3.2 Inorganic chemistry — revision question pack

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

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.

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Answer all questions in the spaces provided.

3.2.1.1 · Classification

Explanation

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.

Tier 1 · Easy

An element has the electron configuration $\mathrm{[Ne]3s^2}$ . State its block in the Periodic Table.
[1 mark]
Total for this question: 1
Element A is in Period 3 and Group 4(14) of the Periodic Table. State the block containing A.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Element X has proton number $34$ and electron configuration $\mathrm{[Ar]3d^{10}4s^2 4p^4}$ . Identify the block to which X belongs and give a reason.
[2 marks]
Total for this question: 2
Element B is the second p-block element in Period 3. Deduce the proton number of B and identify the element.
[2 marks]
Total for this question: 2
An incomplete Period 4 row shows potassium, calcium, ten blank positions and then gallium. Determine the proton-number range covered by the ten blanks, identify their block, and explain why that block contains ten positions.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Elements Y and Z have consecutive proton numbers. Y has configuration $\mathrm{[Ar]4s^2}$ and Z has configuration $\mathrm{[Ar]3d^1 4s^2}$ . State the block of each element and explain why the classifications differ even though both contain $4s$ electrons.
[3 marks]
Total for this question: 3
A student assigns zinc to the s block because the $\mathrm{Zn^{2+}}$ ion has configuration $\mathrm{[Ar]3d^{10}}$ and therefore has no partly filled d subshell. Diagnose the error and state the correct block for zinc.
[3 marks]
Total for this question: 3
A student proposes two rules for assigning a block: (A) use the occupied subshell containing the most electrons; (B) use the subshell written last in an electron configuration. Use $\mathrm{[Ar]3d^{10}4s^2 4p^1}$ to test rule A and $\mathrm{[Ar]3d^2 4s^2}$ to test rule B. Give the correct classifications and rule.
[5 marks]
Total for this question: 5
A long-form Periodic Table shows element 57 followed by fourteen positions numbered 58 to 71 before element 72. The differentiating electrons across positions 58 to 71 enter $4f$ , while the differentiating electron of element 72 enters $5d$ . Deduce the block containing positions 58 to 71, explain why it has fourteen positions, and determine the block of element 72.
[5 marks]
Total for this question: 5
Ions $\mathrm{P^{2+}}$ and $\mathrm{Q^{3+}}$ are isoelectronic and both have configuration $\mathrm{[Ar]3d^5}$ . Determine the proton number, identity and block of P and Q, then justify why the block is the same for both elements.
[6 marks]
Total for this question: 6

3.2.1.2 · Physical properties of Period 3 elements

Explanation

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.

Tier 1 · Easy

Arrange Na, Mg and Cl in order of decreasing atomic radius and explain the order.
[3 marks]
Total for this question: 3
A Period 3 element is a monatomic gas at room temperature and has the lowest melting point in the period. Identify the element and explain the low melting point.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Explain why the first ionisation energy falls from Mg to Al and also falls from P to S, despite the general increase across Period 3.
[4 marks]
Total for this question: 4
Four samples are Na, Mg, Al and Si. Their properties are: W is a good conductor with relative hardness 0.5; X is a good conductor with relative hardness 2.5; Y is a good conductor with relative hardness 2.8; Z is a poor conductor with relative hardness 6.5. Deduce the order W, X, Y and Z in increasing proton number and identify each sample.
[5 marks]
Total for this question: 5
Explain why metallic bonding generally becomes stronger from sodium to aluminium across Period 3. Refer to ion charge, ion size and delocalised electrons.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Three Period 3 elements have melting points of approximately $1410\,{}^\circ\mathrm{C}$ , $115\,{}^\circ\mathrm{C}$ and $-189\,{}^\circ\mathrm{C}$ . Identify the elements as Si, S or Ar and explain the large differences.
[6 marks]
Total for this question: 6
A student says that melting point rises continuously from Na to Si because metallic bonding becomes stronger. Explain what is correct and what is wrong in this statement, referring to Na, Al, Si and P.
[5 marks]
Total for this question: 5
An unknown Period 3 element has metallic radius $0.160\,\mathrm{nm}$ , melting point $650\,{}^\circ\mathrm{C}$ and first ionisation energy $738\,\mathrm{kJ\,mol^{-1}}$ . Identify the element and justify your answer using at least two of these properties.
[5 marks]
Total for this question: 5
Explain why the first ionisation energy of aluminium is higher than that of sodium although it is lower than that of magnesium.
[5 marks]
Total for this question: 5
Period 3 element X has a smaller atomic radius than Y but a lower first ionisation energy. Deduce two possible identity pairs (X, Y) and explain the pair in which both elements lose an electron from the same subshell.
[5 marks]
Total for this question: 5

3.2.2 · Group 2, the alkaline earth metals

Explanation

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.

Tier 1 · Easy

Steam is passed over heated magnesium ribbon. Give the chemical equation and the visible change.
[2 marks]
Total for this question: 2
Compare the reactions of magnesium with cold water and with steam. Give both equations and the visible observations.
[5 marks]
Total for this question: 5

Tier 2 · Standard

A solution may contain both sulfate and carbonate ions. Describe how to test specifically for sulfate ions using barium chloride solution, including the reason for the first reagent and the final observation.
[4 marks]
Total for this question: 4
Design a comparison that demonstrates that magnesium hydroxide is less soluble than barium hydroxide. Give the variables to control, the measurement to make and the expected result.
[5 marks]
Total for this question: 5
Give a different Group 2 compound for each use: neutralising excess stomach acid, treating acidic soil, and providing an X-ray contrast medium. For the contrast medium, explain why the chosen barium compound is safe to swallow.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A flue-gas stream contains $1.28\times10^3\,\mathrm{mol}$ of $\mathrm{SO_2}$ . A power station uses limestone to capture $85.0\%$ of this amount in a $1{:}1$ mole ratio of $\mathrm{CaCO_3}$ to $\mathrm{SO_2}$ . The limestone is $92.0\%$ $\mathrm{CaCO_3}$ by mass. Calculate the mass of limestone required. Use $M_r(\mathrm{CaCO_3})=100.1$ .
[5 marks]
Total for this question: 5
A student is given one colourless solution known to contain either $\mathrm{Mg^{2+}}$ or $\mathrm{Ba^{2+}}$ . Design the shortest reliable test using one added reagent. State a suitable control, the two possible observations and the ionic equation for a positive result.
[6 marks]
Total for this question: 6
A solution contains a total of $0.0800\,\mathrm{mol}$ of $\mathrm{Mg^{2+}}$ and $\mathrm{Ba^{2+}}$ . Excess sulfate ions produce $9.33\,\mathrm{g}$ of dry precipitate. Calculate the percentage of the original Group 2 ions that were $\mathrm{Ba^{2+}}$ and write the precipitation equation with state symbols. Use $M_r(\mathrm{BaSO_4})=233.4$ .
[5 marks]
Total for this question: 5
Calcium and barium are added separately to cold water. Write an equation with state symbols for each reaction and contrast the observations.
[4 marks]
Total for this question: 4
Titanium is extracted by $\mathrm{TiCl_4}$ reacting with magnesium. A batch uses $12.0\,\mathrm{kg}$ of $\mathrm{TiCl_4}$ and the process requires $8.0\%$ excess magnesium. The magnesium feed is $95.0\%$ pure by mass. Calculate the mass of magnesium feed required, given $A_r(\mathrm{Ti})=47.9$ , $A_r(\mathrm{Cl})=35.5$ and $A_r(\mathrm{Mg})=24.3$ .
[6 marks]
Total for this question: 6

3.2.3.1 · Trends in properties

Explanation

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.

Tier 1 · Easy

State and explain the trend in electronegativity down Group 7 from chlorine to iodine.
[2 marks]
Total for this question: 2
Acidified silver nitrate gives a pale precipitate whose colour cannot be judged reliably as white or cream. State the two halide ions that this single result cannot distinguish and name the follow-up step that distinguishes them.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Three colourless solutions contain chloride, bromide and iodide ions, one ion per solution. Describe a test that distinguishes all three, including every precipitate colour and its behaviour with ammonia.
[6 marks]
Total for this question: 6
A solution contains bromide ions. A student acidifies it with dilute hydrochloric acid before adding silver nitrate solution. State the precipitate observations and explain what went wrong.
[4 marks]
Total for this question: 4
Explain why the boiling points increase from $\mathrm{Cl_2}$ to $\mathrm{Br_2}$ to $\mathrm{I_2}$ . Your answer must refer to electron clouds and the intermolecular attraction involved.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Solid sodium iodide is treated with concentrated sulfuric acid. Give the initial acid-base equation, then an equation in which iodide reduces sulfuric acid to hydrogen sulfide. State two observations from the redox reaction and explain why iodide gives deeper reduction than bromide.
[7 marks]
Total for this question: 7
Three solid sodium halides are each warmed separately with concentrated sulfuric acid. Sample 1 gives only misty fumes. Sample 2 gives misty fumes and a red-brown vapour. Sample 3 gives a purple vapour and a gas that smells of rotten eggs. Identify the halide in each sample and explain, in terms of ion size and shielding, why the extent of reduction differs.
[6 marks]
Total for this question: 6
Halogen solutions P, Q and R are $\mathrm{Cl_2}$ , $\mathrm{Br_2}$ and $\mathrm{I_2}$ in an unknown order. Halide solutions X, Y and Z are $\mathrm{Cl^-}$ , $\mathrm{Br^-}$ and $\mathrm{I^-}$ in an unknown order. P reacts with Y and Z but not X; Q reacts only with Z; R reacts with none. Deduce all six identities and write the ionic equation for the reaction between Q and Z.
[6 marks]
Total for this question: 6
In the test for halide ions, explain why silver nitrate solution is the reagent used and why the mixture is acidified, and explain why neither hydrochloric acid nor sulfuric acid can be used for the acidification.
[4 marks]
Total for this question: 4
A $0.0200\,\mathrm{mol}$ mixture of sodium chloride and sodium bromide is dissolved, acidified with nitric acid and treated with excess silver nitrate. The dry precipitate is shaken with dilute ammonia, then filtered and dried. The remaining solid has mass $1.93\,\mathrm{g}$ . Calculate the amount and mole percentage of each halide in the original mixture and give the mass of the initial silver-halide precipitate, given $M_r(\mathrm{AgCl})=143.4$ and $M_r(\mathrm{AgBr})=187.8$ .
[5 marks]
Total for this question: 5

3.2.3.2 · Uses of chlorine and chlorate(I)

Explanation

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.

