Bonding, structure, and the properties of matter

4.2 Bonding, structure, and the properties of matter — revision question pack

18 specification points · notes, questions, answers and worked methods

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.

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4.2.1.1 · Chemical bonds

Explanation

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.

Tier 1 · Easy

State how electrons are shared in covalent bonding and in metallic bonding.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Magnesium reacts with chlorine to form magnesium chloride. Hydrogen atoms join to form hydrogen molecules. Compare how the strong bonds form in these two substances.
[4 marks]
Total for this question: 4
Compare the strong bonding in calcium metal and calcium oxide. Include the particles attracted in each substance.
[4 marks]
Total for this question: 4
A student says that ionic, covalent and metallic bonds all consist of a pair of electrons shared between two atoms. Explain why this description applies only to covalent bonding, and describe ionic and metallic bonding correctly.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Substance P is a giant structure containing positive and negative particles. Substance Q is a giant structure containing one type of atom and mobile outer-shell electrons. Substance R contains separate groups of non-metal atoms joined by shared electron pairs. Identify the bonding in P, Q and R and explain the electrostatic attraction in each bond.
[6 marks]
Total for this question: 6
Sodium metal reacts with chlorine molecules to produce sodium chloride. Describe the strong bonding in each reactant and in the product, including the relevant electrons or charged particles.
[5 marks]
Total for this question: 5
Substance A is a malleable element with atoms in regular layers and conducts electricity as a solid. Substance B contains only non-metals, has a low boiling point and does not conduct. Substance C is formed from a metal and a non-metal, has a high melting point and conducts only when molten. Identify the strong bonding in A, B and C and explain how the evidence supports each choice.
[6 marks]
Total for this question: 6
A bonding investigation gives this dataset for solid R: melting point $805\,^{\circ}\text{C}$ ; solid conductivity trials — no, no, yes; molten conductivity trials — yes, yes. Identify and reject the contradictory datum, then determine the bonding in R and justify your answer.
[6 marks]
Total for this question: 6
A substance is known to have metallic bonding. Choose the two statements that cannot be true and justify both eliminations: its outer-shell electrons are shared in pairs between particular atoms; its outer-shell electrons are delocalised throughout a giant structure; it contains negatively charged metal ions attracted to the electrons. Explain why the remaining statement is correct.
[5 marks]
Total for this question: 5

4.2.1.2 · Ionic bonding

Explanation

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.

Tier 1 · Easy

State the ions formed when potassium, a Group 1 metal, reacts with sulfur, a Group 6 non-metal.
[2 marks]
Total for this question: 2
A student says a Group 2 metal atom gains one electron and forms a negative ion. Give two corrections.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe the electron transfer when magnesium chloride, MgCl₂, forms, and state the charge on every ion produced from one magnesium atom.
[4 marks]
Total for this question: 4
Describe the electron transfer when lithium oxide forms and deduce its formula. Lithium is in Group 1 and oxygen is in Group 6.
[3 marks]
Total for this question: 3
Atom A has the electron arrangement $2,8,1$ and atom B has the electron arrangement $2,7$ . Describe the electron transfer, give the ions formed and write the formula of the compound.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Element X is in Group 2 and element Y is in Group 7. Describe a dot-and-cross diagram for the ionic compound they form and deduce its formula using X and Y.
[5 marks]
Total for this question: 5
A student draws sodium oxide using one Na⁺ ion and one O⁻ ion, with no brackets. Identify the errors and describe a correct dot-and-cross diagram.
[5 marks]
Total for this question: 5
An ionic compound has the formula PQ₂. P is either a Group 1 or Group 2 metal, and Q is either a Group 6 or Group 7 non-metal. Determine the groups of P and Q, and explain the electron transfer when one formula unit forms.
[5 marks]
Total for this question: 5
An ionic compound formed from a metal X and a non-metal Y has the formula XY. X is in Group 1 or Group 2, and Y is in Group 6 or Group 7. Determine both possible pairs of groups and explain why the formula alone does not identify one pair uniquely.
[5 marks]
Total for this question: 5
A dot-and-cross diagram shows one X²⁺ ion with electron arrangement $2,8$ and two Y⁻ ions, each with electron arrangement $2,8$ . Work backwards to determine the original electron arrangement and group of each atom, then give the compound formula.
[6 marks]
Total for this question: 6

4.2.1.3 · Ionic compounds

Explanation

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.

Tier 1 · Easy

Describe the structure and bonding in a solid ionic compound.
[2 marks]
Total for this question: 2
A diagram shows a repeating arrangement of positive and negative ions. State the type of structure and the force holding it together.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A model of an ionic lattice contains $18$ M²⁺ ions and $36$ N⁻ ions. Work out the empirical formula of the compound.
[2 marks]
Total for this question: 2
A neutral lattice section contains $6$ T³⁺ ions. Its negative ions are U²⁻. Calculate the number of U²⁻ ions and deduce the empirical formula.
[3 marks]
Total for this question: 3
Explain why a giant ionic lattice should not be described as a collection of separate ion pairs.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A section of an ionic model contains $8$ A³⁺ ions and $12$ B²⁻ ions. Deduce the empirical formula and give two limitations of using a ball-and-stick model for this lattice.
[4 marks]
Total for this question: 4
Three models represent the same ionic lattice: a dot-and-cross diagram, a flat ball-and-stick diagram and a three-dimensional ball-and-stick diagram. Compare the information they show and give two limitations shared by the ball-and-stick models.
[4 marks]
Total for this question: 4
A model is claimed to show a neutral ionic lattice containing $10$ X²⁺ ions and $15$ Y⁻ ions. Show that the model is not neutral, determine the number of Y⁻ ions needed for neutrality and give the empirical formula.
[4 marks]
Total for this question: 4
An ionic compound has the formula R₃S₂. Each S ion has charge $3-$ . Determine the charge on one R ion and show how the charges in the formula balance.
[5 marks]
Total for this question: 5
A ball-and-stick section contains $12$ R⁺ ions, $6$ S²⁻ ions and $36$ sticks. One student uses the ion counts to write R₂S; another uses the numbers of sticks and ions to propose a different formula. Determine which approach is valid and justify the formula.
[4 marks]
Total for this question: 4

4.2.1.4 · Covalent bonding

Explanation

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.

Tier 1 · Easy

A chlorine molecule contains two chlorine atoms joined by one covalent bond. State what this bond represents.
[1 mark]
Total for this question: 1
State the number of shared electron pairs between the two atoms in an oxygen molecule.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Describe a dot-and-cross diagram for one water molecule, H₂O. Include the bonding pairs and the unshared outer-shell electrons on oxygen.
[4 marks]
Total for this question: 4
Describe a dot-and-cross diagram for ammonia, NH₃. Include all outer-shell electrons on nitrogen.
[4 marks]
Total for this question: 4
A ball-and-stick model of H₂ shows two equal spheres joined by one stick. Describe one feature represented correctly and give two limitations of this model.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A model shows two nitrogen atoms joined together, with each nitrogen also joined to two hydrogen atoms. Deduce the molecular formula, count the covalent bonds shown, and give one limitation of the model.
[3 marks]
Total for this question: 3
Compare dot-and-cross diagrams for hydrogen chloride, HCl, and nitrogen, N₂. State the shared and unshared electron pairs in each molecule.
[5 marks]
Total for this question: 5
Describe a complete dot-and-cross diagram for a chlorine molecule, Cl₂, and explain how the diagram shows that each chlorine atom has a full outer shell.
[4 marks]
Total for this question: 4
A student draws methane, CH₄, with four shared pairs but shows both electrons in every pair as coming from carbon and also gives carbon one unshared pair. Evaluate and correct the electron arrangement.
[5 marks]
Total for this question: 5
A dot-and-cross description of XF₃ shows three shared pairs around central atom X. X supplies one electron to every bond and retains one unshared pair. Determine how many outer-shell electrons X had before bonding and hence state its group.
[6 marks]
Total for this question: 6

4.2.1.5 · Metallic bonding

Explanation

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.

Tier 1 · Easy

Name the two types of charged particle shown in the usual model of metallic bonding.
[2 marks]
Total for this question: 2
State what is meant by a delocalised electron in a metal.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain why the bonding in a metal is strong.
[3 marks]
Total for this question: 3
Describe the particles and attraction represented in a metallic-bonding model of aluminium.
[3 marks]
Total for this question: 3
A model contains positive ions of two different sizes in a regular giant structure, surrounded by delocalised electrons. Identify the material as a metal or an alloy and explain the bonding shown.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A diagram shows identical circles in regular layers with many smaller negative symbols between them. A student says the diagram represents a simple molecule with electrons shared only between neighbouring pairs of atoms. Evaluate the student's statement.
[5 marks]
Total for this question: 5
A student draws a metal as separate pairs of positive ions, with two electrons trapped between each pair. Evaluate the model and describe a more accurate metallic-bonding model.
[4 marks]
Total for this question: 4
Compare the sharing of electrons and the electrostatic attractions in a metal with those in a simple covalent molecule.
[5 marks]
Total for this question: 5
A metallic-bonding model contains 12 ions, each labelled 2+, formed when every atom contributed two outer-shell electrons to the delocalised electron population. Calculate the number of delocalised electrons and explain how the model represents a neutral, strongly bonded metal.
[5 marks]
Total for this question: 5
A metal sheet is bent so that layers of positive ions move relative to one another, but the sheet remains one piece. Explain why this observation supports a delocalised-electron model rather than a model of fixed electron pairs between neighbouring ions.
[5 marks]
Total for this question: 5

4.2.2.1 · The three states of matter

Explanation

The three states of matter are solid, liquid and gas. Melting and freezing occur at the melting point; boiling and condensing occur at the boiling point.
To predict state, compare the given temperature with both change-of-state temperatures: below the melting point the substance is solid, between the points it is liquid, and above the boiling point it is a gas.
Changing state transfers energy because forces between particles must be overcome or form; stronger forces give higher melting and boiling points.
Atoms do not themselves have bulk properties such as being solid.
Higher tier: the simple particle model omits forces and represents every particle as a solid, inelastic sphere.

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 $25\,^{\circ}\mathrm{C}$ .

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.

Tier 1 · Easy

At the boiling point, name the change from liquid to gas and the reverse change from gas to liquid.
[2 marks]
Total for this question: 2
Name the change from solid to liquid and the reverse change from liquid to solid.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Substance S has a melting point of $-15\,^{\circ}\text{C}$ and a boiling point of $84\,^{\circ}\text{C}$ . State its physical state at $-30\,^{\circ}\text{C}$ , $20\,^{\circ}\text{C}$ and $100\,^{\circ}\text{C}$ .
[3 marks]
Total for this question: 3
Compare the particle changes when a liquid freezes and when the liquid boils. Include energy transfer.
[4 marks]
Total for this question: 4
A pure solid is heated at its melting point. Energy continues to enter the substance, but its temperature stays constant until all the solid has melted. Explain this observation using the particle model.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Substance P melts at $40\,^{\circ}\text{C}$ and boils at $85\,^{\circ}\text{C}$ . Substance Q melts at $780\,^{\circ}\text{C}$ and boils at $1400\,^{\circ}\text{C}$ . Compare their states at $100\,^{\circ}\text{C}$ and explain what the data suggest about the forces or bonds between their particles.
[5 marks]
Total for this question: 5
Explain three limitations of representing particles in solids, liquids and gases as solid, inelastic spheres with gaps between them.
[4 marks]
Total for this question: 4
A student says, ‘When a solid melts, each atom becomes liquid and expands.’ Evaluate this statement and describe the particle changes during melting.
[4 marks]
Total for this question: 4
Substance R is solid at $12\,^{\circ}\text{C}$ , liquid at $18\,^{\circ}\text{C}$ and $72\,^{\circ}\text{C}$ , and a gas at $78\,^{\circ}\text{C}$ . Use the observations to give a range for each change-of-state temperature and predict the state at $50\,^{\circ}\text{C}$ .
[5 marks]
Total for this question: 5
A pure substance cools from a gas. Its temperature stays at $91\,^{\circ}\text{C}$ while liquid first forms, then falls to $27\,^{\circ}\text{C}$ and stays there while solid forms. Identify both changes, state the boiling and melting points, and describe the energy transfer and particle change at each plateau.
[6 marks]
Total for this question: 6

4.2.2.2 · State symbols

Explanation

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.

Tier 1 · Easy

State the meaning of the symbols (l) and (aq) in a chemical equation.
[2 marks]
Total for this question: 2
Give the state symbol for liquid bromine and for potassium bromide dissolved in water.
[2 marks]
Total for this question: 2

Tier 2 · Standard

At room conditions, methane, oxygen and carbon dioxide are gases, while the water produced is liquid. Represent the complete combustion of methane by a balanced symbol equation with a state symbol after each formula.
[3 marks]
Total for this question: 3
Zinc metal is placed in copper(II) sulfate solution. Zinc sulfate solution and copper metal form. Write a balanced symbol equation with state symbols.
[4 marks]
Total for this question: 4
Magnesium burns in oxygen to form solid magnesium oxide. Write a balanced symbol equation with state symbols and state why oxygen has the state symbol shown.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A technician mixes solutions of barium chloride and sodium sulfate. The barium sulfate product is a solid precipitate, while sodium chloride stays dissolved. Give, in order, the state symbols for barium chloride, sodium sulfate, barium sulfate and sodium chloride.
[4 marks]
Total for this question: 4
Copper(II) chloride solution reacts with sodium hydroxide solution. Solid copper(II) hydroxide and sodium chloride solution form. Write a balanced symbol equation with state symbols.
[4 marks]
Total for this question: 4
Solid calcium carbonate reacts with hydrochloric acid solution to form calcium chloride solution, liquid water and carbon dioxide gas. Write a balanced symbol equation with a state symbol after every formula.
[4 marks]
Total for this question: 4
A student writes the neutralisation equation as HCl(l) + NaOH(l) → NaCl(l) + H₂O(aq). The acid, alkali and salt are all dissolved in water. Correct every state symbol and explain the distinction between (aq) and (l) in this reaction.
[4 marks]
Total for this question: 4
Hydrogen burns in oxygen in a vessel at 130 °C. Both reactants are gases and the water formed remains as steam. Write the balanced equation with state symbols. The products are then cooled to 25 °C; state the new symbol for the water and explain the change.
[5 marks]
Total for this question: 5

4.2.2.3 · Properties of ionic compounds

Explanation

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.

