3.2 Inorganic chemistry — notes on one page

Every AQA A-level Chemistry 7405 specification point in 3.2, read straight through: 13 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 7405 section 3.2. Review basis: the qualification registry sourced from the AQA A-level Chemistry (7405) specification; registry verification recorded 11 July 2026.

How this checking works

3.2.1.1 · Classification

An element is classified as s, p, d or f block from its position in the Periodic Table, and that position is determined by proton number.
The block corresponds to the subshell receiving the differentiating electron: an outer $s$ subshell gives s block, an outer $p$ subshell gives p block, and filling a $(n-1)d$ subshell gives d block.
Electron configuration therefore explains classification, but any filled inner subshell must not override the differentiating subshell.
For example, $\mathrm{[Ar]3d^{10}4s^2 4p^4}$ is p block despite containing occupied d orbitals.

Worked example

Classify an element with configuration $\mathrm{[Ar]3d^1 4s^2}$ and explain the classification.

1.Identify the subshell receiving the differentiating electron.
2.The added electron occupies $3d$ , not the already occupied $4s$ subshell.

Answer: The element is in the d block.

Common mistakes

Don't classify every element containing d electrons as d block, even when the d subshell is a filled inner shell.
Don't use the highest principal quantum number alone instead of the differentiating subshell.

Exam tip. For a block-classification question, name the differentiating subshell rather than merely quoting the full configuration.

3.2.1.2 · Physical properties of Period 3 elements

Across Period 3 from Na to Ar, atomic radius generally decreases because proton number rises while added electrons enter the same principal shell with similar shielding.
First ionisation energy generally increases, but falls from Mg to Al because Al loses a higher-energy $3p$ electron, and from P to S because repulsion within a paired $3p$ orbital eases removal.
Melting points require structure and bonding: metallic bonding strengthens from Na to Al, giant covalent silicon is very high, while $\mathrm{P_4}$ , $\mathrm{S_8}$ and $\mathrm{Cl_2}$ molecules and monatomic Ar melt by overcoming London forces whose strength depends on electron-cloud size.

The Period 3 melting-point pattern reflects changing metallic, giant covalent and simple molecular or atomic structures.

Worked example

Explain why sulfur has a higher melting point than phosphorus and argon.

1.Sulfur and phosphorus are simple molecular, while argon is monatomic.
2. $\mathrm{S_8}$ has a larger, more polarisable electron cloud than $\mathrm{P_4}$ , and both exceed Ar.
3.Larger temporary dipoles produce stronger London forces.

Answer: Sulfur has the strongest London forces and therefore the highest melting point of the three.

Common mistakes

Don't explain the decreasing atomic radius by increasing shielding, although electrons are added to the same main shell.
Don't treat the Mg-to-Al and P-to-S ionisation-energy dips as random anomalies without using subshell energy or electron pairing.
Don't say covalent bonds inside $\mathrm{S_8}$ molecules break during melting instead of intermolecular forces.

Exam tip. For a Period 3 melting-point comparison, identify each structure before naming the attraction overcome.

3.2.2 · Group 2, the alkaline earth metals

From Mg to Ba, atomic radius increases and first ionisation energy decreases as each element adds an electron shell. All have giant metallic structures; melting-point changes reflect metallic-bond strength and lattice packing rather than a simple monotonic trend.
Reactions with water become more vigorous, forming $\mathrm{M(OH)_2}$ and $\mathrm{H_2}$ , while Mg reacts with steam to form MgO.
Magnesium extracts titanium from $\mathrm{TiCl_4}$ .
Hydroxide solubility increases down the group, whereas sulfate solubility decreases: $\mathrm{Mg(OH)_2}$ and $\mathrm{MgSO_4}$ have medicinal uses, $\mathrm{Ca(OH)_2}$ treats acidic soil, CaO or $\mathrm{CaCO_3}$ removes flue-gas $\mathrm{SO_2}$ , and insoluble $\mathrm{BaSO_4}$ is a safe contrast medium.
Sulfate testing uses acidified $\mathrm{BaCl_2}$ .

Worked example

Describe a test for sulfate ions in a solution that may also contain carbonate ions, including the reason for acidification.

1.Add dilute hydrochloric acid to remove carbonate as $\mathrm{CO_2}$ and water.
2.Add aqueous barium chloride to the acidified sample.
3.A white $\mathrm{BaSO_4}$ precipitate confirms sulfate.

