AQA A-Level Chemistry Revision
Explore AQA A-Level Chemistry revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
AQA · A-Level Chemistry7405Exam years · 2027
Assessment The assessment includes Paper 1, Paper 2 and Paper 3; see the official specification for the full structure.
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Reviewed course guidance
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Topic group
Assessment and qualification identity
- Core
Scope: Core · §2.3 A-level assessment and paper scope; §4.2 A-level assessment-objective weightings; §5.1 A-level award 7405 and QAN
Open revision guidance
What to know
- A-level 7405 is linear: all A-level examinations are taken at the end of the course. Paper 1 and Paper 2 each last 2 hours and are worth 105 marks (35% each); Paper 3 lasts 2 hours, is worth 90 marks (30%), and the three papers scale to 300 marks.
- Paper 1 assesses physical chemistry §§3.1.1–3.1.4, 3.1.6–3.1.8 and 3.1.10–3.1.12, all inorganic chemistry §3.2, and relevant practical skills. Paper 2 assesses physical chemistry §§3.1.2–3.1.6 and 3.1.9, all organic chemistry §3.3, and relevant practical skills.
- Paper 1 and Paper 2 use short- and long-answer questions. Paper 3 assesses any content and practical skills: 40 marks cover practical techniques/data analysis, 20 marks test across the specification and 30 marks are multiple choice. Overall assessment-objective weightings are AO1 knowledge 30%, AO2 application 45% and AO3 analysis/evaluation 25%. At least 20% of marks require mathematical skills and at least 15% assess practical skills.
- The registered award is AQA Advanced Level GCE Chemistry 7405, QAN 601/5731/8. AS Chemistry 7404 is a distinct award (QAN 601/5730/6), even though content is co-taught.
- A-level component AO weightings (approximate): Paper 1 AO1/AO2/AO3 = 30%/48%/22%; Paper 2 = 30%/48%/22%; Paper 3 = 32%/34%/34%. Across the A-level, the overall weightings are 30%/45%/25%. AO1 assesses knowledge and understanding of scientific ideas, processes, techniques and procedures; AO2 assesses applying these in theoretical and practical contexts and when handling qualitative and quantitative data; AO3 assesses analysing, interpreting and evaluating scientific information, ideas and evidence (including issues) to make judgements/reach conclusions and develop/refine practical design and procedures.
What to do in an assessment
- Plan revision against the stated component scope: Paper 1 physical/inorganic, Paper 2 physical/organic, and Paper 3 any content plus practical skills.
- For every paper, show calculations and units and interpret practical data; do not assume practical questions appear only on Paper 3.
- Keep 7405 A-level assessment separate from the AS 7404 paper structure and award; construct sustained, relevant explanations and calculations for extended responses.
- When revising assessment demands, distinguish the per-paper AO table from the overall weighting: learn 30/48/22 for both Papers 1 and 2 and 32/34/34 for Paper 3; practise AO1 recall/explanation, AO2 application in theoretical/practical and qualitative/quantitative contexts, and AO3 evidence evaluation, justified judgement/conclusion and practical-design refinement.
In plain language
Know what each paper assesses: Paper 1 is physical plus inorganic, Paper 2 is physical plus organic, and Paper 3 can draw on the whole A-level course.
Self-check
What are the A-level AO1/AO2/AO3 weightings for Paper 3, and what does AO3 require?
Show the answer
Paper 3 is approximately 32%/34%/34% AO1/AO2/AO3. AO3 requires analysing, interpreting and evaluating scientific information, ideas and evidence, including issues, to reach judgements/conclusions and develop or refine practical design and procedures.
Topic group
Atomic structure and amount of substance
- Core
Open revision guidance
What to know
- Atomic models changed as evidence accumulated: Dalton's indivisible solid-sphere model was revised after Thomson's cathode-ray discovery of electrons (plum-pudding model); Rutherford's alpha-scattering evidence showed a tiny, dense, positively charged nucleus and mostly empty space; Bohr introduced electron shells, and the modern quantum model describes electrons in orbitals. Scientific models change when new evidence requires them.
- The nucleus contains protons and neutrons, surrounded by electrons; relative charge is +1, 0, −1 and relative mass is approximately 1, 1, 1/1836. Atomic number Z is proton number and mass number A is proton plus neutron number; isotopes share Z but differ in neutron number.
- A time-of-flight mass spectrometer ionises particles, accelerates ions to the same kinetic energy, lets them drift, detects arrival times and analyses the data. It gives precise isotopic masses and relative abundances, can identify elements and can determine relative molecular mass.
- Relative atomic mass is the weighted mean of isotopic masses. Relative atomic and molecular masses are defined relative to one twelfth of the mass of a carbon-12 atom; ionic compounds use relative formula mass.
- Electron configurations for atoms and ions up to Z = 36 use shells and s, p and d subshells. First ionisation energy is removal of one electron from each atom in one mole of gaseous atoms to form gaseous 1+ ions; successive ionisation energies remove further electrons from gaseous 1+ etc. ions.
- Across Period 3 and down Group 2, first and successive ionisation-energy patterns and large jumps reveal shell/subshell structure; local deviations reflect the subshell occupied and electron pairing.
What to do in an assessment
- Use A, Z and ionic charge to calculate proton, neutron and electron counts; write ionisation equations with correct gaseous states and charges.
- Explain a time-of-flight spectrum through ionisation, equal kinetic energy, mass-dependent drift time, detection and data conversion; read m/z and relative abundance.
- Calculate relative atomic mass as Σ(isotopic mass × fractional abundance), and report an appropriate significant-figure result.
- Write electron configurations to Z = 36 and explain Period 3/Group 2 ionisation-energy patterns using shell, subshell and electron-pair evidence.
In plain language
Atomic structure explains an element's electron arrangement; isotope patterns in a mass spectrum can be used to identify atoms and calculate their average mass.
Self-check
An element has isotopes of mass 10 (20%) and mass 11 (80%). What is its relative atomic mass?
Show the answer
(10 × 0.20) + (11 × 0.80) = 10.8.
- Core
Open revision guidance
What to know
- Relative atomic and molecular masses are measured against 1/12 of a carbon-12 atom; ionic compounds use relative formula mass. A mole applies to particles, atoms, molecules, ions, formulae and electrons; one mole contains the Avogadro constant (its numerical value is supplied when needed).
- Use n = m/Mr and c = n/V, with solution volume in dm³ and concentration in mol dm⁻³. Use the supplied Avogadro constant to convert between entities and moles; its value is not recalled. The ideal-gas equation is pV = nRT in SI units; R is supplied when required.
- An empirical formula is the simplest whole-number ratio; a molecular formula gives the actual atom numbers and is an integer multiple of the empirical formula. Full and ionic equations must conserve atoms and charge.
- Balanced-equation mole ratios give reacting masses, gas volumes, solution concentrations and titration quantities. Percentage yield = actual/theoretical × 100; atom economy = (Mr of desired product × its equation coefficient)/(sum of reactant Mr values × their equation coefficients) × 100.
- High atom economy can reduce unwanted by-products and improve economic, environmental and ethical outcomes. Experimental calculations also use percentage yield, mean titres, outlier selection and uncertainty; the titre uncertainty includes both burette readings, and results cannot be more precise than the least accurate measurements.
What to do in an assessment
- Convert grams, concentration/volume or gas measurements to moles, use the balanced equation ratio, then convert to the requested quantity and unit.
- For pV = nRT convert pressure to Pa, volume to m³ and temperature to K; rearrange algebraically and use consistent supplied R units.
- Find empirical formula by converting composition to moles and dividing by the smallest value; obtain molecular formula from Mr divided by empirical-formula mass.
- Calculate atom economy from the balanced reactant side, percentage yield from actual and theoretical yield, and titration amounts from concordant titres and the equation ratio; reject justified outliers and estimate uncertainty from both burette readings.
- Balance unfamiliar molecular or ionic equations by conserving each element and total charge; estimate and report uncertainty from measurement precision.
In plain language
Most chemistry calculations become manageable when you convert the measurements into moles, use the balanced equation's ratio, then convert to the requested units.
Self-check
How many moles are present in 25.0 cm³ of 0.100 mol dm⁻³ solution?
Show the answer
Convert 25.0 cm³ to 0.0250 dm³, then n = cV = 0.100 × 0.0250 = 0.00250 mol.
Topic group
Bonding
- Core
Open revision guidance
What to know
- Ionic lattices contain oppositely charged ions; use Periodic Table position to predict simple ionic charges and write formulae for compounds containing ions such as sulfate, hydroxide, nitrate, carbonate and ammonium.