Tier 1 · Easy

State whether an ionic compound usually conducts electricity when solid and when molten.
[2 marks]
Total for this question: 2
Explain why solid potassium chloride does not conduct electricity.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why magnesium oxide has a high melting point.
[3 marks]
Total for this question: 3
An ionic solid is dissolved in water and the resulting solution completes an electrical circuit. Explain this observation.
[3 marks]
Total for this question: 3
A student crushes a solid ionic compound into a fine powder and predicts that it will conduct because the ions have been freed. Evaluate the prediction.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Compound T has a high melting point. It does not conduct electricity as a solid, but it conducts when molten and when dissolved in water. Explain all three observations and identify the likely structure of T.
[6 marks]
Total for this question: 6
Solid J melts at $1015\,^{\circ}\text{C}$ , does not conduct when solid and conducts when molten. Substance K melts at $-28\,^{\circ}\text{C}$ and never conducts. Deduce the likely structure of each substance and explain all the data.
[6 marks]
Total for this question: 6
An ionic compound is tested as a solid, dissolved in water, recovered as solid crystals by evaporating the water, and finally melted. Predict the conductivity at all four stages and explain each prediction.
[6 marks]
Total for this question: 6
A student claims that sodium chloride solution conducts only because water carries the charge. Distilled water and sugar solution do not complete a circuit, but sodium chloride solution and molten sodium chloride do. Evaluate the claim using all the evidence.
[5 marks]
Total for this question: 5
A student says molten sodium chloride conducts because electrons leave chloride ions and travel through the liquid. Evaluate this explanation and describe how charge actually flows in the molten compound and why the solid does not conduct.
[5 marks]
Total for this question: 5

4.2.2.4 · Properties of small molecules

Explanation

Substances made of small molecules are usually gases or liquids with relatively low melting and boiling points. The covalent bonds within each molecule are strong, but the intermolecular forces between separate molecules are weak.
Melting or boiling overcomes these intermolecular forces; it does not break the covalent bonds.
Intermolecular forces increase with molecular size, so larger molecules generally require more energy to separate and have higher melting and boiling points.
Small molecular substances do not conduct electricity because their molecules have no overall electric charge.
In explanations, distinguish clearly between forces between molecules and bonds within molecules, then link the relevant attraction to the observed bulk property.

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.

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.

Tier 1 · Easy

Explain why a substance made of small molecules can have a low boiling point even though its covalent bonds are strong.
[2 marks]
Total for this question: 2
Explain why a pure substance made of neutral small molecules does not conduct electricity.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Molecule V is larger than molecule U. Both substances consist of small molecules. Predict which substance usually has the higher boiling point and explain your answer.
[3 marks]
Total for this question: 3
A liquid made of small molecules boils. Describe which attractions are overcome and which bonds remain intact.
[3 marks]
Total for this question: 3
Molecular substance X melts at −62 °C and boils at −8 °C. State its physical state at 20 °C and explain what its boiling point suggests about the forces between its molecules.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Substances A and B both contain neutral small molecules. A has relative molecular mass $34$ and boils at $22\,^{\circ}\text{C}$ ; B has relative molecular mass $122$ and boils at $168\,^{\circ}\text{C}$ . Explain the difference in boiling point and predict whether either pure substance conducts electricity.
[5 marks]
Total for this question: 5
A student claims that molecular substance M melts easily because its covalent bonds are weak, and that liquid M conducts because its particles can move. Evaluate both claims.
[5 marks]
Total for this question: 5
A sealed sample of a pure small-molecule substance is boiled and then condensed. A student claims that both changes are chemical reactions because bonds must break and reform. Evaluate the claim.
[4 marks]
Total for this question: 4
Equal numbers of molecules of U, V and W require 13 kJ, 29 kJ and 54 kJ respectively to boil under the same conditions. All three substances consist of neutral small molecules. Give the order of likely molecular size and explain what is, and is not, overcome during boiling.
[5 marks]
Total for this question: 5
Molecular substance Z boils at 36 °C, but much more energy is needed to split each Z molecule into its atoms. A student says these observations contradict each other. Explain why both observations are consistent and predict whether pure liquid Z conducts electricity.
[5 marks]
Total for this question: 5

4.2.2.5 · Polymers

Explanation

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.

Tier 1 · Easy

A substance consists of separate, very long carbon-based chains. Name this class of substance and state the type of bond joining atoms within each chain.
[2 marks]
Total for this question: 2
A bonding diagram shows a covalent chain inside brackets followed by the symbol n. Identify the substance type and state what n represents.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A short-chain molecular substance is a liquid at $20\,^\circ\text{C}$ , whereas material P, made from long covalent chains, is solid. Explain why P is solid at this temperature.
[3 marks]
Total for this question: 3
Describe the bonding within a polymer chain and the forces between separate polymer chains.
[3 marks]
Total for this question: 3
A substance consists of separate chain molecules, each containing thousands of covalently bonded atoms. Explain why the substance is a polymer and why it is solid at room temperature.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Material R contains long covalent chains with no covalent bonds from one chain to another. Material S is one continuous covalent network. Identify the polymer and explain two structural differences between R and S.
[4 marks]
Total for this question: 4
Polymer T consists of separate long chains. It is solid at room temperature, softens when heated and becomes solid again on cooling. Explain these observations without stating that covalent bonds break.
[5 marks]
Total for this question: 5
Polymer P is changed into much shorter chain molecules without changing the types of covalent bond within each chain. Predict how its melting point is likely to change and explain why.
[4 marks]
Total for this question: 4
A student interprets a diagram written as [–A–B–]_n as many separate A–B molecules and says that n is the number of atoms in the sample. Evaluate the interpretation and explain why the material can be solid at room temperature.
[5 marks]
Total for this question: 5
Samples P and Q of the same type of polymer are compared. The samples have equal masses, but Q's molecules are on average much shorter than P's. Q melts at a lower temperature than P. Evaluate sample size and molecular size as competing explanations for Q's lower melting point.
[5 marks]
Total for this question: 5

4.2.2.6 · Giant covalent structures

Explanation

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.

Tier 1 · Easy

A diagram shows atoms covalently bonded in a repeating network that extends in every direction. State the type of structure shown.
[1 mark]
Total for this question: 1
Give two examples of substances with giant covalent structures.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Silicon dioxide is solid at $1500\,^\circ\text{C}$ . Explain this observation using its structure and bonding.
[3 marks]
Total for this question: 3
A model shows covalent bonds continuing beyond every edge of the displayed section. Deduce the structure type and predict its melting point.
[3 marks]
Total for this question: 3
Silicon dioxide has the formula SiO₂, but it does not contain separate SiO₂ molecules. Explain what the formula represents and explain its very high melting point.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two covalent substances are heated. U melts at $84\,^\circ\text{C}$ and consists of separate molecules. V is still solid at $1700\,^\circ\text{C}$ and has no separate molecules. Deduce the structure of V and explain the difference in melting behaviour.
[4 marks]
Total for this question: 4
A student says silicon dioxide is made of separate SiO₂ molecules and has a high melting point because the intermolecular forces are strong. Evaluate and correct the explanation.
[4 marks]
Total for this question: 4
Diamond and silicon dioxide contain different elements but both remain solid at very high temperatures. Explain why this similarity does not require the two substances to have identical structures.
[4 marks]
Total for this question: 4
A block of a giant covalent substance is ground into a fine powder. A student predicts a much lower melting point because each grain is now small. Evaluate the prediction using structure and bonding.
[4 marks]
Total for this question: 4
Carbon dioxide and silicon dioxide both have formulas containing one atom of the first element for every two oxygen atoms. Carbon dioxide is a gas at room temperature, whereas silicon dioxide is a high-melting solid. Explain why the same 1:2 formula ratio does not give similar properties.
[6 marks]
Total for this question: 6

4.2.2.7 · Properties of metals and alloys

Explanation

Metals have giant structures of atoms with strong metallic bonding, so most have high melting and boiling points. In a pure metal, equal-sized atoms form regular layers.
These layers can slide over one another when a force is applied, allowing the metal to be bent and shaped.
Pure metals are too soft for many uses, so they are mixed with other elements to make alloys.
Atoms of different sizes distort the regular layers and make it harder for them to slide, so the alloy is harder.
The required explanation is not that an alloy has more bonds; it is the link between different atom sizes, layer distortion and reduced movement.

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.

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.

Tier 1 · Easy

Explain why a pure metal can be bent into shape.
[2 marks]
Total for this question: 2
A metal model contains atoms of several different sizes. State whether it represents a pure metal or an alloy, and predict whether it is harder than the pure metal.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A manufacturer replaces pure metal M with an alloy containing a second element whose atoms have a different size. Explain why the alloy is harder than M.
[3 marks]
Total for this question: 3
A sheet of pure metal is bent without breaking. Explain this behaviour using its arrangement of atoms.
[3 marks]
Total for this question: 3
Explain how a pure metal can have a high melting point but still be bent into shape without breaking.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Pure metal Q bends when a force of $18\,\text{N}$ is applied, but an alloy of Q needs $47\,\text{N}$ . Both have high melting points. Explain both observations in terms of structure and bonding.
[5 marks]
Total for this question: 5
A student says an alloy is harder only because it has stronger bonds, and that a soft pure metal must have a low melting point. Evaluate both claims.
[5 marks]
Total for this question: 5
A clip must resist a force of 50 N in use but must be shaped using a force below 70 N. Pure metal P bends at 12 N, alloy A at 38 N and alloy B at 62 N. Choose the most suitable material and explain the choice in terms of structure.
[5 marks]
Total for this question: 5
Equal-sized samples are tested. Pure metal M bends at 16 N, an alloy containing 4% of a second element bends at 43 N, and an alloy containing 9% bends at 39 N. Evaluate the claim that adding more of the second element always makes the material harder, and explain why both alloys are harder than M.
[5 marks]
Total for this question: 5
A 2 mm diameter rod of pure metal bends at 21 N. A 4 mm diameter rod of an alloy bends at 58 N. A student concludes that the alloy is harder. Evaluate the investigation and describe how to obtain valid structural evidence for the comparison.
[5 marks]
Total for this question: 5

4.2.2.8 · Metals as conductors

Explanation

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.

Tier 1 · Easy

Name the particles that carry electrical charge through a metal wire.
[1 mark]
Total for this question: 1
Name the particles that transfer thermal energy through a metal.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A metal strip completes a circuit, but a solid polymer strip does not. Explain why the metal conducts electricity.
[3 marks]
Total for this question: 3
Solid aluminium conducts electricity, but solid sodium chloride does not. Explain the difference in terms of mobile charged particles.
[4 marks]
Total for this question: 4
A student says that a metal wire conducts because positive metal ions travel from one end of the wire to the other. Explain why the statement is incorrect and describe how the wire conducts.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Copper is used for both electrical wiring and the base of a saucepan. Explain how the same structural feature makes copper suitable for both uses.
[4 marks]
Total for this question: 4
Compare electrical conduction in a solid metal, a solid ionic compound and the same ionic compound when molten. Identify the charge carrier in every conducting sample.
[6 marks]
Total for this question: 6
A metal rod is drawn into a longer, thinner wire without changing its composition. Explain why the wire can still conduct electricity and identify the roles of both particles in the metallic structure.
[4 marks]
Total for this question: 4
Molten sodium chloride conducts electricity but transfers thermal energy much less effectively than liquid copper. Explain the difference by contrasting the mobile-ion mechanism with the delocalised-electron mechanism.
[5 marks]
Total for this question: 5
Loose grains of a metal powder do not complete a circuit between two clips, but the same powder conducts after it is compressed into a continuous pellet. A student says compression must have created delocalised electrons. Evaluate the explanation.
[5 marks]
Total for this question: 5

4.2.3.1 · Diamond

Explanation

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.

Tier 1 · Easy

In diamond, how many bonds per carbon?
[1 mark]
Total for this question: 1
State the structure type of diamond and the number of covalent bonds formed by each carbon atom.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why a diamond-tipped cutting tool can scratch most materials.
[3 marks]
Total for this question: 3
Explain why diamond is an electrical insulator.
[3 marks]
Total for this question: 3
A ball-and-stick model shows a small section of diamond. Give three limitations of using this model to represent a real diamond crystal.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Diamond is proposed for an electrically heated cutting element. Explain its high melting point and hardness, and decide whether it can carry the heating current.
[5 marks]
Total for this question: 5
Methane and diamond both contain covalent bonds, but methane is a gas at room temperature while diamond is a hard solid with a very high melting point. Explain the difference.
[5 marks]
Total for this question: 5
Compare diamond with graphite in terms of the bonds made by each carbon atom, the arrangement of the atoms, hardness and electrical conductivity.
[6 marks]
Total for this question: 6
A complete model section of diamond contains 28 carbon atoms and all four bonds from every atom are contained within the section. Calculate the number of distinct covalent bonds, then use the bonding to predict hardness and electrical conductivity.
[5 marks]
Total for this question: 5
A material is very hard, has a very high melting point and does not conduct electricity. Determine whether the material could be diamond, graphite or a metal, and justify the rejection of each alternative.
[5 marks]
Total for this question: 5

4.2.3.2 · Graphite

Explanation

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.

Tier 1 · Easy

State the number of covalent bonds made by each carbon atom in graphite.
[1 mark]
Total for this question: 1
State the shape of the carbon rings in graphite and whether covalent bonds join adjacent layers.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why graphite can be used as an electrode.
[3 marks]
Total for this question: 3
Explain why powdered graphite can reduce friction between two moving surfaces.
[3 marks]
Total for this question: 3
A complete section of a graphite model contains $24$ carbon atoms. Determine the number of delocalised electrons contributed by these atoms and explain how they allow graphite to conduct electricity.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Graphite is used in a high-temperature electrical contact that must also slide against another surface. Explain three properties that make graphite suitable.
[6 marks]
Total for this question: 6
A student claims that graphite is soft because all its bonds are weak and that a non-metal cannot conduct electricity. Evaluate both claims.
[6 marks]
Total for this question: 6
Compare the bonding and electrical conduction in graphite with those in a metal.
[5 marks]
Total for this question: 5
A carbon allotrope conducts electricity and is soft enough to mark paper. Determine how many covalent bonds each of its carbon atoms forms, and justify that number using both observations.
[5 marks]
Total for this question: 5
A force parallel to graphite's layers makes them slide, but a force that tries to pull carbon atoms apart within one layer meets much greater resistance. Explain the difference and predict whether sliding the layers removes graphite's electrical conductivity.
[5 marks]
Total for this question: 5

4.2.3.3 · Graphene and fullerenes

Explanation

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.