Answer: Acidification prevents a white barium carbonate precipitate from causing a false positive.

Common mistakes

Don't reverse the solubility trends and state that Group 2 sulfates become more soluble down the group.
Don't add barium chloride before acidifying, allowing carbonate to form a misleading white precipitate.
Don't suggest soluble toxic barium compounds for X-ray imaging instead of insoluble $\mathrm{BaSO_4}$ .

Exam tip. For a sulfate-test explanation, name the interfering carbonate ion and state how acid removes it.

3.2.3.1 · Trends in properties

Down Group 7, electronegativity decreases as atomic radius and shielding increase, while boiling point rises because larger $\mathrm{X_2}$ electron clouds form stronger London forces.
Halogen oxidising ability decreases, so a halogen displaces halide ions below it; halide reducing ability increases.
With concentrated sulfuric acid, chloride shows acid–base behaviour, bromide reduces sulfur mainly to $\mathrm{SO_2}$ and iodide can reduce it to sulfur or $\mathrm{H_2S}$ .
Acidified $\mathrm{AgNO_3}$ distinguishes halides: AgCl is white and dissolves in dilute ammonia, AgBr is cream and dissolves in concentrated ammonia, and AgI is yellow and insoluble.
Nitric acid avoids adding halide ions.

Worked example

Chlorine water is added to aqueous potassium bromide. State the observation and write the ionic equation.

1.Chlorine is the stronger oxidising agent and accepts electrons from bromide.
2.Bromide is oxidised to orange bromine while chlorine is reduced to chloride.

Answer: The solution becomes orange: $\mathrm{Cl_2+2Br^-\rightarrow2Cl^-+Br_2}$ .

Common mistakes

Don't state that halogen oxidising ability increases down the group, confusing it with halide reducing ability.
Don't use hydrochloric acid before silver nitrate and introduce chloride ions into the test.
Don't claim AgBr dissolves in dilute ammonia instead of requiring concentrated ammonia.

Exam tip. For a halide-identification question, give precipitate colour and ammonia solubility because colour alone may be ambiguous.

3.2.3.2 · Uses of chlorine and chlorate(I)

Chlorine disproportionates in water to chloride and chlorate(I): $\mathrm{Cl_2+H_2O\rightleftharpoons2H^++Cl^-+ClO^-}$ . In light, chlorine water can instead produce chloride ions and oxygen overall.
Cold, dilute aqueous NaOH gives $\mathrm{NaCl}$ and $\mathrm{NaClO}$ for bleaching and disinfection.
Chlorination kills pathogens and supplies residual protection; controlled health benefits outweigh toxic effects, though risks require assessment.
Decisions about adding chemicals such as fluoride also balance evidence.
Required practical 4 reliably identifies Group 2 and $\mathrm{NH_4^+}$ cations and halide, $\mathrm{OH^-}$ , $\mathrm{CO_3^{2-}}$ and $\mathrm{SO_4^{2-}}$ anions by test-tube reactions.

Worked example

Use oxidation states to show that chlorine reacting with water to form $\mathrm{Cl^-}$ and $\mathrm{ClO^-}$ is disproportionation.

1.Chlorine has oxidation state $0$ in $\mathrm{Cl_2}$ .
2.It changes to $-1$ in $\mathrm{Cl^-}$ and $+1$ in $\mathrm{ClO^-}$ .
3.The same element is both reduced and oxidised.

Answer: The reaction is disproportionation.

Common mistakes

Don't call chlorate(I) chlorine with oxidation state $-1$ instead of $+1$ .
Don't evaluate chlorination using toxicity alone and omit pathogen removal and residual disinfection.
Don't write hot concentrated alkali products when the specified condition is cold dilute aqueous NaOH.

Exam tip. For an 'assess' water-treatment question, link one quantified or chemical benefit to one risk before reaching a balanced conclusion.

3.2.4 · Properties of Period 3 elements and their oxides (A-level only)

Definition. Across Period 3 the elements Na, Mg, Al, Si, P and S react with oxygen to form Na₂O, MgO, Al₂O₃, SiO₂, P₄O₁₀, SO₂ and SO₃. The bonding in the oxides changes from ionic on the left, through amphoteric Al₂O₃, to covalent on the right, and this controls both the melting points and the pH of the solutions formed with water.

