Chemistry › Key ideas

Key ideas in A-level chemistry

Each of these 420 concise statements sets out one idea to get right. They are grouped by unit and link to the lesson that explains each point.

Jump to: Working in the laboratory · Atomic structure · Amount of substance · Bonding and structure · Energetics · Kinetics · Chemical equilibria · Redox · Thermodynamics · Rate equations · Electrochemistry · Acids and bases · Periodicity and Period 3 · Group 2 and Group 7 · Transition metals · Ions in aqueous solution · Introduction to organic chemistry · Alkanes and halogenoalkanes · Alkenes and alcohols · Organic analysis · Carbonyls and carboxylic acids · Aromatic chemistry and amines · Organic synthesis

420 ideas

Working in the laboratoryUnit C0015 ideas

Measurement, uncertainty and significant figures in chemistry

Double the reading uncertainty for a titre or a mass by difference: ±0.10 cm³ for a titre, ±0.02 g for a mass on a balance reading to 0.01 g.
Divide by the measurement, not by the capacity: ±0.10 cm³ on a titre of 24.35 cm³, not on a 50 cm³ burette.
Adding the absolute uncertainties of a mass and a volume is meaningless; multiplied or divided quantities add their percentages.
'Repeat and take a mean' never shrinks the uncertainty of one reading and never removes a systematic error.
A result rounded at each line, or given to more figures than the least precise data, misses the answer mark.

Apparatus, techniques and safety: the twelve required practicals in outline

Measure a volume that enters a calculation with a pipette or a burette; a measuring cylinder is too coarse, however carefully it is read.
Weigh by difference, the container before and after tipping, so that grains left behind on it are not counted.
A stoppered reflux condenser, or cooling water entering at the top, is a fault to name and correct.
An aldehyde refluxed before it is distilled has become the carboxylic acid.
Too much hot solvent, or warm solvent for washing, leaves product dissolved and lowers the yield.

Formulae, equations and state symbols

Bracket a polyatomic ion that appears more than once: calcium hydroxide is Ca(OH)₂, and CaOH₂ is a different formula.
Writing H₂O₂ to balance the oxygen changes the product; only the coefficients may change.
Write H₂O(l) even for a reaction in solution: the water is the solvent, and H₂O(aq) would be water dissolved in water.
An ionic equation with Na⁺ or Cl⁻ still on both sides is unfinished.
Splitting CaCO₃(s), H₂O(l) or CH₃COOH(aq) into ions is wrong: only aqueous ionic compounds and strong acids split.

Atomic structureUnit C0120 ideas

Protons, neutrons, electrons and isotopes

'Different masses' does not define isotopes: two different elements can share a mass number.
Count the electrons in an ion from its protons: a 2+ ion has two fewer, so Mg²⁺ has 10 electrons, not 14.
35.5 is a relative atomic mass, not a mass number; a mass number is always a whole number.
'Same number of protons' does not explain identical chemistry; the electron arrangement does.
A model named without the experiment that replaced it answers only half of a question on how the model changed.

Time-of-flight mass spectrometry and relative atomic mass

'Deflected by a magnetic field' describes a different machine: a time-of-flight instrument accelerates ions with an electric field.
Dividing by 100 when the abundances are relative heights gives an Ar outside the isotopic masses.
Convert the mass to kilograms and divide by L before working out a flight time; a mass in grams makes the time about thirty times too long.
An electrospray peak at Mr + 1 read as Mr names the wrong molecule; subtract the proton.
The detector current measures abundance; m/z comes from the arrival time, not from the size of the current.

Electron configuration: shells, subshells and orbitals

Fill 4s before 3d: potassium is [Ar] 4s¹, not 2, 8, 9 or [Ar] 3d¹.
Fe³⁺ written [Ar] 3d³ 4s² is wrong: an ion loses its 4s electrons first.
An orbital is a region of high probability, not a path an electron follows.
Chromium as 3d⁴ 4s² and copper as 3d⁹ 4s² are the predicted forms, not the real ones.
Superscripts that do not add up to the number of electrons sink the whole configuration, however good the order.

Ionisation energies and the evidence for subshells

Leaving out (g), or writing Na(g) → Na²⁺(g) + 2e⁻ for the second ionisation energy, is the wrong equation.
'More protons' down a group, without distance and shielding outweighing them, explains nothing.
For aluminium, say that the electron removed is in 3p, which is higher in energy than 3s, as well as further out.
For sulfur, place the pair in one 3p orbital before saying that paired electrons repel.
The group is the number of electrons removed before the largest jump, not the number of the ionisation after it.

Amount of substanceUnit C0225 ideas

The mole, the Avogadro constant and molar mass

A mass in kg or mg put straight into n = m ÷ M is out by a factor of a thousand.
Mass ratios read off the equation's coefficients give the wrong mass of product.
Keep at least one figure more than the answer needs until the last line; rounding to two figures between lines can move the third.
Multiply by the number of atoms or ions in the formula when they are asked for, not molecules: 1 mol of CaCl₂ holds 2 mol of Cl⁻ ions.
One heating and one weighing cannot show that all the water has gone.

Gases and the ideal gas equation

A volume left in cm³ in pV = nRT puts the amount out by a factor of a million.
Convert to kelvin before pV = nRT or a ratio of temperatures: 50 °C is not twice as hot as 25 °C, since 323 K is not twice 298 K.
Keep 24.0 dm³ mol⁻¹ for room temperature and pressure; anywhere else, work the volume out from pV = nRT.
Dividing cm³ by 100 to reach m³ gives an answer ten thousand times too large.
Leave liquid water out of a reacting-volume ratio: only the gases count.