Tier 1 · Easy

Describe the relationship between graphene and graphite.
[1 mark]
Total for this question: 1
State two defining features of a fullerene.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An electronic sensor needs a conducting layer that adds very little thickness. Explain why graphene is suitable.
[4 marks]
Total for this question: 4
A carbon structure is a hollow cylinder whose length is much greater than its diameter. Identify the structure and give two application areas.
[3 marks]
Total for this question: 3
A composite needs a reinforcing carbon layer that is strong but adds very little thickness. Explain why graphene is suitable.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Three carbon materials are described: A is a one-atom-thick sheet; B is a hollow sphere containing 60 carbon atoms; C is a hollow cylinder whose length is far greater than its diameter. Identify A, B and C, then give one suitable type of application for A and one for C.
[5 marks]
Total for this question: 5
Compare graphene with Buckminsterfullerene, C₆₀, in terms of shape, scale and carbon-ring arrangement. Give one application area for graphene.
[5 marks]
Total for this question: 5
A student says, ‘Every fullerene contains only hexagonal rings and every fullerene is spherical.’ Evaluate both parts of the statement using Buckminsterfullerene and carbon nanotubes.
[5 marks]
Total for this question: 5
Distinguish a graphite particle made from 15 stacked carbon layers from 15 graphene sheets that have been completely separated. Explain why only the separated sheets provide the one-atom-thick benefit.
[5 marks]
Total for this question: 5
An analysis reports only that a carbon material has chemical formula C. A student claims this is enough to decide whether it is graphene, Buckminsterfullerene or a carbon nanotube. Evaluate the claim and state the structural observation needed to identify each material.
[5 marks]
Total for this question: 5

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

Explanation

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.

Tier 1 · Easy

Particle T is $72\,\text{nm}$ across. A typical atom is $0.24\,\text{nm}$ across. State whether T is nano, fine or coarse, and calculate how many atom diameters span T.
[2 marks]
Total for this question: 2
State whether a particle of diameter $65\,\text{nm}$ and a particle of diameter $900\,\text{nm}$ are nano, fine or coarse particles.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A nano-cube has edge length $5\,\text{nm}$ . Calculate its surface area, volume and surface-area-to-volume ratio.
[3 marks]
Total for this question: 3
A cube of material has side length $80\,\text{nm}$ . It is divided into $64$ equal cubes without losing material. Calculate the side length of each small cube and the factor by which the total surface area increases.
[4 marks]
Total for this question: 4
Cube-shaped particles of the same material have side lengths $2.4\times10^{-7}\,\text{m}$ (particle X) and $8.0\times10^{-9}\,\text{m}$ (particle Y). Convert each side length to nanometres, state whether each particle is nano, fine or coarse, and determine how many times larger Y's surface-area-to-volume ratio is than X's.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Fine particles have side $800\,\text{nm}$ ; nano-sized cubes of the same material have side $40\,\text{nm}$ . Determine how many times larger the nano cubes' surface-area-to-volume ratio is. If effectiveness per gram is proportional to this ratio and $12\,\text{g}$ of the fine particles is effective, estimate the effective mass of nano cubes.
[4 marks]
Total for this question: 4
A cube has a surface-area-to-volume ratio of $0.15:1$ . Calculate its side length, classify it as nano, fine or coarse, and determine how many $0.20\,\text{nm}$ atom diameters fit along one edge. Predict the new ratio if the side length doubles.
[5 marks]
Total for this question: 5
A material is supplied as cubes with side length $1.5\times10^{-6}\,\text{m}$ . A new version must make the same mass $25$ times as effective. Assume effectiveness per gram is proportional to surface-area-to-volume ratio and $A:V=6/a$ . Work out the required side length in metres and nanometres, calculate the new ratio and state whether the new cubes are nanoparticles.
[5 marks]
Total for this question: 5
Model a cube-shaped particle with edge length $2.4\,\text{nm}$ using atoms whose diameter is $0.24\,\text{nm}$ . Estimate the number of atoms in the particle and comment on the statement that nanoparticles may contain a few hundred atoms. State the packing assumption used.
[5 marks]
Total for this question: 5
Particle A is measured as $99 \pm 3\,\text{nm}$ , particle B as $2492 \pm 12\,\text{nm}$ and particle C as $2520 \pm 7\,\text{nm}$ . Use the full measurement ranges to decide which particles fall definitely within the nano, fine or coarse size ranges.
[6 marks]
Total for this question: 6

4.2.4.2 · Uses of nanoparticles (chemistry only)

Explanation

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.

Tier 1 · Easy

Give two application areas in which nanoparticles are used.
[2 marks]
Total for this question: 2
Give one reason to research new nanoparticle applications and one reason to test each application before widespread use.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A catalyst works equally well when a factory replaces $8.0\,\text{g}$ of bulk material with $0.40\,\text{g}$ of nanoparticles. Suggest two advantages and one possible disadvantage of the change.
[4 marks]
Total for this question: 4
A wound dressing containing $2.0\,\text{g}$ of larger particles kills $72\%$ of tested bacteria. A nanoparticle dressing containing $0.20\,\text{g}$ kills $90\%$ . Traces of the nanoparticles reach waste water, but no environmental harm has been demonstrated. Suggest two advantages and one concern about the nanoparticle dressing.
[4 marks]
Total for this question: 4
A water-treatment catalyst gives the same effect using one twenty-fifth of the mass when it is supplied as nanoparticles. Suggest two advantages of using the nanoparticles and one possible risk.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A company compares two sun creams. Product J uses larger particles, blocks $96\%$ of ultraviolet radiation and leaves a visible layer. Product K uses nanoparticles, blocks $94\%$ , is transparent and needs one quarter as much active material. Tests detect K's particles inside $2\%$ of sampled skin cells, but no harm has been established. Evaluate which product the company should develop.
[6 marks]
Total for this question: 6
A manufacturer compares conductive inks for an electronic display. Ink B uses $6.0\,\text{g}$ of larger silver particles at £ $1.20$ per gram. Ink N uses $0.80\,\text{g}$ of silver nanoparticles at £ $5.00$ per gram. Both give the same electrical performance, but N forms a thinner transparent track. Waste tests detect nanoparticles leaving the factory; their long-term effect is unknown. Evaluate which ink should be developed.
[6 marks]
Total for this question: 6
A conventional medicine needs a 100 mg dose, is effective for 70% of patients and produces side effects in 8%. A nanoparticle delivery system needs a 20 mg dose, is effective for 86% and produces side effects in 12%. Nanoparticles were detected in the liver during a short trial, but long-term harm has not been established. Evaluate whether the nanoparticle system should be adopted.
[6 marks]
Total for this question: 6
A conventional deodorant uses 3.6 g of active material, reduces measured odour by 84% and causes skin irritation in 2% of users. A nanoparticle version uses 0.45 g, reduces odour by 81% and causes irritation in 7%. Nanoparticles are detected in waste water, but their environmental effect is unknown. Evaluate which product should be developed.
[6 marks]
Total for this question: 6
A nanoparticle cosmetic uses half as much pigment and gives a more even finish than a conventional product. A two-day study of 12 people found no irritation. A separate six-month study of 1200 people found irritation in 3% and detected nanoparticles in blood samples, without showing that they caused illness. Evaluate the strength of the evidence and suggest a next step.
[6 marks]
Total for this question: 6

Answer key

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4.2.1.1 · Chemical bonds

Tier 1 · Easy

Mark scheme for 4.2.1.1 Tier 1 · Easy
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01.1	A covalent bond contains a pair of electrons shared between atoms. In metallic bonding, outer-shell electrons are delocalised through the giant structure.	Contrast localised and delocalised sharing. A covalent pair belongs to two bonded atoms, whereas metallic electrons are shared throughout the metal.	2
Total Question 1			2

Tier 2 · Standard

Mark scheme for 4.2.1.1 Tier 2 · Standard
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01.1	In magnesium chloride, electrons are transferred from magnesium atoms to chlorine atoms. The resulting oppositely charged ions attract electrostatically, forming ionic bonds. In hydrogen, two atoms share a pair of electrons. The shared pair is attracted to both nuclei, forming a covalent bond.	Treat each substance separately. Magnesium is a metal and chlorine is a non-metal, so describe electron transfer and attraction between the ions formed. Hydrogen contains only non-metal atoms, so describe a shared pair of electrons and its attraction to both nuclei.	4
Total Question 1			4
02.1	Calcium has metallic bonding. Positive calcium ions are attracted to delocalised electrons. Calcium oxide has ionic bonding. Positive calcium ions and negative oxide ions attract electrostatically.	Treat the element and compound separately. A metal contains positive ions and delocalised electrons. A compound of a metal with a non-metal contains oppositely charged ions. In each case, name the particles joined by the electrostatic attraction.	4
Total Question 2			4
03.1	A covalent bond is a shared pair of electrons between atoms. Ionic bonding is the electrostatic attraction between oppositely charged ions. Metallic bonding is the electrostatic attraction between positive metal ions and delocalised electrons. The metallic electrons are shared through the giant structure rather than being localised between one pair of atoms.	Test the statement against the particles in each bonding model. Only a covalent bond has one electron pair shared by two atoms. Ionic and metallic bonding are both electrostatic attractions, but the attracted particles are different.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.1.1 Tier 3 · Hard
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01.1	P has ionic bonding: oppositely charged ions attract. Q has metallic bonding: positive metal ions are attracted to delocalised electrons. R has covalent bonding: each shared electron pair is attracted to the nuclei of the bonded atoms.	Match each particle description to a bonding model. Oppositely charged ions indicate ionic bonding. Mobile delocalised electrons in a giant metal structure indicate metallic bonding. Shared electron pairs between non-metal atoms indicate covalent bonding. Then state the relevant electrostatic attraction rather than stopping at the bond name.	6
Total Question 1			6
02.1	Sodium has metallic bonding between positive metal ions and delocalised electrons. A chlorine molecule has a covalent bond formed by a shared pair of electrons. An electron transfers from each sodium atom to a chlorine atom when sodium chloride forms. The product contains positive sodium ions and negative chloride ions. Electrostatic attraction between the oppositely charged ions is ionic bonding.	Follow the particles through the reaction. Sodium begins as a giant metallic structure, while each chlorine molecule contains a shared electron pair. Electron transfer produces Na⁺ and Cl⁻; their opposite charges then attract in the ionic product.	5
Total Question 2			5
03.1	A has metallic bonding. Its solid electrical conduction is explained by mobile delocalised electrons in a giant metallic structure. B has covalent bonding within small molecules. Its low boiling point is consistent with weak forces between neutral molecules, which have no mobile charged particles. C has ionic bonding because it is formed from a metal and a non-metal and has a high melting point. Its ions are fixed when solid but mobile when molten, so the molten substance conducts.	Use all the evidence rather than one property alone. A conducting element fits metallic bonding. A low-boiling substance made only from non-metals fits small covalent molecules. A high-melting metal–non-metal compound whose liquid conducts fits a giant ionic lattice.	6
Total Question 3			6
04.1	The ‘yes’ result for solid conductivity contradicts the two repeated ‘no’ results and should be rejected as anomalous. The melting point is high, showing that much energy is needed to overcome strong attractions in a giant structure. The accepted solid results show that R has no mobile charged particles when solid. The repeated molten results show that charged particles become mobile when R melts. This combined evidence supports ionic bonding in R. R contains oppositely charged ions that are fixed in the solid lattice but free to move when molten.	Check repeated measurements before using the property pattern. Two solid trials agree and one conflicts, so reject the isolated conducting result. The remaining high melting point, non-conducting solid and conducting liquid form the ionic pattern: strong attractions hold a giant lattice together, fixed ions cannot carry charge in the solid, and mobile ions carry charge in the melt.	6
Total Question 4			6
05.1	The outer-shell electrons cannot be shared in pairs between particular atoms. Shared pairs between specific atoms describe covalent bonding, not metallic bonding. The substance cannot contain negatively charged metal ions. Metal atoms lose their outer-shell electrons, so the metal ions left in the lattice are positively charged. The delocalised statement is correct: the outer-shell electrons move freely throughout the giant structure and are attracted to the positive metal ions, holding it together.	Test each statement against the metallic bonding model. Fixed shared pairs belong to covalent bonding, so that statement fails. Metals form positive ions by losing electrons, so negative metal ions are impossible. What remains is the model itself: a giant structure of positive ions with delocalised outer-shell electrons.	5
Total Question 5			5