- A single covalent bond is one shared electron pair, multiple bonds contain multiple pairs, and a dative covalent bond has both bonding electrons from one donor atom. Metallic bonding is attraction between delocalised electrons and positive ions in a lattice.
- The four crystal structures are ionic, metallic, giant covalent and molecular. Diamond and graphite are giant covalent; ice and iodine are molecular; magnesium is metallic; sodium chloride is ionic. Their melting point, conductivity and solubility follow from structure and bonding.
- For 2, 3, 4, 5 and 6 electron pairs around a central atom, the electron-pair arrangements are linear (180°), trigonal planar (120°), tetrahedral (109.5°), trigonal bipyramidal (90°, 120° and 180°) and octahedral (90° and 180°). Replacing bonding pairs with lone pairs gives bent H₂O (about 104.5°), trigonal-pyramidal NH₃ (about 107°), seesaw (<90° and <120°), T-shaped (about 90°), linear (180°), square-pyramidal (about 90°) and square-planar (90°) molecular shapes; lone-pair repulsions are stronger and distort bond angles.
- Electronegativity differences produce polar bonds; molecular dipoles also depend on geometry. Intermolecular forces are permanent dipole–dipole, induced dipole–dipole (London/dispersion) and hydrogen bonding, where H bonded to N, O or F interacts with a lone pair. Their relative strength affects melting/boiling points; hydrogen bonding explains water's anomalous properties and ice's open, low-density structure.
What to do in an assessment
- Use ion charges to construct neutral formulae and represent covalent bonds by lines and coordinate bonds by arrows from the electron-pair donor.
- Draw ionic, metallic, giant-covalent and molecular particle diagrams with specified particle numbers, and relate melting, conductivity and state-change energy to bonds or intermolecular forces overcome.
- Apply VSEPR to up to six electron pairs, state molecular shape and bond angles, and explain angle changes from lone-pair repulsion.
- Use δ+ and δ− to mark bond polarity, decide whether bond dipoles cancel, and explain intermolecular forces and physical trends for familiar and unfamiliar molecules.
In plain language
A substance's properties follow from its structure: identify the particles, the bonds within them and the attractions between them.
Self-check
Why is the H–O–H bond angle smaller than the tetrahedral angle?
Show the answer
Water has two lone pairs on oxygen; lone-pair repulsion is stronger than bonding-pair repulsion and compresses the H–O–H angle.
Topic group
Energetics
- Core
Scope: Core · §3.1.4 Energetics: enthalpy definitions; calorimetry; Hess's law; mean bond enthalpies
Open revision guidance
What to know
- A reaction may be endothermic or exothermic; ΔH is the heat change at constant pressure. A standard enthalpy change uses 100 kPa and a stated temperature, commonly 298 K. Standard enthalpy of formation forms one mole of a compound from its elements in their standard states; standard enthalpy of combustion burns one mole completely in oxygen.
- Calorimetry uses q = mcΔT, where m is the mass that changes temperature, c its specific heat capacity and ΔT the temperature change. A molar reaction enthalpy is obtained by relating q to the reacting amount and applying the reaction sign. The data booklet supplies c when needed.
- Hess's law states that a reaction enthalpy is independent of route. Use formation or combustion data in cycles and reverse or scale ΔH with the equation. A mean bond enthalpy is an average for a bond in different gaseous environments; Σ(bonds broken) − Σ(bonds formed) estimates gaseous reaction ΔH.
- Calorimetry examples in the specification include dissolving salts, neutralisation, metal displacement and alcohol combustion; Hess-cycle contexts include thermal decomposition of sodium hydrogencarbonate and hydration of magnesium or copper sulfate.
What to do in an assessment
- Define ΔcH° and ΔfH° with one mole and standard states explicit; assign negative ΔH to heat released and positive ΔH to heat absorbed.
- Calculate q with consistent mass, heat capacity and temperature units, divide by the correct moles, and explain heat loss, incomplete reaction, solution-density assumptions and instrument resolution.
- Construct Hess cycles from formation or combustion data, reverse/scale equations consistently and solve the signed enthalpy sum.
- Calculate approximate gaseous ΔH using bond enthalpies and explain why mean-bond estimates differ from Hess values based on exact compounds.
In plain language
Measure or calculate the heat change, connect it to the amount reacting, and keep track of whether energy enters or leaves the chemical system.
Self-check
A reaction warms its surroundings. What is the sign of the reaction enthalpy change?
Show the answer
Negative: the reaction releases heat and is exothermic.
Topic group
Kinetics, equilibrium and redox
- Core
Scope: Core · §§3.1.5–3.1.7 Kinetics, chemical equilibria/Le Chatelier/Kc, oxidation states and redox equations
Open revision guidance
What to know
- Activation energy is the minimum energy for a reaction; most collisions do not react because they lack sufficient energy or suitable orientation. A Maxwell–Boltzmann curve plots molecular energy against population; increasing temperature broadens the distribution and increases the fraction above Ea.
- Rate is change in concentration per time. Higher temperature increases successful collisions; higher concentration or gas pressure increases collision frequency. A catalyst provides a route of lower activation energy without changing its chemical composition or amount. For a gas reaction, a catalyst lowers the activation-energy threshold: on a Maxwell–Boltzmann graph at fixed temperature, draw the catalyst Ea line to the left of the uncatalysed Ea line; the distribution curve itself is unchanged, but the larger area to the right of the lower threshold represents a greater fraction of molecules with sufficient energy.
- At dynamic equilibrium in a closed homogeneous system, forward and reverse rates are equal and concentrations remain constant. Le Chatelier predicts the shift opposing a change in concentration, pressure or temperature; an industrial process may use compromise temperature/pressure to balance yield, rate, cost and safety. A catalyst changes neither equilibrium position nor Kc.
- For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d/([A]^a[B]^b), with equilibrium concentrations in mol dm⁻³ and stoichiometric exponents; derive any Kc units from the expression. Concentration and catalyst do not change Kc; temperature can: heating increases Kc for an endothermic forward reaction and decreases it for an exothermic forward reaction. Redox means electron transfer: oxidation is electron loss, reduction electron gain; oxidising agents accept electrons and reducing agents donate them. Free elements have oxidation state 0, monatomic ions have oxidation state equal to charge, the total in a compound is 0, and common values include Group 1 +1, Group 2 +2, H +1 and O −2 (with known exceptions).
What to do in an assessment
- Draw Maxwell–Boltzmann distributions at different temperatures, label Ea and explain the increased area beyond Ea, not just the higher mean energy. To show catalyst action, keep the same Maxwell–Boltzmann curve and temperature, draw a second lower Ea line to its left, shade/compare the areas to the right of both lines, and explain that the larger catalysed area means more molecules can react per unit time; do not shift or redraw the distribution curve.
- Explain concentration/pressure changes by collision frequency and catalytic effects by lower Ea; identify what variables a rate experiment controls.
- Predict equilibrium shifts from the imposed change, count gas moles and use the direction of the enthalpy change; explain industrial compromise conditions.
- Build the Kc expression from a balanced equation, omit pure solids/liquids, calculate unknown equilibrium concentrations or Kc and explain qualitatively why temperature changes Kc.
- Assign oxidation states from formulae using the rules for atoms, ions and compounds (including the total oxidation state equalling the species charge), identify oxidation/reduction and agents, write electron-balanced half-equations and combine them into an overall redox equation.
- For a catalyst question using a Maxwell–Boltzmann graph, state that the catalyst provides an alternative lower-Ea route, mark the lower threshold at the same temperature, and compare the fractions above Ea rather than claiming that the catalyst changes molecular energies or the distribution.
In plain language
Rates describe how quickly composition changes; equilibrium describes a balanced pair of ongoing reactions; redox tracks electron transfer.
Self-check
At fixed temperature, how should a Maxwell–Boltzmann graph show a catalyst, and why does the reaction rate increase?
Show the answer
Keep the distribution curve unchanged and draw the catalysed Ea threshold to the left of the uncatalysed threshold. A larger fraction of molecules then has energy at least equal to Ea, so there are more successful collisions per unit time.
Topic group
A-level physical chemistry
- Core
Scope: Core · §§3.1.8–3.1.10 Born–Haber cycles, entropy/Gibbs free energy, rate equations/determination and Kp (A-level only)
Open revision guidance
What to know
- Born–Haber cycles apply Hess's law to ionic solids. Define lattice enthalpy consistently as either lattice dissociation (separating a solid into gaseous ions) or lattice formation (forming the solid from gaseous ions); define formation, atomisation (forming one mole of gaseous atoms), bond enthalpy (breaking one mole of gaseous bonds), ionisation energy and electron affinity. Compare cycle values with a perfect ionic model to infer covalent character. Hydration enthalpy is the enthalpy change when gaseous ions become aqueous ions; combine it with lattice enthalpy to calculate enthalpy of solution.