Empirical and molecular formulae

Dividing the percentages by each other, not by Ar, gives a mass ratio, not a formula.
Rounding 1.5 or 1.33 to the nearest whole number gives a different compound.
A molecular formula written without the empirical formula mass and multiplier has no working to credit.
Find oxygen in a combustion analysis by subtracting the masses of C and H from the sample's mass; part of the oxygen in the CO₂ and H₂O came from the air.
Give any missing percentage to oxygen when the elements named fall short of 100.

Concentration, volumetric solutions and titration calculations

A volume left in cm³ in n = cV gives an amount a thousand times too large.
A rough titre, or one outside 0.10 cm³ of the others, in the mean moves the answer.
Scale up from the pipetted portion to the whole flask: 25.0 cm³ taken from 250.0 cm³ is a tenth of it.
Leaving out the 2 for a diprotic acid or a carbonate halves or doubles the answer.
In a back titration, subtract the acid left over from the acid added to find the acid that reacted.

Reacting masses, percentage yield and atom economy

Divide each reactant's moles by its coefficient before comparing them to find the limiting reagent.
A theoretical yield worked from the reactant in excess is too large.
Leaving out a coefficient in atom economy halves or doubles the answer.
Read a yield over 100% as a wet or impure product, not as extra product.
Calling a high-yield process 'efficient' says nothing about its atom economy or solvent.

Bonding and structureUnit C0325 ideas

Ionic bonding and ionic lattices

'Molecules of sodium chloride' or 'intermolecular forces' in a salt is a wrong structure.
Say that the ions move when a molten salt conducts; no electrons carry the current through it.
Compare lattices by charge and ionic radius together: MgO is stronger than NaCl on both counts.
A full outer shell is not why the ions form; the energy released as the lattice forms pays for them.
A dot-and-cross diagram with no brackets or charges on the ions is not credited.

Covalent, dative covalent and metallic bonding

'Shared electrons' for a covalent bond is not the definition: it is a shared pair.
Draw every lone pair on the central atom: nitrogen in NH₃ keeps one, and leaving it off shows six electrons round N.
Claiming a dative bond is weaker, or can be told apart once made, is wrong.
Name the positive ions as well as the delocalised electrons in metallic bonding: the bond is the attraction between them.
Melting points of metals across a period need charge, electron number and size together.

The four crystal structures and their properties

Covalent bonds breaking when iodine or ice melts is wrong: only intermolecular forces are overcome.
Intermolecular forces in diamond, graphite or silicon dioxide do not exist to overcome.
Graphite 'has free electrons' without saying they are delocalised in the layers is incomplete.
Use conductivity to separate ionic, metallic and giant covalent (macromolecular) solids, since all three melt high: only the metal conducts as a solid, and the ionic one only when molten.
Ice floating is explained by density and the open lattice, not by trapped air.

Shapes of molecules and ions

'Bonds repel' or 'atoms repel' in place of electron pairs is not the theory.
Calling NH₃ tetrahedral names the pairs, not the atoms, and loses the shape mark.
Add the ion's charge to the electron count before halving: NH₄⁺ has 8 electrons round N, four bonding pairs and no lone pair.
Count a double bond as one region: CO₂ is linear and SO₂ bent.
A lone pair placed axial in a trigonal bipyramid gives SF₄ a trigonal pyramid in place of its seesaw.

Electronegativity, polarity and intermolecular forces

Check the shape before calling a molecule polar: CCl₄ and BF₃ have polar bonds and no overall dipole.
'Hydrogen bonding' in HCl or CH₃Cl, with no H on N, O or F, is wrong.
End a drawn hydrogen bond on the lone pair, not on the other molecule's hydrogen, and show δ+ and δ− at each end.
Explaining water's boiling point by its strong O–H bonds confuses bonds with forces.
Name the kind of force in every comparison: induced dipole–dipole forces (van der Waals forces), permanent dipole–dipole forces or hydrogen bonds.

EnergeticsUnit C0420 ideas

Enthalpy change, standard conditions and the definitions

Give ΔH a sign every time: '178 kJ mol⁻¹' for a decomposition says nothing about which way the heat went, and the scheme wants +178.
A ΔH arrow drawn from the peak, or an Ea arrow from the axis, measures the wrong gap: both start on the reactant line.
'Under standard conditions' written in place of 'standard states' loses the third mark of a definition; it says nothing about states.
Check a neutralisation definition against sulfuric acid: one mole of it forms two moles of water, so 'per mole of acid' gives double the value.
Doubling a combustion equation to clear the ½O₂ burns two moles, and the ΔH written beside it is no longer ΔcH°.

Measuring an enthalpy change: calorimetry

Use the total volume for m when two solutions are mixed: 25 cm³ with 25 cm³ is 50 g, since both solutions warm.
Divide q by the moles of the limiting reagent, or of the water formed; the reactant in excess gives a value too small.
A ΔH left in J mol⁻¹ and labelled kJ mol⁻¹, or given positive when the solution warmed, loses the answer mark.
A cooling line taken back to time zero instead of the time of mixing reads ΔT for a moment before anything was mixed.
'Heat loss' alone for a spirit burner leaves two reasons unclaimed: fuel evaporating from the wick and incomplete combustion.

Hess's law and enthalpy cycles

A formation cycle drawn with its arrows pointing down, or a combustion cycle with them up, gives the right size with the opposite sign.
3CO₂ entered as −394 instead of 3 × (−394) loses the method mark for that sum.
Write (−1182) − (−1157) with its brackets before working it out: subtracting a negative is where the sign goes.
Scale each measured step by the amount the target equation needs; a step used once where two moles are needed halves that arrow.
'Heat loss' as the reason an enthalpy change is found indirectly is rejected; the reason is that the reaction cannot be carried out cleanly.