4.2.1.2 · Ionic bonding

Tier 1 · Easy

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01.1	K⁺ S²⁻	A Group 1 atom loses one electron, so potassium forms K⁺. A Group 6 atom gains two electrons to complete its outer shell, so sulfur forms S²⁻.	2
Total Question 1			2
02.1	A Group 2 metal atom loses two electrons. It forms a positive ion with charge $2+$ .	Metals lose outer-shell electrons. A Group 2 atom loses both outer electrons, leaving an ion with two more protons than electrons and therefore charge $2+$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.2 Tier 2 · Standard
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01.1	The magnesium atom loses two outer-shell electrons. Each of two chlorine atoms gains one electron. One Mg²⁺ ion and two Cl⁻ ions form.	Magnesium is in Group 2, so it loses two electrons and forms Mg²⁺. Chlorine is in Group 7, so each atom gains one electron and forms Cl⁻. Two chlorine atoms are therefore needed to receive the two transferred electrons.	4
Total Question 1			4
02.1	Two lithium atoms each lose one electron, forming two Li⁺ ions. One oxygen atom gains two electrons, forming one O²⁻ ion. The formula is Li₂O.	The oxygen atom needs two electrons, but each Group 1 lithium atom supplies one. Two Li⁺ ions balance one O²⁻ ion, so the neutral compound has the formula Li₂O.	3
Total Question 2			3
03.1	A transfers one electron to B. A forms A⁺ with electron arrangement $2,8$ . B forms B⁻ with electron arrangement $2,8$ . The ions combine in a $1:1$ ratio, so the formula is AB.	A loses its single outer-shell electron and B gains that electron. The resulting $1+$ and $1-$ charges balance in equal numbers, so one ion of each type is required.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.1.2 Tier 3 · Hard
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01.1	X loses two electrons to form X²⁺. Two Y atoms each gain one electron to form two Y⁻ ions. Each ion is shown in brackets with a complete outer shell and its charge. Dots and crosses distinguish electrons originally from X and Y, including the transferred electrons. The formula is XY₂.	Use the group numbers to obtain X²⁺ and Y⁻. One X atom supplies two electrons, so two Y atoms are needed. In the diagram, use dots and crosses to distinguish the original and transferred outer-shell electrons, bracket every ion, and label the charges. The $1:2$ ion ratio gives XY₂.	5
Total Question 1			5
02.1	Two sodium atoms are needed, not one. Each sodium atom transfers one electron to the oxygen atom. The diagram shows two Na⁺ ions and one O²⁻ ion. Each ion is in square brackets with a complete outer shell and its charge. Dots and crosses show which atom supplied each outer-shell electron.	Oxygen needs two electrons for a full outer shell, while each Group 1 sodium atom supplies one. The correct charge balance is $2\times(+1)+(-2)=0$ , giving Na₂O. Show both transferred electrons on the bracketed O²⁻ ion.	5
Total Question 2			5
03.1	P is in Group 2 and forms P²⁺. Q is in Group 7 and forms Q⁻. One P atom loses two outer-shell electrons. Each of two Q atoms gains one electron. One P²⁺ ion and two Q⁻ ions have zero total charge, giving PQ₂.	Work backwards from the $1:2$ ion ratio. Of the permitted groups, a $2+$ metal ion is balanced by two $1-$ non-metal ions. Then account for both transferred electrons atom by atom.	5
Total Question 3			5
04.1	One possibility is Group 1 X with Group 7 Y. X loses one electron and Y gains one electron, giving X⁺ and Y⁻ in a $1:1$ ratio. The other possibility is Group 2 X with Group 6 Y. X loses two electrons and Y gains two electrons, giving X²⁺ and Y²⁻ in a $1:1$ ratio. Both charge combinations produce the formula XY, so more information is needed to choose between them.	List the permitted ion charges, then find the pairs whose charges cancel in equal numbers. Charges of $1+$ and $1-$ give a $1:1$ ratio, but charges of $2+$ and $2-$ also give a $1:1$ ratio. The empirical formula records only the simplest ratio, not the size of each ion charge.	5
Total Question 4			5
05.1	The original X atom had electron arrangement $2,8,2$ . It lost two electrons to form X²⁺. X is therefore a Group 2 metal. Each original Y atom had electron arrangement $2,7$ . Each Y atom gained one electron, so Y is a Group 7 non-metal. One X²⁺ balances two Y⁻ ions, so the formula is XY₂.	Reverse the electron changes shown by the ions. Add the two lost electrons back to neutral X to obtain $2,8,2$ . Remove the gained electron from each neutralised Y shell to obtain $2,7$ . These arrangements place X in Group 2 and Y in Group 7, and charge balance gives one X ion for two Y ions.	6
Total Question 5			6

4.2.1.3 · Ionic compounds

Tier 1 · Easy

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01.1	It has a giant lattice of oppositely charged ions. Strong electrostatic attractions act between the ions in all directions.	Name both the scale of the structure and the force holding it together: a giant lattice, not separate molecules, with electrostatic attraction between opposite charges in every direction.	2
Total Question 1			2
02.1	A giant ionic lattice. Strong electrostatic attraction between oppositely charged ions.	A repeating array of ions is a giant lattice. The positive and negative ions are held together by electrostatic forces acting between opposite charges.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.3 Tier 2 · Standard
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01.1	MN₂	The ion count ratio is $18:36$ . Divide both numbers by $18$ to obtain $1:2$ , so the simplest empirical formula is MN₂. The charges also balance: one $2+$ ion is matched by two $1-$ ions.	2
Total Question 1			2
02.1	The positive charge is $6\times3=18$ . $9$ U²⁻ ions provide total charge $-18$ . The ratio $6:9$ simplifies to $2:3$ , so the empirical formula is T₂U₃.	Neutrality requires equal positive and negative charge. Six T³⁺ ions give $+18$ , so nine U²⁻ ions are needed. Simplify T:U from $6:9$ to $2:3$ .	3
Total Question 2			3
03.1	The positive and negative ions form one repeating giant structure. Each ion is attracted to oppositely charged ions around it. The strong electrostatic attractions act in all directions through the lattice, not only within separate pairs.	Focus on the extent and direction of the bonding. An ionic diagram may show only a small section, but the alternating arrangement and attractions repeat throughout the giant lattice.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.1.3 Tier 3 · Hard
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01.1	The ion ratio $8:12$ simplifies to $2:3$ . The empirical formula is A₂B₃. The model may not show the ions at their relative sizes or separations. Sticks can suggest directional physical bonds that are not present.	Simplify the count ratio $8:12$ by dividing by $4$ , giving $2:3$ and hence A₂B₃. The charges balance because two A³⁺ ions give $+6$ and three B²⁻ ions give $-6$ . For limitations, compare the simplified representation with a giant lattice whose electrostatic attractions act in all directions.	4
Total Question 1			4
02.1	A dot-and-cross diagram can distinguish the origins of outer-shell electrons. A three-dimensional ball-and-stick model shows the spatial arrangement more clearly than a flat diagram. The balls may not show the relative sizes or separations of the ions accurately. The sticks may incorrectly suggest directional physical links, or the models show only a small part of the giant lattice.	Match each representation to its purpose: electron origins for dot-and-cross and spatial arrangement for a three-dimensional model. Then compare the artificial sizes, separations, sticks and finite displayed section with the real giant lattice.	4
Total Question 2			4
03.1	The total positive charge is $10 × (+2) = +20$ and the total negative charge shown is $15 × (-1) = -15$ . The displayed model has an overall charge of $+5$ , so it is not neutral. $20$ Y⁻ ions are needed to balance the $10$ X²⁺ ions. The simplest ratio X:Y is $10:20=1:2$ , so the empirical formula is XY₂.	Check the asserted model constructively by totaling both charges. Correct the negative-ion count until it supplies charge $-20$ , then simplify the corrected ion ratio.	4
Total Question 3			4
04.1	R₃S₂ contains three R ions and two S ions in its simplest ratio. The two S³⁻ ions have a total charge of $2 × (-3) = -6$ . The three R ions must therefore have a total charge of $+6$ for the compound to be neutral. The charge on one R ion is $+6 ÷ 3 = +2$ . R forms R²⁺ ions, and $3 × (+2) + 2 × (-3) = 0$ .	Read the subscripts as ion counts and work backwards from neutrality. Two ions carrying charge $3-$ contribute $-6$ altogether, so the three R ions must contribute $+6$ . Dividing that required total by three gives charge $2+$ on each R ion.	5
Total Question 4			5
05.1	The ion-count ratio is $12:6=2:1$ , so it gives R₂S. The total positive charge is $12 × (+1) = +12$ and the total negative charge is $6 × (-2) = -12$ , confirming neutrality. The number of sticks must not be used to determine the empirical formula. Each ion attracts several oppositely charged neighbours in the three-dimensional lattice, and the sticks are only a model of those attractions.	Use the relative numbers and charges of the ions, not the drawing aids. Simplifying $12:6$ gives $2:1$ , and the charge totals cancel. Sticks do not represent separate molecules or a count that belongs in the formula because ionic attractions extend in all directions.	4
Total Question 5			4

4.2.1.4 · Covalent bonding

Tier 1 · Easy

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01.1	A shared pair of electrons.	Use the definition of a covalent bond: the single bond between the two non-metal atoms represents one pair of electrons shared by both atoms.	1
Total Question 1			1
02.1	Two shared pairs of electrons.	Each oxygen atom needs two more electrons for a full outer shell, so the atoms share two pairs. This is a double covalent bond.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.2.1.4 Tier 2 · Standard
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01.1	Oxygen is in the centre with one hydrogen on each side. There is one shared electron pair between oxygen and each hydrogen. Oxygen also has two unshared pairs of outer-shell electrons. Dots and crosses distinguish which atom supplied each electron.	Oxygen has six outer-shell electrons and each hydrogen has one. Oxygen shares one electron with each hydrogen, making two O–H shared pairs. Four oxygen electrons remain as two lone pairs. Use different symbols for electrons from oxygen and hydrogen.	4
Total Question 1			4
02.1	Nitrogen is central with three hydrogen atoms around it. There is one shared electron pair in each of three N–H bonds. Nitrogen has one unshared pair of outer-shell electrons. Dots and crosses distinguish the electrons supplied by nitrogen and hydrogen.	Nitrogen has five outer-shell electrons. It contributes one electron to each of three shared pairs and retains two electrons as one unshared pair. Each hydrogen contributes its one electron to a bonding pair.	4
Total Question 2			4
03.1	It correctly represents two hydrogen atoms joined by one covalent bond. The stick does not show that the bond is a shared pair of electrons. The sizes of the atoms or the distance between them may not be to scale.	Separate the information encoded by the model from its conventions. The connectivity is useful, but a solid stick is not a literal bond and model dimensions are not necessarily realistic.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.1.4 Tier 3 · Hard
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01.1	The molecular formula is N₂H₄. Five covalent bonds are shown. The sticks do not show electron pairs, or the model may not show realistic atom sizes, distances or bond angles.	Count atoms rather than bonds to obtain two nitrogen atoms and four hydrogen atoms, so the formula is N₂H₄. Count the one N–N connection and four N–H connections to obtain five bonds. A ball-and-stick model simplifies electron density, scale and geometry.	3
Total Question 1			3
02.1	HCl has one shared pair between H and Cl. The chlorine atom in HCl has three unshared pairs. N₂ has three shared pairs between the nitrogen atoms. Each nitrogen atom in N₂ has one unshared pair. Dots and crosses distinguish the atom from which each electron originated.	Hydrogen needs one more electron and chlorine needs one, giving one shared pair; chlorine's other six outer electrons form three unshared pairs. Each nitrogen needs three more electrons, so the two atoms share three pairs and each retains one unshared pair.	5
Total Question 2			5
03.1	The two chlorine atoms share one pair of electrons, forming one covalent bond. One electron in the shared pair comes from each chlorine atom. Each chlorine atom also has three unshared pairs of outer-shell electrons. Counting the shared pair, each chlorine atom has access to eight outer-shell electrons.	Start with seven outer electrons on each chlorine atom. Put one electron from each atom into the shared pair, then arrange the remaining six electrons on each atom as three unshared pairs.	4
Total Question 3			4
04.1	Each C–H bond should contain one electron from carbon and one from hydrogen. Methane has four shared electron pairs, one in each C–H bond. Carbon has no unshared outer-shell pair in methane. The four shared pairs give carbon access to eight outer-shell electrons. Each hydrogen has access to two electrons through its shared pair.	Account for the original outer electrons before checking the completed shells. Carbon contributes one electron to each of four bonds and each hydrogen contributes one to its bond. All four carbon electrons are therefore in shared pairs, leaving no lone pair on carbon, while sharing completes carbon's and hydrogen's outer shells.	5
Total Question 4			5
05.1	The diagram contains three shared pairs around X. X contributed one electron to each shared pair. Therefore three of X's original outer-shell electrons are used in bonds. The one unshared pair contains two more electrons from X. X originally had $3 + 2 = 5$ outer-shell electrons. An atom with five outer-shell electrons is in Group 5.	Count only the electrons that the symbols show came from X. Its contribution to the three bonding pairs accounts for three electrons, and its lone pair accounts for two. The total of five is the original outer-shell count, which gives the main-group number.	6
Total Question 5			6

4.2.1.5 · Metallic bonding

Tier 1 · Easy

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01.1	Positive metal ions. Delocalised electrons.	The outer-shell electrons are no longer attached to individual atoms. The model therefore shows a regular array of positive metal ions surrounded by mobile, negatively charged delocalised electrons.	2
Total Question 1			2
02.1	An electron that is free to move through the whole metallic structure rather than belonging to one atom.	Delocalised means that the electron is not fixed to a particular metal ion or pair of atoms; it is shared across the giant structure.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.2.1.5 Tier 2 · Standard
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01.1	Metal ions are positively charged. Delocalised electrons are negatively charged. There is a strong electrostatic attraction between the ions and delocalised electrons throughout the giant structure.	Identify the opposite charges and then state their attraction. Because the ions and delocalised electrons extend through a giant structure, the electrostatic attraction acts throughout the metal.	3
Total Question 1			3
02.1	Positive aluminium ions are arranged in a regular giant structure. Outer-shell electrons are delocalised through the structure. Strong electrostatic attraction acts between the positive ions and the delocalised electrons.	Identify both charged components of the model, then name the attraction between them. The electrons are shared through the metal rather than attached to individual ions.	3
Total Question 2			3
03.1	The material is an alloy because it contains metal ions of different sizes. Its outer-shell electrons are delocalised through the giant structure. The metal ions are positively charged. Strong electrostatic attraction between the positive ions and delocalised electrons forms metallic bonds.	Use the two ion sizes to identify a mixture of metallic elements. The bonding remains metallic: the attraction is between the positive metal ions and the shared, mobile population of electrons.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.1.5 Tier 3 · Hard
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01.1	The diagram represents a giant metallic structure, not a simple molecule. The circles represent positive metal ions in a regular arrangement. The negative symbols represent delocalised electrons. The electrons move through and are shared across the whole structure, not only one neighbouring pair. Metallic bonding is the electrostatic attraction between the positive ions and delocalised electrons.	Use every feature of the diagram: regular repeated particles show a giant lattice, while many small negative symbols between them show delocalised electrons. Correct both parts of the claim, then give the electrostatic definition of metallic bonding.	5
Total Question 1			5
02.1	A metal is a giant structure, not separate pairs. The positive metal ions form a regular arrangement. The outer-shell electrons are delocalised and free to move through the whole structure, not trapped between pairs. Metallic bonding is the electrostatic attraction between the positive ions and delocalised electrons.	Correct both scale and electron location. The structure repeats throughout the metal, and the electron population is shared across it. The strong attraction is between opposite charges, not a localised shared pair as in a covalent bond.	4
Total Question 2			4
03.1	In a metal, outer-shell electrons are delocalised through the whole giant structure. The delocalised electrons are attracted to positive metal ions. In a simple covalent molecule, each bonding pair is shared between two particular atoms. A shared pair is attracted to the nuclei of the two bonded atoms. Both involve electrostatic attraction, but metallic sharing is extended whereas each covalent pair is localised.	Organise the comparison around electron location and the positive particles attracting those electrons. Do not describe metallic bonding as separate electron pairs between neighbouring ions.	5
Total Question 3			5
04.1	The model contains $12 × 2 = 24$ delocalised electrons. The 12 ions have a total charge of $+24$ and the 24 electrons have a total charge of $-24$ , so the model is neutral overall. The positive metal ions form a regular giant structure. The electrons are delocalised through the whole structure rather than fixed between ion pairs. Strong electrostatic attraction between the positive ions and delocalised electrons holds the structure together.	Multiply the number of ions by the number of electrons contributed per atom. The 24 negative electron charges balance the total ion charge of $12 × 2 = +24$ . Then interpret the two charged components as a giant lattice held together by attraction throughout the structure.	5
Total Question 4			5
05.1	The metal is a giant structure containing layers of positive ions. Bending changes which ions are next to one another as the layers move. The electrons are delocalised and can redistribute through the whole structure. Electrostatic attraction between the ions and delocalised electrons is maintained after the layers move. Fixed local electron pairs would be tied to particular neighbours and would not explain bonding being maintained as neighbours change.	Follow the particles during deformation. The neighbouring ions change, so a model based on permanent local pairs is unsuitable. A mobile population of delocalised electrons can remain attracted to the positive ions before, during and after the layers shift, maintaining metallic bonding across the sheet.	5
Total Question 5			5