- Entropy change measures dispersal of energy/matter. Calculate ΔS° from absolute entropies, including stoichiometric coefficients. ΔG = ΔH − TΔS (consistent units, T in K); ΔG ≤ 0 means feasible under those conditions. A ΔG-versus-T line gives intercept ΔH and slope −ΔS.
- A rate equation rate = k[A]^m[B]^n is determined experimentally; AQA restricts orders m and n to 0, 1 or 2. The overall order controls the units of k. k varies with temperature according to k = Ae^(−Ea/RT), equivalently ln k = −Ea/(RT) + ln A.
- Initial-rate comparisons and concentration–time curves establish orders; rate data then give k and its units. For a zero-order reaction, the concentration–time gradient gives −k for a reactant or +k for a product. Orders can inform a possible mechanism and rate-determining/limiting step but do not by themselves prove a complete mechanism.
- For aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp = (pC)^c(pD)^d/((pA)^a(pB)^b), using equilibrium partial pressures at constant temperature. Partial pressure = mole fraction × total pressure. Temperature changes Kp; pressure shifts equilibrium position but does not change Kp, and a catalyst only changes how rapidly equilibrium is reached.
What to do in an assessment
- Draw a Born–Haber or solution cycle, define each enthalpy arrow, keep sign/direction consistent, and calculate an unknown lattice, hydration, formation or other enthalpy.
- Calculate ΔS and ΔG with matched J/kJ units, rearrange the Gibbs equation for an unknown or feasibility temperature, and interpret slope/intercept of ΔG against T.
- Use initial rates or rate–concentration data to determine orders, derive the rate equation, calculate rate or k and units, and use a tangent for instantaneous rate.
- Linearise Arrhenius data as ln k against 1/T, identify slope −Ea/R and intercept ln A, and apply the supplied R and data consistently.
- For Kp, calculate mole fractions and partial pressures, form the correct expression, solve equilibrium quantities and distinguish temperature effects on Kp from pressure effects on position.
In plain language
These A-level-only ideas connect energy dispersal to feasibility, experiments to rate laws, and gas composition to pressure-based equilibrium calculations.
Self-check
Can a catalyst change the equilibrium constant Kp?
Show the answer
No. A catalyst can make equilibrium establish faster but does not change the equilibrium constant; only temperature changes its value.
- Core
Scope: Core · §3.1.11 Electrode potentials, cell notation and commercial electrochemical cells (A-level only)
Open revision guidance
What to know
- Write electrode half-equations in the IUPAC reduction convention and represent cells with a single line at phase boundaries and a double line for the salt bridge (for example Zn|Zn²⁺||Cu²⁺|Cu). Standard electrode potentials are measured relative to the standard hydrogen electrode at 298 K, 100 kPa and 1.00 mol dm⁻³; the Nernst equation is not required.
- The electrochemical series compares standard reduction potentials. E°cell = E°(more positive reduction couple) − E°(less positive reduction couple); a positive value predicts the feasible direction under standard conditions, not the reaction rate.
- The specified lithium-cell half-equations are Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻ and Li → Li⁺ + e⁻. In an alkaline H₂/O₂ fuel cell, oxidation is 2H₂ + 4OH⁻ → 4H₂O + 4e⁻ and reduction is O₂ + 2H₂O + 4e⁻ → 4OH⁻, giving 2H₂ + O₂ → 2H₂O. Non-rechargeable, rechargeable and fuel cells differ in reversibility.
- A cell generates current because oxidation and reduction occur at separate electrodes and electrons travel through an external circuit. Commercial benefits, resource/safety risks and environmental effects must be assessed together.
What to do in an assessment
- Use E° data to predict simple redox direction, write and apply cell notation, calculate EMF and identify electron-flow direction.
- Explain the role of standard conditions and why changing concentration or temperature changes measured potential; do not apply an unrequired Nernst equation.
- Use supplied electrode data to deduce reactions and EMF for non-rechargeable/rechargeable cells, and explain how an alkaline hydrogen–oxygen fuel cell produces current.
- Evaluate a cell's application using evidence for benefits, limitations, hazards and lifecycle impacts rather than treating voltage as the only criterion.
In plain language
Electrode potentials let you compare redox couples and predict the voltage and favoured direction of a cell reaction.
Self-check
A cell has E°cell = +0.42 V for a proposed forward reaction. What does this imply under standard conditions?
Show the answer
The forward reaction is thermodynamically feasible under standard conditions; it does not tell you how fast it occurs.
- Core
Scope: Core · §3.1.12 Brønsted–Lowry equilibria, pH, Kw, Ka, titration curves/indicators and buffer action (A-level only)
Open revision guidance
What to know
- Brønsted–Lowry acids donate protons and bases accept them; acid–base equilibria transfer protons and conjugate pairs differ by H⁺. Strong acids/bases ionise almost completely; weak acids/bases dissociate only slightly.
- pH = −log₁₀[H⁺], with [H⁺] in mol dm⁻³. Calculate [H⁺] from pH and pH of strong acids from concentration. Kw = [H⁺][OH⁻] for slightly dissociated water and varies with temperature; use it to calculate pH of a strong base.
- For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA] and pKa = −log₁₀Ka. Use Ka, concentration and pH relationships for weak acids; convert Ka ↔ pKa.
- Typical curves are required for every strong/weak monoprotic acid/base pairing: strong/strong has a steep change around equivalence pH 7; weak-acid/strong-base equivalence is above 7; strong-acid/weak-base equivalence is below 7; weak/weak has no comparably steep vertical region. Select an indicator whose transition interval lies in the steep pH region.
- A buffer resists pH changes on dilution or small additions of acid/base. An acidic buffer contains a weak acid and its salt; a basic buffer contains a weak base and its salt. Buffer action follows removal of added H⁺ or OH⁻ by the conjugate pair; buffers have important industrial and biological applications.
What to do in an assessment
- Identify conjugate acid/base pairs and construct Ka expressions using equilibrium concentrations.
- Calculate pH/[H⁺], strong-acid and strong-base pH, weak-acid pH and Ka/pKa with correct logarithms, powers of ten, units and significant figures.
- Use titration stoichiometry on experimental data; sketch and explain all strong/weak monoprotic curve types and select an indicator from the steep change.
- Explain qualitatively how each acidic or basic buffer component removes added acid/base and calculate acidic-buffer pH using Ka or pH = pKa + log10([salt]/[acid]), making the appropriate approximation.
In plain language
Acid-base calculations combine proton transfer with equilibrium; a pH curve shows how the mixture changes as titrant is added.
Self-check
What is the pH of a solution with [H+] = 1.0 × 10⁻³ mol dm⁻³?
Show the answer
pH = −log10(1.0 × 10⁻³) = 3.00.
Topic group
Inorganic chemistry
- Core
Scope: Core · §§3.2.1–3.2.3 Periodic classification and Period 3 properties; Group 2; Group 7 trends, tests and chlorine uses
Open revision guidance
What to know
- The Periodic Table is ordered by proton number. An element is s, p, d or f block according to its position. Across Na–Ar, atomic radius generally falls and first ionisation energy rises with local subshell/pairing deviations; explain melting points from each element's structure and bonding.
- For Mg–Ba, atomic radius increases and first ionisation energy generally falls down Group 2; explain melting-point pattern using metallic structure/bonding. Mg–Ba react with water; Mg also reacts with steam. Magnesium reduces TiCl₄ in titanium extraction.
- Down Mg–Ba, hydroxide solubility increases (Mg(OH)₂ is sparingly soluble) while sulfate solubility decreases (BaSO₄ is insoluble). Mg(OH)₂ is used medicinally; Ca(OH)₂ adjusts agricultural soil pH; CaO/CaCO₃ remove SO₂ from flue gas. Acidified BaCl₂ tests for sulfate while acid prevents carbonate interference; BaSO₄ is used in medicine.
- For Group 7, electronegativity decreases down the group as the bonding pair is farther from the nucleus and more shielded; boiling point rises with molecular size and London forces. Oxidising ability of halogens decreases down the group while reducing ability of halide ions increases.
- Halogen displacement and reaction of solid sodium halides with concentrated H₂SO₄ reveal those trends. Acidified AgNO₃ identifies halides: AgCl is white, AgBr cream and AgI yellow; dilute/concentrated ammonia solubility distinguishes the precipitates. Dilute nitric acid removes carbonate interference without adding halide ions; ammonia compares silver-halide solubility.