Melting points must be explained by structure and bonding, never by 'stronger bonds' alone. Na₂O and MgO are ionic with giant lattices, so their melting points are high, and MgO is higher than Na₂O because Mg²⁺ has a greater charge and a smaller radius, giving stronger electrostatic attraction.
Al₂O₃ is ionic with covalent character. SiO₂ is macromolecular, or giant covalent, so melting it breaks many strong covalent bonds and its melting point is very high.
P₄O₁₀, SO₂ and SO₃ are simple molecular, so only weak van der Waals forces between molecules are overcome and their melting points are low. Sodium reacts vigorously with cold water to give sodium hydroxide and hydrogen; magnesium reacts very slowly with cold water but rapidly with steam to give magnesium oxide and hydrogen.
Na₂O dissolves to give NaOH at about pH 13-14; MgO is sparingly soluble and gives roughly pH 9; Al₂O₃ and SiO₂ are insoluble and give pH 7, though Al₂O₃ is amphoteric, dissolving in both acids and bases; P₄O₁₀ gives phosphoric(V) acid H₃PO₄ at pH 0-1; and SO₂ and SO₃ give sulfurous and sulfuric acids, SO₃ giving the lower pH.
The mark scheme is strict about attributing structures to the oxides rather than to the elements.

Across Period 3, oxide structure and acid–base character change from mainly ionic and basic to molecular and acidic.

Worked example

Classify $\mathrm{Al_2O_3}$ and $\mathrm{SiO_2}$ , then give one reaction of each with hydroxide ions.

1. $\mathrm{Al_2O_3}$ is amphoteric and forms tetrahydroxoaluminate with base.
2. $\mathrm{SiO_2}$ is acidic and forms silicate with base.

Answer: $\mathrm{Al_2O_3+2OH^-+3H_2O\rightarrow2[Al(OH)_4]^-}$ ; $\mathrm{SiO_2+2OH^-\rightarrow SiO_3^{2-}+H_2O}$ .

Mark-scheme wording to use

“M₁ SiO₂ is macromolecular, giant covalent or a giant molecule; do not allow only giant, giant atomic or giant ionic. M₂ strong covalent bonds between atoms require much energy to break” — AQA 7405, November 2021 Paper 1, Q03.3
“If the structures are attributed to the elements rather than the oxides, M₁ and M₂ are lost” — AQA 7405, June 2018 Paper 3, Q02.4
“P₄O₁₀ + 6H₂O → 4H₃PO₄. Allow multiples, fractions and an ionic form of the acid products. Do not allow P₂O₅ as the phosphorus(V) oxide formula” — AQA 7405, November 2020 Paper 3, Q02.2

What the examiner wants. 'Macromolecular' or 'giant covalent' in full — 'giant' alone is rejected — and covalent bonds described as being between ATOMS within SiO₂, never between molecules.

Common mistakes

Don't call $\mathrm{Al_2O_3}$ only basic and omit its reaction with bases.
Don't explain the high melting point of molecular $\mathrm{P_4O_{10}}$ by breaking covalent bonds rather than intermolecular forces.
Don't state that sulfur burns directly to $\mathrm{SO_3}$ as the principal product instead of $\mathrm{SO_2}$ .

Exam tip. For an oxide-trend question, pair every melting-point explanation with the correct structure and bonding.

3.2.5.1 · General properties of transition metals (A-level only)

For Ti to Cu, transition-metal characteristics arise from an incomplete d sub-level in atoms or ions. These characteristics are complex formation, coloured ions, variable oxidation states and catalytic activity.
A ligand is a molecule or ion that donates an electron pair to the central metal atom or ion, forming a co-ordinate bond.
A complex consists of that central metal surrounded by ligands.
Co-ordination number counts co-ordinate bonds to the centre, not the number of ligand particles: a bidentate ligand supplies two donor atoms.
Applying the definition excludes any d-block element whose relevant ions are only $d^0$ or $d^{10}$ .

Worked example

For $\mathrm{[Cr(C_2O_4)_3]^{3-}}$ , determine chromium's oxidation state and co-ordination number. Ethanedioate is bidentate.

1.Three $\mathrm{C_2O_4^{2-}}$ ligands contribute $-6$ , so $x-6=-3$ and $x=+3$ .
2.Three bidentate ligands each form two co-ordinate bonds.