4.2.2.1 · The three states of matter

Tier 1 · Easy

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01.1	Liquid to gas: boiling. Gas to liquid: condensing.	Use the named changes at the boiling point: heating a liquid produces a gas by boiling, while cooling a gas produces a liquid by condensing.	2
Total Question 1			2
02.1	Solid to liquid: melting. Liquid to solid: freezing.	At the melting point, heating changes a solid into a liquid. Cooling reverses the change, so a liquid becomes a solid by freezing.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.2.1 Tier 2 · Standard
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01.1	At $-30\,^{\circ}\text{C}$ : solid. At $20\,^{\circ}\text{C}$ : liquid. At $100\,^{\circ}\text{C}$ : gas.	$-30\,^{\circ}\text{C}$ is below the melting point, so S is solid. $20\,^{\circ}\text{C}$ lies between the melting and boiling points, so S is liquid. $100\,^{\circ}\text{C}$ is above the boiling point, so S is a gas.	3
Total Question 1			3
02.1	During freezing, particles become fixed in a regular arrangement. Energy is transferred from the substance to the surroundings during freezing. During boiling, particles become widely separated as a gas forms. Energy is transferred to the substance to overcome forces between particles during boiling.	Track one sample in opposite energy-transfer directions. Freezing releases energy as particles become fixed in a solid arrangement. Boiling requires energy to overcome attractions so that particles can separate into the gas state.	4
Total Question 2			4
03.1	The energy is used to overcome forces between the particles. The particles change from fixed positions to being able to move past one another. The energy is not increasing the particles' kinetic energy during the change, so the temperature stays constant.	At a change of state, track where the transferred energy goes. During melting it changes the particle arrangement by overcoming attractions rather than raising the temperature.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.2.1 Tier 3 · Hard
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01.1	P is a gas at $100\,^{\circ}\text{C}$ . Q is a solid at $100\,^{\circ}\text{C}$ . Q has much stronger forces or bonds between its particles than P. More energy is needed to overcome the attractions in Q, giving it higher melting and boiling points.	$100\,^{\circ}\text{C}$ is above P's boiling point, so P is a gas, but it is below Q's melting point, so Q is solid. Higher change-of-state temperatures mean that more energy must be transferred to separate or rearrange the particles, so the attractions in Q are stronger.	5
Total Question 1			5
02.1	The model does not show forces between particles. It represents every particle as a sphere even though particles can have different structures or shapes. It represents particles as solid and inelastic, which is an oversimplification.	Compare the drawing conventions with what the model omits or simplifies. The required limitations concern absent forces, universal spherical shape and the assumption that each particle is a solid inelastic object.	4
Total Question 2			4
03.1	Individual atoms do not have the bulk property of being solid or liquid. The particles themselves do not expand during melting. Energy is transferred to overcome some of the forces holding the particles in fixed positions. The particles remain close together but become able to move past one another in the liquid.	Distinguish properties of the bulk material from properties of one particle. Melting changes the particles' arrangement and movement; it does not turn each particle into a tiny liquid object.	4
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04.1	The melting point is above $12\,^{\circ}\text{C}$ . The melting point is at or below $18\,^{\circ}\text{C}$ , so it lies in the interval above $12\,^{\circ}\text{C}$ to $18\,^{\circ}\text{C}$ . The boiling point is above $72\,^{\circ}\text{C}$ . The boiling point is at or below $78\,^{\circ}\text{C}$ , so it lies in the interval above $72\,^{\circ}\text{C}$ to $78\,^{\circ}\text{C}$ . At $50\,^{\circ}\text{C}$ , R is a liquid because this temperature is above the melting range and below the boiling range.	Bracket each change using the nearest observations on either side. A solid at $12\,^{\circ}\text{C}$ but a liquid at $18\,^{\circ}\text{C}$ places the melting point between those temperatures. A liquid at $72\,^{\circ}\text{C}$ but a gas at $78\,^{\circ}\text{C}$ similarly brackets the boiling point. $50\,^{\circ}\text{C}$ lies safely between the two ranges.	5
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05.1	At $91\,^{\circ}\text{C}$ the gas condenses, so the boiling point is $91\,^{\circ}\text{C}$ . During condensation, gas particles become close together in a liquid. Energy is transferred from the substance during condensation. At $27\,^{\circ}\text{C}$ the liquid freezes, so the melting point is $27\,^{\circ}\text{C}$ . During freezing, particles become fixed in a regular arrangement. Energy is transferred from the substance during freezing.	A constant-temperature section during cooling identifies a change of state. The higher plateau is condensation at the boiling point; the lower plateau is freezing at the melting point. In both changes energy leaves the substance, while the particles become closer or more ordered rather than gaining kinetic energy.	6
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4.2.2.2 · State symbols

Tier 1 · Easy

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01.1	(l) means liquid. (aq) means dissolved in water.	Read each symbol as a physical state. The symbol (aq) is reserved for an aqueous solution, so it must not be treated as another symbol for a pure liquid.	2
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02.1	Bromine: (l). Potassium bromide solution: (aq).	A pure liquid receives (l). A substance dissolved in water is aqueous and receives (aq), even though the solution itself is liquid.	2
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Tier 2 · Standard

Mark scheme for 4.2.2.2 Tier 2 · Standard
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01.1	CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).	Balance carbon and hydrogen first, then oxygen. Translate the stated physical state of each substance directly: methane, oxygen and carbon dioxide receive (g), while water receives (l). Coefficients do not change state symbols.	3
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02.1	Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).	Translate each stated substance and state directly. The formula count is already $1:1:1:1$ : one Zn, one Cu and one SO₄ group occur on each side, so the equation is balanced.	4
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03.1	2Mg(s) + O₂(g) → 2MgO(s). The equation is balanced with two magnesium atoms and two oxygen atoms on each side. Oxygen is a gas under the reaction conditions, so its state symbol is (g).	Write Mg, O₂ and MgO first. A coefficient of 2 before MgO balances oxygen, and a coefficient of 2 before Mg balances magnesium. Then attach the physical states.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.2.2 Tier 3 · Hard
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01.1	Barium chloride: (aq). Sodium sulfate: (aq). Barium sulfate: (s). Sodium chloride: (aq).	Both named reactant solutions receive (aq). The barium sulfate is stated to be a solid precipitate, so it receives (s). Sodium chloride remains dissolved in water, so it receives (aq), not (l).	4
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02.1	CuCl₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaCl(aq).	Use CuCl₂, NaOH, Cu(OH)₂ and NaCl. A coefficient of 2 before NaOH supplies two hydroxide groups and two sodium atoms; a coefficient of 2 before NaCl then balances sodium and chlorine. Both reactants and sodium chloride are aqueous, while the precipitate is solid.	4
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03.1	CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). The coefficient 2 before HCl supplies two H atoms and two Cl atoms. Ca, C, O, H and Cl are balanced atom by atom. The state symbols are (s), (aq), (aq), (l) and (g), respectively.	Balance calcium and the carbonate products first, then use two HCl to supply the two chlorides in CaCl₂ and the two hydrogens in H₂O. Finally apply the states stated in the prompt.	4
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04.1	The corrected equation is HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). HCl, NaOH and NaCl are aqueous because each is dissolved in water. The water product is liquid, so its symbol is (l). The symbol (aq) means dissolved in water, whereas (l) means the substance itself is in the liquid state.	Use the physical descriptions rather than assuming every fluid is a liquid substance. Solutions receive (aq), including the dissolved salt product. The water produced is the liquid substance itself, so it receives (l). The formula equation is already balanced in a 1:1:1:1 ratio.	4
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05.1	The balanced equation is 2H₂(g) + O₂(g) → 2H₂O(g). The coefficient 2 before H₂ balances four hydrogen atoms on each side. Hydrogen and oxygen have the state symbol (g) because both reactants are gases. At 130 °C the water product has the state symbol (g) because it is steam. After cooling to 25 °C the water has symbol (l) because the steam condenses to liquid water.	Balance the formulas before adding state symbols. Two hydrogen molecules react with one oxygen molecule to make two water molecules. The state of the product depends on the stated temperature: it is gaseous steam in the hot vessel and liquid after cooling and condensation.	5
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4.2.2.3 · Properties of ionic compounds

Tier 1 · Easy

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01.1	Solid: does not conduct. Molten: conducts.	In the solid, the ions are held in fixed lattice positions and cannot carry charge through the material. Melting frees the ions to move, so the liquid conducts.	2
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02.1	Its ions are held in fixed positions in the lattice. The charged particles cannot move through the solid to carry charge.	Having charged ions is not sufficient for conduction. In the solid lattice, those ions cannot move, so there is no mobile charge carrier.	2
Total Question 2			2

Tier 2 · Standard

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01.1	Magnesium oxide has a giant ionic lattice. There are strong electrostatic attractions between oppositely charged ions in all directions. A large amount of energy is needed to overcome the many strong ionic bonds.	Build the explanation from structure to bonding to energy: identify the giant lattice, state the attraction between opposite charges, then connect the many strong attractions to the large energy transfer required for melting.	3
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02.1	The ionic solid separates into positive and negative ions in the water. The ions are free to move through the solution. Moving ions carry electrical charge through the solution.	Dissolving removes the fixed lattice arrangement. Both types of ion become mobile in the aqueous solution and can transport charge between the electrodes.	3
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03.1	The powder will not conduct electricity. Crushing breaks the solid into smaller pieces but the ions remain fixed in ionic lattices. Charge can flow only when the ions become mobile, for example after melting or dissolving the compound in water.	Particle size is not the deciding factor for ionic conduction. Test whether charged ions can move through the sample; crushing does not give lattice ions that mobility.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.2.3 Tier 3 · Hard
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01.1	T has a giant ionic lattice. Strong electrostatic attractions act between oppositely charged ions in all directions. Much energy is needed to overcome these attractions, so the melting point is high. In the solid, ions are fixed and cannot carry charge through the lattice. When molten or dissolved, ions are free to move and carry charge.	The combined pattern is diagnostic of an ionic compound. Use the giant lattice and strong attractions to explain the melting point. Then compare mobility: ions are fixed in the solid but mobile in the melt and aqueous solution. Charge is carried by ions, not by delocalised electrons.	6
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02.1	J has a giant ionic lattice. Many strong electrostatic attractions require much energy to overcome, giving J a high melting point. J's ions are fixed in the solid, so it does not conduct. The ions become mobile when J melts and carry charge. K consists of small neutral molecules with weak intermolecular forces, giving a low melting point. K has no mobile charged particles, so it does not conduct.	Use the change in conductivity to identify J as ionic, then link lattice attractions to its high melting point. K's low melting point and absence of conduction fit small neutral molecules held together by weak intermolecular forces.	6
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03.1	The original solid does not conduct because its ions are fixed in the lattice. The aqueous solution conducts because the dissolved ions are free to move. The recovered solid crystals do not conduct because the ions are fixed again. The molten compound conducts because its ions are mobile. The charge carriers in the conducting stages are ions, not electrons. Evaporation removes water but does not change the ionic compound into a molecular substance.	Follow ion mobility through the physical changes. Dissolving and melting make ions mobile; crystallising restores a rigid lattice. The substance remains ionic throughout.	6
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04.1	Distilled water not conducting shows that water alone is not the charge carrier in this test. Sugar solution not conducting shows that dissolved neutral molecules do not provide mobile charged particles. Sodium chloride solution contains positive and negative ions that are free to move through the water. Molten sodium chloride also conducts because its ions are free to move even though no water is present. The evidence therefore supports mobile ions, not water molecules, carrying charge in the solution.	Use the distilled water and sugar solution as comparisons. They separate the effect of water and dissolving from the effect of producing mobile ions. The molten sample is decisive because it conducts without any water, leaving mobile ions as the explanation common to both conducting samples.	5
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05.1	The charge carriers in molten sodium chloride are ions, not electrons released from chloride ions. Positive sodium ions are free to move through the liquid. Negative chloride ions are also free to move through the liquid. The oppositely charged ions move in opposite directions and both movements transfer charge. In solid sodium chloride the ions are fixed in lattice positions, so charge cannot flow.	Identify the charged particles already present in the ionic compound. Melting frees both kinds of ion to move without requiring electron transfer or new particles. Their motion carries charge through the liquid, whereas the same ions are immobilised by the solid lattice.	5
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4.2.2.4 · Properties of small molecules