- Chlorine reacts with water to form chloride and chlorate(I), or chloride and oxygen; cold dilute NaOH produces a chlorate(I)-containing solution. Chlorination disinfects water, and society weighs toxicity against the health benefit.
What to do in an assessment
- Explain periodic radius, ionisation-energy and melting-point patterns through proton number, shielding, distance, subshell, structure and bonding; explain halogen electronegativity and boiling-point trends through shielding/distance and London forces.
- Write balanced equations and compare Mg–Ba reactions with water, Mg with steam and Mg reduction of TiCl₄; explain hydroxide/sulfate solubility trends and uses.
- Describe and interpret halogen displacement and concentrated-sulfuric-acid tests; distinguish halides using acidified AgNO₃ then ammonia.
- Explain why AgNO₃ is used, why it is acidified and why ammonia is added, linking each reagent to interference or precipitate identification.
- Write chlorine-water and chlorine/NaOH equations; identify aqueous chlorate(I) (hypochlorite) uses such as bleaching and disinfection. State the specified conclusion: when chlorine is used in a controlled water-treatment dose, its health benefit from killing pathogens outweighs its toxic effects; include the hazard of chlorine and the need for controlled use.
In plain language
Element trends make sense when linked to electron structure, bonding, and how strongly particles attract one another.
Self-check
Will bromine displace chloride ions from aqueous sodium chloride?
Show the answer
No. Bromine is less reactive than chlorine and cannot oxidise chloride ions to chlorine.
- Core
Open revision guidance
What to know
- Na and Mg react with water. In Period 3, the specified reactions with oxygen form Na₂O, MgO, Al₂O₃, SiO₂, P₄O₁₀, SO₂ and SO₃. Explain highest-oxide melting-point trends through ionic, giant covalent or molecular structure/bonding.
- Na₂O and MgO form alkaline solutions in water; Al₂O₃ is amphoteric, SiO₂ is an acidic oxide but insoluble in water, and P₄O₁₀/SO₂/SO₃ form acidic solutions. P₄O₁₀ + 6H₂O → 4H₃PO₄; the acid has O=P(OH)₃ structure and its H₂PO₄⁻, HPO₄²⁻ and PO₄³⁻ ions have tetrahedral P–O frameworks with one, two and three deprotonated O atoms. SO₂ + H₂O forms H₂SO₃ (O=S(OH)₂) and HSO₃⁻/SO₃²⁻, with a trigonal-pyramidal S centre. SO₃ + H₂O → H₂SO₄ (O=S(=O)(OH)₂), with HSO₄⁻/SO₄²⁻ ions having tetrahedral S–O frameworks. Draw the Lewis structures, showing O–H groups and deprotonated O atoms, and explain oxide pH/reactivity through bonding.
- Ti–Cu transition-metal characteristics arise from an incomplete d subshell in atoms or ions: complex formation, coloured ions, variable oxidation states and catalysis. A ligand donates an electron pair; coordination number counts coordinate bonds.
- H₂O, NH₃ and Cl⁻ are monodentate; NH₃/H₂O are similar in size and uncharged. Their substitution can keep coordination number or be incomplete. Larger Cl⁻ can change coordination number. Ethane-1,2-diamine and C₂O₄²⁻ are bidentate; EDTA⁴⁻ is multidentate. Chelation is favoured by the entropy/enthalpy balance. Haem binds O₂ at Fe(II); CO competes for that site.
- Small ligands commonly give octahedral complexes, where monodentate ligands can show cis–trans isomerism (for example, cis/trans-[Co(NH₃)₄Cl₂]⁺: like ligands adjacent at 90° or opposite at 180°). Octahedral complexes with bidentate ligands can be optical isomers; [Co(en)₃]³⁺ forms a pair of non-superimposable mirror images. Larger Cl⁻ commonly gives tetrahedral complexes. Square-planar complexes such as [Pt(NH₃)₂Cl₂] show cis–trans isomerism; cisplatin is the cis isomer. [Ag(NH₃)₂]⁺ is linear. Visible absorption promotes d electrons; ΔE = hν = hc/λ, and oxidation state, coordination and ligand affect colour. Colorimetry relates absorbance to concentration.
- Transition ions show variable oxidation states: acidified zinc reduces vanadate(V) through V(IV), V(III) and V(II); ligand and pH affect redox potentials. Tollens' reagent oxidises aldehydes while [Ag(NH₃)₂]⁺ is reduced. Fe²⁺/ethanedioate and MnO₄⁻ redox titrations use mole ratios.
- Transition-metal catalysts may be heterogeneous (surface active sites; poisoning; supports) or homogeneous (intermediate species). A cheap support spreads catalyst into small particles, maximising exposed active surface area per amount of catalyst and minimising catalyst cost. V₂O₅ in the Contact process cycles through V₂O₄ (V₂O₅ + SO₂ → V₂O₄ + SO₃; V₂O₄ + ½O₂ → V₂O₅); Fe is a heterogeneous Haber-process catalyst. In the homogeneous I⁻/S₂O₈²⁻ reaction, S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺, then 2Fe³⁺ + 2I⁻ → I₂ + 2Fe²⁺. Mn²⁺ autocatalyses ethanedioate/MnO₄⁻ through Mn³⁺ intermediates: MnO₄⁻ + 8H⁺ + 4Mn²⁺ → 5Mn³⁺ + 4H₂O; 2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂.
- Aqueous ions include [M(H₂O)₆]²⁺ (M = Fe, Cu) and [M(H₂O)₆]³⁺ (M = Al, Fe). The 3+ aqua ions are more acidic because greater charge/size ratio polarises O–H bonds. With OH⁻, Cu²⁺ gives a blue precipitate, Fe²⁺ a green precipitate (which oxidises brown), Fe³⁺ a red-brown precipitate and Al³⁺ a white precipitate that dissolves in excess hydroxide. NH₃ initially gives corresponding hydroxide precipitates; excess NH₃ dissolves the Cu²⁺ precipitate to a deep-blue complex but not Fe/Al hydroxides. CO₃²⁻ gives Cu/Fe(II) carbonate precipitates and Al/Fe(III) hydroxide precipitates with CO₂. Al(OH)₃ dissolves in both acids and bases (amphoteric).
What to do in an assessment
- Write equations for Na/Mg reactions with water, the specified Period 3 element/oxide reactions, oxide hydration and oxide reactions with supplied acids/bases; relate structure/bonding to melting points and pH.
- Represent complexes, count coordination number and explain ligand substitution, coordination changes, incomplete exchange and the chelate effect.
- Draw octahedral, tetrahedral, square-planar and linear complexes; identify/draw cis–trans and optical isomers and distinguish cisplatin's cis structure.
- Use ΔE = hν = hc/λ to connect visible absorption and complex colour; use a calibration graph for colorimetric concentration.
- Explain vanadium reduction and Tollens' reaction, and write catalytic steps for the Contact-process V(V)/V(IV) cycle, Fe²⁺/Fe³⁺ iodide–peroxodisulfate cycle and Mn²⁺/Mn³⁺ ethanedioate–permanganate autocatalysis; calculate Fe²⁺/C₂O₄²⁻–MnO₄⁻ and other redox-titration amounts from balanced ratios.
- Describe transition-ion tests with OH⁻, NH₃ and CO₃²⁻, record precipitate/colour/dissolution, and use charge density to explain aqua-ion acidity and amphoterism.
In plain language
A-level inorganic chemistry uses reactions and observations to connect periodic trends, metal-ion structure and the identity of dissolved ions.
Self-check
What does a ligand contribute when it bonds to a metal ion?
Show the answer
A lone pair of electrons, forming a coordinate (dative covalent) bond.
Topic group
Organic chemistry
- Core
Scope: Core · §§3.3.1–3.3.2 Organic formulae/nomenclature, mechanisms, isomerism; crude oil, cracking, combustion and alkane chlorination
Open revision guidance
What to know
- Organic compounds use empirical, molecular, general, structural, displayed and skeletal formulae. A homologous series shares a functional group; IUPAC naming/drawing is required for chains and rings up to six carbon atoms. Structural isomers differ in connectivity; stereoisomers differ in 3D arrangement. E–Z isomerism arises from restricted rotation and uses Cahn–Ingold–Prelog priorities.