Tier 1 · Easy

Write the equation for the reaction of chlorine with cold, dilute aqueous sodium hydroxide and state one use of the solution formed.
[2 marks]
Total for this question: 2
Chlorine water is left in bright light. State the visible change and write the ionic equation for the reaction that produces oxygen.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Use oxidation states to show that the formation of chloride and chlorate(I) ions when chlorine reacts with water is disproportionation.
[4 marks]
Total for this question: 4
Evaluate the use of chlorine in drinking-water treatment. Give two benefits and one risk, then state why chlorination is used despite that risk.
[4 marks]
Total for this question: 4
For $\mathrm{Cl_2+H_2O\rightleftharpoons2H^++Cl^-+ClO^-}$ , predict how adding dilute acid to one sample of chlorine water and adding aqueous sodium hydroxide to another affect the equilibrium amount of $\mathrm{ClO^-}$ . State what happens to the equilibrium constant if temperature is unchanged.
[4 marks]
Total for this question: 4

Tier 3 · Hard

$1.42\,\mathrm{g}$ of chlorine reacts completely with excess cold, dilute aqueous sodium hydroxide. The final solution has volume $250\,\mathrm{cm^3}$ . Calculate the concentration of $\mathrm{NaClO}$ formed. Use $M_r(\mathrm{Cl_2})=71.0$ .
[4 marks]
Total for this question: 4
Required practical 4: two colourless solutions contain $\mathrm{CO_3^{2-}}$ and $\mathrm{OH^-}$ , one ion in each. Design the shortest test using dilute hydrochloric acid and limewater. Include a reagent blank, every observation and both ionic equations.
[7 marks]
Total for this question: 7
Required practical 4: a solution known to contain ammonium ions is tested. A student adds aqueous sodium hydroxide at room temperature and holds dry red litmus paper in the gas space, then reports that no ammonium ions are present. Give the correct procedure and observation, write the ionic equation, and explain the two procedural faults.
[6 marks]
Total for this question: 6
A treatment works processes $2.50\times10^6\,\mathrm{dm^3}$ of water each day and adds chlorine at $0.800\,\mathrm{mg\,dm^{-3}}$ . Reactions with impurities consume $65.0\%$ of the added chlorine. A residual concentration of at least $0.250\,\mathrm{mg\,dm^{-3}}$ is needed for disinfection, while concentrations above $0.400\,\mathrm{mg\,dm^{-3}}$ exceed the works' by-product control limit. Calculate the daily chlorine mass and residual concentration, then evaluate this dose against both limits.
[5 marks]
Total for this question: 5
Write separate reduction and oxidation half-equations in alkaline solution for chlorine reacting with cold, dilute aqueous sodium hydroxide. Write the overall equation with electrons cancelled, then determine the mole ratio of chloride to chlorate(I) formed.
[5 marks]
Total for this question: 5

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

Explanation

Na and Mg react with water.
Burning Period 3 elements gives the specified $\mathrm{Na_2O}$ , MgO, $\mathrm{Al_2O_3}$ , $\mathrm{SiO_2}$ , $\mathrm{P_4O_{10}}$ , $\mathrm{SO_2}$ and $\mathrm{SO_3}$ chemistry.
Highest-oxide melting points reflect a change from giant ionic lattices through giant covalent $\mathrm{SiO_2}$ to molecular phosphorus and sulfur oxides.
With water, ionic oxides form alkaline solutions while $\mathrm{P_4O_{10}}$ , $\mathrm{SO_2}$ and $\mathrm{SO_3}$ form $\mathrm{H_3PO_4}$ , $\mathrm{H_2SO_3}$ and $\mathrm{H_2SO_4}$ and their phosphate, sulfite and sulfate anions.
Basic oxides react with acids, acidic oxides with bases, and amphoteric $\mathrm{Al_2O_3}$ with both.

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

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.

Tier 1 · Easy

Write equations for the formation from the elements of magnesium oxide and phosphorus(V) oxide, $\mathrm{P_4O_{10}}$ .
[2 marks]
Total for this question: 2
A Period 3 oxide dissolves in water to give a colourless solution of pH 13. Identify the oxide as $\mathrm{Na_2O}$ , $\mathrm{MgO}$ or $\mathrm{SiO_2}$ and write the equation for its reaction with water.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Explain why $\mathrm{Al_2O_3}$ and $\mathrm{SiO_2}$ have high melting points but $\mathrm{P_4O_{10}}$ has a much lower melting point.
[5 marks]
Total for this question: 5
Two white solids are $\mathrm{Al_2O_3}$ and $\mathrm{SiO_2}$ . Suggest one reagent that distinguishes them, state both observations and write the ionic equation for the reaction that occurs.
[5 marks]
Total for this question: 5
Oxides A, B and C are $\mathrm{MgO}$ , $\mathrm{SiO_2}$ and $\mathrm{SO_3}$ in an unknown order. A has a high melting point and conducts electricity when molten. B has a high melting point and has no mobile charged particles. C has a low melting point and consists of discrete molecules. Identify A, B and C and explain A's molten conductivity.
[4 marks]
Total for this question: 4

Tier 3 · Hard

For each oxide $\mathrm{Al_2O_3}$ , $\mathrm{SiO_2}$ and $\mathrm{SO_3}$ , state whether it is basic, amphoteric or acidic. Give one ionic equation showing the reaction of $\mathrm{Al_2O_3}$ with a base, one equation for $\mathrm{SiO_2}$ with a base, and the equation for $\mathrm{SO_3}$ with water.
[7 marks]
Total for this question: 7
An unnamed Period 3 element X burns in oxygen to form a white oxide. The oxide is sparingly soluble in water. Predict the approximate pH of the resulting solution, deduce the identity of X, and write equations for the combustion and the reaction of the oxide with water.
[6 marks]
Total for this question: 6
$0.620\,\mathrm{g}$ of a Period 3 element E, with $A_r=31.0$ , gains $0.800\,\mathrm{g}$ when burned completely in excess oxygen. Deduce the empirical formula of the oxide, identify E and give the molecular formula of the oxide. State its structure type and write its reaction with water.
[6 marks]
Total for this question: 6
A $2.81\,\mathrm{g}$ mixture contains only $\mathrm{Na_2O}$ and $\mathrm{MgO}$ . It reacts exactly with a hydrochloric acid portion measuring $100.0\,\mathrm{cm^3}$ ; the acid concentration is $1.00\,\mathrm{mol\,dm^{-3}}$ . Calculate the percentage by mass of $\mathrm{Na_2O}$ in the mixture, given $M_r(\mathrm{Na_2O})=62.0$ and $M_r(\mathrm{MgO})=40.3$ .
[6 marks]
Total for this question: 6
Separate $0.0120\,\mathrm{mol}$ samples of $\mathrm{P_4O_{10}}$ and $\mathrm{SO_3}$ react completely with excess water. Each resulting acid is then neutralised completely by aqueous sodium hydroxide. Write the water-reaction and complete-neutralisation equations for both oxides, and calculate the ratio of sodium hydroxide amounts required for the phosphorus oxide sample and the sulfur oxide sample.
[5 marks]
Total for this question: 5

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

Explanation

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.

Tier 1 · Easy

Define the term ligand.
[2 marks]
Total for this question: 2
Which of $\mathrm{NH_3}$ , $\mathrm{CH_4}$ , $\mathrm{H_2O}$ and $\mathrm{BF_3}$ can act as ligands? Explain your choices.
[4 marks]
Total for this question: 4

Tier 2 · Standard

For the complex ion $\mathrm{[Cr(C_2O_4)_3]^{3-}}$ , determine the oxidation state of chromium and the co-ordination number. Each ethanedioate ion is bidentate.
[3 marks]
Total for this question: 3
A vanadium(III) complex has co-ordination number 6 and contains four water ligands and one bidentate ethanedioate ion, $\mathrm{C_2O_4^{2-}}$ . Deduce the formula and overall charge of the complex.
[4 marks]
Total for this question: 4
Metal X forms a blue $\mathrm{X^{2+}}$ ion, a green $\mathrm{X^{3+}}$ ion and the complex $\mathrm{[X(NH_3)_6]^{2+}}$ . It also speeds up a reaction without being consumed. State four characteristic transition-metal properties evidenced by these observations.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Sc, Fe and Zn are d-block elements. Their common ions include $\mathrm{Sc^{3+}}$ with $d^0$ , $\mathrm{Fe^{2+}}$ with $d^6$ , and $\mathrm{Zn^{2+}}$ with $d^{10}$ . Use the definition of a transition metal to identify which of these three is a transition metal and justify every exclusion.
[4 marks]
Total for this question: 4
Copper forms $\mathrm{Cu^+}$ with configuration $d^{10}$ and $\mathrm{Cu^{2+}}$ with configuration $d^9$ . A student argues that copper is not a transition metal because one of its ions has a full d subshell. Evaluate this argument using the definition of a transition metal.
[3 marks]
Total for this question: 3
In the complex $\mathrm{[M(L)_2Cl_2]^+}$ , M is in oxidation state $+3$ and L is bidentate. Deduce the charge on L, explain why co-ordination number 6 does not mean that six ligand particles are present, and state the shape of the complex.
[5 marks]
Total for this question: 5
Chromium atoms have configuration $\mathrm{[Ar]3d^5 4s^1}$ and manganese atoms have configuration $\mathrm{[Ar]3d^5 4s^2}$ . Determine the configurations of $\mathrm{Cr^{2+}}$ and $\mathrm{Mn^{2+}}$ , and justify whether each element is a transition metal.
[5 marks]
Total for this question: 5
A cobalt complex contains $\mathrm{Co^{3+}}$ , four $\mathrm{NH_3}$ ligands and one $\mathrm{C_2O_4^{2-}}$ ligand. Deduce its overall charge, number of ligand particles, co-ordination number and shape.
[5 marks]
Total for this question: 5

3.2.5.2 · Substitution reactions (A-level only)

Explanation

$\mathrm{H_2O}$ , $\mathrm{NH_3}$ 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.

Tier 1 · Easy

State whether each of $\mathrm{NH_3}$ , ethane-1,2-diamine and EDTA is a monodentate, bidentate or multidentate ligand.
[3 marks]
Total for this question: 3
Explain why carbon monoxide is toxic by referring to ligand substitution in haemoglobin.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write an equation for the reaction of $\mathrm{[Cu(H_2O)_6]^{2+}}$ with excess ammonia and explain why the co-ordination number is unchanged.
[4 marks]
Total for this question: 4
Concentrated hydrochloric acid is added to a pink solution of $\mathrm{[Co(H_2O)_6]^{2+}}$ . Write the ligand-substitution equation, state the final solution colour and give the change in co-ordination number.
[5 marks]
Total for this question: 5
For $\mathrm{[Co(H_2O)_6]^{2+}+4Cl^-\rightleftharpoons[CoCl_4]^{2-}+6H_2O}$ , predict the colour change when the blue equilibrium mixture is diluted with water. Explain the shift and state whether the equilibrium constant changes at constant temperature.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The substitution $\mathrm{[M(H_2O)_6]^{2+}+3en\rightleftharpoons [M(en)_3]^{2+}+6H_2O}$ is strongly favoured, where en is neutral ethane-1,2-diamine. Explain the chelate effect using enthalpy, entropy and particle numbers.
[5 marks]
Total for this question: 5
A neutral ligand L undergoes the substitution $\mathrm{[Co(H_2O)_6]^{3+}+2L\rightleftharpoons[CoL_2]^{3+}+6H_2O}$ . The co-ordination number remains 6. Deduce the denticity of L and explain how the equation supports your answer.
[3 marks]
Total for this question: 3
For an unnamed chelate-substitution reaction, $\Delta H=+8.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=+60.0\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Calculate $\Delta G$ at $298\,\mathrm{K}$ and state what the sign of $\Delta G$ shows. Explain why this result does not show that the substitution is fast.
[5 marks]
Total for this question: 5
A blood sample contains $4.00\times10^{-5}\,\mathrm{mol}$ of haem oxygen-binding sites. Carbon monoxide occupies $18.0\%$ of the sites, and each unblocked site can bind one $\mathrm{O_2}$ molecule. Write an equation representing carbon monoxide displacing oxygen, then calculate the maximum amount of oxygen that the remaining sites can carry and the mass of oxygen the sample can no longer carry, given $M_r(\mathrm{O_2})=32.0$ .
[5 marks]
Total for this question: 5
Two substitutions start from $\mathrm{[M(H_2O)_6]^{2+}}$ . Reaction A uses one hexadentate ligand $\mathrm{Y^{4-}}$ to form $\mathrm{[MY]^{2-}}$ and six water molecules. Reaction B uses two neutral tridentate ligands L to form $\mathrm{[ML_2]^{2+}}$ and six water molecules. Assuming the metal-ligand bond enthalpy changes are similar, determine the particle-number change for each reaction and deduce which has the more favourable entropy contribution.
[5 marks]
Total for this question: 5

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

Explanation

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.