Answer: Chromium is in oxidation state $+3$ and the co-ordination number is $6$ .

Common mistakes

Don't define every d-block element as a transition metal without checking for an incomplete d sub-level in an ion.
Don't count three bidentate ligands as co-ordination number three instead of six.
Don't describe a ligand as accepting a lone pair from the metal rather than donating one.

Exam tip. For a complex-ion calculation, determine oxidation state from charge separately from co-ordination number.

3.2.5.2 · Substitution reactions (A-level only)

$\mathrm{H_2O}$ , $\mathrm{NH_3}$ and $\mathrm{Cl^-}$ are monodentate ligands. Similar-sized uncharged water and ammonia exchange without changing co-ordination number, although substitution may stop at $\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}$ .
Larger chloride can replace water with a co-ordination-number change.
Ethane-1,2-diamine and ethanedioate are bidentate; EDTA is multidentate.
Their replacement of monodentate ligands is the chelate effect: metal–ligand bond enthalpy changes broadly balance, while producing more particles makes $\Delta S$ positive.
Haem is an Fe(II) multidentate complex that binds oxygen; carbon monoxide is toxic because it substitutes for oxygen and prevents transport.

Worked example

Explain why $\mathrm{[M(H_2O)_6]^{2+}+3en\rightleftharpoons[M(en)_3]^{2+}+6H_2O}$ favours the chelate complex.

1.Six metal–oxygen bonds are replaced by six metal–nitrogen bonds, so the enthalpy change is relatively small.
2.Four reactant particles form seven product particles.
3.The particle increase gives positive $\Delta S$ , making $-T\Delta S$ favourable.

Answer: The positive entropy contribution favours the chelated product.

Common mistakes

Don't explain the chelate effect only as stronger bonds and omit the increase in particle number and entropy.
Don't assume replacing water with larger chloride must leave the co-ordination number unchanged.
Don't state that carbon monoxide destroys haem rather than replacing co-ordinated oxygen at Fe(II).

Exam tip. For a chelate-effect explanation, count particles on both sides and compare the numbers and types of metal–ligand bonds.

3.2.5.3 · Shapes of complex ions (A-level only)

Small ligands such as water and ammonia commonly form six-coordinate octahedral complexes; larger chloride ligands commonly form four-coordinate tetrahedral complexes.
Four-coordinate complexes may instead be square planar, while $\mathrm{[Ag(NH_3)_2]^+}$ in Tollens' reagent is linear and two-coordinate.
Octahedral complexes with suitable monodentate ligands and square-planar complexes can show cis–trans isomerism; cisplatin is the cis form of $\mathrm{[Pt(NH_3)_2Cl_2]}$ .
Octahedral complexes with bidentate ligands can show optical isomerism when their two arrangements are non-superimposable mirror images.
Drawings must preserve three-dimensional geometry, not merely change page orientation.

Square-planar cis and trans platinum complexes contrast with the linear silver complex in Tollens' reagent.

Worked example

Describe the two stereoisomers of square-planar $\mathrm{[Pt(NH_3)_2Cl_2]}$ and identify cisplatin.

1.Place all four ligands in one plane around platinum.
2.Adjacent chloride ligands give the cis isomer; opposite chloride ligands give the trans isomer.

Answer: Cisplatin is the cis isomer with adjacent chloride ligands.

Common mistakes

Don't draw a tetrahedral arrangement for cisplatin instead of square planar.
Don't call two rotated drawings optical isomers without checking that they are non-superimposable mirror images.
Don't assign $\mathrm{[Ag(NH_3)_2]^+}$ an octahedral shape despite co-ordination number two.

Exam tip. For a complex-shape question, state both co-ordination number and shape before drawing stereoisomers.

3.2.5.4 · Formation of coloured ions (A-level only)

A transition-metal ion appears coloured because it absorbs selected visible wavelengths and transmits or reflects the remainder. Absorption promotes a d electron from a ground state to an excited state, with the energy gap $\Delta E=h\nu=hc/\lambda$ .
Changing ligand, oxidation state or co-ordination number changes this gap and therefore the absorbed wavelength and observed colour.
Spectroscopy uses visible absorption quantitatively.
In colorimetry, a blank sets zero, a suitable filter or wavelength maximises useful absorption, and standards provide an absorbance–concentration calibration graph.
An unknown concentration is interpolated only within the calibrated range, with any preparation dilution reversed afterwards.