- In mechanisms, a curly arrow starts at an electron pair or covalent bond and shows its movement. Methane chlorination: initiation Cl₂ → 2Cl• (UV); propagation Cl• + CH₄ → HCl + CH₃• and CH₃• + Cl₂ → CH₃Cl + Cl•; termination includes Cl• + Cl• → Cl₂, CH₃• + Cl• → CH₃Cl and CH₃• + CH₃• → C₂H₆. Every step is a balanced equation; radical mechanisms mark unpaired electrons with dots and do not use curly arrows.
- Crude oil/petroleum is a mixture mainly of alkanes; fractional distillation separates fractions by boiling range. Cracking breaks C–C bonds: thermal cracking at high pressure/high temperature yields many alkenes; catalytic cracking uses hot zeolite at slight pressure and yields motor fuels/aromatics. Economic reason: match supply to demand for shorter useful molecules.
- Alkanes burn completely or incompletely. Engines produce NOx, CO, carbon and unburned hydrocarbons; catalytic converters remove gaseous pollutants. Sulfur-containing fuel forms SO₂, which is captured by CaO or CaCO₃. Methane chlorination is free-radical substitution: UV initiates Cl radicals, propagation transfers H/Cl, termination joins radicals.
What to do in an assessment
- Translate between all listed formula representations; name and draw IUPAC chains/rings up to six carbons, including chain, position and functional-group isomers.
- Assign CIP priorities and draw E/Z isomers; distinguish structural isomerism from stereoisomerism.
- Draw non-radical mechanisms with curly arrows from electron source to destination; write, balance and classify radical initiation, both propagation steps and possible termination steps for alkane chlorination.
- Explain fraction separation by boiling range, cracking choices from product demand, combustion products/pollutants, catalytic-converter or flue-gas treatment, and methane initiation/propagation/termination.
In plain language
Organic chemistry is easier when structures are drawn consistently and each reaction is linked to the functional group and electron movement.
Self-check
Is fractional distillation of crude oil a chemical reaction?
Show the answer
No. It separates hydrocarbons by boiling range without changing their molecular identities.
- Core
Open revision guidance
What to know
- Halogenoalkanes have polar C–X bonds and undergo nucleophilic substitution with OH⁻, CN⁻ and NH₃. C–X bond enthalpy influences hydrolysis rate. With hydroxide, substitution competes with elimination: hydroxide can be nucleophile or base, so conditions/products distinguish the pathways. For example, with 2-bromopropane, aqueous hydroxide substitutes Br to form propan-2-ol, whereas hot ethanolic hydroxide favours elimination to propene. In the concerted elimination, a hydroxide lone pair forms a bond to a β-hydrogen, the C–H electron pair forms the C=C bond as the adjacent C–Br bond breaks, and its electron pair leaves with bromide; show all three curly arrows for this pathway.
- Ozone forms naturally in the upper atmosphere and is beneficial because it absorbs ultraviolet radiation. UV breaks C–Cl bonds in CFCs, forming chlorine radicals; Cl• + O₃ → ClO• + O₂ and ClO• + O₃ → 2O₂ + Cl• regenerate Cl• and catalyse ozone loss. Research evidence from different groups supported legislation restricting CFC solvents/refrigerants and the development of chlorine-free replacements.
- Alkenes are unsaturated and have a high-electron-density C=C bond. In HBr addition, the polarised H–Br bond is attacked at H by the π pair; the H–Br pair leaves with Br, producing a carbocation that Br⁻ attacks. With H₂SO₄, protonation forms a carbocation, HSO₄⁻ attacks to give an alkyl hydrogensulfate, and hydrolysis gives an alcohol. In Br₂ addition, the π bond polarises Br₂; electron-pair arrows show C–Br bond formation and Br–Br bond breaking, followed by attack by Br⁻ at the other carbon. For unsymmetrical alkenes, major/minor products follow the relative stability of primary, secondary and tertiary carbocations. Bromine is a test for unsaturation.
- Addition polymers form from alkenes/substituted alkenes. Know repeat-unit ↔ monomer conversions, IUPAC naming, poly(chloroethene) (PVC) uses/plasticiser effects, and why polyalkenes are unreactive; attractions between polymer molecules are intermolecular forces.
- Alcohols are industrially made by hydration of an alkene with steam and an acid catalyst: the alkene π pair accepts H⁺ to form the more stable carbocation, water donates a lone pair to that C⁺, and deprotonation forms the alcohol and regenerates H⁺ (for example, CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH). Ethanol fermentation uses aqueous glucose, yeast enzymes, warm conditions (typically about 30–40 °C) and exclusion of oxygen; conditions balance enzyme activity against denaturation and contamination. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂; ethanol is separated by fractional distillation and used as a biofuel. The carbon-neutral argument compares photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂) with combustion (C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O), but must also include cultivation, transport, processing emissions and ethical land/food use.
- Primary alcohol oxidation gives aldehyde or carboxylic acid depending on distillation/reflux and oxidant; secondary gives ketone; tertiary alcohols resist oxidation. Acidified dichromate(VI) is suitable. Alcohol dehydration is acid-catalysed elimination; cyclohexanol can yield cyclohexene.
What to do in an assessment
- For 2-bromopropane, distinguish the conditions and products: aqueous OH⁻ gives substitution to propan-2-ol; hot ethanolic OH⁻ favours elimination to propene. Draw the elimination mechanism as one concerted step with curly arrows from the hydroxide lone pair to a β-H, from the C–H bond to form C=C, and from the C–Br bond to Br⁻; identify OH⁻ acting as a base. For substitution, show the nucleophile attacking carbon and the C–Br pair leaving to Br⁻; identify OH⁻ acting as a nucleophile.
- Write both ozone catalytic steps and explain chlorine regeneration; assess evidence-led policy and replacement-compound trade-offs.
- Show each electrophilic-addition mechanism with correctly placed curly arrows and intermediates for HBr, H₂SO₄ and Br₂; justify regioisomer proportions from the relative stability of primary, secondary and tertiary carbocations, and use bromine-water decolourisation for unsaturation.
- Draw addition-polymer repeat units from monomer and chain fragments in both directions, name the polymer and explain properties/uses and plasticiser effects.
- Write the equation and show protonation, water attack and deprotonation steps for acid-catalysed alkene hydration; write the glucose-fermentation equation, justify fermentation conditions, purify ethanol by fractional distillation and evaluate the photosynthesis/combustion carbon-balance argument and its limitations.
- Write alcohol oxidation equations and explain aldehyde-versus-acid isolation; use Tollens' or Fehling's to distinguish aldehydes from ketones; outline acid-catalysed dehydration.
In plain language
Follow the functional group: a carbon–halogen bond can be substituted or eliminated, a carbon–carbon double bond adds across, and an alcohol can be oxidised or dehydrated.
Self-check
Show the electron-pair movement when hot ethanolic hydroxide reacts with 2-bromopropane by elimination, and state the product.
Show the answer
In one concerted step, an OH⁻ lone pair forms an O–H bond to a β-hydrogen; the C–H bond pair forms the C=C bond while the C–Br bond pair leaves with Br⁻. The organic product is propene: CH₃CHBrCH₃ + OH⁻ → CH₃CH=CH₂ + H₂O + Br⁻.
- Core
Scope: Core · §3.3.6 Functional-group tests, precise-mass spectrometry and infrared spectroscopy
Open revision guidance
What to know
- The written course requires identification of alcohols, aldehydes, alkenes and carboxylic acids using specified organic reactions. Bromine water is decolourised by an alkene; carboxylic acids release CO₂ from carbonate; aldehydes reduce Fehling's solution/Tollens' reagent while ketones do not; primary/secondary alcohols are oxidised by acidified dichromate(VI), with tertiary alcohols not readily oxidised.
- A mass spectrum can determine molecular formula when precise atomic masses and precise molecular mass are used, not merely nominal Mr. IR bonds absorb at characteristic wavenumbers; fingerprinting compares spectra to identify a molecule and impurities.
- Use the AQA Chemistry Data Booklet for diagnostic bonds/functional groups. IR absorption by CO₂, methane and water vapour connects molecular vibrations to global warming.
What to do in an assessment
- Choose and carry out a separate appropriate test for the specified functional group; state reagent/conditions and the positive observation, and infer only what the result supports.
- Determine a molecular formula from precise mass data by testing integer atom combinations against exact mass; check consistency with formula and ion peak.
- Use diagnostic IR absorptions and fingerprint matching to identify functional groups, propose structures and spot impurities; explain the greenhouse connection for CO₂, methane and water vapour.
In plain language
A reliable identification combines tests and spectra: each supplies evidence, and the structure should explain all of it.
Self-check
Why should an infrared spectrum not usually be used as the only evidence for a complete structure?