Tier 1 · Easy

State the shapes of $\mathrm{[Ag(NH_3)_2]^+}$ and $\mathrm{[CoCl_4]^{2-}}$ .
[2 marks]
Total for this question: 2
State the shape of $\mathrm{[Fe(H_2O)_6]^{2+}}$ and the bond angle between adjacent metal-oxygen bonds.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$\mathrm{[Pt(NH_3)_2Cl_2]}$ forms two stereoisomers. Name the type of stereoisomerism, describe the ligand arrangement in each isomer, and identify cisplatin.
[4 marks]
Total for this question: 4
The octahedral ion $\mathrm{[Co(NH_3)_4Cl_2]^+}$ forms two geometrical isomers. Describe the positions of the chloride ligands in each isomer and name the isomerism.
[3 marks]
Total for this question: 3
State which of octahedral $\mathrm{[MA_4B_2]}$ , octahedral $\mathrm{[MA_5B]}$ and tetrahedral $\mathrm{[MA_2B_2]}$ show cis-trans isomerism, and justify each answer.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An octahedral ion $\mathrm{[M(en)_3]^{3+}}$ contains three identical bidentate ligands. State its co-ordination number, the number and type of stereoisomers it forms, and explain the origin of the stereoisomerism.
[5 marks]
Total for this question: 5
Cisplatin has formula $\mathrm{[Pt(NH_3)_2Cl_2]}$ and has a distinct cis isomer and trans isomer. Explain why this evidence requires a square-planar arrangement rather than a tetrahedral arrangement around platinum.
[4 marks]
Total for this question: 4
A student claims to have drawn the two optical isomers of octahedral $\mathrm{[M(en)_3]^{3+}}$ . Drawing B is obtained only by rotating drawing A. Drawing C is a mirror image of A, but in drawing C each chelate ring is formed from one nitrogen of one en and one nitrogen of a different en. Identify both errors and state what a valid optical-isomer pair must show.
[5 marks]
Total for this question: 5
State the total number of stereoisomers of octahedral $\mathrm{[M(C_2O_4)_2(H_2O)_2]^{n-}}$ , describe the ligand arrangement in each, and justify whether each is chiral or achiral, without drawing the structures.
[5 marks]
Total for this question: 5

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

Explanation

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.

Tier 1 · Easy

Explain why a transition-metal ion can appear coloured when white light passes through its solution.
[3 marks]
Total for this question: 3
Explain why aqueous $\mathrm{Zn^{2+}}$ ions are colourless.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The energy gap between two d-electron levels is $3.70\times10^{-19}\,\mathrm{J}$ . Calculate the wavelength of light absorbed. Take $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and Planck's constant as $6.63\times10^{-34}\,\mathrm{J\,s}$ .
[3 marks]
Total for this question: 3
Two complexes contain the same metal ion in the same oxidation state and have the same co-ordination number, but one contains water ligands and the other ammonia ligands. Explain qualitatively why their solutions can have different colours.
[4 marks]
Total for this question: 4
A coloured complex has maximum absorbance at $625\,\mathrm{nm}$ . A student proposes using a $450$ , $625$ or $750\,\mathrm{nm}$ filter, zeroing with pure water even though the samples contain an additional reagent that absorbs weakly at $625\,\mathrm{nm}$ , and handling marked cuvettes without wiping them. Identify the filter and both procedural errors.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A calibration is linear: solutions of a coloured ion at $1.00$ , $2.00$ , $3.00$ and $4.00\,\mathrm{mmol\,dm^{-3}}$ give absorbances $0.120$ , $0.240$ , $0.360$ and $0.480$ . A ten-fold diluted sample gives absorbance $0.330$ . Determine the concentration of the original sample in $\mathrm{mol\,dm^{-3}}$ .
[4 marks]
Total for this question: 4
Complexes R, S and T absorb most strongly at $650\,\mathrm{nm}$ , $520\,\mathrm{nm}$ and $430\,\mathrm{nm}$ respectively. Deduce the order of their d-electron energy gaps from largest to smallest and explain the order.
[3 marks]
Total for this question: 3
Repeated colorimetry gives standards $(c/\mathrm{mmol\,dm^{-3}},A)$ of $(0,0.000)$ , $(1.00,0.150)$ , $(2.00,0.300)$ , $(3.00,0.450)$ and $(4.00,0.560)$ . The final value is reproducible. An unknown has $A=0.375$ . Determine which calibration range to use and the unknown concentration, then explain why the $4.00\,\mathrm{mmol\,dm^{-3}}$ standard should not be used in the linear fit.
[5 marks]
Total for this question: 5
Four cobalt complexes are U, $\mathrm{[Co(H_2O)_6]^{2+}}$ ; V, $\mathrm{[Co(NH_3)_6]^{2+}}$ ; W, $\mathrm{[CoCl_4]^{2-}}$ ; and X, $\mathrm{[Co(NH_3)_6]^{3+}}$ . Determine which comparison isolates the effect of ligand identity, which isolates oxidation state, and whether these four complexes contain a comparison that isolates co-ordination number. Explain why U and W alone cannot establish the effect of co-ordination number on colour.
[5 marks]
Total for this question: 5
Complex A absorbs most strongly at $480\,\mathrm{nm}$ and complex B at $620\,\mathrm{nm}$ . Calculate the d-electron energy gap for each complex, the difference between the gaps per photon, and the difference per mole of photons. State which complex has the larger gap, given $h=6.63\times10^{-34}\,\mathrm{J\,s}$ , $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[5 marks]
Total for this question: 5

3.2.5.5 · Variable oxidation states (A-level only)

Explanation

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.

Tier 1 · Easy

State the oxidation state and colour of vanadium in $\mathrm{VO^{2+}}$ , formed during reduction of acidified vanadate(V).
[2 marks]
Total for this question: 2
During the reduction of acidified vanadate(V) by zinc, a solution has changed from yellow to blue and then to green. State the oxidation state and formula of the vanadium species in the green solution.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write the balanced ionic equation for the reaction of acidified manganate(VII) ions with ethanedioate ions, $\mathrm{C_2O_4^{2-}}$ .
[4 marks]
Total for this question: 4
Write two reduction half-equations for the yellow-to-blue and blue-to-green stages when acidified $\mathrm{VO_2^+}$ is reduced. Include the formula and colour of each vanadium species.
[6 marks]
Total for this question: 6
A blue solution containing $\mathrm{M^{3+}}$ becomes green after a reducing agent forms $\mathrm{M^{2+}}$ . Adding ligand L to the green solution then produces a yellow $\mathrm{M^{2+}}$ complex. Explain which change demonstrates variable oxidation states, which is ligand substitution, and why colour change alone cannot distinguish the two processes.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $0.455\,\mathrm{g}$ iron supplement is dissolved, then diluted to $200\,\mathrm{cm^3}$ . Titration of a $20.0\,\mathrm{cm^3}$ aliquot uses $17.30\,\mathrm{cm^3}$ of $0.00220\,\mathrm{mol\,dm^{-3}}$ acidified $\mathrm{KMnO_4}$ . Calculate the percentage by mass of iron in the supplement. Use $A_r(\mathrm{Fe})=55.8$ and $\mathrm{MnO_4^-{:}Fe^{2+}}=1{:}5$ .
[6 marks]
Total for this question: 6
An aldehyde is warmed with Tollens' reagent, $\mathrm{[Ag(NH_3)_2]^+}$ . State the exact observation, write the overall ionic equation in alkaline solution, and explain the redox changes. Use RCHO for the aldehyde.
[7 marks]
Total for this question: 7
The following are standard electrode potentials: $\mathrm{M^{3+}+e^-\rightarrow M^{2+}}$ , $E^\circ=+0.62\,\mathrm{V}$ ; $\mathrm{[ML]^{3+}+e^-\rightarrow[ML]^{2+}}$ , $E^\circ=+0.20\,\mathrm{V}$ ; $\mathrm{N^{3+}+e^-\rightarrow N^{2+}}$ , $E^\circ=+0.45\,\mathrm{V}$ . Determine the spontaneous reaction and $E^\circ_{\mathrm{cell}}$ when $\mathrm{M^{3+}}$ , $\mathrm{M^{2+}}$ , $\mathrm{N^{3+}}$ and $\mathrm{N^{2+}}$ are mixed under standard conditions. Then do the same when $\mathrm{[ML]^{3+}}$ , $\mathrm{[ML]^{2+}}$ , $\mathrm{N^{3+}}$ and $\mathrm{N^{2+}}$ are mixed under standard conditions. Explain the reversal.
[6 marks]
Total for this question: 6
$1.250\,\mathrm{g}$ of hydrated ethanedioic acid, $\mathrm{H_2C_2O_4\cdot xH_2O}$ , is dissolved and made up to $250.0\,\mathrm{cm^3}$ . A $25.00\,\mathrm{cm^3}$ aliquot requires $19.80\,\mathrm{cm^3}$ of $0.0200\,\mathrm{mol\,dm^{-3}}$ acidified potassium manganate(VII). Write the ionic equation and determine $x$ , given $M_r(\mathrm{H_2C_2O_4})=90.0$ and $M_r(\mathrm{H_2O})=18.0$ .
[6 marks]
Total for this question: 6
Excess zinc reduces $0.0180\,\mathrm{mol}$ of yellow acidified $\mathrm{VO_2^+}$ completely to vanadium(II). Each zinc atom forms $\mathrm{Zn^{2+}}$ . Calculate the minimum mass of zinc required and state the formula and colour of each vanadium species encountered after the yellow starting ion, given $A_r(\mathrm{Zn})=65.4$ .
[6 marks]
Total for this question: 6

3.2.5.6 · Catalysts (A-level only)

Explanation

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.

Tier 1 · Easy

State the three surface stages in heterogeneous catalysis and explain how a catalyst poison lowers the rate.
[4 marks]
Total for this question: 4
Iron is a solid catalyst in a reaction between gaseous nitrogen and hydrogen. State the type of catalysis and where the reaction occurs.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write two equations showing how $\mathrm{V_2O_5}$ catalyses oxidation of $\mathrm{SO_2}$ in the Contact process, and show that the catalyst is regenerated.
[4 marks]
Total for this question: 4
A $\mathrm{Mn^{2+}}$ catalyst provides this route for the reaction between peroxodisulfate and iron(II) ions: $\mathrm{S_2O_8^{2-}+2Mn^{2+}\rightarrow2SO_4^{2-}+2Mn^{3+}}$ and $\mathrm{2Mn^{3+}+2Fe^{2+}\rightarrow2Mn^{2+}+2Fe^{3+}}$ . Combine the steps to show that $\mathrm{Mn^{2+}}$ is regenerated and absent from the overall equation.
[4 marks]
Total for this question: 4
Equal masses of active metal are tested as an unsupported lump, dispersed on a support, and dispersed on the same support after exposure to a poison. Their relative initial rates are $1.0$ , $8.5$ and $0.4$ . Explain both rate changes in terms of active sites.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The reaction $\mathrm{S_2O_8^{2-}+2I^-\rightarrow 2SO_4^{2-}+I_2}$ is catalysed by $\mathrm{Fe^{2+}}$ . Write two ionic equations for a catalytic route through $\mathrm{Fe^{3+}}$ , then show that $\mathrm{Fe^{2+}}$ is absent from the overall equation.
[6 marks]
Total for this question: 6
Design an experiment to test whether added $\mathrm{Mn^{2+}}$ catalyses the reaction between acidified manganate(VII) and ethanedioate ions. State the independent and dependent variables, a control experiment, three controlled variables and the expected observations.
[8 marks]
Total for this question: 8
State whether each claim about catalysis is true or false and give the correct statement for each false claim: (1) an intermediate must appear in the overall equation; (2) a catalyst increases the equilibrium constant; (3) a catalyst lowers activation energy only for the forward reaction; (4) a homogeneous catalyst works by adsorption; (5) a homogeneous catalyst forms an intermediate species with a reactant; (6) a support increases the surface area available per unit mass of metal.
[6 marks]
Total for this question: 6
The uncatalysed reaction is $\mathrm{S_2O_8^{2-}+2I^-\rightarrow2SO_4^{2-}+I_2}$ . Standard reduction potentials are $E^\circ(\mathrm{S_2O_8^{2-}/SO_4^{2-}})=+2.01\,\mathrm{V}$ , $E^\circ(\mathrm{Co^{3+}/Co^{2+}})=+1.82\,\mathrm{V}$ , $E^\circ(\mathrm{I_2/I^-})=+0.54\,\mathrm{V}$ and $E^\circ(\mathrm{V^{3+}/V^{2+}})=-0.26\,\mathrm{V}$ . Determine whether the cobalt or vanadium couple can provide a two-step homogeneous catalytic route. Write the spontaneous cobalt steps before calculating their cell potentials, and explain the rejection of the other couple.
[6 marks]
Total for this question: 6
An unsupported catalyst contains $2.40\,\mathrm{kg}$ of active metal and produces product at $18.0\,\mathrm{mol\,min^{-1}}$ . A supported catalyst contains $0.300\,\mathrm{kg}$ of the same metal and produces $9.00\,\mathrm{mol\,min^{-1}}$ before poisoning reduces its rate to $2.25\,\mathrm{mol\,min^{-1}}$ . Calculate the catalytic activity per unit mass of metal for the unsupported and unpoisoned supported catalysts, compare them quantitatively, calculate the percentage activity lost on poisoning, and explain the cost implication for the poisoned catalyst bed.
[5 marks]
Total for this question: 5