Mean bond enthalpies

Copy every bond enthalpy as a positive number; the sign of ΔH comes from the subtraction.
Ethanol counted with six C–H bonds has one too many; the sixth hydrogen is on the oxygen.
Take a double bond once at the C=C value, never as two single bonds or at the C–C value.
'Heat loss' as the reason a bond enthalpy answer misses the data book is rejected; nothing was measured.
Comparing a bond enthalpy answer with a ΔcH° that forms liquid water ignores the state difference; bond enthalpies are for gases only.

KineticsUnit C0515 ideas

Collision theory, activation energy and the rate of reaction

Put time into every collision answer, more frequent collisions or more collisions per second; 'more collisions' alone gives no rate.
An answer that names only the energy condition is half an answer: a collision with enough energy in the wrong orientation is a bounce.
Start an Ea arrow on the reactant line and end it at the peak; one from the axis or from the products measures nothing.
'Collide harder' or 'more energy' in an answer about concentration or surface area describes temperature and is rejected.
Take tangent points at least half the grid apart; two points close together magnify a grid misreading.

The Maxwell–Boltzmann distribution, temperature and catalysts

An axis labelled 'speed', 'rate' or 'temperature', or a tail that lands on the axis, loses the sketch mark.
'Average energy' written at the peak is wrong: the mean lies to the right of the most probable energy.
A hotter curve drawn as tall as the cooler one shows extra molecules; the peak must be lower as well as further right.
Draw one curve on a catalyst diagram and mark two values of Ea; a second curve says the catalyst changed the molecules' energies.
Give the first temperature mark to more molecules with at least Ea; more frequent collisions are true but the smaller effect.

Measuring the rate of a reaction

A choice of method with no calculation behind it is a guess: find the mass or volume of gas and compare it with the balance or the syringe.
Rule the tangent to the whole curve; a rate from the first two readings uses the least certain points.
A narrower flask, or a different volume over the cross, changes the amount of sulfur at the end point, and that run no longer compares.
Record the mixture's temperature, not the bath's: a hot run that has cooled plots too far right.
'Wear eye protection' answers the acid, not the gas: the hazard mark needs sulfur dioxide named and ventilation.

Chemical equilibriaUnit C0615 ideas

Dynamic equilibrium, Le Chatelier's principle and industrial compromise

Count only moles of gas for a pressure shift; a solid or liquid counted among them can pick the other side.
Write 'shifts in the endothermic direction'; 'the reaction uses up the heat' names no direction.
'A high temperature gives a high yield' for an exothermic reaction is not credited however well the compromise is argued.
A catalyst said to 'increase the yield' is wrong; it shortens the time to reach equilibrium and leaves the position alone.
Restating Le Chatelier's principle without the direction of the shift and its effect on the yield earns none of the three marks.

The equilibrium constant Kc

An expression with the reactants on top is not credited, even with every power right.
Amounts in moles go straight into Kc only when the powers on top and bottom add up to the same number; otherwise divide by the volume first.
Double the change for HI: the 2 in 2HI makes its change twice that of I₂.
'Kc increases' when a reactant is added is rejected even with the right shift; Kc is unchanged and the ratio is momentarily below it.
Write 'no units' when the units cancel, rather than leaving the space blank.

Kp: equilibria in the gas phase

Work mole fractions from the equilibrium total; the starting amounts make them add up to more than 1.
Write p(NH₃), never [NH₃], in a Kp expression: square brackets make it a Kc expression.
Converting Kp from atm⁻² to kPa⁻² by dividing by 101 once is a hundred times out; the factor carries the power of the units.
An amount in moles, or a pressure from before the reaction, substituted into Kp is not a partial pressure at equilibrium.
'Kp increases with pressure' is rejected; the position shifts and the constant does not.

RedoxUnit C0715 ideas

Oxidation states and the rules for assigning them

Put the sign before the number in an oxidation state, +2; 2+ is the notation for an ionic charge.
A mean of +1 for nitrogen in NH₄NO₃ is wrong: split the formula into its two ions and give −3 and +5.
Hydrogen set at +1 in NaH breaks the sum rule; the metal's rule ranks higher, so hydrogen is −1.
'Manganese' as the oxidising agent is not credited; the agent is the species that accepts the electrons, MnO₄⁻.
Read the Roman numeral as the oxidation state, not a count of atoms: copper(I) oxide is Cu₂O.

Half-equations and building redox equations

Add the electrons last, once the atoms balance and the charge on each side is known.
Count every H on a side at step 3, the H in NH₄⁺ as well as the water's.
Electrons left over in an overall equation mean the multipliers were wrong, and the equation mark is not awarded.
'Iodine is oxidised' names the product; the species oxidised is the iodide ion, −1 to 0.
'Chlorine is oxidised and reduced' alone earns only part of the credit; give 0 to −1 and 0 to +1 and say which is which.

Redox titrations: manganate(VII) and thiosulfate

Read the top of the meniscus at both ends in a manganate(VII) burette; the purple hides the bottom.
'Clear' for the flask before the end point is not accepted; the word is colourless.
Add starch at pale straw; added at the start it holds iodine in the complex and makes the titre too large.
Leaving out the scale factor from the 25.0 cm³ portion to the 250.0 cm³ flask gives a tenth of the mass.
A mean that includes the rough titre, or a titre outside 0.10 cm³ of the others, is not awarded the mark.

ThermodynamicsUnit C0815 ideas

Lattice enthalpy and Born–Haber cycles

A sign that does not match the direction the question names is wrong, however right the size: formation negative, dissociation positive.
Atomisation of chlorine is per mole of atoms, ½Cl₂(g) → Cl(g); writing Cl₂(g) → 2Cl(g) doubles the value.
Every species between the elements and the solid needs (g), and every level after an ionisation carries its electrons.
Count each step as often as the formula needs it: undoubled chlorine steps for MgCl₂ give −2752 kJ mol⁻¹ in place of −2525.
The gap between experiment and theory is covalent character from polarisation, never experimental error.