Show the answer
Many structures share functional-group absorptions; molecular formula, mass-spectral evidence and chemical tests help distinguish candidates.
Topic group
A-level organic chemistry
- Core
Open revision guidance
What to know
- A molecule with one chiral centre can form non-superimposable enantiomers with opposite effects on plane-polarised light; a 1:1 racemate is optically inactive. Aldehydes oxidise to acids; NaBH₄ reduces aldehydes to primary alcohols and ketones to secondary alcohols. KCN followed by dilute acid adds HCN across the carbonyl to form a hydroxynitrile; attack on either side of the planar C=O can produce enantiomers. KCN is hazardous.
- Carboxylic acids are weak acids and release CO₂ from carbonates. Acid-catalysed reaction with alcohols forms esters; common ester uses include solvents (for example ethyl ethanoate), plasticisers, perfumes and food flavourings. Ester hydrolysis gives alcohol plus acid (acidic conditions) or carboxylate salt (alkaline). Oils/fats are glycerol esters; alkaline hydrolysis gives soap and glycerol, and transesterification with methanol gives biodiesel methyl esters. Acyl chlorides and acid anhydrides undergo nucleophilic addition–elimination with water, alcohols, ammonia and primary amines to form carboxylic acids, esters and amides. Ethanoic anhydride has manufacturing advantages over ethanoyl chloride for aspirin.
- Benzene is planar with intermediate C–C bond lengths; delocalised p electrons stabilise it relative to hypothetical cyclohexa-1,3,5-triene. Hydrogenation enthalpy evidence explains the extra stability and preference for substitution over addition, which preserves the delocalised ring. Nitration uses concentrated nitric/sulfuric acids to generate NO₂⁺; Friedel–Crafts acylation uses an acyl chloride and AlCl₃.
- Amines arise from ammonia with alkyl/aryl substitutions; primary aliphatic amines are prepared from halogenoalkanes plus ammonia or reduction of nitriles, and aromatic amines from reduction of nitro compounds (used in dye manufacture). Amines are weak bases and nucleophiles; alkyl groups make the nitrogen lone pair more available than in ammonia, while delocalisation into an aromatic ring makes it less available, so primary aliphatic amines are stronger bases than ammonia and primary aromatic amines are weaker. In substitution, an N lone pair attacks the carbon bonded to halogen as the C–X bond pair leaves; deprotonation gives the amine, and further alkylation can form secondary/tertiary amines and quaternary ammonium salts used as cationic surfactants. For acyl-chloride addition–elimination, the N lone pair attacks the carbonyl carbon, the C=O π pair moves to O, O reforms C=O as chloride leaves, then deprotonation gives an amide.
- Condensation polymers arise from dicarboxylic acids + diols, dicarboxylic acids + diamines or amino acids; polyester and polyamide links and repeat units include Terylene, nylon 6,6 and Kevlar. Terylene is used in textile fibres/fabrics, nylon in fibres, fabrics and ropes, and Kevlar in strong protective materials. Polyesters have dipole attractions and polyamides can hydrogen-bond between chains. Polyalkenes are inert/non-biodegradable; polyesters/polyamides hydrolyse and are biodegradable. Compare disposal and recycling.
- Amino acids are amphoteric zwitterions; proteins are amino-acid sequences with peptide links and primary, α-helix/β-sheet secondary and tertiary structures maintained in part by H-bonds and S–S bonds. Hydrolysis gives amino acids; TLC/ninhydrin or UV and Rf identify them. Enzyme active sites are stereospecific and inhibitors can block them.
- DNA contains phosphate, 2-deoxyribose and A/C/G/T bases in covalently linked sugar-phosphate strands; complementary strands form a hydrogen-bonded double helix. Cisplatin binds guanine through ligand replacement and inhibits cancer-cell DNA replication; drug benefits must be weighed against adverse effects. Computers can support the design of enzyme-inhibitor drugs by modelling candidate binding to an enzyme's stereospecific active site.
What to do in an assessment
- Draw one-centre enantiomers and racemates; write carbonyl-reduction equations using [H], outline NaBH₄ reduction with H⁻ as nucleophile, write HCN hydroxynitrile equations, outline KCN/dilute-acid nucleophilic addition and explain why attack on either face can produce enantiomers.
- Write esterification, ester hydrolysis, soap/biodiesel and acylation equations/mechanisms; compare acylating-agent manufacturing hazards/advantages.
- Use thermochemical evidence to explain benzene stability; generate NO₂⁺ and outline electrophilic substitution for nitration and AlCl₃-catalysed acylation.
- Write amine preparation and base reactions; draw curly-arrow substitution mechanisms for ammonia/amine with halogenoalkanes, including deprotonation and possible further alkylation; draw nucleophilic addition–elimination for ammonia/primary amine with an acyl chloride, showing attack, tetrahedral intermediate, chloride loss and deprotonation; explain base-strength order through nitrogen lone-pair availability.
- Draw polyester/polyamide repeating units and infer monomers; link Terylene, nylon 6,6 and Kevlar structures/properties to typical uses; explain hydrolysis vs inert polyalkenes and evaluate disposal choices.
- Draw zwitterions and acid/base forms, peptides of up to three residues and hydrolysis products; identify protein structural levels, explain links, calculate amino-acid Rf and describe stereospecific enzyme inhibition.
- Draw nucleotide/DNA building blocks, explain complementary base hydrogen bonding and cisplatin's guanine substitution; discuss both therapeutic benefit and adverse effects.
In plain language
The A-level-only organic families extend functional-group chemistry to aromatic substitution, nitrogen bases, macromolecules and stereochemistry.
Self-check
What is the key difference between addition and condensation polymerisation?
Show the answer
Addition polymerisation joins unsaturated monomers without eliminating a small molecule; condensation polymerisation forms links while eliminating a small molecule.
- Core
Scope: Core · §§3.3.14–3.3.16 Organic synthesis; NMR spectroscopy; chromatography (A-level only)
Open revision guidance
What to know
- Plan organic syntheses using reactions in this specification, with routes up to four steps. Green-chemistry aims include avoiding solvents and hazardous starting materials, reducing steps and increasing atom economy.
- NMR confirms structures using ¹H and ¹³C environments; ¹³C spectra are simpler. Chemical shift δ depends on environment; ¹H integration gives relative proton numbers. Samples use deuterated solvent or CCl₄. TMS is a suitable standard because its 12 equivalent protons give one sharp signal at δ = 0 and it is chemically unreactive.
- For aliphatic ¹H NMR, the n+1 rule gives doublets, triplets or quartets from n equivalent adjacent protons. Use supplied data-booklet shifts rather than memorised unsupported values.
- TLC uses a solid-coated plate and solvent moving up; column chromatography uses a solid-packed column and solvent moving down; GC passes pressurised gas through a heated column containing a solid or liquid-coated solid. Separation balances solubility in the mobile phase and retention in the stationary phase. Rf and retention times are compared with standards; GC can be coupled to mass spectrometry.
What to do in an assessment
- Devise a route of no more than four syllabus-supported transformations, naming reagents, conditions and intermediate structures; justify solvent minimisation, safer starting materials, fewer stages and atom economy.
- Use ¹H/¹³C shifts to propose structures/part structures, integrate ¹H peaks, apply n+1 only to adjacent non-equivalent aliphatic protons, and combine spectra with formula/mass evidence.
- Explain the choice and operation of TLC, column chromatography or GC; measure Rf from baseline and solvent-front distances and compare Rf/retention times with standards.
- Describe GC–MS as separation followed by mass analysis of components, and use combined evidence rather than one peak or chromatographic value alone.
In plain language
Synthesis plans explain how each chemical transformation reaches the target; spectroscopy and chromatography test whether the proposed product fits the evidence.
Self-check
A TLC spot travels 3.0 cm and the solvent front 7.5 cm. What is its Rf?
Show the answer
Rf = 3.0 ÷ 7.5 = 0.40; it has no unit.
Topic group
Mathematics and practical skills
- Core
Scope: Core · §6.1–6.5 Mathematical requirements MS 0.0–4.3; §8.1 apparatus/techniques AT a–l; §8.3 written practical skills PS 1.1–4.1
Open revision guidance
What to know
- At least 20% of A-level Chemistry marks require mathematical skills at least equivalent to higher-tier GCSE mathematics. MS 0.0–0.4 covers appropriate units (including rate/Kc constants and J↔kJ), ordinary/standard form, ratios/percentages, estimates and calculator powers/exponentials/logarithms, including Avogadro, pH/pKa and approximate buffer calculations.