Tier 1 · Easy

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01.1	Only weak intermolecular forces need to be overcome when it boils. The strong covalent bonds within the molecules are not broken.	Separate the two scales of attraction. Covalent bonds hold atoms together inside each molecule, but boiling moves whole molecules apart by overcoming the much weaker forces between them.	2
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02.1	The molecules have no overall electric charge. There are no mobile charged particles to carry charge.	Electrical conduction needs charged particles that can move. Neutral molecules provide neither ions nor delocalised electrons.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.2.4 Tier 2 · Standard
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01.1	V usually has the higher boiling point. Larger molecules have stronger intermolecular forces. More energy is needed to overcome these forces between V molecules.	Apply the specified trend: increasing molecular size strengthens the forces between molecules. Stronger intermolecular forces require a larger energy transfer to overcome, so the boiling point is higher.	3
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02.1	Weak intermolecular forces between molecules are overcome. The molecules move apart to form a gas. The strong covalent bonds within each molecule remain intact.	Boiling changes the spacing between complete molecules. It therefore overcomes forces between molecules rather than the bonds holding atoms together inside each molecule.	3
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03.1	X is a gas at 20 °C because this temperature is above its boiling point. Its low boiling point suggests that the intermolecular forces are weak. Only a modest energy transfer is required to separate the molecules.	Locate the stated temperature above the boiling point, then connect a low boiling point to the energy needed to overcome attractions between separate molecules.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.2.4 Tier 3 · Hard
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01.1	B is made of larger molecules than A. B therefore has stronger intermolecular forces. More energy is needed to overcome the forces between B molecules, so B has the higher boiling point. Neither pure substance conducts electricity because the molecules have no overall charge.	Use relative molecular mass as evidence that B has larger molecules. Link larger molecules to stronger intermolecular forces and hence a greater energy requirement for boiling. For conductivity, use the stated neutral molecular particles: there are no mobile charged particles in either pure substance.	5
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02.1	The covalent bonds inside M's molecules are strong. M melts easily because only weak intermolecular forces are overcome. The covalent bonds remain unbroken when M melts. Movement alone does not cause electrical conduction. M's molecules are neutral, so liquid M has no mobile charged particles and does not conduct.	Correct the scale of each attraction first: melting separates molecules without breaking their internal covalent bonds. Then apply the charge-carrier test; mobile neutral molecules cannot carry an electric current.	5
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03.1	Boiling and condensing are physical changes, not chemical reactions. Boiling overcomes weak intermolecular forces between molecules. The strong covalent bonds within each molecule are not broken, so the molecules remain chemically unchanged. During condensation, intermolecular attractions form again as energy is transferred to the surroundings.	Distinguish bonds within molecules from forces between molecules. Reversible changes of state alter the intermolecular attractions and particle spacing without making a new substance.	4
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04.1	The likely size order is U smallest, then V, then W largest. Larger molecules usually have stronger intermolecular forces. W needs the most energy because its intermolecular forces are strongest, while U's are weakest. Boiling overcomes the intermolecular forces between molecules. The strong covalent bonds within each molecule are not broken.	Because the samples contain equal numbers of molecules and are compared under the same conditions, use the energy needed to separate them as evidence for intermolecular-force strength. The specification trend links stronger intermolecular forces to larger molecules. Keep those attractions separate from the covalent bonds inside every molecule.	5
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05.1	Boiling Z requires only the weak intermolecular forces between molecules to be overcome. The covalent bonds within each Z molecule remain intact during boiling. Splitting a molecule into atoms requires strong covalent bonds to be broken, so it needs much more energy. The two observations involve different attractions and therefore do not contradict each other. Pure liquid Z does not conduct because its neutral molecules provide no mobile charged particles.	Separate a change in physical state from breaking a molecule apart chemically. The low boiling point measures the forces between intact molecules, while atomising a molecule measures its strong internal covalent bonds. Electrical conduction is a third question: neutral molecules cannot carry a current through the liquid.	5
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4.2.2.5 · Polymers

Tier 1 · Easy

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01.1	Polymer. Covalent bonds join the atoms within a chain.	Use the phrase ‘separate, very long chains’ to identify a polymer. Then distinguish the bonds inside a molecule from the forces between molecules: the atoms along each chain are linked by strong covalent bonds.	2
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02.1	The substance is a polymer. n represents a large number of repeating units.	A bracketed covalent unit repeated many times is the standard representation of a polymer molecule. The symbol n indicates that the unit occurs a large number of times.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.2.5 Tier 2 · Standard
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01.1	P has very large polymer molecules. The intermolecular forces between its molecules are relatively strong. At $20\,^\circ\text{C}$ there is insufficient energy to overcome these forces, so P remains solid.	Build a structure-to-property chain. Long covalent chains mean very large molecules; these have relatively strong intermolecular forces. Link that explicitly to the state: room-temperature energy does not overcome enough of those forces for the molecules to move past one another.	3
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02.1	Atoms within each chain are joined by strong covalent bonds. Each chain is a very large molecule. Relatively strong intermolecular forces act between separate polymer molecules.	Keep the two structural levels separate. Covalent bonds join atoms inside one long molecule, while intermolecular forces act from one molecule or chain to another.	3
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03.1	It is a polymer because each molecule is a very long chain containing many covalently bonded atoms. There are intermolecular forces between the separate polymer molecules. These intermolecular forces are relatively strong, so enough energy is needed to separate the chains that the substance is solid at room temperature.	Use two structural scales. Strong covalent bonds make each very large chain molecule, while the combined attractions between long, separate molecules determine the room-temperature state.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.2.5 Tier 3 · Hard
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01.1	R is the polymer. R consists of separate very large molecules or chains. Forces between R's chains are intermolecular forces rather than covalent bonds. S is a giant covalent structure in which covalent bonds connect atoms throughout one continuous network.	First decide whether the covalent bonding stops at the edge of a molecule. It does in R, leaving separate chains, so R is the polymer. In S the covalent bonds continue throughout the structure, so S is a giant covalent network rather than a collection of polymer molecules.	4
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02.1	T contains very large polymer molecules. Relatively strong intermolecular forces act between the chains. At room temperature these forces prevent the chains moving past one another easily, so T is solid. Heating supplies enough energy to overcome some intermolecular forces, allowing the chains to move and the material to soften. On cooling, intermolecular forces form between the chains again; the covalent bonds within each chain remain intact.	Use intermolecular forces to explain changes in chain mobility. Heating affects attractions between the separate molecules, not the strong covalent framework within each chain. Cooling reverses the energy transfer and restricts chain movement again.	5
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03.1	The melting point is likely to decrease. The shorter molecules have weaker intermolecular forces between them. Less energy is therefore needed to overcome the attractions between molecules. The covalent bonds within each chain do not need to break when the substance melts.	Keep intramolecular bonding separate from intermolecular attraction. Shortening the molecules reduces the attractions between different chains, while leaving the type of strong bond inside each chain unchanged.	4
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04.1	The brackets show a repeating unit within a polymer chain, not a separate A–B molecule. The symbol n is a large number of repeating units in the long chain. Atoms within each polymer molecule are joined by strong covalent bonds. Different long polymer molecules have relatively strong intermolecular forces between them. Enough energy is needed to overcome these interchain forces that the polymer can remain solid at room temperature.	Interpret the brackets as a structural repeat rather than a boundary around a small molecule. Repeating the unit many times makes one very large covalently bonded molecule. The bulk sample contains many such chains, and their relatively strong intermolecular attractions explain its room-temperature state.	5
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05.1	Sample size cannot be the explanation: the samples have equal masses, and melting point does not depend on the amount of substance heated. Q's molecules are much shorter, so there are fewer points of contact between neighbouring molecules. The total intermolecular forces between Q's shorter molecules are therefore weaker than between P's very large molecules. Less energy is needed to overcome the weaker intermolecular forces and separate Q's molecules, giving the lower melting point. The evidence therefore supports molecular size, not sample size, as the explanation.	Test each proposed cause against the information given. Equal masses, and the fact that a melting point is a property of the substance rather than the sample, eliminate sample size. Polymer properties depend on the relatively strong intermolecular forces between their very large molecules, so shorter molecules mean weaker total attractions and a lower melting point.	5
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4.2.2.6 · Giant covalent structures

Tier 1 · Easy

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01.1	A giant covalent structure.	Look for a network with no boundary around an individual molecule. Because the covalent bonding continues throughout the sample, the structure is giant covalent.	1
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02.1	Any two from diamond, graphite and silicon dioxide (silica).	The specified examples are the two giant forms of carbon, diamond and graphite, and the giant network compound silicon dioxide.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.2.6 Tier 2 · Standard
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01.1	Silicon dioxide has a giant covalent structure. There are many strong covalent bonds between its atoms. A large amount of energy is needed to overcome these bonds, giving it a very high melting point.	Name the structure, identify the strong bonds that extend throughout it, and then link bond strength to energy. Since $1500\,^\circ\text{C}$ is still below the melting point, the structure remains solid.	3
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02.1	It is a giant covalent structure. It has a very high melting point. Many strong covalent bonds throughout the network require a large energy transfer to overcome.	Bonds that continue beyond the displayed section show that the drawing is part of one giant network rather than a complete molecule. Melting that network requires many strong covalent bonds to be overcome.	3
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03.1	SiO₂ represents a $1:2$ ratio of silicon atoms to oxygen atoms in a giant structure. All the atoms are joined in a giant covalent network. There are many strong covalent bonds throughout the structure. Melting requires enough energy to overcome many of these strong bonds.	Interpret the formula as the simplest atom ratio in an extended structure. Its melting point is controlled by the strong covalent bonds in that network, not by forces between molecules.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.2.6 Tier 3 · Hard
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01.1	V has a giant covalent structure. Strong covalent bonds extend throughout V's network. Many strong covalent bonds must be overcome to melt V, requiring a large amount of energy. Melting U only overcomes intermolecular forces between its separate molecules, so U melts at a much lower temperature.	Use ‘no separate molecules’ and the extreme temperature to infer a giant network for V. Compare what must be overcome: covalent bonds throughout V but only intermolecular forces between U's molecules. This accounts for the large difference in energy needed.	4
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02.1	Silicon dioxide has a giant covalent structure, not separate molecules. Silicon and oxygen atoms are connected by covalent bonds throughout a continuous network. Many strong covalent bonds must be overcome to melt the structure. A large amount of energy is therefore needed; intermolecular forces are not the explanation.	Do not infer molecular structure from the empirical formula. Replace the claimed molecules and intermolecular forces with the continuous giant network and its many strong covalent bonds.	4
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03.1	Both substances have giant covalent structures. In each substance, atoms are linked by strong covalent bonds throughout a continuous network. Many strong covalent bonds must be overcome for either substance to melt, requiring a large amount of energy. The atoms and detailed arrangements can differ while the giant covalent structure type still gives both substances very high melting points.	Compare the shared structural feature that controls melting rather than claiming every atom or bond arrangement is the same. A continuous network of strong bonds is sufficient for the common bulk property.	4
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04.1	Grinding makes the pieces smaller but does not change the giant covalent structure within each grain. Atoms in every grain remain linked by strong covalent bonds in a continuous network. Many strong covalent bonds still have to be overcome for the powder to melt. The melting point should therefore remain very high rather than falling greatly just because the grains are smaller.	Separate bulk particle size from atomic bonding. Grinding creates more grains and new surfaces, but it does not turn the material into small molecules. Each grain retains an extended network of strong covalent bonds, so the reason for the high melting point remains.	4
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05.1	CO₂ consists of separate small molecules. Only weak intermolecular forces between CO₂ molecules are overcome when it changes state. SiO₂ is a giant covalent structure and does not contain separate SiO₂ molecules. Strong covalent bonds link silicon and oxygen atoms throughout the continuous network. Many strong covalent bonds must be overcome to melt silicon dioxide, requiring much more energy. A chemical formula can show an atom ratio without specifying whether the substance is molecular or giant.	Use the formula only as a ratio, then identify the different structural scales. Carbon dioxide has intact molecules separated by weak attractions. Silicon dioxide is one extended network whose strong internal bonds must be overcome. The structure, not the shared numerical ratio, controls the melting behaviour.	6
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4.2.2.7 · Properties of metals and alloys

Tier 1 · Easy

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01.1	Its atoms are arranged in layers. The layers can slide over one another when a force is applied.	Identify the relevant structural feature, which is the layered arrangement of atoms, and link movement of those layers to the observed change of shape.	2
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02.1	It represents an alloy. It is harder than the pure metal.	A pure metal contains equal-sized atoms in regular layers. Different sizes identify an alloy and distort the layers, making them harder to slide.	2
Total Question 2			2

Tier 2 · Standard

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01.1	The different-sized atoms distort the regular layers in M's structure. The distorted layers cannot slide over one another as easily. More force is therefore needed to change the alloy's shape, so it is harder.	Follow the causal sequence demanded by AQA: different atom sizes, then distortion of the layers, then reduced sliding. Finish by connecting reduced sliding to increased hardness.	3
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02.1	Equal-sized atoms are arranged in regular layers. The layers can slide over one another when a force is applied. The sliding changes the shape of the sheet without separating all the atoms.	Use the regular-layer model for a pure metal. Bending moves one layer relative to another; it does not require the entire giant structure to split apart.	3
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03.1	The metal has a giant structure with strong metallic bonding. Melting requires enough energy to overcome many metallic bonds, giving a high melting point. The atoms in a pure metal are arranged in regular layers. The layers can slide over one another when a force is applied, so the metal can be bent.	Explain the two properties with different aspects of the same structure. Bond strength controls melting, whereas the regular layered arrangement controls how readily the solid changes shape.	4
Total Question 3			4

Tier 3 · Hard

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01.1	Both materials have giant metallic structures with strong metallic bonding. A large amount of energy is needed to overcome the strong metallic bonding, so their melting points are high. Pure Q has regular layers of atoms that can slide over one another. Different-sized atoms in the alloy distort these layers, making sliding more difficult. Therefore the alloy needs the larger force to bend it.	Treat the two properties separately. Explain melting using strong metallic bonding in a giant structure. Explain bending using layer movement: regular layers slide in pure Q, while different-sized alloy atoms distort the layers and oppose sliding, consistent with $47\,\text{N}>18\,\text{N}$ .	5
Total Question 1			5
02.1	The explanation for alloy hardness is distortion of the layers, not simply stronger bonds. Different-sized atoms in an alloy make the layers harder to slide. A pure metal can be soft because its regular layers slide easily. The pure metal still has a giant structure with strong metallic bonding. Much energy may be needed to overcome the metallic bonds, so softness does not by itself mean a low melting point.	Separate resistance to shape change from melting. Hardness depends on how easily layers slide, whereas melting depends on the energy needed to overcome metallic bonding in the giant structure.	5
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03.1	Alloy B is the most suitable material. It does not bend under the 50 N force used in service because it requires 62 N. It can still be shaped below 70 N. Different-sized atoms in the alloy distort the regular layers. The distortion makes it more difficult for the layers to slide than in pure P or alloy A.	Apply both numerical constraints before using the particle model. Only B has a bending force between 50 N and 70 N; its greater hardness is explained by distorted layers.	5
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04.1	Both alloys are harder than pure M because each needs a larger force to bend. Atoms of the second element have a different size from atoms of M. The different-sized atoms distort the regular layers found in the pure metal. The distortion makes it more difficult for the layers to slide. The 9% alloy bends at 39 N compared with 43 N for the 4% alloy, so these data do not support the claim that more always means harder.	First compare each alloy with the pure-metal control, then compare the two alloys with each other. The particle model explains the overall increase through layer distortion. However, the second comparison runs against the word ‘always’, because the sample with the larger percentage bends under a smaller force.	5
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05.1	The alloy may be harder because different-sized atoms can distort its layers and hinder sliding. The test does not by itself prove this because the rods have different diameters. A thicker rod could require more force even if its material were not harder. The samples should have the same dimensions and be tested in the same way. If the equal-sized alloy then needs more force, the result supports the distorted-layer explanation.	Separate the expected chemistry from the quality of the evidence. Alloy structure gives a plausible prediction, but diameter is a confounding variable. A fair comparison keeps size and test conditions constant so that a force difference can be attributed to the different atomic arrangements.	5
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4.2.2.8 · Metals as conductors