Enthalpies of solution and hydration

Count each ion as often as the formula holds it: CaCl₂ takes the chloride's hydration twice, and once is 364 kJ mol⁻¹ out.
A hydration definition that starts from anything but gaseous ions describes a different quantity.
Take the lattice dissociation value, +787, into a solution cycle; lattice formation plus hydration gives a huge negative number.
Hydration is explained by the ion attracting the oppositely charged end of the water dipole; 'charge density' alone explains nothing.
'Endothermic, so it cannot dissolve' is wrong: sodium chloride dissolves at +4 kJ mol⁻¹ (the cycle gives +17) because the entropy gain outweighs it.

Entropy and Gibbs free-energy change

Leaving oxygen out of a ΔS sum because 'elements are zero' confuses entropy with enthalpy of formation; every substance has an entropy.
ΔS left in J K⁻¹ mol⁻¹ inside ΔG = ΔH − TΔS makes the TΔS term a thousand times too large.
Write 'feasible' for ΔG ≤ 0, not 'spontaneous', which suggests the reaction happens at once.
A positive ΔG read as 'no reaction' is wrong; K is small, not zero, and some product is present at equilibrium.
Give the activation energy as the reason diamond does not turn to graphite: ΔG is −2.9 kJ mol⁻¹, so the change is feasible.

Rate equationsUnit C0915 ideas

Rate equations, orders of reaction and the rate constant

An order written without its reactant, or a rate equation without k, is incomplete: rate = k[A]ᵐ[B]ⁿ, every term named.
A coefficient of 2 in the equation says nothing about the order; only the rate equation does.
Each concentration factor is raised to the power of its own order, then the factors are multiplied, never added.
Units of k worked out for the wrong overall order are wrong even when the value is right: s⁻¹ for first, mol⁻¹ dm³ s⁻¹ for second.
Raising a concentration raises the rate and leaves k alone; only temperature or a catalyst changes k.

Finding orders from initial rates, half-lives and graphs

Divide out the effect of the known order first when two concentrations change between a pair of runs.
1/t measures the rate only if every run makes the same fixed amount of product.
A chord between two points gives an average rate; the rate at a moment needs the tangent.
One half-life shows nothing about the order; two or more equal ones show first order.
A straight line in the RP7 run shows zero order in iodine only: propanone and acid were not varied in it.

The Arrhenius equation and the rate-determining step

Convert to kelvin before plotting ln k against 1/T; a temperature in °C throws every point off the line.
Dropping the minus sign of the gradient gives a negative Ea, which is not credited; Ea = −gradient × R.
'The first step' as the definition of the rate-determining step is wrong as often as it is right; it is the slowest step.
Replace an intermediate in a rate equation with the reactants that formed it.
A slow step holding one A and one B cannot give rate = k[A]²; the slow step must hold exactly the particles the rate equation names.

ElectrochemistryUnit C1015 ideas

Electrode potentials and the standard hydrogen electrode

Saying the salt bridge carries electrons is rejected: only ions move through it.
A definition of E° without the hydrogen electrode, or without the values of the standard conditions, is incomplete.
Doubling a half-equation to balance electrons never doubles its E°.
A wire in place of the bridge adds two new electrode potentials to the reading.
Put the oxidised forms next to the double line, Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s); the reduced forms sit at the outer ends.

The electrochemical series and predicting feasibility

Name the oxidised form as the oxidising agent: Cl₂ oxidises, Cl⁻ does not.
Put the half-cell being oxidised on the left, so that E°right − E°left gives E°cell its proper sign.
Multiplying E° by the coefficient used to balance electrons gives a wrong E°cell.
'Not feasible' is the wrong explanation for a feasible reaction that is slow; name the activation energy.
Name the equilibrium and the way it shifts when a change of concentration moves an electrode potential.

Commercial cells, rechargeable cells and fuel cells

A fuel cell described as 'recharged' has been confused with a rechargeable cell.
Use hydroxide ions, not H⁺, in the alkaline cell's half-equations; H₂ → 2H⁺ + 2e⁻ belongs to the acidic cell.
Giving Li⁺ an (aq) state symbol ignores the organic solvent in the electrolyte.
Reverse both half-equations for charging, so that the electrons still have somewhere to go.
'No pollution' is true only at the point of use; the hydrogen's source decides the rest.

Acids and basesUnit C1120 ideas

Brønsted–Lowry acids and bases, pH and strong acids

'Hydrogen donor' in place of 'proton donor' does not define an acid.
Calling a dilute strong acid 'weak' confuses strength with concentration.
Add the two volumes before dividing the excess: 25.0 cm³ mixed with 20.0 cm³ is 45.0 cm³ of solution.
Count two H⁺ per H₂SO₄ when a question treats it as fully dissociated; one halves [H⁺].
Using the leftover [OH⁻] in −log gives pOH, not pH; Kw is needed first.

Weak acids, Ka and pKa; Kw and strong bases

Taking pH 7 as neutral at a temperature other than 298 K puts the neutral point in the wrong place.
Convert [OH⁻] to [H⁺] with Kw before taking −log; [OH⁻] straight into −log gives pOH.
Applying [H⁺] = √(Ka × [HA]) to a strong acid or a buffer uses an approximation that does not hold there.
Keep [H⁺] unrounded until it is squared; rounding first can move the third figure of Ka.
Reading a larger pKa as a stronger acid turns the scale upside down.