- MS 1.1–1.3 covers significant figures/least-accurate measurement, weighted means and outlier-aware mean titres, and uncertainty from the two burette readings in a titre. MS 2.1–2.5 covers symbols (=, <, <<, >>, >, ∝, ~ and ⇌), equation rearrangement, numerical substitution with units, solving equations (including Hess/rate calculations) and logarithms for pH/pKa.
- MS 3.1–3.5 covers translation between graphical, numerical and algebraic forms (including spectra/orders), plotting concentration–time data with best-fit curves, gradients/intercepts, rate from graph gradient and tangents. MS 4.1–4.3 covers shape/bond angles, 2D/3D structure and isomer drawing, and symmetry/chirality.
- AT a–l: measure mass/time/liquid and gas volume/temperature; heat with water/electric/sand bath; measure pH by chart, meter or probe/logger; use burette/pipette, reflux/distillation glassware on stands/clamps, qualitative tests and fluted/reduced-pressure filtration; prepare solutions in volumetric flasks; use indicators; recrystallise solids and purify liquids with a separating funnel; measure melting points; use TLC/paper chromatography; set up cells/measure voltage; handle hazardous materials; measure rates by initial and continuous methods.
- At least 15% of A-level marks assess practical skills in written papers: PS 1.1–1.2 solve practical problems/apply science; PS 2.1–2.4 evaluate design, present data, evaluate uncertainty/errors and identify controlled variables; PS 3.1–3.3 plot/interpret, process/analyse and judge accuracy/precision; PS 4.1 understand equipment/techniques.
- The endorsement in §8.4 is a separate teacher-observed practical competency outcome and does not contribute to the A-level grade. Practical methods and skills that can be assessed in written papers remain in scope.
What to do in an assessment
- Convert units, use standard form and ratios, rearrange equations, substitute values with units, apply logarithms and report significant figures limited by measurement precision.
- Calculate means from appropriate concordant results, identify outliers, estimate uncertainty including two burette readings, and explain accuracy vs precision.
- Plot well-labelled graphs, translate representations, calculate slopes/intercepts and tangents, and derive rates/orders from suitable data.
- Predict molecular geometry and stereoisomerism from 2D/3D diagrams and explain bond angles, chirality and symmetry.
- For a practical question, state apparatus and technique, independent/dependent/control variables, repeat/measurement method, risk controls, data presentation, uncertainty and a conclusion supported by evidence.
- Distinguish the written-exam practical skills from the separately observed endorsement; do not omit written practical planning, calculations or evaluation.
In plain language
Exams test chemistry and the quantitative and experimental reasoning used to make measurements, interpret evidence and judge how strong a conclusion is.
Self-check
Why should a titre use the same volume unit as the concentration equation n = cV?
Show the answer
Because n = cV requires V in dm³ when c is in mol dm⁻³; convert the measured cm³ value before substitution.
Topic group
Required practical activities
- Core
Scope: Core · §8.2 Required practical activity 1: make up a volumetric solution and carry out a simple acid–base titration
Open revision guidance
What to know
- A volumetric solution has a known concentration. Quantitative transfer, rinsing and a calibrated volumetric flask reduce loss and volume error. A burette delivers a measured titre; a pipette transfers a fixed aliquot.
- The indicator endpoint is an observable colour change close to the equivalence point. Use concordant titres and the balanced equation to determine the unknown concentration.
What to do in an assessment
- Dissolve the weighed solute, transfer quantitatively to a volumetric flask, rinse all material into the flask, make up to the calibration mark and invert to mix.
- Rinse the burette with titrant and the pipette with analyte; record initial and final burette readings, perform a rough titration, then repeat accurately to obtain concordant titres.
- Use mean concordant titre, aliquot amount and the balanced equation ratio to calculate the unknown concentration; report units and appropriate significant figures.
In plain language
Prepare a solution with an accurately known concentration, then measure how much it takes to react with a fixed volume of another solution.
Self-check
Why rinse a pipette with the solution it will transfer?
Show the answer
Residual water would dilute the aliquot and make its amount/concentration different from the intended value.
- Core
Scope: Core · §8.2 Required practical activity 2: measurement of an enthalpy change
Open revision guidance
What to know
- The temperature change of a known mass of solution estimates heat transfer using q = mcΔT. The measured heat is converted to a molar enthalpy change using the reacting amount and balanced equation.
- Insulation, a lid, prompt mixing and consistent temperature readings reduce heat exchange and improve comparability.
What to do in an assessment
- Measure reactant quantities and initial temperature, combine in an insulated calorimeter, stir consistently and record the temperature change.
- Use solution mass and the stated/supplied specific heat capacity to calculate q; convert to kJ and divide by the correct amount in moles.
- Assign the sign for the reaction, identify heat losses and other assumptions, and suggest an improvement tied to the limitation.
In plain language
Measure how much the reacting mixture warms or cools, convert that temperature change to energy, and express it per mole.
Self-check
If an exothermic reaction warms the solution, what sign should its ΔH have?
Show the answer
Negative, because the reacting system releases heat to the surroundings.
- Core
Scope: Core · §8.2 Required practical activity 3: investigation of how the rate of a reaction changes with temperature
Open revision guidance
What to know
- Temperature affects rate because particles have a higher mean kinetic energy and a greater fraction of collisions exceed activation energy. A valid comparison changes temperature while controlling the other conditions.
- Rate can be compared using a timed endpoint or continuous measurement, depending on the reaction and method.
What to do in an assessment
- Set a measured reaction temperature using a water bath and allow reactants to reach it before mixing.
- Keep concentrations, volumes, apparatus and the chosen endpoint consistent; start timing at mixing and repeat across temperatures.
- Plot a suitable rate measure against temperature, identify anomalous results and explain the trend using collision theory.
In plain language
Change only the reaction temperature, measure the same rate indicator each time, and use collision theory to explain the result.
Self-check
Why pre-equilibrate the reactants to the chosen water-bath temperature?
Show the answer
So the reaction begins at the intended temperature rather than warming or cooling during the measurement.
- Core
Open revision guidance
What to know
- The activity covers Group 2 and NH4+ cations, and Group 7 halide, OH−, CO3²− and SO4²− anions. Group 2 flame tests distinguish Mg2+ (no characteristic colour), Ca2+ (brick-red), Sr2+ (crimson) and Ba2+ (apple-green); hydroxide and sulfate precipitation/solubility patterns provide additional evidence where specified. NH4+ warmed with alkali releases ammonia, which turns damp red litmus blue.
- Acidified silver nitrate gives AgCl (white), AgBr (cream) or AgI (yellow); ammonia distinguishes their solubilities. Carbonate plus dilute acid effervesces CO2, which turns limewater milky. Acidified barium chloride gives white BaSO4 with sulfate; acidification removes carbonate interference. Hydroxide is identified by an alkaline indicator/pH response.
- A reliable qualitative test uses clean apparatus, small controlled quantities, and observations recorded before drawing an inference.
What to do in an assessment
- Divide samples into separate portions so one reagent does not contaminate a later test; add reagents in the prescribed order and note precipitate, flame colour, gas or dissolution.
- Warm a separate ammonium-ion portion with alkali and test evolved gas with damp red litmus; use the stated halide precipitate/ammonia sequence; test carbonate by acid and gas confirmation; test sulfate with acidified BaCl2.
- For an unknown, compare the complete pattern of results with the known-ion tests and state both supporting and excluding observations; use Group 2 flame colours and hydroxide/sulfate solubility trends where required.
- Use eye protection, handle corrosive reagents carefully and follow the laboratory's safety and disposal instructions.
In plain language
Test separate small portions with the right reagents, record exactly what happens, and identify an ion from the pattern of results.
Self-check
Why should separate portions of an unknown be used for different anion tests?
Show the answer
To prevent reagents from the previous test contaminating the sample and changing the next observation.
- Core
Scope: Core · §8.2 Required practical activity 5: distillation of a product from a reaction
Open revision guidance
What to know
- Distillation separates a volatile product from a reaction mixture by vaporisation and condensation. The thermometer monitors vapour temperature near the still head; cooling water passes through the condenser.
- The product's boiling behaviour and collection range are evidence about identity and purity, but distillation alone does not prove purity.
What to do in an assessment
- Assemble the specified distillation apparatus securely with the condenser water inlet low and outlet high; ensure the system is not sealed.
- Heat the reaction mixture in a controlled way, observe the temperature and collect the fraction in the specified range.
- Explain separation using volatility/boiling point and identify how contamination, leaks, excessive heating or poor temperature measurement affect yield or purity.
In plain language
Heat the mixture so the more volatile product enters the vapour, then cool that vapour back into a collected liquid.