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

Explanation

The specified aqua ions are $\mathrm{[Fe(H_2O)_6]^{2+}}$ , $\mathrm{[Cu(H_2O)_6]^{2+}}$ , $\mathrm{[Al(H_2O)_6]^{3+}}$ and $\mathrm{[Fe(H_2O)_6]^{3+}}$ . Their hydroxide precipitates are green, blue, white and brown respectively.
A $3+$ aqua ion is more acidic than a comparable $2+$ ion because greater charge-to-size ratio polarises ligand O–H bonds more strongly.
Excess hydroxide dissolves amphoteric aluminium hydroxide; excess ammonia dissolves copper(II) hydroxide as a deep-blue ammine complex.
Carbonate precipitates metal carbonates from the $2+$ ions but deprotonates acidic $3+$ aqua ions, giving hydroxide precipitate and $\mathrm{CO_2}$ .
Required practical 11 identifies transition-metal ions by these test-tube reactions.

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.

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.

Tier 1 · Easy

State the observations when aqueous sodium hydroxide is added separately to solutions containing $\mathrm{Cu^{2+}}$ and $\mathrm{Fe^{3+}}$ ions.
[2 marks]
Total for this question: 2
Aqueous ammonia is added dropwise and then in excess to separate solutions of $\mathrm{Al^{3+}}$ and $\mathrm{Cu^{2+}}$ . State the complete observation sequence for each ion.
[4 marks]
Total for this question: 4

Tier 2 · Standard

A blue precipitate forms when a small amount of ammonia is added to $\mathrm{[Cu(H_2O)_6]^{2+}}$ . State what happens in excess ammonia, name the product ion's colour, and write the ligand-substitution equation.
[5 marks]
Total for this question: 5
Solutions A and B contain $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$ , one ion in each. State how aqueous sodium hydroxide distinguishes them and write both ionic precipitation equations with state symbols.
[6 marks]
Total for this question: 6
Separate $10.0\,\mathrm{cm^3}$ samples of $0.0500\,\mathrm{mol\,dm^{-3}}$ $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$ are precipitated completely with $0.100\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide. Calculate the minimum hydroxide volume for each sample and state both precipitate colours.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Solutions of $\mathrm{[Fe(H_2O)_6]^{2+}}$ and $\mathrm{[Fe(H_2O)_6]^{3+}}$ are treated with aqueous carbonate ions. Predict the different observations, write an ionic equation for each reaction, and explain why carbon dioxide forms with only one of the ions.
[8 marks]
Total for this question: 8
Required practical 11: four labelled solutions contain $\mathrm{Al^{3+}}$ , $\mathrm{Fe^{2+}}$ , $\mathrm{Cu^{2+}}$ and $\mathrm{Fe^{3+}}$ , one ion in each. Design the shortest sodium-hydroxide sequence that identifies all four. Include controls, exact observations and ionic equations for both stages involving aluminium.
[9 marks]
Total for this question: 9
Explain, in terms of the charge and size of the metal ion, why $\mathrm{[M(H_2O)_6]^{3+}}$ is more acidic than $\mathrm{[M(H_2O)_6]^{2+}}$ . Write the proton-transfer equation showing hydrolysis of each aqua ion.
[6 marks]
Total for this question: 6
A solution contains only $\mathrm{Al^{3+}}$ and $\mathrm{Cu^{2+}}$ . Give a method using aqueous sodium hydroxide, filtration and dilute acid to separate the ions and recover a hydroxide precipitate of each metal. Give the key observations and equations for copper(II) precipitation, aluminium hydroxide dissolving in excess hydroxide, and aluminium hydroxide reforming.
[6 marks]
Total for this question: 6
A solution contains $0.0100\,\mathrm{mol}$ in total of $\mathrm{Al^{3+}}$ and $\mathrm{Cu^{2+}}$ . Complete precipitation, with no hydroxide left over, requires a $26.0\,\mathrm{cm^3}$ sodium hydroxide portion whose concentration is $1.00\,\mathrm{mol\,dm^{-3}}$ . Calculate the amount and mole percentage of each metal ion, then state the observation when excess sodium hydroxide is added.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.2.1.1 · Classification

Tier 1 · Easy

Mark scheme for 3.2.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	s block	The differentiating electrons occupy an $s$ subshell, so the element is in the s block.	1
02.1	p block	A Group 4(14) element is positioned in the p-block region of the Periodic Table, so A is a p-block element.	1

Tier 2 · Standard

Mark scheme for 3.2.1.1 Tier 2 · Standard
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01.1	X is a p-block element because its differentiating electrons occupy the $4p$ subshell.	Read the occupied subshell at the end of the configuration. Although X contains a filled $3d$ subshell, the differentiating electrons are in $4p$ , so its position is in the p block.	2
02.1	Proton number 14; silicon, Si.	The Period 3 p block begins with aluminium at proton number 13. The second position is therefore silicon, with proton number 14.	2
03.1	The blank positions run from proton number $21$ to proton number $30$ . They form the d block. This region corresponds to progressive occupation of the five $3d$ orbitals. Each orbital holds two electrons, so the d block has $5\times2=10$ positions.	Calcium has proton number $20$ and gallium has proton number $31$ , so the ten consecutive positions between them are $21$ through $30$ . This region corresponds to progressive occupation of the five $3d$ orbitals. Two electrons fit in each orbital, accounting for the ten-element width of the Period 4 d block.	4

Tier 3 · Hard

Mark scheme for 3.2.1.1 Tier 3 · Hard
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01.1	Y is s block; Z is d block; the differentiating electron in Z enters the $3d$ subshell.	Y ends the filling of the $4s$ subshell, so it occupies the s block. The next proton and electron in Z begin occupation of $3d$ . Block classification follows the differentiating subshell, making Z d block despite its retained $4s^2$ electrons.	3
02.1	Zinc is a d-block element. The student has used the electron configuration of an ion to assign the block of the element; a full $3d^{10}$ subshell in $\mathrm{Zn^{2+}}$ does not move zinc out of the d-block region of the Periodic Table.	Block classification belongs to the element and follows its position in the Periodic Table. The $d^{10}$ configuration quoted is for $\mathrm{Zn^{2+}}$ , so it cannot be used to reassign zinc to the s block.	3
03.1	Rule A wrongly assigns $\mathrm{[Ar]3d^{10}4s^2 4p^1}$ to the d block because $3d$ contains the most electrons. $\mathrm{[Ar]3d^{10}4s^2 4p^1}$ is p block. Rule B wrongly assigns $\mathrm{[Ar]3d^2 4s^2}$ to the s block because $4s$ is written last. $\mathrm{[Ar]3d^2 4s^2}$ is d block. The correct rule is to identify the subshell receiving the differentiating electron.	Apply only rule A to the first configuration: it points to $3d$ , but the element is p block because its differentiating electron enters $4p$ . Apply only rule B to the second configuration: it points to $4s$ , but the element is d block because its differentiating electrons enter $3d$ . Block classification follows the differentiating subshell.	5
04.1	Positions 58 to 71 form the f block. An f subshell contains seven orbitals. Each orbital can hold two electrons. Progressive occupation therefore gives $7\times2=14$ f-block positions. Element 72 is d block because its differentiating electron enters $5d$ .	Use the stated differentiating subshell rather than the highest occupied shell. Filling the seven $4f$ orbitals with two electrons each accounts for fourteen consecutive f-block positions. Once that subshell is complete, the stated differentiating electron for element 72 enters $5d$ , so element 72 resumes the d-block sequence.	5
05.1	Each ion contains $18+5=23$ electrons. P has proton number $23+2=25$ . P is manganese, Mn. Q has proton number $23+3=26$ and is iron, Fe. Both elements are d-block elements. Their neutral atoms are reached in the sequence where differentiating electrons occupy the $3d$ subshell; ionic charge does not reassign an element's block.	Count the electrons in the common ion configuration first: argon contributes 18 and $3d^5$ contributes five. A $2+$ ion has two fewer electrons than protons, giving P proton number 25, manganese. A $3+$ ion has three fewer electrons than protons, giving Q proton number 26, iron. Block is a property of each element's Periodic Table position, and both positions belong to the sequence in which $3d$ is being filled.	6

3.2.1.2 · Physical properties of Period 3 elements

Tier 1 · Easy

Mark scheme for 3.2.1.2 Tier 1 · Easy
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01.1	$\mathrm{Na>Mg>Cl}$ ; nuclear charge increases across the period while shielding is similar, so attraction to the outer shell increases.	All three atoms add their outer electrons to the third shell. Proton number rises from Na to Cl without an additional inner shell, so effective nuclear attraction increases and pulls the outer shell closer. Therefore the decreasing order is $\mathrm{Na>Mg>Cl}$ .	3
02.1	Argon; only weak London forces act between its atoms, so little energy is needed to overcome these attractions.	The only monatomic Period 3 gas is argon. It has no metallic lattice, covalent network or molecules; melting overcomes weak London forces between individual Ar atoms.	3

Tier 2 · Standard

Mark scheme for 3.2.1.2 Tier 2 · Standard
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01.1	Al loses a higher-energy, more shielded $3p$ electron rather than Mg's $3s$ electron. S has a pair of electrons in one $3p$ orbital, so electron-electron repulsion makes one easier to remove than an unpaired $3p$ electron from P.	Compare the subshells first: Mg ends $3s^2$ but Al ends $3p^1$ , and a $3p$ electron is higher in energy and slightly more shielded. Then compare orbital occupancy: P is $3p^3$ with one electron per orbital, while S is $3p^4$ and contains a repelling pair. Each effect lowers the energy needed for the stated removal.	4
02.1	$\mathrm{W<X<Y<Z}$ in increasing proton number: W is Na, X is Mg, Y is Al and Z is Si.	The poor conductor and hardest sample is giant covalent silicon. The three conductors are the metals; their increasing hardness places the very soft sodium first, then magnesium, then aluminium. Their proton numbers are 11, 12, 13 and 14 respectively.	5
03.1	The metal-ion charge increases from $+1$ to $+3$ . The metal ions become smaller across the period. The number of delocalised electrons supplied per atom increases from one to three. The electrostatic attraction between the positive ions and delocalised electrons therefore becomes stronger.	Across Na, Mg and Al, nuclear charge rises and the metallic ions have increasing charge and decreasing radius. Each atom also contributes more electrons to the delocalised sea. Greater charge density and more delocalised electrons strengthen the electrostatic attraction in the giant metallic lattice.	4