A calibration line converts the measured absorbance of an unknown into concentration.

Worked example

A d-electron energy gap is $3.70\times10^{-19}\,\mathrm{J}$ . Calculate the absorbed wavelength using $h=6.63\times10^{-34}\,\mathrm{J\,s}$ and $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ .

1.Rearrange $\Delta E=hc/\lambda$ to $\lambda=hc/\Delta E$ .
2.Substitute the three values with SI units.
3.Convert metres to nanometres.

Answer: $\lambda=5.38\times10^{-7}\,\mathrm{m}=538\,\mathrm{nm}$ .

Common mistakes

Don't say the observed colour is the wavelength absorbed rather than the light transmitted or reflected.
Don't attribute every colour change to concentration instead of considering ligand, oxidation state and co-ordination number.
Don't divide by a dilution factor when recovering the original concentration instead of multiplying.

Exam tip. For a colorimetry calculation, read the diluted concentration from the calibration line before applying the stated dilution factor.

3.2.5.5 · Variable oxidation states (A-level only)

Transition elements show variable oxidation states. Zinc reduces vanadate(V) in acid through vanadium(IV), vanadium(III) and vanadium(II), producing characteristic colour changes.
A transition-metal redox potential depends on pH and ligand because these conditions change the relative stability of oxidation states.
Tollens' reagent, $\mathrm{[Ag(NH_3)_2]^+}$ , is reduced to metallic silver by an aldehyde and distinguishes aldehydes from ketones.
Quantitative redox titrations include $\mathrm{Fe^{2+}}$ and $\mathrm{C_2O_4^{2-}}$ with acidified $\mathrm{MnO_4^-}$ ; balanced half-equations determine ratios used for iron-tablet, steel, hydrated-salt, ethanedioic-acid or hydrogen-peroxide calculations.
Accurate aliquot scaling is essential.

Worked example

$20.0\,\mathrm{cm^3}$ of $\mathrm{Fe^{2+}}$ solution requires $16.40\,\mathrm{cm^3}$ of $0.00250\,\mathrm{mol\,dm^{-3}}$ acidified $\mathrm{MnO_4^-}$ . Calculate $[\mathrm{Fe^{2+}}]$ .

1.Moles $\mathrm{MnO_4^-}=0.00250\times0.01640=4.10\times10^{-5}$ .
2. $\mathrm{MnO_4^-{:}Fe^{2+}}=1{:}5$ , so moles $\mathrm{Fe^{2+}}=2.05\times10^{-4}$ .
3.Divide by $0.0200\,\mathrm{dm^3}$ .

Answer: $[\mathrm{Fe^{2+}}]=1.03\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ .

Common mistakes

Don't use a $1{:}1$ manganate(VII)-to-iron(II) ratio instead of the balanced $1{:}5$ ratio.
Don't describe Tollens' reagent as oxidising ketones to give a silver mirror.
Don't treat electrode potential as independent of pH and ligand for transition-metal couples.

Exam tip. In a redox-titration calculation, write the balanced ionic equation or explicit mole ratio before scaling an aliquot.

3.2.5.6 · Catalysts (A-level only)

A heterogeneous catalyst is in a different phase from reactants; reaction occurs at surface active sites through adsorption, reaction and desorption. A support maximises surface area and reduces cost, while poisons block sites and lower efficiency.
Iron catalyses the Haber process and $\mathrm{V_2O_5}$ the Contact process.
A homogeneous catalyst shares the reactant phase and forms an intermediate.
Variable oxidation states allow $\mathrm{Fe^{2+}/Fe^{3+}}$ to catalyse iodide–peroxodisulfate through two electron-transfer steps and $\mathrm{Mn^{2+}}$ to autocatalyse ethanedioate–manganate(VII) through manganese intermediates.
Catalyst equations must show the intermediate oxidation state and regenerate the catalyst so it cancels from the overall reaction.

A catalyst supplies an alternative route with lower activation energy while leaving reactant and product energies unchanged.

Worked example

Show with two equations how $\mathrm{V_2O_5}$ catalyses the Contact-process oxidation of $\mathrm{SO_2}$ .

1. $\mathrm{V_2O_5+SO_2\rightarrow V_2O_4+SO_3}$ reduces the catalyst.
2. $\mathrm{V_2O_4+\tfrac12O_2\rightarrow V_2O_5}$ regenerates it.
3.Adding the steps cancels both vanadium oxides.