Self-check
Why must a distillation apparatus not be sealed?
Show the answer
Heating a sealed apparatus can cause pressure to build and create a serious hazard; vapour must not be trapped.
- Core
Scope: Core · §8.2 Required practical activity 6: tests for alcohol, aldehyde, alkene and carboxylic acid
Open revision guidance
What to know
- The required tests cover alcohol, aldehyde, alkene and carboxylic-acid functional groups. Bromine water is decolourised by C=C; carboxylic acid plus carbonate releases CO2; aldehydes reduce Tollens' reagent to silver or Fehling's solution to brick-red Cu2O while ketones do not; acidified dichromate(VI) changes orange to green when it oxidises primary/secondary alcohols, while tertiary alcohols are not readily oxidised.
- A positive test is a specific observed change under stated conditions, and a negative result alone may not identify a unique compound. A purposeful sequence of small-scale tests preserves sample and avoids cross-contamination.
What to do in an assessment
- Use fresh portions of the unknown for each test; state reagent, conditions and observation, including bromine-water decolourisation, carbonate effervescence, silver mirror/brick-red precipitate, or dichromate colour change as appropriate.
- Combine independent tests to distinguish candidate functional groups; do not infer an aldehyde from oxidation alone when a ketone or other oxidisable compound remains possible.
- Use eye protection and the relevant precautions for corrosive, oxidising or flammable reagents and organic products.
In plain language
Use a different, controlled test for each possible functional group and combine the observed results to identify the unknown.
Self-check
Why is one positive functional-group test not always enough to identify an unknown molecule?
Show the answer
More than one candidate molecule may contain that group; additional tests or spectral evidence are needed to distinguish structures.
- Core
Scope: Core · §8.2 Required practical activity 7: rate measured by an initial-rate method and continuous monitoring
Open revision guidance
What to know
- An initial-rate method estimates rate near the start, often by timing a defined endpoint at different initial concentrations. Continuous monitoring records a changing measurable property throughout a reaction.
- Rate is change in concentration or measurable quantity per time; a tangent to a concentration–time curve gives instantaneous rate at that point.
What to do in an assessment
- For initial rates, keep conditions constant, vary one reactant concentration, record the same endpoint time and calculate a consistent rate estimate.
- For continuous monitoring, calibrate or zero the sensor/apparatus, start recording promptly after mixing, and plot the measured quantity against time.
- Use repeated measurements, suitable gradients and units; explain how mixing delay, endpoint judgement and temperature affect uncertainty.
In plain language
One method compares how quickly different starting mixtures reach the same endpoint; another follows the reaction continuously as it happens.
Self-check
On a concentration–time curve, how is the instantaneous rate at a chosen time found?
Show the answer
Draw a tangent at that time and calculate its gradient, with the sign interpreted for reactant disappearance or product formation.
- Core
Scope: Core · §8.2 Required practical activity 8: measurement of the EMF of an electrochemical cell
Open revision guidance
What to know
- A cell's EMF is measured as the potential difference with negligible current. Two half-cells are connected by an external circuit and an ionic pathway; a high-resistance voltmeter minimises current.
- The electrode materials, solutions and conditions determine the measured cell potential; polarity and cell notation must be reported consistently.
What to do in an assessment
- Set up the specified half-cells with clean electrodes and the prescribed ionic connection; connect a high-resistance voltmeter.
- Record the stable potential and identify which electrode is positive under the chosen meter connections; repeat if appropriate.
- Compare the result with a prediction only when concentration, temperature and standard-state assumptions are stated.
In plain language
Connect two redox half-cells and measure the voltage without drawing significant current from the cell.
Self-check
Why use a high-resistance voltmeter to measure EMF?
Show the answer
It draws negligible current, so the measured potential is close to the cell's open-circuit EMF.
- Core
Scope: Core · §8.2 Required practical activity 9: weak acid with strong base and strong acid with weak base
Open revision guidance
What to know
- The activity compares a weak acid reacting with a strong base and a strong acid reacting with a weak base. pH depends on acid/base strength, concentration and the composition of the mixture during titration.
- A calibrated pH probe yields a curve from which the steep region, equivalence volume and buffer region can be interpreted.
What to do in an assessment
- Calibrate the pH meter/probe as instructed, measure a known starting volume, add titrant in measured increments and mix before each pH reading.
- Use smaller additions near the steep section and record pH against cumulative titrant volume for each acid/base pairing.
- Plot both curves with labelled axes and compare initial pH, buffer behaviour, steep region, equivalence point and indicator suitability.
In plain language
Add a strong or weak acid/base in measured portions, record pH each time, and compare how the two titration curves differ.
Self-check
Why take smaller titrant additions near the steep part of a pH curve?
Show the answer
A small volume change can cause a large pH change there, so finer measurements locate the steep region and equivalence volume more accurately.
- Core
Open revision guidance
What to know
- The activity includes preparation and purification of a pure organic solid followed by a purity test, and preparation of a pure organic liquid. Suitable purification depends on physical properties and the product state.
- Recrystallisation removes soluble and insoluble impurities by controlled dissolution, hot filtration and crystal formation; liquid products may be isolated by distillation. A melting point can help assess a solid's purity.
What to do in an assessment
- Follow the specified synthesis and safety instructions; separate the product from the reaction mixture using a method appropriate to solid or liquid product.
- For a solid, purify by an appropriate crystallisation procedure, collect and dry crystals, then determine melting point and compare with a reliable reference when provided.
- For a liquid, distil and collect the appropriate fraction; record yield, appearance and relevant boiling behaviour, and evaluate sources of product loss or contamination.
In plain language
Make the organic product, separate it from the reaction mixture, purify it using a method suited to its properties, and collect evidence about purity.
Self-check
Why should crystals be dry before measuring their melting point?
Show the answer
Residual solvent can contaminate the sample and distort its measured melting behaviour.
- Core
Scope: Core · §8.2 Required practical activity 11: test-tube reactions to identify transition-metal ions in aqueous solution
Open revision guidance
What to know
- The assessed aqueous ions are [Fe(H2O)6]2+ (pale green), [Cu(H2O)6]2+ (blue), [Al(H2O)6]3+ (colourless) and [Fe(H2O)6]3+ (yellow). With OH−, respectively, they form green, blue, white and red-brown hydroxide precipitates; only the Al(OH)3 precipitate dissolves in excess OH−.
- NH3 initially forms the corresponding hydroxide precipitates; excess NH3 dissolves only the Cu2+ precipitate, giving a deep-blue ammine complex. CO3²− gives green Fe(II) and blue-green Cu(II) carbonate precipitates, but the acidic 3+ aqua ions give Al(OH)3 or Fe(OH)3 precipitates plus effervescence of CO2. These reagent-specific outcomes, solution colour and response to excess reagent distinguish the ions.
What to do in an assessment
- Place measured portions of each solution in clean labelled tubes; add OH−, NH3 or CO3²− dropwise and in excess where directed.
- Record initial colour, precipitate colour, dissolution in excess and any effervescence; use the full response to distinguish the four ions.
- Write ionic equations for supported precipitation or ligand-substitution changes and follow local reagent disposal instructions.
In plain language
Add reagents in a controlled sequence and identify a transition-metal ion from its full colour and precipitation pattern.
Self-check
Why record a result both before and after adding excess reagent?
Show the answer
Some hydroxide precipitates dissolve in excess ligand/base or change through ligand substitution, and that second observation helps distinguish ions.
- Core
Scope: Core · §8.2 Required practical activity 12: separation of species by thin-layer chromatography
Open revision guidance
What to know
- A TLC plate is the stationary phase and the solvent is the mobile phase. Components move different distances because they have different attractions to each phase. Rf = distance travelled by substance ÷ distance travelled by solvent front.
- The baseline is drawn in pencil above the solvent level. A UV lamp or suitable locating agent can reveal colourless separated spots.
What to do in an assessment
- Apply a small concentrated spot to the pencil baseline and place the plate in a covered chamber with solvent below the baseline.
- Allow solvent to rise, remove the plate before the front reaches the top, immediately mark the solvent front, and visualise spots safely.
- Measure each spot from the baseline and calculate its Rf using the solvent-front distance; compare only under matching stationary and mobile phase conditions.
In plain language
A solvent carries mixture components up a plate at different rates, separating them into spots that can be compared and measured.
Self-check
A spot travels 2.4 cm and its solvent front travels 6.0 cm. What is Rf?
Show the answer
Rf = 2.4 ÷ 6.0 = 0.40; Rf is a ratio without units.
Earlier topic checklist
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Chemistry topic list
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