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AQA GCSE Chemistry revision notes

Bonding, structure, and the properties of matter

Section 4.2
18 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8462 section 4.2

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Sodium is a metal and chlorine is a non-metal, so sodium chloride contains oppositely charged ions and has ionic bonding.
  2. 2.Chlorine atoms are non-metals joined by shared electron pairs, so chlorine has covalent bonding.
  3. 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 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

Your progress and exam materials
4.2.1.2

Ionic bonding

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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+1+ and 2+2+ ions, while Group 6 and Group 7 non-metals form 22- and 11- 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. 1.A magnesium atom loses two outer-shell electrons to form Mg2+^{2+}.
  2. 2.Two chlorine atoms each gain one electron to form two Cl^- ions.
  3. 3.The charges balance: one 2+2+ ion is matched by two 11- ions.

Answer: One Mg2+^{2+} ion and two Cl^- ions form, giving MgCl2_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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.2.1.3

Ionic compounds

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 1212 X2+^{2+} ions and 2424 Y^- ions. Deduce the empirical formula.

  1. 1.Count the ions to obtain the ratio X:Y =12:24=12:24.
  2. 2.Divide both numbers by 1212 to obtain the simplest ratio 1:21:2.
  3. 3.Check charge balance: 1(2+)+2(1)=01(2+)+2(1-)=0.

Answer: The empirical formula is XY2_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

ORIGINAL

Describe the structure and bonding in a solid ionic compound.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A model of an ionic lattice contains 1818 M2+ ions and 3636 N ions. Work out the empirical formula of the compound.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A section of an ionic model contains 88 A3+ ions and 1212 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

4.2.1.4

Covalent bonding

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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, CH4_4.

  1. 1.Place one carbon atom at the centre with four hydrogen atoms around it.
  2. 2.Carbon shares one pair of electrons with each hydrogen atom.
  3. 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

ORIGINAL

A chlorine molecule contains two chlorine atoms joined by one covalent bond. State what this bond represents.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.2.1.5

Metallic bonding

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Magnesium forms a giant regular structure of positive metal ions.
  2. 2.Its outer-shell electrons are delocalised through the whole structure.
  3. 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

ORIGINAL

Name the two types of charged particle shown in the usual model of metallic bonding.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain why the bonding in a metal is strong.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

4.2.2.1

The three states of matter

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 20C-20\,^{\circ}\mathrm{C} and boils at 70C70\,^{\circ}\mathrm{C}. State its physical state at 25C25\,^{\circ}\mathrm{C}.

  1. 1.25C25\,^{\circ}\mathrm{C} is above the melting point, so the substance is not solid.
  2. 2.25C25\,^{\circ}\mathrm{C} is below the boiling point, so the substance is not a gas.
  3. 3.The temperature lies between the two points.

Answer: The substance is liquid at 25C25\,^{\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

Substance S has a melting point of 15C-15\,^{\circ}\text{C} and a boiling point of 84C84\,^{\circ}\text{C}. State its physical state at 30C-30\,^{\circ}\text{C}, 20C20\,^{\circ}\text{C} and 100C100\,^{\circ}\text{C}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Substance P melts at 40C40\,^{\circ}\text{C} and boils at 85C85\,^{\circ}\text{C}. Substance Q melts at 780C780\,^{\circ}\text{C} and boils at 1400C1400\,^{\circ}\text{C}. Compare their states at 100C100\,^{\circ}\text{C} and explain what the data suggest about the forces or bonds between their particles.

[5 marks]

Total for this question: 5

4.2.2.2

State symbols

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Magnesium is stated to be solid, so write Mg(s).
  2. 2.Both solutions are aqueous, so write HCl(aq) and MgCl2_2(aq).
  3. 3.Hydrogen is released as a gas, so write H2_2(g), then balance the equation.

Answer: Mg(s) + 2HCl(aq) → MgCl2_2(aq) + H2_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

ORIGINAL

State the meaning of the symbols (l) and (aq) in a chemical equation.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.2.2.3

Properties of ionic compounds

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

State whether an ionic compound usually conducts electricity when solid and when molten.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Explain why magnesium oxide has a high melting point.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

4.2.2.4

Properties of small molecules

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Boiling separates whole molecules from one another.
  2. 2.Only the weak intermolecular forces between molecules are overcome.
  3. 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Substances A and B both contain neutral small molecules. A has relative molecular mass 3434 and boils at 22C22\,^{\circ}\text{C}; B has relative molecular mass 122122 and boils at 168C168\,^{\circ}\text{C}. Explain the difference in boiling point and predict whether either pure substance conducts electricity.

[5 marks]

Total for this question: 5

4.2.2.5

Polymers

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.The polymer contains very large molecules formed from long chains.
  2. 2.Intermolecular forces between these large molecules are relatively strong.
  3. 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A short-chain molecular substance is a liquid at 20C20\,^\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

Tier 3 · Hard

ORIGINAL

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

4.2.2.6

Giant covalent structures

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 1500C1500\,^{\circ}\mathrm{C}. Explain this observation.