Titration curves and choosing an indicator

Putting every equivalence point at pH 7 ignores the salt formed.
Level a curve for an added weak base off near pH 11; pH 12.5 is where a strong base ends.
Read pKa at half the equivalence volume, where [HA] = [A⁻], not at the equivalence volume itself.
An indicator whose range only touches the vertical section changes colour before or after equivalence.
Naming an indicator for a weak acid–weak base titration is wrong: none gives a sharp end point.

Buffer solutions and buffer calculations

'Keeps the pH constant' describes no real buffer; a buffer resists change.
Pair the weak acid with its conjugate base; hydrochloric acid with sodium chloride is no buffer, since Cl⁻ cannot take up added H⁺.
Take [A⁻] from the salt inside a buffer; setting [H⁺] = [A⁻] gives the pH of the acid alone.
Adding the moles of acid to the wrong reservoir moves the pH the wrong way.
An account of buffer action with no equation, or no mention of large reservoirs, is incomplete.

Periodicity and Period 3Unit C1215 ideas

Periodicity: trends across Period 3 and the blocks of the table

File an element by the subshell its last electron entered: selenium's went into 4p, so it is p block, full 3d or not.
'More shielding across the period' is not credited: the inner shells are the same ten electrons in every atom, so the shielding is similar.
Melt chlorine by overcoming the induced dipole–dipole forces (van der Waals forces) between Cl₂ molecules; the Cl–Cl bond stays whole.
Compare S₈ with P₄, not S with P: the size of the molecule is the whole comparison.
Explaining the fall at aluminium by a change in radius is wrong: the radius falls at every step, and the 3p electron is the reason.

The Period 3 elements: reactions with oxygen and water

Describe what is seen, never the product's name: fizzing, a white solid, a blue flame.
Steam gives MgO, cold water gives Mg(OH)₂; swapping them loses the equation.
SO₃ written as the product of burning sulfur is wrong in any amount of oxygen: combustion stops at SO₂, and SO₃ needs V₂O₅ at about 450 °C.
Give P₄O₁₀ when the molecular formula is asked for; P₂O₅ is only the empirical formula.
The pH of a solution is a number or a narrow range, never just 'alkaline'.

The Period 3 oxides: structure, bonding and acid–base character

Classify an oxide by its reactions with acids and bases, not by the pH it leaves in water.
Putting Al₂O₃ above MgO on melting point because Al³⁺ carries more charge ignores its covalent character: MgO melts highest.
Written to NaAl(OH)₄, the aluminate equation needs 3H₂O on the left, or it does not balance.
Giving SO₂ and SO₃ the same pH in water is wrong: H₂SO₃ is a weak acid at pH 2 to 3, H₂SO₄ a strong one at pH 0 to 1.
Put every acidic hydrogen of H₃PO₄ on an oxygen; none sits on the phosphorus.

Group 2 and Group 7Unit C1320 ideas

Group 2: trends down the group and reactions with water

More protons down the group is not more grip: distance and shielding win, and the answer must say so.
Name magnesium as the melting-point exception and do not try to explain it by ion size.
Magnesium with cold water gives Mg(OH)₂; with steam, MgO. The two equations are not interchangeable.
Calling magnesium the oxidising agent in TiCl₄ + 2Mg because it is oxidised reverses the terms: it is oxidised, so it is the reducing agent.
Argon is there to keep oxygen and nitrogen off both hot metals, not to cool or dilute anything.

Group 2 compounds: solubility, uses and the sulfate test

'The hydroxides get more soluble' with no end member named is half an answer: Mg(OH)₂ sparingly soluble, Ba(OH)₂ soluble.
Acidify a sulfate test with hydrochloric or nitric acid, never sulfuric, and say the acid removes carbonate.
Explain a barium meal by the sulfate's insolubility in stomach acid as well as water: barium carbonate dissolves there and releases toxic Ba²⁺.
Calcium is the ion to confirm with a second test: its precipitates are slight with both reagents.
CO₂ written as a product of CaO + SO₂ is wrong: only the CaCO₃ route releases carbon dioxide.

Group 7: trends down the group and displacement reactions

'The Br–Br bond breaks on boiling' describes a different change: only the induced dipole–dipole forces (van der Waals forces) between molecules are overcome.
Oxidising power is about gaining an electron; do not borrow the Group 2 argument about losing one.
A displacement equation with K⁺ still on both sides is unfinished: cancel the spectators and check that the charges balance.
An orange tube is not a reaction until it has changed colour; compare it with a control.
Give every colour with its solvent: bromine and iodine look alike in water.

Halide ions, the silver nitrate test, concentrated sulfuric acid, and chlorine in water

Chloride with concentrated sulfuric acid is not redox: nothing changes oxidation state.
Pair every observation with its product: brown fumes are Br₂, the bad-egg smell is H₂S.
Acidify silver nitrate with nitric acid; hydrochloric adds chloride and sulfuric adds sulfate.
Confirm the halide with ammonia; white, cream and yellow alone are too close to call.
Call a reaction disproportionation only after giving both chlorine oxidation-state changes.

Transition metalsUnit C1425 ideas

Transition metal characteristics and complex formation

A definition of a transition metal that mentions only the atom, not a stable ion, lets zinc in.
Writing Fe³⁺ as [Ar] 3d³ 4s² takes the electrons from the wrong subshell: 4s goes first.
'A ligand bonds to the metal' is half a definition: say it donates a lone pair, and that the bond formed is coordinate.
Three ethane-1,2-diamine ligands give coordination number 6; counting ligands instead of bonds gives 3.
Explain carbon monoxide's toxicity by its much stronger bond to the Fe²⁺ of haem, released only very slowly, not by 'blocking the blood'.