Tier 3 · Hard

Mark scheme for 3.2.1.2 Tier 3 · Hard
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01.1	$1410\,{}^\circ\mathrm{C}$ : Si; $115\,{}^\circ\mathrm{C}$ : S; $-189\,{}^\circ\mathrm{C}$ : Ar. Si has a giant covalent lattice, $\mathrm{S_8}$ molecules have London forces between molecules, and monatomic Ar has still weaker London forces.	Assign the giant covalent structure first: many strong Si-Si covalent bonds must be broken, giving the highest value. Sulfur exists as relatively large $\mathrm{S_8}$ molecules, so its London forces are appreciable but far weaker than covalent bonds. Argon is monatomic with the least polarizable electron cloud of the two simple species, so it has the weakest attractions and the lowest melting point.	6
02.1	Metallic bonding generally strengthens from Na to Al, but Si is giant covalent rather than metallic. Its high melting point involves breaking many covalent bonds. Phosphorus is simple molecular $\mathrm{P_4}$ , so its melting point is much lower because only London forces between molecules are overcome.	Separate the trend into structural regions. The metallic explanation applies only from sodium to aluminium. Silicon changes to a giant covalent network, and phosphorus changes again to simple molecules, so no single metallic-bonding argument can describe the whole sequence.	5
03.1	The element is magnesium, Mg. Its metallic radius of $0.160\,\mathrm{nm}$ is smaller than sodium's but larger than those of the remaining Period 3 elements. Its first ionisation energy of $738\,\mathrm{kJ\,mol^{-1}}$ is higher than sodium's; the next element, aluminium, has the characteristic decrease caused by removal of a $3p$ electron. Its $650\,{}^\circ\mathrm{C}$ melting point is consistent with metallic magnesium and is much higher than sodium's because magnesium forms stronger metallic bonding; the melting point alone would not distinguish Mg from Al.	Use the properties together rather than relying on one trend. The metallic radius places the element near the left of Period 3 but not at sodium. The $738\,\mathrm{kJ\,mol^{-1}}$ first ionisation energy is the Period 3 value for Mg, before the Mg-to-Al decrease. The relatively high melting point also fits stronger metallic bonding in Mg than in Na, although melting point alone would not distinguish Mg from Al. These independent matches identify magnesium.	5
04.1	Aluminium has a greater nuclear charge than sodium. The added electrons from sodium to aluminium enter the same principal energy level, so the increase in shielding is small. The stronger attraction between the aluminium nucleus and its outer electron means more energy is needed to remove it than from sodium. Magnesium loses a $3s$ electron, whereas aluminium loses a $3p$ electron. The aluminium $3p$ electron is higher in energy and better shielded than magnesium's $3s$ electron, so it is easier to remove and aluminium has the lower first ionisation energy of the two.	Treat the two comparisons separately. From sodium to aluminium, increasing nuclear charge across the same shell dominates the small shielding increase, so first ionisation energy rises overall. The magnesium-to-aluminium dip has a different cause: aluminium's outer electron occupies the higher-energy, better-shielded $3p$ subshell rather than magnesium's $3s$ subshell.	5
05.1	One possible pair is X = Al and Y = Mg. Al is to the right of Mg in Period 3, so Al has the smaller atomic radius, but Al has the lower first ionisation energy. The other possible pair is X = S and Y = P. S is to the right of P, so S has the smaller atomic radius, but S has the lower first ionisation energy. For the S/P pair, both elements lose a $3p$ electron, but in sulfur two electrons are paired in one $3p$ orbital and their mutual repulsion makes one easier to remove than an unpaired $3p$ electron in phosphorus.	Radius generally decreases from left to right, so X must lie to the right of Y. The only adjacent Period 3 positions where first ionisation energy falls instead of rising are Mg to Al and P to S, giving (Al, Mg) and (S, P). The subshell change explains the first dip; paired-electron repulsion in a sulfur $3p$ orbital explains the second.	5

3.2.2 · Group 2, the alkaline earth metals

Tier 1 · Easy

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01.1	$\mathrm{Mg(s)+H_2O(g)\rightarrow MgO(s)+H_2(g)}$ ; magnesium burns or glows brightly and a white solid forms.	Steam oxidises Mg to MgO and is reduced to hydrogen. One Mg atom and one water molecule balance the equation; the white product and bright reaction are the expected observations.	2
02.1	With cold water, magnesium reacts very slowly: a few bubbles of hydrogen form and a white coating or suspension of $\mathrm{Mg(OH)_2}$ appears. $\mathrm{Mg(s)+2H_2O(l)\rightarrow Mg(OH)_2(s)+H_2(g)}$ . With steam, magnesium burns or glows brightly and forms white $\mathrm{MgO}$ and hydrogen. $\mathrm{Mg(s)+H_2O(g)\rightarrow MgO(s)+H_2(g)}$ .	Cold water produces the hydroxide slowly, whereas steam produces the oxide in a much more vigorous reaction. Both equations conserve magnesium, oxygen and hydrogen atoms.	5

Tier 2 · Standard

Mark scheme for 3.2.2 Tier 2 · Standard
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01.1	Add dilute hydrochloric acid to remove carbonate, then add barium chloride solution; a white precipitate of $\mathrm{BaSO_4}$ confirms sulfate.	Acid first converts carbonate to carbon dioxide and water: $\mathrm{CO_3^{2-}+2H^+\rightarrow CO_2+H_2O}$ . Adding $\mathrm{BaCl_2(aq)}$ then supplies $\mathrm{Ba^{2+}}$ , and sulfate gives the insoluble white solid by $\mathrm{Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)}$ .	4
02.1	Add equal excess masses of $\mathrm{Mg(OH)_2}$ and $\mathrm{Ba(OH)_2}$ to equal volumes of water. Stir for the same time at the same temperature, filter, then measure the pH of each filtrate. The barium hydroxide filtrate has the higher pH because more hydroxide has dissolved.	Using excess solid ensures each filtrate is saturated. Control water volume, solid particle size, stirring time and temperature, then use pH as evidence for the relative dissolved hydroxide concentration.	5
03.1	$\mathrm{Mg(OH)_2}$ is used to neutralise excess stomach acid. $\mathrm{Ca(OH)_2}$ is used to neutralise acidic soil. $\mathrm{BaSO_4}$ is used as the X-ray contrast medium. $\mathrm{BaSO_4}$ is extremely insoluble, so very few toxic $\mathrm{Ba^{2+}}$ ions are released.	Use hydroxides for the two neutralisation applications: magnesium hydroxide is the medicinal antacid and calcium hydroxide is spread on acidic soil. Barium sulfate is suitable for imaging because its very low solubility prevents an appreciable concentration of soluble barium ions forming in the body.	4

Tier 3 · Hard

Mark scheme for 3.2.2 Tier 3 · Hard
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01.1	$1.18\times10^2\,\mathrm{kg}$ of limestone	The amount captured is $(1.28\times10^3)(0.850)=1.088\times10^3\,\mathrm{mol}$ . The $1{:}1$ ratio requires the same amount of $\mathrm{CaCO_3}$ , whose mass is $(1.088\times10^3)(100.1)=1.089\times10^5\,\mathrm{g}$ . This is only $92.0\%$ of the limestone, so the required mass is $(1.089\times10^5)/0.920=1.18\times10^5\,\mathrm{g}=1.18\times10^2\,\mathrm{kg}$ .	5
02.1	Add dilute sulfuric acid or a soluble sulfate solution. A white precipitate identifies $\mathrm{Ba^{2+}}$ ; no precipitate identifies $\mathrm{Mg^{2+}}$ under the same conditions. Compare with known $\mathrm{Ba^{2+}}$ and $\mathrm{Mg^{2+}}$ controls. $\mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)}$ .	One sulfate addition is discriminating because $\mathrm{BaSO_4}$ is insoluble while $\mathrm{MgSO_4}$ is soluble. Known-ion controls verify that the reagent forms the expected white solid and that absence of a precipitate is meaningful.	6
03.1	$n(\mathrm{BaSO_4})=9.33/233.4=0.03997\,\mathrm{mol}$ . The $1{:}1$ precipitation ratio gives $0.03997\,\mathrm{mol}$ of $\mathrm{Ba^{2+}}$ . The percentage of Group 2 ions that were $\mathrm{Ba^{2+}}$ is $49.97\%$ , or $50.0\%$ to three significant figures. $\mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)}$ .	Only the barium ions are removed quantitatively because barium sulfate is insoluble while magnesium sulfate remains soluble. Convert the precipitate mass to moles, use the one-to-one ionic equation to obtain the original barium amount, then calculate $(0.03997/0.0800)\times100=49.97\%$ and round only at the end.	5
04.1	$\mathrm{Ca(s)+2H_2O(l)\rightarrow Ca(OH)_2(s)+H_2(g)}$ . $\mathrm{Ba(s)+2H_2O(l)\rightarrow Ba(OH)_2(aq)+H_2(g)}$ . Calcium hydroxide is sparingly soluble, so the calcium reaction produces a white cloudy suspension. Barium hydroxide is soluble, so the barium reaction produces a colourless solution, with more vigorous effervescence than calcium.	Both metals form the hydroxide and hydrogen. Calcium hydroxide is only sparingly soluble, so enough remains dispersed as a white solid to make the mixture cloudy; it must not be described as insoluble. Barium hydroxide is soluble, so the liquid remains colourless. Barium is more reactive down Group 2 and therefore gives faster, more vigorous hydrogen effervescence.	4
05.1	$\mathrm{TiCl_4+2Mg\rightarrow Ti+2MgCl_2}$ . $M_r(\mathrm{TiCl_4})=47.9+4(35.5)=189.9$ . $n(\mathrm{TiCl_4})=12000/189.9=63.19\,\mathrm{mol}$ . The stoichiometric magnesium amount is $2(63.19)=126.4\,\mathrm{mol}$ , with mass $3.071\,\mathrm{kg}$ . Allowing $8.0\%$ excess requires $3.071\times1.080=3.317\,\mathrm{kg}$ of pure magnesium. The magnesium feed mass is $3.317/0.950=3.49\,\mathrm{kg}$ to three significant figures.	Balance the extraction equation before using its $1{:}2$ mole ratio. Convert the titanium(IV) chloride mass to moles with $M_r=189.9$ , double this amount for magnesium, and convert to mass. Apply the excess factor to the pure magnesium requirement, then divide by the feed's mass fraction because only $95.0\%$ of the feed is magnesium.	6

3.2.3.1 · Trends in properties

Tier 1 · Easy

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01.1	Electronegativity decreases because atomic radius and electron shielding increase down the group.	Each element down the group has an additional occupied electron shell. The bonding pair is farther from the nucleus and more shielded, so the nucleus attracts it less strongly and electronegativity decreases.	2
02.1	The ions could be chloride, $\mathrm{Cl^-}$ , or bromide, $\mathrm{Br^-}$ . Add dilute ammonia: white $\mathrm{AgCl}$ dissolves, whereas cream $\mathrm{AgBr}$ remains insoluble.	White silver chloride and cream silver bromide can be difficult to separate by colour alone. Their different solubilities in dilute ammonia provide the discriminating observation.	3

Tier 2 · Standard

Mark scheme for 3.2.3.1 Tier 2 · Standard
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01.1	Acidify with dilute nitric acid and add silver nitrate: chloride gives white AgCl, bromide cream AgBr and iodide yellow AgI. AgCl dissolves in dilute ammonia, AgBr dissolves in concentrated ammonia, and AgI remains insoluble.	Nitric acid removes interfering ions without adding a halide, and $\mathrm{Ag^+}$ precipitates each silver halide. Record white, cream and yellow in the order AgCl, AgBr and AgI. If colours are uncertain, add dilute ammonia first, then concentrated ammonia: successive dissolution identifies AgCl then AgBr, leaving AgI.	6
02.1	The bromide ions form a cream $\mathrm{AgBr}$ precipitate, but chloride ions introduced by the hydrochloric acid also form a white $\mathrm{AgCl}$ precipitate. The observed solid is therefore a mixture and its colour gives an unreliable halide result. Dilute nitric acid should have been used instead.	Hydrochloric acid contaminates the sample with the ion being tested for: $\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}$ . Nitric acid acidifies without adding a halide ion, so only the cream silver bromide should form.	4
03.1	The molecules contain progressively more electrons down the group. Their electron clouds become larger and more polarisable. Larger temporary dipoles are formed, so van der Waals' forces become stronger. More energy is required to overcome the intermolecular forces, so boiling point rises.	All three substances are non-polar simple molecular halogens, so boiling separates molecules by overcoming van der Waals' forces. Down the group the larger electron cloud is more easily distorted, producing stronger temporary dipole-induced dipole attractions and therefore a higher boiling point.	4

