other carbon atoms, creating layers of hexagonal rings. There are no covalent bonds between adjacent layers, so weak forces between the layers allow them to slide over one another.
This makes graphite soft and useful where layers must move. Strong covalent bonds within the layers require much energy to overcome, giving graphite a high melting point.
The fourth outer electron from each carbon atom is delocalised.
These electrons can move through the structure and carry charge, so graphite conducts electricity in a similar way to metals.
Explanations must keep bonding within layers separate from forces between layers.

Graphite has strong covalent bonds within layers, weak forces between layers and mobile delocalised electrons.

Worked example

Explain why graphite conducts electricity and can act as a lubricant.

1.One electron from each carbon atom is delocalised and free to move.
2.The mobile electrons carry electrical charge through graphite.
3.There are no covalent bonds between layers, so weak forces allow the layers to slide.

Answer: Delocalised electrons make graphite conductive, while sliding layers make it suitable as a lubricant.

Common mistakes

Don't say graphite is soft because covalent bonds within its layers are weak.
Don't claim carbon atoms move between layers to carry charge.
Don't state that graphite has no strong covalent bonds.

Exam tip. Use ‘within layers’ for strong bonds and ‘between layers’ for weak forces; those phrases prevent contradictory explanations.

4.2.3.3 · Graphene and fullerenes

Graphene is a single layer of graphite: a one-atom-thick sheet of carbon atoms joined in hexagonal rings. Its strong covalent bonds, delocalised electrons and very small thickness make it useful in electronics and composites.
Fullerenes are hollow molecules made only of carbon. Their structures are based mainly on hexagonal rings but may also contain rings of five or seven carbon atoms.
Buckminsterfullerene, C $_{60}$ , is spherical. Carbon nanotubes are cylindrical fullerenes with very high length-to-diameter ratios.
Their properties support uses in nanotechnology, electronics and materials.
Diagrams and descriptions may require graphene, spherical fullerenes and nanotubes to be distinguished by shape and bonding.

Recognising graphene, Buckminsterfullerene and a carbon nanotube.

Worked example

A sensor needs an electrically conducting layer that adds very little thickness. Explain why graphene is suitable.

1.Graphene is a single layer of graphite and is only one atom thick.
2.It therefore adds very little thickness to the sensor.
3.Graphene contains delocalised electrons that move and carry charge.

Answer: Graphene is both extremely thin and electrically conducting, so it suits the sensor.

Common mistakes

Don't describe graphene as several graphite layers rather than one layer.
Don't draw a carbon nanotube as a solid rod instead of a hollow cylindrical fullerene.
Don't state that every fullerene contains only hexagonal rings.

Exam tip. For ‘recognise the structure’, use the defining shape: sheet for graphene, hollow sphere for C $_{60}$ and hollow cylinder for a nanotube.

4.2.4.1 · Sizes of particles and their properties (chemistry only)

Nanoscience concerns structures from $1$ to $100\,\mathrm{nm}$ , of the order of a few hundred atoms. Fine particles have diameters from $100$ to $2500\,\mathrm{nm}$ , while coarse particles span $2500$ to $10\,000\,\mathrm{nm}$ .
Nano dimensions may be compared with typical atomic and molecular dimensions using ratios and standard form.
For a cube of side $a$ , surface area is $6a^2$ , volume is $a^3$ , and the surface-area-to-volume ratio is $6/a$ .
Reducing the side by a factor of $10$ increases this ratio by a factor of $10$ .
A high ratio can give nanoparticles different properties from the bulk material and make smaller quantities effective.

Smaller cubes have a larger surface-area-to-volume ratio than a larger cube.

Worked example

A cube-shaped nanoparticle has side length $5\,\mathrm{nm}$ . Calculate its surface area, volume and surface-area-to-volume ratio.

1.Surface area $=6a^2=6\times(5\,\mathrm{nm})^2=150\,\mathrm{nm}^2$ .
2.Volume $=a^3=(5\,\mathrm{nm})^3=125\,\mathrm{nm}^3$ .
3.Surface area : volume $=150:125=1.2:1$ .

Answer: Surface area $150\,\mathrm{nm}^2$ ; volume $125\,\mathrm{nm}^3$ ; surface-area-to-volume ratio $1.2:1$ .

Common mistakes

Don't calculate a cube’s surface area as $a^2$ instead of $6a^2$ .
Don't compare total surface area alone instead of surface area relative to volume.
Don't classify a $2500\,\mathrm{nm}$ particle as nano-sized.

Exam tip. For a surface-area-to-volume calculation, show $6a^2$ , $a^3$ and the simplified ratio as separate lines.

4.2.4.2 · Uses of nanoparticles (chemistry only)

Nanoparticles have applications in medicine, electronics, cosmetics and sun creams, deodorants and catalysts, and new uses remain an active area of research. Their high surface-area-to-volume ratio can make a smaller quantity effective, which may reduce material use, cost or waste.
Evaluation questions provide information about a specified application and require advantages to be weighed against disadvantages.
Possible health and environmental risks must be considered because the small particles may interact with organisms differently from bulk material.
Evidence must control the conclusion: finding that particles enter cells does not by itself prove harm, while absence of proven harm does not prove safety.
A justified judgement should compare performance, quantity, cost and uncertainty.

Worked example

A nanoparticle catalyst gives the same reaction rate as a bulk catalyst while using one twentieth of the mass. Evaluate one benefit and one possible risk.

1.Equal reaction rate shows that performance is maintained.
2.Using one twentieth of the mass reduces raw-material use and may reduce cost or waste.
3.Small particles may create uncertain health or environmental risks, so exposure and long-term effects require testing.

Answer: The nanoparticles reduce material use without reducing performance, but further evidence is needed before possible risks can be judged.

Common mistakes

Don't list a benefit and risk without using the data supplied in the question.
Don't claim nanoparticles are definitely harmful merely because a possible risk exists.
Don't claim nanoparticles are safe because no harm has yet been demonstrated.

Exam tip. For ‘evaluate’, compare the supplied evidence on both sides and finish with a conditional, justified conclusion.