Ligands, the shapes of complexes, isomerism and ligand substitution

Put four chloride ligands round Cu²⁺, not six: they are larger, so the complex is tetrahedral.
Copper with excess ammonia keeps two water ligands: [Cu(NH₃)₆]²⁺ is not the product.
'The new bonds are stronger' does not explain the chelate effect; the rise in the number of particles does.
Keep cis–trans isomerism for square planar and octahedral complexes; every corner of a tetrahedron is next to every other.
Bond ammonia to the metal through nitrogen, whose lone pair forms the coordinate bond.

Colour in transition metal complexes and colorimetry

A blue solution absorbs orange and red light; 'it absorbs blue' is the opposite of the truth.
Convert λ to metres before ΔE = hc/λ; left in nanometres it gives an answer 10⁹ times too small.
'Larger ΔE, longer wavelength' reverses hc/λ: a larger gap absorbs a shorter wavelength.
'Zn²⁺ is not a transition metal ion' does not explain its lack of colour; the full 3d subshell does.
Choose the complementary filter, the colour the solution absorbs; a filter of the solution's own colour lets the light through.

Variable oxidation states: vanadium, chromium and manganate(VII) titrations

The first green when zinc reduces vanadium(V) is a yellow and blue mixture; calling it V³⁺ is wrong.
Take E°cell as E°(reduced) − E°(oxidised), the species reduced first.
Hydrochloric acid in a manganate(VII) titration gives a titre that is too large, because chloride is oxidised.
The mole ratio for ethanedioate is 2 : 5, not 1 : 5; using the iron ratio doubles the amount of ethanedioate calculated.
Give the reason with 'easier to oxidise in alkali': the E° of the couple is less positive at high pH.

Catalysis: heterogeneous, homogeneous and autocatalytic

'A catalyst takes no part in the reaction' is wrong: it reacts and is regenerated.
Write 'adsorbed' for reactants held on a catalyst surface; 'absorbed' describes a sponge.
'It lowers the activation energy' alone does not explain how V₂O₅ works; the two equations and the change of oxidation state do.
Explain the faster Fe²⁺ route by charge: the ions in each of its steps are oppositely charged, where S₂O₈²⁻ and I⁻ repel.
An autocatalysis curve drawn steepest at the start is the shape of an ordinary reaction.

Ions in aqueous solutionUnit C1510 ideas

Metal-aqua ions and the acidity of their solutions

'A salt solution is neutral' fails for every 3+ aqua ion: they are weak acids.
Lower the complex's charge by one for each proton lost in a hydrolysis equation.
'Fe³⁺ is more charged' is one link of three; the polarised O–H bond and the release of H⁺ are the other two.
The brown solid from iron(III) and carbonate is the hydroxide; Fe₂(CO₃)₃ is never formed.
Square [H⁺] when working out Ka from pH; without the square the value is out by orders of magnitude.

Reactions of aqua ions with hydroxide, ammonia and carbonate

'Goes blue' is not an observation: say precipitate or solution, and its colour.
Ammonia equations balance with NH₄⁺ on the right; water there leaves them unbalanced.
Aluminium hydroxide does not dissolve in excess ammonia; only excess sodium hydroxide dissolves it.
[Cu(NH₃)₆]²⁺ is the wrong product: copper keeps two water ligands in excess ammonia.
Iron(II) recorded after standing looks like iron(III); describe its precipitate as it first forms.

Introduction to organic chemistryUnit C1615 ideas

Naming and representing organic molecules

Count every route through the carbons before naming: the chain drawn straight across the page is not always the longest.
C₄H₁₀ is a molecular formula; a question asking for a structural formula wants CH₃CH₂CH₂CH₃, and the molecular one is rejected.
Leave the hydrogens off a skeletal formula; drawn on each carbon, they make it a displayed formula.
'2-Ethylbutane' is hexane misnamed: the longest chain was missed because the ethyl branch continues it.
Numbering from the wrong end gives 3-methylbutane; the name with the lowest locant, 2-methylbutane, is the only one credited.

Structural isomerism and E/Z isomerism

Compare what is bonded to what before counting isomers; one chain bent two ways on paper is one compound.
Define a stereoisomer as the same structural formula with a different arrangement of atoms in space; 'different arrangement' alone is too little.
Giving but-1-ene E and Z isomers is wrong: one carbon of its C=C carries two hydrogens.
'The molecule cannot rotate' is not the reason: rotation about C=C would break the π bond, and that is the sentence credited.
Priority goes by the atomic number of the atom bonded to the C=C carbon, not the size of the group: Br beats CH₂CH₃.

Reaction types, curly arrows and how a mechanism is written

'Addition' for bromoethane with hydroxide is wrong: a group is replaced, so the reaction is a substitution.
Draw homolytic fission with two fishhook arrows, one electron to each atom; one double-headed arrow gives ions.
Define a nucleophile as an electron pair donor; 'attracted to a positive charge' describes one without defining it.
Start the arrow on the nucleophile's lone pair and end it on the δ+ carbon.
A carbocation without its + sign, or a leaving halide without its − sign, loses the charge mark; count the charges in and out.

Alkanes and halogenoalkanesUnit C1720 ideas

Alkanes: fractional distillation, cracking and combustion

Explain a longer alkane's higher boiling point by stronger induced dipole–dipole forces (van der Waals forces) between molecules; no covalent bond breaks on boiling.
A fraction is a range of boiling points, not one compound: 'petrol boils at 40 °C' is not credited.
Balance a cracking equation in carbon and hydrogen: C₁₀H₂₂(g) → C₈H₁₈(g) + C₂H₄(g), not C₈H₁₆(g) + C₂H₄(g).
A catalytic converter removes CO, NO and unburnt hydrocarbons; an answer that has it removing CO₂ is rejected.
Name the products of incomplete combustion, carbon monoxide or carbon with water; 'less energy released' answers a different question.