Answer: Overall: $\mathrm{SO_2+\tfrac12O_2\rightarrow SO_3}$ .

Common mistakes

Don't include the catalyst in the overall equation instead of cancelling regenerated species.
Don't describe a catalyst poison as raising activation energy rather than blocking active sites.
Don't call $\mathrm{Mn^{2+}}$ an added catalyst in the ethanedioate reaction although it forms as a product and autocatalyses.

Exam tip. For a catalytic-cycle question, add the elementary equations and demonstrate explicitly that catalyst and intermediate cancel.

3.2.6 · Reactions of ions in aqueous solution (A-level only)

Definition. In aqueous solution transition and Group 3 metals form metal-aqua ions: [M(H₂O)₆]²⁺ limited to Fe and Cu, and [M(H₂O)₆]³⁺ limited to Al and Fe. The acidity of [M(H₂O)₆]³⁺ is greater than that of [M(H₂O)₆]²⁺ because the 3+ ion has a higher charge to size ratio.

The higher charge density of the 3+ ion polarises the O-H bonds of the coordinated water more strongly, so a proton is released more readily and the solution is more acidic. Test-tube reactions with three bases are examinable, and each behaves differently.
Sodium hydroxide and ammonia both act as bases and remove protons, giving the neutral hydroxide precipitate M(H₂O)₃(OH)₃ or M(H₂O)₄(OH)₂: Fe(II) gives a green precipitate, Cu(II) a blue precipitate, Fe(III) a brown precipitate and Al(III) a white one. In excess sodium hydroxide only the aluminium hydroxide dissolves, because it is amphoteric and forms [Al(OH)₄]⁻;
in excess ammonia only the copper hydroxide dissolves, forming the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺ by ligand substitution.
Sodium carbonate distinguishes the 2+ ions from the 3+ ions: the weakly acidic 2+ aqua ions merely form the metal carbonate precipitate, whereas the strongly acidic 3+ aqua ions are acidic enough to release carbon dioxide, so effervescence accompanies the hydroxide precipitate.
Every equation must be balanced with the full complex ion formulae, and every answer needs a colour and a state. This is Required practical 11.

Selected transition-metal aqua ions form characteristic hydroxide precipitates with aqueous hydroxide.

Worked example

Predict the different reactions of aqueous carbonate with $\mathrm{[Fe(H_2O)_6]^{2+}}$ and $\mathrm{[Fe(H_2O)_6]^{3+}}$ .

1.The $2+$ ion forms green $\mathrm{FeCO_3}$ precipitate without gas.
2.The more acidic $3+$ aqua ion transfers protons to carbonate.
3.This produces brown iron(III) hydroxide and effervescence of $\mathrm{CO_2}$ .

Answer: $\mathrm{Fe^{2+}}$ : green carbonate precipitate; $\mathrm{Fe^{3+}}$ : brown hydroxide precipitate plus carbon dioxide.

Mark-scheme wording to use

“M₁ [Al(H₂O)₆]₃+ + 3NH₃ → Al(H₂O)₃(OH)₃ + 3NH₄+. M₂ white precipitate or white solid. Do not accept effervescence or that the precipitate dissolves in excess NH₃” — AQA 7405, June 2023 Paper 1, Q07.5
“M₁ a blue-green, green-blue or green precipitate or solid; do not accept blue” — AQA 7405, June 2025 Paper 1, Q03.3
“M₁ white precipitate/solid. M₂ effervescence/bubbles/fizzing. M₃ balanced equation; accept multiples and balanced spectator-ion versions” — AQA 7405, June 2018 Paper 1, Q07.3

What the examiner wants. A colour AND a state for every observation, and the exact behaviour in excess reagent. The schemes are colour-strict — 'blue' is rejected where 'blue-green' is required — and dissolving in excess is credited only for the right cation and the right reagent.

Common mistakes

Don't predict a stable iron(III) carbonate precipitate instead of hydroxide and carbon dioxide.
Don't state that $2+$ aqua ions are more acidic because they have lower charge.
Don't claim excess aqueous ammonia dissolves aluminium hydroxide in the same way as excess hydroxide.

Exam tip. For an ion-identification question, give reagent, initial precipitate colour and any change in excess reagent.