Tier 3 · Hard

Mark scheme for 3.2.3.1 Tier 3 · Hard
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01.1	$\mathrm{NaI+H_2SO_4\rightarrow NaHSO_4+HI}$ and $\mathrm{8HI+H_2SO_4\rightarrow 4I_2+H_2S+4H_2O}$ . Purple iodine vapour or a black solid and the rotten-egg smell of $\mathrm{H_2S}$ ; iodide is a stronger reducing agent because its outer electron is more shielded and farther from the nucleus.	First transfer one proton from sulfuric acid to iodide to form HI. For the deepest reduction, sulfur changes from $+6$ in $\mathrm{H_2SO_4}$ to $-2$ in $\mathrm{H_2S}$ and gains eight electrons; eight iodide ions each lose one electron, forming four $\mathrm{I_2}$ . Balance H and O with four waters. Down the group the larger, more shielded halide ion loses an electron more readily, so iodide reduces sulfuric acid further than bromide.	7
02.1	Sample 1 is sodium chloride: misty fumes of hydrogen chloride only, because chloride is not a strong enough reducing agent to reduce sulfuric acid. Sample 2 is sodium bromide: misty fumes of hydrogen bromide and red-brown bromine vapour, because bromide reduces sulfur from $+6$ to $+4$ , forming sulfur dioxide. Sample 3 is sodium iodide: purple iodine vapour and hydrogen sulfide, which smells of rotten eggs, because iodide reduces sulfur all the way from $+6$ to $-2$ . The reducing power increases down the group because the halide ion becomes larger and more shielded, so its outer electron is lost more readily.	Match each observation set to the halide by the extent of reduction: chloride gives no redox, bromide reduces sulfuric acid to sulfur dioxide, and iodide reduces it further to hydrogen sulfide. Then explain the trend using ionic radius and shielding: the larger, more shielded ion gives up an electron more readily, so reduction of sulfur goes further down the group.	6
03.1	P is $\mathrm{Cl_2}$ , Q is $\mathrm{Br_2}$ and R is $\mathrm{I_2}$ . X is $\mathrm{Cl^-}$ , Y is $\mathrm{Br^-}$ and Z is $\mathrm{I^-}$ . $\mathrm{Br_2+2I^-\rightarrow2Br^-+I_2}$ .	The strongest oxidising halogen displaces two halides, so P is chlorine. The intermediate halogen displaces only iodide, so Q is bromine, while R must be iodine. The halide resisted by every halogen is chloride, the one displaced only by chlorine is bromide, and the one displaced by both chlorine and bromine is iodide. Substituting Q and Z gives the balanced bromine-iodide equation.	6
04.1	$\mathrm{Ag^+}$ forms insoluble silver halide precipitates while the nitrate ions remain in solution with the other ions. Carbonate and hydroxide ions would also precipitate with $\mathrm{Ag^+}$ , so the sample is acidified to remove them. Hydrochloric acid cannot be used because it introduces chloride ions and would give a false positive. Sulfuric acid cannot be used because silver sulfate is sparingly soluble and would give a white precipitate.	Choose the reagent by what must precipitate: silver ions capture halides as insoluble silver halides while nitrate stays dissolved. Acidification exists to destroy carbonate and hydroxide, which would otherwise give silver carbonate or silver oxide precipitates that could be mistaken for halides. The acid itself must not introduce a precipitating anion: chloride from hydrochloric acid is a false positive, and sulfate gives sparingly soluble silver sulfate.	4
05.1	Dilute ammonia dissolves $\mathrm{AgCl}$ , so the remaining solid is $\mathrm{AgBr}$ and $n(\mathrm{Br^-})=1.93/187.8=0.01028\,\mathrm{mol}$ . $n(\mathrm{Cl^-})=0.0200-0.01028=0.00972\,\mathrm{mol}$ . The original mixture is $51.4\%$ bromide by amount. The original mixture is $48.6\%$ chloride by amount. The initial precipitate mass is $1.93+(0.00972)(143.4)=3.32\,\mathrm{g}$ to three significant figures.	Use the selective solubility step before doing any mass calculation: dilute ammonia removes silver chloride but leaves silver bromide. Convert the residual silver bromide mass to moles, which equals the original bromide amount by the one-to-one precipitation ratio. Subtract from the stated total halide amount to obtain chloride, calculate both percentages, then add the calculated silver chloride mass to the residual mass.	5

3.2.3.2 · Uses of chlorine and chlorate(I)

Tier 1 · Easy

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01.1	$\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}$ ; the solution is used as a bleach or disinfectant.	One chlorine atom is reduced to chloride and the other is oxidised to chlorate(I), so equal amounts of NaCl and NaClO form. Balance sodium, hydrogen and oxygen with two NaOH and one water.	2
02.1	The pale green chlorine water becomes colourless and oxygen is formed: $\mathrm{2Cl_2(aq)+2H_2O(l)\rightarrow4H^+(aq)+4Cl^-(aq)+O_2(g)}$ .	Light promotes the reaction of dissolved chlorine with water to give chloride ions and oxygen. The equation has four chlorine atoms, four hydrogen atoms, two oxygen atoms and zero total charge on each side.	3

Tier 2 · Standard

Mark scheme for 3.2.3.2 Tier 2 · Standard
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01.1	Chlorine changes from $0$ in $\mathrm{Cl_2}$ to $-1$ in $\mathrm{Cl^-}$ and to $+1$ in $\mathrm{ClO^-}$ , so the same element is both reduced and oxidised.	Assign chlorine oxidation state $0$ in the element. In chloride it is $-1$ , which is reduction, while in chlorate(I) the oxygen is $-2$ and the ion charge is $-1$ , making chlorine $+1$ , which is oxidation. Both changes occur in $\mathrm{Cl_2+H_2O\rightleftharpoons 2H^++Cl^-+ClO^-}$ , meeting the definition of disproportionation.	4
02.1	Chlorine kills pathogenic microorganisms and leaves residual protection against later contamination. Chlorine is toxic and chlorination can form harmful by-products. Controlled doses are used because the reduction in water-borne disease outweighs these risks.	A balanced evaluation distinguishes the direct health benefit and continuing protection from the toxicity or by-product risk, then makes the specified evidence-based judgement.	4
03.1	Adding dilute acid increases $\mathrm{H^+}$ concentration and shifts the equilibrium left, decreasing the amount of $\mathrm{ClO^-}$ . Aqueous sodium hydroxide removes $\mathrm{H^+}$ and shifts the equilibrium right, increasing the amount of $\mathrm{ClO^-}$ . The equilibrium constant is unchanged because the temperature is unchanged.	Apply Le Chatelier's principle to the position of equilibrium. Added acid supplies a product, so the system responds to the left. Hydroxide neutralises a product, so the system responds to the right and replaces it. These concentration changes alter equilibrium composition but not the equilibrium constant at fixed temperature.	4

Tier 3 · Hard

Mark scheme for 3.2.3.2 Tier 3 · Hard
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01.1	$0.0800\,\mathrm{mol\,dm^{-3}}$	The amount of chlorine is $1.42/71.0=0.0200\,\mathrm{mol}$ . The equation $\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}$ gives a $1{:}1$ ratio, so $0.0200\,\mathrm{mol}$ of NaClO forms. Convert the volume to $0.250\,\mathrm{dm^3}$ and calculate $c=n/V=0.0200/0.250=0.0800\,\mathrm{mol\,dm^{-3}}$ .	4
02.1	Add dilute hydrochloric acid to separate portions. Carbonate effervesces; pass the colourless gas into limewater, which turns milky. Hydroxide gives no gas or visible change. A blank containing the same acid but no sample should not produce gas or turn limewater milky. $\mathrm{CO_3^{2-}(aq)+2H^+(aq)\rightarrow CO_2(g)+H_2O(l)}$ and $\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}$ .	Gas formation makes the first acid addition discriminating. Limewater confirms that the gas from carbonate is carbon dioxide, while the reagent blank checks that the apparatus and acid do not create the positive result.	7
03.1	Warm gently to release ammonia gas. The ammonia turns damp red litmus paper blue. $\mathrm{NH_4^+(aq)+OH^-(aq)\rightarrow NH_3(g)+H_2O(l)}$ . The solution was not warmed, so too little ammonia is released for a reliable result. The red litmus was dry. Ammonia must dissolve in the water on the paper to give an alkaline solution before litmus can respond.	Add aqueous sodium hydroxide and warm gently so that any ammonium ions release ammonia. Test the gas with damp red litmus paper, which turns blue. Room-temperature testing may not release enough ammonia, while dry litmus cannot show the alkaline response because ammonia must dissolve in water before it produces hydroxide ions.	6
04.1	The daily chlorine mass is $(2.50\times10^6)(0.800)=2.00\times10^6\,\mathrm{mg}=2.00\,\mathrm{kg}$ . The fraction remaining is $1-0.650=0.350$ . The residual concentration is $(0.800)(0.350)=0.280\,\mathrm{mg\,dm^{-3}}$ . $0.280\,\mathrm{mg\,dm^{-3}}$ is above the minimum disinfecting concentration. $0.280\,\mathrm{mg\,dm^{-3}}$ is below the by-product limit, so the dose meets both stated requirements under these conditions.	Multiply flow volume by dose to obtain the total mass added, keeping milligrams until the final conversion to kilograms. The residual is the unconsumed $35.0\%$ of the input concentration. Compare the unrounded residual separately with the lower disinfection threshold and the upper by-product threshold before reaching the conclusion.	5
05.1	$\mathrm{Cl_2+2e^-\rightarrow2Cl^-}$ . $\mathrm{Cl_2+4OH^-\rightarrow2ClO^-+2H_2O+2e^-}$ . Adding and dividing by two gives $\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$ . The chloride-to-chlorate(I) mole ratio is $1{:}1$ . One chlorine atom is reduced from $0$ to $-1$ while another is oxidised from $0$ to $+1$ .	Balance the reduction to chloride with two electrons. For oxidation to chlorate(I), balance oxygen and hydrogen using hydroxide and water, placing two electrons on the product side. The electron numbers already match, so add the half-equations and divide every coefficient by two. The product coefficients give the required one-to-one ratio and show disproportionation explicitly.	5

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

Tier 1 · Easy

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01.1	$\mathrm{2Mg+O_2\rightarrow 2MgO}$ and $\mathrm{P_4+5O_2\rightarrow P_4O_{10}}$ .	Use the specified oxide formula in each case. Two Mg atoms balance one oxygen molecule for MgO, while ten oxygen atoms in $\mathrm{P_4O_{10}}$ require five $\mathrm{O_2}$ molecules.	2
02.1	$\mathrm{Na_2O}$ ; $\mathrm{Na_2O(s)+H_2O(l)\rightarrow2NaOH(aq)}$ .	A strongly alkaline solution indicates the soluble basic oxide sodium oxide. Magnesium oxide is sparingly soluble, giving pH about 9, and silicon dioxide does not react with water.	3

Tier 2 · Standard

Mark scheme for 3.2.4 Tier 2 · Standard
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01.1	$\mathrm{Al_2O_3}$ has a giant ionic lattice with strong electrostatic attractions, $\mathrm{SiO_2}$ is giant covalent with many strong covalent bonds, and molecular $\mathrm{P_4O_{10}}$ has only intermolecular forces between molecules.	Identify each structure before naming the force overcome. Melting $\mathrm{Al_2O_3}$ requires overcoming strong attraction between oppositely charged ions, and melting $\mathrm{SiO_2}$ requires breaking many Si-O covalent bonds. In $\mathrm{P_4O_{10}}$ , covalent bonds within each molecule remain intact and only the much weaker forces between molecules are overcome.	5
02.1	Add warm dilute hydrochloric acid. $\mathrm{Al_2O_3}$ dissolves to give a colourless solution, whereas $\mathrm{SiO_2}$ remains as a white solid. $\mathrm{Al_2O_3(s)+6H^+(aq)\rightarrow2Al^{3+}(aq)+3H_2O(l)}$ .	Amphoteric aluminium oxide reacts with warm dilute acid, but acidic silicon dioxide does not. One acid addition therefore distinguishes the two without needing a second reagent.	5
03.1	A is $\mathrm{MgO}$ . B is $\mathrm{SiO_2}$ . C is $\mathrm{SO_3}$ . Molten $\mathrm{MgO}$ conducts because its ions are free to move and carry charge.	The oxide that conducts only after its giant lattice melts must be ionic magnesium oxide. Giant covalent silicon dioxide has no mobile ions or delocalised electrons, while sulfur trioxide is the low-melting molecular oxide. In molten MgO, mobile $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ ions carry current.	4