Tier 1 · Easy

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01.1	Delocalised electrons.	In a solid metal the positive ions do not travel along the wire. The mobile charge carriers are the delocalised electrons.	1
Total Question 1			1
02.1	Delocalised electrons.	The mobile delocalised electrons move through the metallic structure and transfer energy from hotter regions to cooler regions.	1
Total Question 2			1

Tier 2 · Standard

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01.1	The metal contains delocalised electrons. These electrons are free to move through the metallic structure. Their movement carries electrical charge through the strip.	State which charged particles are present, establish that they are mobile, and then connect their motion to charge transfer. The comparison does not require claiming that the polymer has mobile ions or electrons.	3
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02.1	Aluminium contains delocalised electrons. These electrons can move through the solid metal and carry charge. Sodium chloride contains ions held in fixed lattice positions when solid. Its ions cannot move through the solid to carry charge.	Identify the charge carrier in each structure. A solid metal already contains mobile electrons, whereas the charged ions in a solid ionic lattice remain fixed.	4
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03.1	The positive metal ions remain in fixed positions in the solid structure. Delocalised electrons are free to move through the metal. These electrons carry electrical charge through the wire.	Identify which charged particles are mobile in a solid metal. The regular ion structure stays in place; the delocalised electrons provide the current.	3
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Tier 3 · Hard

Mark scheme for 4.2.2.8 Tier 3 · Hard
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01.1	Copper contains delocalised electrons. The delocalised electrons can move through the metallic structure. They carry electrical charge through a wire. They also transfer thermal energy rapidly through the saucepan base.	Identify the shared cause before treating the two uses. Mobile delocalised electrons account for both properties: charge transport explains electrical conduction, while energy transfer by those electrons explains thermal conduction.	4
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02.1	The solid metal conducts. Delocalised electrons are free to move through the metal and carry charge. The solid ionic compound does not conduct. Its ions are charged but fixed in the lattice. The molten ionic compound conducts because its ions are free to move. Both positive and negative ions carry charge through the molten compound.	Do not use one charge carrier for both bonding types. Metals conduct through mobile delocalised electrons even while solid. Ionic compounds require mobile ions, which become available only after melting or dissolving.	6
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03.1	The wire still contains a giant structure of positive metal ions and delocalised electrons. The delocalised electrons remain free to move through the reshaped metal. The moving electrons carry electrical charge through the wire. The positive metal ions form the structure and attract the delocalised electrons, but the ions do not flow along the wire.	Changing the sample's shape does not remove its charge carriers or change its bonding type. State separately what moves to carry charge and what remains as the positive ion structure.	4
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04.1	Molten sodium chloride contains mobile positive and negative ions that carry electrical charge through the liquid. Sodium chloride is ionic, so the only mobile charged particles in the melt are ions, not electrons. The molten ionic compound has no delocalised electrons, so it lacks the fast electron energy-transfer route. Liquid copper contains delocalised electrons that move through the metallic structure and carry electrical charge. Those mobile electrons also transfer thermal energy rapidly from hotter regions to cooler regions, so copper conducts heat better.	Separate electrical charge transport from the efficiency of thermal transfer. Mobile ions are sufficient for current in the molten ionic compound, but they do not provide the fast electron-based energy-transfer route present in copper. Delocalised electrons account for both electrical conduction and copper's much greater thermal conductivity.	5
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05.1	Each metal grain already contains delocalised electrons before compression. The electrons can move through the metallic structure within each separate grain. Air gaps or poor contacts between loose grains interrupt a continuous conducting path. Compression brings grains into contact and creates a path across the pellet. Conduction then occurs because existing delocalised electrons carry charge through the connected metal, not because new electrons were created.	Distinguish the charge carriers inside each grain from the route across the whole sample. Metallic bonding and delocalised electrons exist in both tests. Compression changes contact between grains, allowing those mobile electrons to pass through a continuous network from one clip to the other.	5
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4.2.3.1 · Diamond

Tier 1 · Easy

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01.1	Four covalent bonds.	Recall that diamond uses all four outer electrons of each carbon atom to make four covalent bonds in its giant structure.	1
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02.1	Diamond has a giant covalent structure. Each carbon atom forms four covalent bonds.	Diamond is one continuous carbon network. Four bonds from every carbon connect the network in a rigid arrangement.	2
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Tier 2 · Standard

Mark scheme for 4.2.3.1 Tier 2 · Standard
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01.1	Diamond has a giant covalent structure. Each carbon atom is joined by four strong covalent bonds in a rigid network. A large force is needed to deform or break this network, so diamond is very hard.	Translate the use into the required property, hardness. Then link hardness to the rigid three-dimensional network of four strong covalent bonds around each carbon atom.	3
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02.1	Each carbon atom uses all four outer electrons in covalent bonds. Diamond therefore has no delocalised electrons. There are no mobile charged particles to carry electrical charge.	Link the four bonds per carbon to electron availability. With every outer electron held in a covalent bond, no electron can move through the structure as a charge carrier.	3
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03.1	The sticks do not represent the shared electron pairs in covalent bonds. The atom sizes and distances between atoms may not be to scale. The model shows only a finite section even though the giant covalent network continues throughout the crystal.	Check what the model simplifies about bonds, scale and extent. A useful connectivity diagram is not a literal, complete picture of the electron distribution or whole crystal.	3
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Tier 3 · Hard

Mark scheme for 4.2.3.1 Tier 3 · Hard
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01.1	Diamond has a giant covalent structure with four covalent bonds from each carbon atom. Many strong covalent bonds must be overcome to melt it, so its melting point is very high. The rigid network of strong bonds makes diamond hard. All four outer electrons of each carbon atom are used in covalent bonds, so there are no delocalised electrons. Diamond cannot carry the heating current because it does not conduct electricity.	Separate the three requested properties. Use strong covalent bonds throughout the network for melting point, rigidity for hardness, and the absence of mobile delocalised electrons for electrical conduction. The last point rules out diamond as the current-carrying element.	5
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02.1	Methane consists of separate small molecules. Only weak intermolecular forces are overcome when methane changes state. Diamond has a giant covalent structure. Each carbon atom has four strong covalent bonds in a rigid network. Many strong covalent bonds require much energy to overcome, making diamond hard and giving it a very high melting point.	The word covalent does not by itself determine bulk properties. Compare the separate molecules in methane with the continuous bonded network in diamond, then identify the different attractions that must be overcome.	5
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03.1	Each carbon atom in diamond forms four covalent bonds, while each carbon atom in graphite forms three. Diamond is a three-dimensional giant covalent network; graphite contains layers of hexagonal rings. Diamond is hard because strong covalent bonds extend throughout its rigid structure. Graphite is soft because there are no covalent bonds between its layers, so the layers can slide. Diamond does not conduct because it has no delocalised electrons. Graphite conducts because one electron from each carbon atom is delocalised and can carry charge.	Build the comparison from carbon's bonding in each allotrope, then link each structural difference to the relevant property. Do not describe graphite's covalent bonds within layers as weak.	6
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04.1	Counting four bond connections per atom gives $28 × 4 = 112$ bond connections. Each bond joins two atoms and has been counted twice, so there are $112 ÷ 2 = 56$ distinct covalent bonds. The bonds form a rigid giant covalent network, making diamond hard. All four outer electrons from each carbon atom are used in covalent bonds. There are no delocalised electrons to carry charge, so diamond does not conduct electricity.	Multiply atoms by bonds per atom, then correct the double count because every bond has two ends. Interpret the high connectivity as a rigid network. The same four-bond electron accounting leaves no mobile electron population, giving the conductivity prediction.	5
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05.1	The material could be diamond. In diamond each carbon atom forms four strong covalent bonds in a giant structure, which explains the hardness and the very high melting point. Diamond has no delocalised electrons or other free charged particles, so it does not conduct electricity. It cannot be graphite: graphite conducts electricity because one delocalised electron per carbon atom can move, and it is soft because its layers slide over each other. It cannot be a metal: metals conduct electricity even when solid because their delocalised electrons are free to move.	Match each observation to the structure that allows it. Hardness and a very high melting point need a rigid giant structure of strong covalent bonds, and the absence of conduction needs every outer electron held in bonds — both point to diamond. Graphite and metals are eliminated because each contains delocalised electrons that would make the material conduct.	5
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4.2.3.2 · Graphite

Tier 1 · Easy

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01.1	Three covalent bonds.	Each graphite carbon bonds to three neighbouring carbon atoms in a hexagonal layer, leaving one electron delocalised.	1
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02.1	The carbon atoms form hexagonal rings. There are no covalent bonds between adjacent layers.	Graphite is built from sheets of joined carbon hexagons. Strong covalent bonding stops at each layer, leaving only weak forces between neighbouring layers.	2
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Tier 2 · Standard

Mark scheme for 4.2.3.2 Tier 2 · Standard
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01.1	One electron from each carbon atom is delocalised. The delocalised electrons can move through the structure. They carry electrical charge, so graphite conducts electricity.	An electrode must conduct. Identify the delocalised electron supplied by each carbon atom, state that these electrons are mobile, and connect their movement to charge transfer.	3
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02.1	Graphite consists of layers of carbon atoms. There are no covalent bonds between adjacent layers, only weak forces. The layers can slide over one another, so graphite acts as a lubricant.	Focus on attractions between layers rather than the strong bonds inside a layer. Weak interlayer forces allow relative movement and reduce resistance between the surfaces.	3
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03.1	$24$ delocalised electrons are contributed. Each carbon atom in graphite contributes one delocalised electron. The delocalised electrons can move through the structure and carry electrical charge.	Use the one-electron-per-carbon relationship, so 24 × 1 = 24. Then link electron mobility, not layer movement, to electrical conduction.	3
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Tier 3 · Hard

Mark scheme for 4.2.3.2 Tier 3 · Hard
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01.1	Graphite has strong covalent bonds between carbon atoms within each layer. Many strong bonds require a large amount of energy to overcome, so graphite has a high melting point. One electron per carbon atom is delocalised. The delocalised electrons move and carry charge, so graphite conducts electricity. There are no covalent bonds between adjacent layers, only weak forces. The layers can slide over one another, so the contact can move with low friction.	Match one structural feature to each demand. Strong bonds within layers give thermal stability; delocalised electrons give electrical conduction; the absence of covalent bonds between layers allows them to slide. Keep the within-layer and between-layer bonding explanations distinct.	6
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02.1	Graphite has strong covalent bonds within each carbon layer. There are no covalent bonds between layers; weak forces allow the layers to slide, making graphite soft. The strong bonds within layers require much energy to overcome, so graphite has a high melting point. Each carbon atom forms three covalent bonds. One electron from each carbon atom is delocalised. The delocalised electrons move and carry charge, so graphite conducts despite being a non-metal.	Resolve the apparent contradictions using different structural features. Softness concerns weak forces between layers, while melting concerns strong bonds within them. Electrical conduction comes from the fourth, delocalised outer electron on each carbon.	6
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03.1	Both graphite and metals contain delocalised electrons that can move and carry charge. In graphite, each carbon atom is covalently bonded to three others within layers. There are no covalent bonds between adjacent graphite layers. In a metal, positive metal ions form a giant structure. Metallic bonding is the electrostatic attraction between those positive ions and the delocalised electrons.	Start with the shared reason for conduction, then distinguish the structures holding the atoms or ions together. Graphite is not metallic simply because it conducts.	5
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04.1	Each carbon atom forms three covalent bonds. Three bonds leave one outer electron per carbon atom that is not used in bonding. That spare electron is delocalised and free to move, which is why the allotrope conducts electricity. Bonding to only three neighbours produces layers of hexagonal rings with no covalent bonds between the layers. The layers can slide over each other, which is why the material is soft enough to mark paper.	Work backwards from the observations. Conduction needs a mobile charge carrier, so at least one outer electron per atom must be delocalised rather than bonded. Softness needs layers that slide, so there must be no covalent bonds between layers. Both conditions are met when each carbon bonds to exactly three others, matching graphite.	5
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05.1	There are no covalent bonds between adjacent graphite layers. The layers can therefore slide over one another relatively easily. Within a layer, each carbon atom is joined to three others by strong covalent bonds. Pulling atoms apart within a layer requires strong covalent bonds to be overcome, so it meets greater resistance. Sliding does not remove the delocalised electrons within the layers, so graphite can still conduct electricity.	Match each force direction to the bonding it challenges. Shearing moves whole layers where no covalent bonds join one layer to the next. Pulling within a sheet acts against its strong carbon–carbon network. Neither simple layer sliding nor the absence of interlayer bonds removes the delocalised electrons that carry charge.	5
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4.2.3.3 · Graphene and fullerenes

Tier 1 · Easy

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01.1	Graphene is a single layer of graphite.	Treat graphite as a stack of carbon layers. One isolated layer from that stack is graphene.	1
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02.1	It is a molecule made only of carbon atoms. It has a hollow shape.	Fullerenes are discrete hollow carbon molecules. Their cages are based on rings of carbon atoms rather than a flat infinite sheet.	2
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Tier 2 · Standard

Mark scheme for 4.2.3.3 Tier 2 · Standard
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01.1	Graphene is a single layer of graphite and is only one atom thick. It therefore adds very little thickness to the sensor. Graphene contains delocalised electrons. The electrons can move and carry charge, so graphene conducts electricity.	Extract the two design requirements: small thickness and electrical conduction. Link the first to graphene being a single atomic layer, and the second to its mobile delocalised electrons.	4
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02.1	It is a carbon nanotube. Any two from nanotechnology, electronics and materials.	The cylindrical shape and very high length-to-diameter ratio identify a nanotube. Its specified application families include nanotechnology, electronic devices and materials.	3
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03.1	Graphene is only one atom thick, so it adds very little thickness. Each carbon atom is joined to other carbon atoms by strong covalent bonds. The extended sheet is therefore strong and can reinforce the composite.	Match each design requirement to a structural feature: one atomic layer supplies thinness, while strong covalent bonding across that layer supplies strength.	3
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Tier 3 · Hard