  1. 1.Silicon dioxide has a giant covalent structure.
  2. 2.Strong covalent bonds link atoms throughout the continuous network.
  3. 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 1500C1500\,^{\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

Silicon dioxide is solid at 1500C1500\,^\circ\text{C}. Explain this observation using its structure and bonding.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Two covalent substances are heated. U melts at 84C84\,^\circ\text{C} and consists of separate molecules. V is still solid at 1700C1700\,^\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

4.2.2.7

Properties of metals and alloys

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.The pure metal has regular layers of equal-sized atoms.
  2. 2.Different-sized atoms in the alloy distort these layers.
  3. 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

ORIGINAL

Explain why a pure metal can be bent into shape.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Pure metal Q bends when a force of 18N18\,\text{N} is applied, but an alloy of Q needs 47N47\,\text{N}. Both have high melting points. Explain both observations in terms of structure and bonding.

[5 marks]

Total for this question: 5

4.2.2.8

Metals as conductors

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Copper contains delocalised electrons that can move through its metallic structure.
  2. 2.The electrons carry electrical charge through a wire.
  3. 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

ORIGINAL

Name the particles that carry electrical charge through a metal wire.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.2.3.1

Diamond

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Each carbon atom forms four strong covalent bonds in a rigid giant structure.
  2. 2.The rigid network is difficult to deform, so diamond is hard.
  3. 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

ORIGINAL

In diamond, how many bonds per carbon?

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain why a diamond-tipped cutting tool can scratch most materials.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

4.2.3.2

Graphite

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.One electron from each carbon atom is delocalised and free to move.
  2. 2.The mobile electrons carry electrical charge through graphite.
  3. 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

ORIGINAL

State the number of covalent bonds made by each carbon atom in graphite.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain why graphite can be used as an electrode.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

4.2.3.3

Graphene and fullerenes

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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, C60_{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. 1.Graphene is a single layer of graphite and is only one atom thick.
  2. 2.It therefore adds very little thickness to the sensor.
  3. 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 C60_{60} and hollow cylinder for a nanotube.

Tier 1 · Easy

ORIGINAL

Describe the relationship between graphene and graphite.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

An electronic sensor needs a conducting layer that adds very little thickness. Explain why graphene is suitable.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

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

4.2.4.1

Sizes of particles and their properties (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Nanoscience concerns structures from 11 to 100nm100\,\mathrm{nm}, of the order of a few hundred atoms. Fine particles have diameters from 100100 to 2500nm2500\,\mathrm{nm}, while coarse particles span 25002500 to 10000nm10\,000\,\mathrm{nm}.
  • Nano dimensions may be compared with typical atomic and molecular dimensions using ratios and standard form.
  • For a cube of side aa, surface area is 6a26a^2, volume is a3a^3, and the surface-area-to-volume ratio is 6/a6/a.
  • Reducing the side by a factor of 1010 increases this ratio by a factor of 1010.
  • 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 5nm5\,\mathrm{nm}. Calculate its surface area, volume and surface-area-to-volume ratio.

  1. 1.Surface area =6a2=6×(5nm)2=150nm2=6a^2=6\times(5\,\mathrm{nm})^2=150\,\mathrm{nm}^2.
  2. 2.Volume =a3=(5nm)3=125nm3=a^3=(5\,\mathrm{nm})^3=125\,\mathrm{nm}^3.
  3. 3.Surface area : volume =150:125=1.2:1=150:125=1.2:1.

Answer: Surface area 150nm2150\,\mathrm{nm}^2; volume 125nm3125\,\mathrm{nm}^3; surface-area-to-volume ratio 1.2:11.2:1.

Common mistakes

  • Don't calculate a cube’s surface area as a2a^2 instead of 6a26a^2.
  • Don't compare total surface area alone instead of surface area relative to volume.
  • Don't classify a 2500nm2500\,\mathrm{nm} particle as nano-sized.

Exam tip

For a surface-area-to-volume calculation, show 6a26a^2, a3a^3 and the simplified ratio as separate lines.

Tier 1 · Easy

ORIGINAL

Particle T is 72nm72\,\text{nm} across. A typical atom is 0.24nm0.24\,\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

Tier 2 · Standard

ORIGINAL

A nano-cube has edge length 5nm5\,\text{nm}. Calculate its surface area, volume and surface-area-to-volume ratio.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Fine particles have side 800nm800\,\text{nm}; nano-sized cubes of the same material have side 40nm40\,\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 12g12\,\text{g} of the fine particles is effective, estimate the effective mass of nano cubes.

[4 marks]

Total for this question: 4

4.2.4.2

Uses of nanoparticles (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Equal reaction rate shows that performance is maintained.
  2. 2.Using one twentieth of the mass reduces raw-material use and may reduce cost or waste.
  3. 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

ORIGINAL

Give two application areas in which nanoparticles are used.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A catalyst works equally well when a factory replaces 8.0g8.0\,\text{g} of bulk material with 0.40g0.40\,\text{g} of nanoparticles. Suggest two advantages and one possible disadvantage of the change.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A company compares two sun creams. Product J uses larger particles, blocks 96%96\% of ultraviolet radiation and leaves a visible layer. Product K uses nanoparticles, blocks 94%94\%, is transparent and needs one quarter as much active material. Tests detect K's particles inside 2%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

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