Tier 3 · Hard

Mark scheme for 3.2.4 Tier 3 · Hard
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01.1	$\mathrm{Al_2O_3}$ is amphoteric, $\mathrm{SiO_2}$ and $\mathrm{SO_3}$ are acidic. $\mathrm{Al_2O_3+2OH^-+3H_2O\rightarrow 2[Al(OH)_4]^-}$ ; $\mathrm{SiO_2+2OH^-\rightarrow SiO_3^{2-}+H_2O}$ ; $\mathrm{SO_3+H_2O\rightarrow H_2SO_4}$ .	Classify amphoteric aluminium oxide by its ability to react with both acids and bases. Balance its base reaction by forming two tetrahydroxoaluminate ions. Silicon dioxide is an acidic giant oxide and forms silicate with hydroxide. Sulfur trioxide is an acidic molecular oxide and hydrates directly to sulfuric acid.	7
02.1	The solution has pH about 9 and X is magnesium. $\mathrm{2Mg(s)+O_2(g)\rightarrow2MgO(s)}$ and $\mathrm{MgO(s)+H_2O(l)\rightarrow Mg(OH)_2(s)}$ .	A sparingly soluble Period 3 oxide that produces a mildly alkaline solution is magnesium oxide. The dissolved magnesium hydroxide supplies hydroxide ions, giving a pH of about 9. Both equations balance magnesium, oxygen and hydrogen atoms.	6
03.1	$n(\mathrm{E})=0.620/31.0=0.0200\,\mathrm{mol}$ and $n(\mathrm{O})=0.800/16.0=0.0500\,\mathrm{mol}$ . The ratio $\mathrm{E{:}O}=2{:}5$ , so the empirical formula is $\mathrm{E_2O_5}$ . E is phosphorus and the molecular oxide is $\mathrm{P_4O_{10}}$ . $\mathrm{P_4O_{10}}$ is simple molecular. $\mathrm{P_4O_{10}+6H_2O\rightarrow4H_3PO_4}$ .	Use the mass gain as the oxygen mass. The amounts are $0.0200\,\mathrm{mol}$ of E and $0.0500\,\mathrm{mol}$ of O atoms, giving the whole-number ratio $2{:}5$ . Relative atomic mass $31.0$ identifies phosphorus. The empirical unit $\mathrm{P_2O_5}$ corresponds to molecular $\mathrm{P_4O_{10}}$ , whose discrete molecules react with six waters to form four phosphoric acid molecules.	6
04.1	$\mathrm{Na_2O+2HCl\rightarrow2NaCl+H_2O}$ . $\mathrm{MgO+2HCl\rightarrow MgCl_2+H_2O}$ . $n(\mathrm{HCl})=(1.00)(0.1000)=0.100\,\mathrm{mol}$ , so the total oxide amount is $0.0500\,\mathrm{mol}$ . If $x$ is the amount of $\mathrm{Na_2O}$ , then $62.0x+40.3(0.0500-x)=2.81$ . $x=0.0366\,\mathrm{mol}$ , giving $m(\mathrm{Na_2O})=2.27\,\mathrm{g}$ . The mixture is $(2.27/2.81)\times100=80.8\%$ $\mathrm{Na_2O}$ by mass.	Both oxides consume two moles of hydrochloric acid per mole of oxide, so the acid amount fixes their combined amount. Let the sodium oxide amount be $x$ and the magnesium oxide amount be $0.0500-x$ . Use the measured total mass to form and solve the mass equation, then convert the sodium oxide amount to mass and divide by the sample mass.	6
05.1	$\mathrm{P_4O_{10}+6H_2O\rightarrow4H_3PO_4}$ . $\mathrm{H_3PO_4+3NaOH\rightarrow Na_3PO_4+3H_2O}$ . $\mathrm{SO_3+H_2O\rightarrow H_2SO_4}$ . $\mathrm{H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O}$ . The samples require $0.144\,\mathrm{mol}$ and $0.0240\,\mathrm{mol}$ of NaOH respectively, so the ratio is $6{:}1$ .	One mole of $\mathrm{P_4O_{10}}$ forms four moles of triprotic phosphoric acid, so complete neutralisation consumes twelve moles of NaOH per mole of oxide. One mole of $\mathrm{SO_3}$ forms one mole of diprotic sulfuric acid and consumes two moles of NaOH. Multiplying each factor by $0.0120\,\mathrm{mol}$ gives $0.144\,\mathrm{mol}$ and $0.0240\,\mathrm{mol}$ , a six-to-one ratio.	5

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

Tier 1 · Easy

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01.1	A molecule or ion that donates a pair of electrons to a central metal atom or ion to form a co-ordinate bond.	Include both essential ideas: the ligand supplies an electron pair, and that pair forms a co-ordinate bond to the metal centre.	2
02.1	$\mathrm{NH_3}$ and $\mathrm{H_2O}$ can act as ligands because each has a lone pair that can be donated to a metal ion. $\mathrm{CH_4}$ has no lone pair, and electron-deficient $\mathrm{BF_3}$ accepts rather than donates an electron pair.	Apply the ligand definition to each molecule: a ligand must be able to donate a lone pair to form a co-ordinate bond. Nitrogen in ammonia and oxygen in water have available lone pairs; methane does not, while boron trifluoride is an electron-pair acceptor.	4

Tier 2 · Standard

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01.1	Chromium is $+3$ ; co-ordination number $=6$ .	Let the chromium oxidation state be $x$ . Three $\mathrm{C_2O_4^{2-}}$ ligands contribute $-6$ , so $x-6=-3$ and $x=+3$ . Each of the three ligands makes two co-ordinate bonds, giving $3\times2=6$ .	3
02.1	$\mathrm{[V(H_2O)_4(C_2O_4)]^+}$ . The four water ligands are neutral and ethanedioate contributes $-2$ , so $+3-2=+1$ overall.	Four monodentate water ligands contribute four co-ordinate bonds and one bidentate ethanedioate contributes two, matching co-ordination number 6. Combining the vanadium oxidation state with the ligand charges gives the formula $\mathrm{[V(H_2O)_4(C_2O_4)]^+}$ .	4
03.1	Formation of coloured ions. Variable oxidation states, shown by $\mathrm{X^{2+}}$ and $\mathrm{X^{3+}}$ . Formation of complex ions with ligands. Catalytic activity.	Map each observation to a characteristic property. The blue and green ions show colour; the two charges show more than one oxidation state; ammonia surrounding X shows complex formation; and increasing rate without overall consumption identifies catalysis.	4

Tier 3 · Hard

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01.1	Fe is a transition metal; $\mathrm{Fe^{2+}}$ has an incomplete d sub-level. Sc and Zn are excluded because their stated ions are respectively $d^0$ and $d^{10}$ .	Apply the definition to an ion rather than merely checking that the element lies in the d block. An incomplete d sub-level contains between one and nine electrons, so $d^6$ qualifies. Empty $d^0$ and full $d^{10}$ sub-levels do not.	4
02.1	The argument is incorrect. Copper is a transition metal because it forms at least one ion with a partially filled d subshell: $\mathrm{Cu^{2+}}$ is $d^9$ . The $d^{10}$ configuration of $\mathrm{Cu^+}$ does not disqualify copper.	The definition requires the element to form at least one stable ion with an incomplete d sub-level; it does not require every ion to have one. Copper satisfies the definition through $\mathrm{Cu^{2+}}$ .	3
03.1	L is neutral. There are four ligand particles: two L and two $\mathrm{Cl^-}$ . Each bidentate L occupies two co-ordination sites, while each chloride occupies one, giving $2(2)+2(1)=6$ sites from only four ligand particles. The complex is octahedral.	Let the charge on L be $q$ . Charge balance gives $+3+2q-2=+1$ , so $q=0$ . The two bidentate L particles form four co-ordinate bonds and the two monodentate chloride particles form two more. Co-ordination number counts donor atoms bonded to M, not ligand particles, so four particles give co-ordination number 6. Six-coordinate complexes are octahedral.	5
04.1	Electrons are removed from $4s$ before $3d$ when these ions form. $\mathrm{Cr^{2+}}$ has configuration $\mathrm{[Ar]3d^4}$ . $\mathrm{Mn^{2+}}$ has configuration $\mathrm{[Ar]3d^5}$ . Both ions have incomplete d sub-levels. Chromium and manganese therefore both meet the definition of a transition metal.	Remove the $4s$ electron before a $3d$ electron. Chromium loses its $4s$ electron and one $3d$ electron to form $\mathrm{[Ar]3d^4}$ ; manganese loses both $4s$ electrons to form $\mathrm{[Ar]3d^5}$ . An element is a transition metal if it forms at least one stable ion with an incomplete d sub-level, so both derived ions provide the required evidence.	5
05.1	Each ammonia ligand is neutral and ethanedioate has charge $2-$ , so the overall charge is $+3+(-2)=+1$ . The complex formula is $\mathrm{[Co(NH_3)_4(C_2O_4)]^+}$ . There are five ligand particles: four ammonia molecules and one ethanedioate ion. The four monodentate ammonia ligands provide four donor atoms and the bidentate ethanedioate provides two, so the co-ordination number is 6. A co-ordination number of 6 gives an octahedral complex.	Bookkeep charge and donor atoms separately. The four neutral ammonia ligands do not alter the cobalt(III) charge, while one ethanedioate ion contributes $-2$ , leaving charge $+1$ . Co-ordination number counts the six donor atoms bonded to cobalt, not the five ligand particles, and six-coordinate complexes are octahedral.	5

3.2.5.2 · Substitution reactions (A-level only)

Tier 1 · Easy

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01.1	$\mathrm{NH_3}$ is monodentate; ethane-1,2-diamine is bidentate; EDTA is multidentate.	Count the donor atoms that can bind to one metal centre. Ammonia donates through one nitrogen, ethane-1,2-diamine through two nitrogens, and EDTA through several donor atoms.	3
02.1	Carbon monoxide replaces oxygen co-ordinately bonded to $\mathrm{Fe^{2+}}$ in haemoglobin, so less oxygen can be transported in the blood.	Treat oxygen and carbon monoxide as competing ligands for the haem iron(II) centre. Replacement of bound oxygen reduces the number of oxygen-carrying sites.	2

Tier 2 · Standard

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01.1	$\mathrm{[Cu(H_2O)_6]^{2+}+4NH_3\rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}+4H_2O}$ ; both sides have six donor atoms bonded to Cu.	Excess ammonia replaces four water ligands but the product retains two waters. All ligands shown are monodentate, so the reactant has six Cu-O bonds and the product has four Cu-N plus two Cu-O bonds; both co-ordination numbers are six.	4
02.1	$\mathrm{[Co(H_2O)_6]^{2+}(aq)+4Cl^-(aq)\rightleftharpoons[CoCl_4]^{2-}(aq)+6H_2O(l)}$ . The solution changes from pink to blue and the co-ordination number changes from 6 to 4.	Four larger chloride ligands replace six water ligands. The charges balance at $-2$ on both sides, and the tetrahedral tetrachlorocobaltate(II) product is blue.	5
03.1	The mixture becomes pink as more $\mathrm{[Co(H_2O)_6]^{2+}}$ forms. Dilution favours the side with more dissolved particles, so equilibrium shifts left. The equilibrium constant does not change because temperature is constant.	Dilution lowers all dissolved concentrations. The equilibrium responds toward the left-hand side, which has more dissolved solute particles, producing the pink hexaaquacobalt(II) complex and consuming blue tetrachlorocobaltate(II). This changes the equilibrium position, not the value of the equilibrium constant at fixed temperature; pure liquid water is omitted from the equilibrium expression.	4