Free-radical substitution

Cl₂(g) → Cl⁺ + Cl⁻ for initiation is heterolysis and is rejected: ultraviolet light gives two Cl•(g) by homolytic fission.
Have the chlorine radical take a hydrogen, giving HCl(g) and •CH₃(g); Cl• + CH₄ → CH₃Cl + H• is not a propagation step.
Write the two propagation steps, not the overall CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), which is their sum.
Put the dot on the carbon of •CH₃, not floating after the formula; a dot on another atom is a different species.
'Excess chlorine' to make chloromethane is backwards: excess methane limits further substitution, excess chlorine gives CH₂Cl₂, CHCl₃ and CCl₄.

Halogenoalkanes: nucleophilic substitution and rates of hydrolysis

A nucleophile drawn attacking the halogen, or arriving without its lone pair, loses the first arrow's mark.
Name the reagent with its condition, KCN and not HCN, ammonia in excess, or the product changes or a mixture forms.
'C–F is the most polar, so it reacts fastest' reverses the trend: rate follows bond enthalpy, and C–I is the weakest bond.
The silver nitrate comparison needs ethanol as the solvent and the same temperature and concentrations; without ethanol the method mark goes.
Finish the ammonia mechanism with a second ammonia molecule taking H⁺ from the N⁺, giving the amine and NH₄⁺.

Elimination from halogenoalkanes, and CFCs and the ozone layer

KOH without 'in ethanol' and 'heated under reflux' is read as substitution: the solvent decides which reaction happens.
Aim the hydroxide's arrow at a hydrogen on the carbon next to the C–X in elimination; an arrow to the carbon is substitution.
Hydroxide acts as a base here; 'nucleophile' in an elimination answer loses the role mark.
Two alkenes from 2-bromobutane is one short: but-2-ene forms as both E and Z isomers, so three are credited.
End an ozone mechanism with a step that gives Cl• back; one that stops at ClO• shows no catalyst.

Alkenes and alcoholsUnit C1820 ideas

Alkenes: structure, electrophilic addition and Markovnikov's rule

The π electrons, not the σ pair, are the part that attacks: the first arrow starts on the double bond.
Every curly arrow starts on a bond or a lone pair; one that starts on an atom or on a + charge loses its mark.
Br₂ is not polar; an answer that says it is gets no credit. Its dipole is induced by the π electrons.
Explain a major product by the stability of the carbocation, not of the product and not by Markovnikov's rule alone.
Start the second arrow on the bromide ion's lone pair and end it on the carbocation's positive carbon.

Addition polymers

A polymer drawn with a C=C still in the chain is wrong: every double bond opened.
A side group drawn into the backbone loses the repeating-unit mark.
A monomer named from the polymer with the brackets left in is not a monomer: restore the double bond.
PVC's rigidity comes from permanent dipoles between C–Cl bonds, not from the chlorine 'bonding' chains together.
Poly(ethene) is unreactive because its bonds are strong and non-polar, not because it is 'stable'.

Alcohols: hydration, fermentation and classification

Classify by the carbon carrying the OH, never by the length or branching of the chain.
Boiling is about hydrogen bonds between molecules; no O–H bond breaks.
A fermentation condition stated without its reason earns half the answer.
The hydration conditions are a compromise: say what a lower temperature would gain and what it would cost.
'Carbon neutral' without the three equations and a fossil-fuel input is an assertion, not an evaluation.

Oxidation and elimination of alcohols

'Turns green' without the starting orange is half an observation.
Distil with the alcohol in excess for an aldehyde; reflux with the oxidising agent in excess for a carboxylic acid.
A tertiary alcohol is not oxidised because the C–OH carbon has no H, not because it is 'too stable'.
A sealed apparatus is never right: heated glass under pressure bursts.
Forgetting the E/Z pair undercounts the alkenes from an unsymmetrical alcohol.

Organic analysisUnit C1920 ideas

Test-tube reactions for functional groups, and high-resolution mass spectrometry

A test needs reagent, condition and observation together: 'Tollens' reagent' alone, without 'warm' and 'silver mirror', scores one mark of three.
A green dichromate result taken as proof of an alcohol is wrong: an aldehyde turns it green too.
Record effervescence, and limewater turning milky to name the gas; 'a gas is given off' is a conclusion.
Use ¹H = 1.0078 and ¹²C = 12.0000 for precise masses; the periodic-table Ar values give a mass that matches nothing.
Naming a compound from a precise mass alone overreaches: isomers share it, and a test or a spectrum decides between them.

Infrared spectroscopy

Labelling every dip is wasted work; assign the strong ones above 1500 cm⁻¹.
Look past the C–H absorptions, which almost every organic compound shows, to the bonds that identify the group.
An acid's O–H called an alcohol's, or the other way round, misreads the whole spectrum.
Infrared read from the table cannot split an aldehyde from a ketone, or two esters from each other.
Greenhouse gases absorb what the Earth emits, not what the Sun sends.

Proton and carbon-13 NMR spectroscopy

Counting hydrogens instead of environments gives too many peaks.
Read only the number of peaks and their shifts from a carbon-13 spectrum; its peak heights are not a ratio.
An integration ratio not scaled to the formula gives the wrong number of hydrogens.
Count the hydrogens on the neighbouring carbons for n + 1, not those on the carbon itself.
An O–H counted as a neighbour gives a splitting the spectrum does not show.

Chromatography: TLC, column and gas

'The polar one moves less' is incomplete without saying it is more strongly adsorbed onto the stationary phase.
Measure Rf to the centre of the spot and from the baseline, not from the edge of the plate.
An ink baseline, or one below the solvent, ruins the plate before it is run.
Peak heights understate a broad peak: compare areas.
An Rf from another day's plate cannot identify anything.