Mark scheme for 4.2.3.3 Tier 3 · Hard
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01.1	A is graphene. B is Buckminsterfullerene or C₆₀. C is a carbon nanotube. A is suitable for electronics or as part of a composite. C is suitable for nanotechnology, electronics or materials.	Use the defining shapes first: one layer identifies graphene, the 60-atom hollow sphere identifies Buckminsterfullerene, and the high-aspect-ratio cylinder identifies a nanotube. Then select applications from the specified families rather than inventing a shape-dependent use that has not been supported.	5
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02.1	Graphene is a single, one-atom-thick layer of graphite. Its carbon atoms form a flat sheet of hexagonal rings. C₆₀ is a separate hollow spherical molecule containing 60 carbon atoms. Its cage contains hexagonal and pentagonal carbon rings. Graphene is used in electronics or composites.	Compare one extended layer with one discrete cage. Graphene stays flat and one atom thick, while pentagonal rings help the mainly hexagonal C₆₀ network close into a hollow sphere.	5
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03.1	Fullerene structures are based mainly on hexagonal carbon rings, but they may also contain rings with five or seven carbon atoms. Buckminsterfullerene, C₆₀, is a hollow spherical molecule. Carbon nanotubes are also fullerenes, but they are cylindrical rather than spherical. A carbon nanotube has a very high length-to-diameter ratio. Therefore neither the claim about only hexagons nor the claim that every fullerene is spherical is correct.	Test each absolute claim against the specification. Use the possible five- and seven-membered rings to reject the first claim, then contrast spherical C₆₀ with cylindrical nanotubes to reject the second.	5
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04.1	The intact particle is graphite because its 15 carbon layers remain stacked together. That particle is about 15 atomic layers thick, so it is not a one-atom-thick sheet. Complete separation produces 15 individual graphene sheets. Each separated sheet consists of one carbon layer and is therefore only one atom thick. Only the separated sheets can add a single atomic layer to a coating or device; the stacked particle adds all 15 layers together.	Count connected layers rather than only the total number present. A stack remains a multilayer graphite particle even though each layer has graphene's internal structure. The one-atom-thick advantage applies only after the weak attractions between layers have been overcome and every layer exists as a separate sheet.	5
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05.1	The formula C is not enough because all three materials contain only carbon atoms. A flat sheet that is one atom thick identifies graphene. A hollow spherical molecule containing 60 carbon atoms identifies Buckminsterfullerene, C₆₀. A hollow cylinder with a very high length-to-diameter ratio identifies a carbon nanotube. Shape, scale and atomic arrangement are therefore needed in addition to elemental composition.	Separate composition from structure. An elemental formula cannot encode whether the carbon atoms form a sheet, sphere or cylinder. Match each candidate to its defining geometry and, for C₆₀, its molecule size before judging the student's claim.	5
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4.2.4.1 · Sizes of particles and their properties (chemistry only)

Tier 1 · Easy

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01.1	T is a nanoparticle. $300$ atom diameters span T.	The size $72\,\text{nm}$ lies inside the nano interval $1$ – $100\,\text{nm}$ . Because both dimensions use nanometres, compare them directly: $72/0.24=300$ , so T spans about $300$ atom diameters.	2
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02.1	$65\,\text{nm}$ : nanoparticle. $900\,\text{nm}$ : fine particle.	Nanoparticles span $1$ to $100\,\text{nm}$ . Fine particles are larger, from $100$ to $2500\,\text{nm}$ , so the two diameters fall in different ranges.	2
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Tier 2 · Standard

Mark scheme for 4.2.4.1 Tier 2 · Standard
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01.1	Surface area $=150\,\text{nm}^2$ . Volume $=125\,\text{nm}^3$ . Surface-area-to-volume ratio $=1.2:1$ .	A cube has six square faces, so $A=6a^2=6\times5^2=150\,\text{nm}^2$ . Its volume is $V=a^3=5^3=125\,\text{nm}^3$ . Therefore $A:V=150:125=1.2:1$ .	3
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02.1	$64=4^3$ , so four small cubes fit along each original edge. Each small cube has side length $80/4=20\,\text{nm}$ . Original surface area $=6\times80^2=38\,400\,\text{nm}^2$ . Total new surface area $=64\times6\times20^2=153\,600\,\text{nm}^2$ , an increase by a factor of $4$ .	Because $64$ equal cubes form a $4\times4\times4$ arrangement, each edge is quartered. Calculate the original area and the sum of all 64 smaller surface areas; $153\,600/38\,400=4$ .	4
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03.1	X: $2.4\times10^{-7}\,\text{m}=240\,\text{nm}$ , a fine particle. Y: $8.0\times10^{-9}\,\text{m}=8.0\,\text{nm}$ , a nanoparticle. For cubes the surface-area-to-volume ratio is $6/a$ , so it increases by the inverse side-length factor $240/8.0=30$ . Y's surface-area-to-volume ratio is $30$ times larger than X's.	Convert using $1\,\text{nm}=10^{-9}\,\text{m}$ , so multiply each metre value by $10^{9}$ . Compare the results with the ranges: nano $1$ – $100\,\text{nm}$ , fine $100$ – $2500\,\text{nm}$ . Because $A:V=6/a$ for a cube, the smaller cube's ratio is larger by the factor $240/8.0=30$ .	4
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Tier 3 · Hard

Mark scheme for 4.2.4.1 Tier 3 · Hard
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01.1	The nano cubes' surface-area-to-volume ratio is $20$ times larger. Estimated nano mass $=0.60\,\text{g}$ .	For similar cubes, $A:V=6/a$ , so the ratio increases by the inverse side-length factor: $800/40=20$ . Under the stated proportional-effectiveness assumption, divide the required mass by $20$ : $12/20=0.60\,\text{g}$ .	4
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02.1	For a cube, $6/a=0.15$ , so $a=40\,\text{nm}$ . A $40\,\text{nm}$ structure is a nanoparticle. $40/0.20=200$ atom diameters fit along one edge. Doubling the side length halves the surface-area-to-volume ratio. The new ratio is $0.075:1$ .	Rearrange the cube relationship $A:V=6/a$ to obtain $a=6/0.15=40\,\text{nm}$ . This lies in the $1$ – $100\,\text{nm}$ nano range. Divide by the atom diameter, then use the inverse relationship between side length and surface-area-to-volume ratio.	5
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03.1	For similar cubes, $A:V$ is inversely proportional to side length, so a $25$ -fold increase requires the side length to be divided by $25$ . $a=(1.5\times10^{-6})/25=6.0\times10^{-8}\,\text{m}$ . $6.0\times10^{-8}\,\text{m}=60\,\text{nm}$ . The new ratio is $6/(6.0\times10^{-8})=1.0\times10^8\,\text{m}^{-1}$ . $60\,\text{nm}$ is within the $1$ – $100\,\text{nm}$ range, so the new cubes are nanoparticles.	Work backwards from the required improvement. Because $6/a$ increases when $a$ decreases, divide the original side by $25$ , convert the result using $1\,\text{nm}=10^{-9}\,\text{m}$ , then compare it with the nanoparticle range.	5
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04.1	Assume the atoms are packed as equal cubes in a simple cubic arrangement with one atom diameter per position. The number of atoms along one edge is $2.4 ÷ 0.24 = 10$ . A three-dimensional cube therefore contains about $10 × 10 × 10$ atoms. The estimate is $10^3 = 1000$ atoms. $1000$ is within one order of magnitude of a few hundred, so the estimate supports the statement's rough scale; real shape and packing would change the value.	Under the cube-packing assumption, first divide the particle edge by one atom diameter without rounding: $2.4/0.24=10$ atoms per edge. Extend the count through three dimensions, giving $10^3=1000$ . Treat this as an estimate because atoms are not cubes and surface packing is incomplete.	5
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05.1	A could be from $99 - 3 = 96\,\text{nm}$ to $99 + 3 = 102\,\text{nm}$ . A crosses the $100\,\text{nm}$ boundary, so it could be nano or fine and cannot be assigned definitely to either range. B could be from $2492 - 12 = 2480\,\text{nm}$ to $2492 + 12 = 2504\,\text{nm}$ . B crosses the $2500\,\text{nm}$ boundary, so it could be fine or coarse and cannot be assigned definitely to either range. C could be from $2520 - 7 = 2513\,\text{nm}$ to $2520 + 7 = 2527\,\text{nm}$ . C's whole range is above $2500\,\text{nm}$ and below $10\,000\,\text{nm}$ , so C is definitely coarse.	Subtract and add each uncertainty before comparing with the category boundaries. A range that lies on both sides of a boundary cannot support a definite class. Only C has its complete possible range inside one category, the coarse interval from $2500\,\text{nm}$ to $10\,000\,\text{nm}$ .	6
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4.2.4.2 · Uses of nanoparticles (chemistry only)

Tier 1 · Easy

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01.1	Any two from medicine, electronics, cosmetics, sun creams, deodorants and catalysts.	Select two distinct application areas from the specification list. Do not give two products from the same area if the question asks for two areas.	2
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02.1	Nanoparticles may provide useful new properties or uses in areas such as medicine, electronics or catalysis. Testing is needed because possible health or environmental risks may not yet be known.	Balance potential benefit with uncertainty. Research can discover useful applications, while safety testing checks for possible effects on organisms and the environment.	2
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Tier 2 · Standard

Mark scheme for 4.2.4.2 Tier 2 · Standard
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01.1	The nanoparticles have a high surface-area-to-volume ratio, so less catalyst is needed for the same effect. Using less material may reduce raw-material use or cost. Less catalyst may also reduce waste or the mass that must be handled. A possible disadvantage is uncertain health or environmental risk if nanoparticles enter organisms or the environment.	Use the supplied equal-performance data to justify the benefits rather than assuming better performance. Credit the smaller mass through its high surface-area-to-volume ratio, then state a resource, cost or waste advantage. Balance this with a possible risk caused by the particles' small size.	4
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02.1	It kills a greater percentage of the tested bacteria: $90\%$ rather than $72\%$ . It uses one tenth of the mass of active material. Using less material may reduce resource use, cost or waste. Nanoparticles entering waste water create a possible environmental risk, although the data do not demonstrate harm.	Use both performance and mass data for the advantages. Treat detection in waste water as evidence of exposure, not proof of damage, so phrase the disadvantage as an uncertain environmental risk.	4
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03.1	Their high surface-area-to-volume ratio means a much smaller mass is effective. Using less catalyst may reduce material use, waste or cost. The catalyst can provide the same treatment effect as the larger particles. Nanoparticles might enter organisms or the environment, and their long-term effects may be unknown.	Use the supplied performance comparison rather than assuming a new chemical property. Credit the reduced quantity and a practical consequence, then state a possible exposure risk without claiming that harm is proven.	4
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Tier 3 · Hard

Mark scheme for 4.2.4.2 Tier 3 · Hard
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01.1	J blocks slightly more ultraviolet radiation: $96\%$ compared with $94\%$ . K is transparent, which may make it more acceptable to users. K needs only one quarter as much active material, so it may reduce resource use, cost or waste. Detection inside some cells shows exposure is possible and creates a potential health concern. The data do not show that K causes harm, so the risk remains uncertain rather than proven. A justified conclusion may choose either product, but should weigh performance and user/resource benefits against the uncertainty and recommend further safety testing before K is widely used.	Compare every relevant datum: protection, appearance, quantity required and cell exposure. Distinguish evidence of entry into cells from evidence of harm. Reach a conditional judgement; for example, K has strong practical and resource advantages, but development should include further safety testing because the long-term risk is unresolved.	6
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02.1	Both inks give the same electrical performance. B uses $6.0/0.80=7.5$ times as much silver as N. B costs £ $7.20$ in silver per display, while N costs £ $4.00$ . N also gives a thinner transparent track, which is useful for a display. Nanoparticles in waste create a possible environmental risk, but the long-term effect is uncertain rather than proven harmful. N has the stronger material, cost and design case, but development should depend on controlling releases and obtaining further environmental evidence.	Calculate both material costs: $6.0\times1.20=7.20$ and $0.80\times5.00=4.00$ . Combine these values with mass, performance and transparency. Balance the advantages against uncertain environmental exposure before giving a conditional conclusion.	6
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03.1	The nanoparticle system uses one fifth of the dose, so less active material is needed. Its observed effectiveness is higher: 86% compared with 70%. Its observed side-effect rate is also higher: 12% compared with 8%. Detection in the liver shows exposure but does not by itself prove that the particles cause harm. The short trial leaves uncertainty about long-term risks, so larger and longer studies are needed. Adoption could be justified only if further evidence shows that the improved effectiveness and lower dose outweigh the side effects and possible long-term risk.	Compare benefit, quantity and observed risk separately. Preserve the evidence boundary: detection is not proof of harm. A justified conclusion should therefore be conditional on stronger long-term safety evidence.	6
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04.1	The nanoparticle version uses one eighth as much active material because $3.6 ÷ 0.45 = 8$ . It is slightly less effective, reducing odour by 81% rather than 84%. Its observed irritation rate is 5 percentage points higher, at 7% rather than 2%. Using less material could reduce resource use, transport mass or waste. Detection in waste water shows environmental exposure, while the unknown effect leaves a possible risk. The conventional product is the safer current choice unless further tests show that the material saving outweighs the higher irritation rate and environmental uncertainty.	Compare quantity, performance and risk separately. The mass calculation gives an eightfold reduction, but the observed benefit falls by 3 percentage points and irritation rises by 5 percentage points. Detection establishes exposure rather than harm, so the final judgement should remain conditional on better environmental and safety evidence.	6
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05.1	The nanoparticle cosmetic has potential benefits because it uses less pigment and gives a more even finish. The 12-person study is weak evidence of safety because it is small and lasts only two days. The 1200-person, six-month study provides stronger evidence because it is larger and longer. 3% of 1200 is 36 people, so the larger study recorded irritation in 36 participants. Detection in blood shows exposure but does not prove that nanoparticles caused illness. Widespread use should wait while longer safety and cause-and-effect studies investigate the irritation and possible effects after particles enter the body.	Weigh the stated product benefits against the quality and meaning of the safety evidence. Sample size and duration make the second study more informative, and $0.03 × 1200 = 36$ quantifies its irritation result. Keep detection separate from proven harm, then suggest research that addresses the remaining long-term and causal uncertainty.	6
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