Carbonyls and carboxylic acidsUnit C2025 ideas

Optical isomerism

Compare whole groups, not first atoms: CH₂CH₃ and CH₂CH₂CH₃ are different groups.
A CH₂, a CH₃ or a C=O carbon is never a chiral centre.
Two flat drawings at right angles show no chirality; one wedge and one dashed bond each, as mirror images.
Enantiomers rotate the plane of polarised light; they do not 'bend' or 'reflect' it.
A racemate still holds chiral molecules; only the mixture is inactive.

Aldehydes and ketones: nucleophilic addition and the tests that tell them apart

Hydride is H⁻ with a lone pair, not a hydrogen atom, and NaBH₄ leaves a C=C bond alone.
The cyanide arrow starts on the carbon's lone pair, never on the nitrogen.
A test needs the reagent, 'warm', and what both an aldehyde and a ketone show.
'Positive result' is not an observation: write silver mirror or brick-red precipitate.
No chiral product carbon, no racemate: propanone with cyanide gives one compound only.

Carboxylic acids, esters and biodiesel

Weak means slightly dissociated, not dilute: a weak acid still reacts completely with a base.
The water in esterification takes its O from the acid, not the alcohol.
Read an ester's formula from the C=O: that side is the acid and is named second.
Alkaline hydrolysis gives the carboxylate salt, not the acid, until a strong acid is added.
A triglyceride needs three NaOH or three CH₃OH, and gives one glycerol.

Acylation: acyl chlorides and acid anhydrides

An acyl chloride is not a chloroalkane: the nucleophile adds to C=O before Cl⁻ leaves.
With ammonia or an amine, two molecules react: one as nucleophile, one to take up the HCl.
The N- in N-methylethanamide puts the methyl group on nitrogen, not on the chain.
The intermediate carries two charges: O⁻ and a + on the nucleophile's N or O.
Atom economy favours the chloride for aspirin; cost and safety choose the anhydride.

Condensation polymers and the disposal of polymers

A repeating unit with a closed end, or with an extra OH or H left on, is not a repeating unit.
Polyesters have no N–H, so their chains are held by dipoles, not hydrogen bonds.
Biodegradable does not mean it breaks down in landfill: there is too little air and water.
Photodegradable plastics need light; buried, they last as long as any other.
Count 2n water molecules for n of each monomer, or 2n − 1 if the question asks for the exact figure.

Aromatic chemistry and aminesUnit C2120 ideas

Benzene: its structure and the evidence for delocalisation

'Stronger double bonds' is never the reason for anything benzene does: it has no C=C bonds.
Describe p orbitals overlapping sideways above and below the plane of the ring; 'a ring of electrons' or 'resonance' leaves that out.
A delocalisation energy given as −152 kJ mol⁻¹, or as 'more exothermic', inverts the argument: benzene releases less, so it is 152 kJ mol⁻¹ more stable.
A less exothermic hydrogenation means a lower starting energy, not weaker bonds.
Name substituted benzenes with numbers for the lowest locants, never with ortho, meta or para.

Electrophilic substitution: nitration and Friedel–Crafts acylation

An arrow that starts on one carbon, or on a Kekulé double bond, does not show the delocalised electrons moving.
'Heat' or dilute nitric acid loses the conditions mark: both acids concentrated, about 50 °C, and above 55 °C a second nitro group goes on.
The horseshoe covers five carbons, not six, with + inside and both H and the new group on the sixth.
The arrow from the ring ends on the positive atom: N of NO₂⁺, the carbonyl C of RCO⁺.
A catalyst must be shown coming back: H⁺ + HSO₄⁻, or H⁺ + AlCl₄⁻.

Amines: preparation, basicity and nucleophilic reactions

Count carbon groups on the nitrogen, not on the carbon: propan-2-amine is primary.
A quaternary ion has no lone pair, so it is neither a base nor a nucleophile.
Base strength is about the lone pair's availability, never the size of the group.
The amine product is a better nucleophile than ammonia, so it reacts on unless ammonia is in excess.
The nitrile route adds a carbon atom: count the chain before choosing the halogenoalkane.

Amino acids, proteins, enzymes and DNA

Leave glycine out of any list of chiral amino acids: its central carbon carries two hydrogens.
A zwitterion drawn in strong acid is wrong: at low pH both groups carry H, NH₃⁺ and COOH; in alkali neither does.
Two different amino acids give two different dipeptides: write both, with the peptide link the right way round.
Secondary structure is held by hydrogen bonds between peptide links, not between side chains.
Cisplatin drawn bonded to a phosphate or a sugar is at the wrong site: it bonds to a nitrogen on guanine.

Organic synthesisUnit C2215 ideas

Planning a synthetic route

A step with no conditions, or 'the nitrile' for a product, loses its mark.
[O], [H] and plain 'dichromate' are not reagents; name the substance.
Cyanide adds to C=O, never to C–OH: add it before reducing, not after.
Aqueous KOH substitutes, ethanolic KOH eliminates; write which.
Ammonia on a halogenoalkane gives a mixture, so it is not a route to pure amine.

Preparing, purifying and testing an organic solid and liquid

A sealed reflux apparatus or a Bunsen flame under organic vapour is unsafe.
Extra hot solvent lowers the yield, not the purity.
A warm wash dissolves crystals; the wash is a little ice-cold solvent.
A wet product weighs too much and melts low: dry it before both.
The limiting reactant comes from moles, never from volumes or masses.

Putting the analysis together: structure from spectra

Fewer carbon peaks than carbons needs the equivalent carbons named.
An integration ratio of 3 : 1 in an eight-hydrogen molecule is 6H and 2H.
A C=O peak does not separate aldehyde from ketone; the 9 to 10 ppm proton does.
A broad O–H at 2500 to 3000 cm⁻¹ is a carboxylic acid, not an alcohol.
A structure that contradicts one reading is wrong, however well it fits the rest.