Tier 3 · Hard

Mark scheme for 3.2.5.2 Tier 3 · Hard
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01.1	Similar numbers and types of metal-ligand bonds are broken and formed, so $\Delta H$ is small; four particles form seven particles, so $\Delta S$ is positive and the products are favoured.	Count six M-O bonds broken and six M-N bonds formed, so the bond-enthalpy contributions broadly balance. Count independently moving species: one aqua complex plus three en molecules gives four particles, while one chelate complex plus six waters gives seven. The increase in disorder makes $T\Delta S$ favourable in $\Delta G=\Delta H-T\Delta S$ , shifting equilibrium toward the chelate.	5
02.1	L is tridentate. Six water ligands are replaced by two L molecules, so each L supplies three donor atoms and forms three co-ordinate bonds.	The six released water molecules show that all six original co-ordination sites have been replaced. Dividing those six sites between two identical ligand molecules gives $6/2=3$ donor atoms per L. The displayed equation is balanced in charge because L is neutral and both complexes have charge $3+$ .	3
03.1	$T\Delta S=(298)(60.0)=1.788\times10^{4}\,\mathrm{J\,mol^{-1}}=17.88\,\mathrm{kJ\,mol^{-1}}$ . $\Delta G=8.0-17.88=-9.9\,\mathrm{kJ\,mol^{-1}}$ to two significant figures; accept $-9.88\,\mathrm{kJ\,mol^{-1}}$ . The negative $\Delta G$ means the substitution is thermodynamically feasible under the stated conditions. Feasibility does not determine rate; the reaction could still have a large activation energy and be slow.	Convert the entropy term before combining units: $60.0\,\mathrm{J\,K^{-1}\,mol^{-1}}=0.0600\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ . Then $\Delta G=\Delta H-T\Delta S=8.0-(298)(0.0600)=-9.88\,\mathrm{kJ\,mol^{-1}}$ , reported as $-9.9\,\mathrm{kJ\,mol^{-1}}$ because $\Delta H$ is given to two significant figures. A negative value is a thermodynamic statement only; kinetics depends on the activation barrier.	5
04.1	$\mathrm{HbO_2+CO\rightarrow HbCO+O_2}$ . Carbon monoxide occupies $(4.00\times10^{-5})(0.180)=7.20\times10^{-6}\,\mathrm{mol}$ of sites. The unblocked sites can carry $4.00\times10^{-5}-7.20\times10^{-6}=3.28\times10^{-5}\,\mathrm{mol}$ of $\mathrm{O_2}$ . The amount of $\mathrm{O_2}$ no longer carried equals the blocked-site amount, $7.20\times10^{-6}\,\mathrm{mol}$ . The mass of $\mathrm{O_2}$ no longer carried is $(7.20\times10^{-6})(32.0)=2.30\times10^{-4}\,\mathrm{g}$ .	The substitution equation shows carbon monoxide replacing bound oxygen at haem iron. Calculate the blocked-site amount from the stated fraction and subtract it from the total to find the remaining oxygen capacity. Each blocked site represents one oxygen molecule no longer carried, so convert the blocked-site amount, not the remaining capacity, to the requested lost oxygen mass.	5
05.1	Both products retain co-ordination number 6 because each incoming ligand set supplies six donor atoms. Six metal-oxygen bonds are replaced by six metal-donor bonds in each reaction, so the enthalpy contributions are broadly comparable. Reaction A changes from two dissolved reactant particles to seven product particles, an increase of five. Reaction B changes from three dissolved reactant particles to seven product particles, an increase of four. Reaction A has the larger positive entropy change and therefore the more favourable $-T\Delta S$ contribution.	Count donor atoms separately from freely moving particles. Both ligand sets occupy all six sites, and the same number of metal-ligand bonds is exchanged. For entropy, reaction A has one aqua complex plus one Y ion on the left and one complex plus six waters on the right. Reaction B begins with one complex plus two L molecules. The larger particle increase in A gives the stronger entropy advantage under the stated enthalpy assumption.	5

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

Tier 1 · Easy

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01.1	$\mathrm{[Ag(NH_3)_2]^+}$ is linear; $\mathrm{[CoCl_4]^{2-}}$ is tetrahedral.	The silver complex has two co-ordinate bonds arranged linearly. Four large chloride ligands around cobalt favour a tetrahedral arrangement.	2
02.1	Octahedral; $90^\circ$ .	Six monodentate water ligands give co-ordination number 6 and an octahedral arrangement. Adjacent ligand positions subtend $90^\circ$ at the metal ion.	2

Tier 2 · Standard

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01.1	Cis-trans isomerism in a square-planar complex; like ligands are adjacent in cis and opposite in trans; cisplatin is the cis isomer.	Place the four ligands in one plane around Pt. If the two chloride ligands occupy neighbouring positions the isomer is cis; if their positions are $180^\circ$ apart it is trans. The medicinal compound cisplatin is the adjacent-ligand form.	4
02.1	In the cis isomer the chloride ligands are adjacent; in the trans isomer they are opposite. This is cis-trans isomerism.	Keep all six ligand positions octahedral. Placing the two identical chloride ligands at neighbouring positions gives cis; placing them $180^\circ$ apart gives trans.	3
03.1	Octahedral $\mathrm{[MA_4B_2]}$ shows cis-trans isomerism because the two B ligands can be adjacent or opposite. Octahedral $\mathrm{[MA_5B]}$ does not show cis-trans isomerism because there is only one B ligand and all six positions are equivalent. Tetrahedral $\mathrm{[MA_2B_2]}$ does not show cis-trans isomerism because all four positions are equivalent and there are no opposite ligand positions.	Cis-trans isomerism requires distinct adjacent and opposite arrangements. An octahedron supplies both arrangements when two ligands differ from the other four. A single B ligand cannot define an adjacent-or-opposite pair, and a tetrahedron has no opposite sites, so the other two complexes each have only one arrangement.	4

Tier 3 · Hard

Mark scheme for 3.2.5.3 Tier 3 · Hard
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01.1	Co-ordination number $6$ ; it forms two optical isomers that are non-superimposable mirror images.	Each en ligand uses two donor atoms, so three ligands make six co-ordinate bonds and an octahedral ion. The three chelate rings can wind around the metal centre in two opposite handed arrangements. These arrangements are mirror images but cannot be superimposed, so they are a pair of optical isomers.	5
02.1	A square-planar $\mathrm{MA_2B_2}$ complex has two different arrangements: the matching ligands can be adjacent in the cis isomer or opposite in the trans isomer. In a tetrahedral $\mathrm{MA_2B_2}$ complex all four positions are equivalent, so it has only one arrangement and cannot show cis-trans isomerism.	Use the observed number of geometrical isomers to test the two candidate shapes. Only the square plane provides both neighbouring and opposite pairs of positions; a tetrahedron has no opposite ligand positions.	4
03.1	A and B are the same structure because rotation does not create a new isomer. C is not a valid $\mathrm{[M(en)_3]^{3+}}$ structure because each bidentate en ligand must keep its two donor atoms joined within the same ligand. A valid pair must preserve three intact bidentate ligands and octahedral co-ordination number 6. The two valid structures must be mirror images that cannot be superimposed.	First preserve connectivity: each en molecule forms two bonds through its own two nitrogen atoms, so donor atoms cannot be re-paired between ligand molecules. Then compare three-dimensional handedness. A rotated drawing is superimposable on the original; optical isomers require intact octahedral mirror-image arrangements that remain non-superimposable after every rotation.	5
04.1	There are three stereoisomers in total. One is the trans form, in which the two water ligands occupy opposite positions. The trans form is achiral because it is superimposable on its mirror image. The cis arrangement, with adjacent water ligands, exists as two optical isomers. These two cis forms are a pair of non-superimposable mirror images because the two bidentate ethanedioate ligands give the cis octahedral arrangement opposite handednesses.	First classify the relative positions of the two identical water ligands: they can be trans or cis. The trans geometry has an internal symmetry relationship and is superimposable on its mirror image, so it contributes one achiral stereoisomer. The two chelate rings make the cis geometry handed, so its mirror images are non-superimposable and contribute two optical isomers. Hence the total is one trans plus two cis, or three.	5

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

Tier 1 · Easy

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01.1	Some visible wavelengths are absorbed to promote d electrons to a higher energy level, and the remaining wavelengths are transmitted.	Link the electronic change to the observation: a visible photon matching the d-level energy gap is absorbed, so the transmitted light lacks those wavelengths and is seen as a colour.	3
02.1	$\mathrm{Zn^{2+}}$ has a full $d^{10}$ subshell, so no d electron can be promoted between split d energy levels by absorbing visible light.	Colour requires absorption of a visible wavelength for a transition between d energy levels. With all five d orbitals filled in $\mathrm{Zn^{2+}}$ , that d-electron excitation is not available, so visible light is not selectively absorbed.	2

Tier 2 · Standard

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01.1	$5.38\times10^{-7}\,\mathrm{m}$ or $538\,\mathrm{nm}$	Rearrange $\Delta E=hc/\lambda$ to $\lambda=hc/\Delta E$ . Substitution gives $\lambda=(6.63\times10^{-34})(3.00\times10^8)/(3.70\times10^{-19})=5.38\times10^{-7}\,\mathrm{m}$ , which is $538\,\mathrm{nm}$ .	3
02.1	Changing the ligand changes the energy difference between the d-electron levels. A different visible wavelength is therefore absorbed to excite a d electron, so different wavelengths remain transmitted or reflected and the observed colour changes.	The controlled oxidation state and co-ordination number leave ligand identity as the changed factor. Link that change to the energy gap, the absorbed wavelength and finally the light that reaches the observer.	4
03.1	Use the $625\,\mathrm{nm}$ filter because absorbance, and therefore sensitivity, is greatest there. The blank must contain the solvent and the additional reagent but no coloured complex. A pure-water blank would not remove the weak absorbance due to the additional reagent, so the complex's absorbance would be overestimated systematically. Wipe the cuvette because fingerprints or droplets absorb or scatter light and can make the measured absorbance too high.	Use the absorption maximum so a change in complex concentration produces the largest useful signal. A valid blank differs from a sample only by omission of the coloured complex, so it subtracts the additional reagent's weak absorbance as well as the solvent and cuvette contributions. Clean optical faces are also essential because contamination changes transmitted intensity independently of analyte concentration.	4

Tier 3 · Hard

Mark scheme for 3.2.5.4 Tier 3 · Hard
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01.1	$2.75\times10^{-2}\,\mathrm{mol\,dm^{-3}}$	The calibration gradient is $0.120$ absorbance per $1.00\,\mathrm{mmol\,dm^{-3}}$ . The diluted concentration is $0.330/0.120=2.75\,\mathrm{mmol\,dm^{-3}}$ . Undo the ten-fold dilution to obtain $27.5\,\mathrm{mmol\,dm^{-3}}=2.75\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ .	4
02.1	T > S > R. Photon energy is inversely proportional to wavelength, so the shortest absorbed wavelength corresponds to the largest gap.	Use $\Delta E=hc/\lambda$ qualitatively: with the same constants, decreasing wavelength increases the energy absorbed. Therefore $430\,\mathrm{nm}$ gives the largest gap and $650\,\mathrm{nm}$ the smallest.	3
03.1	Use the linear range from $0$ to $3.00\,\mathrm{mmol\,dm^{-3}}$ . The first four points have gradient exactly $0.150$ absorbance per $\mathrm{mmol\,dm^{-3}}$ . $c=0.375/0.150=2.50\,\mathrm{mmol\,dm^{-3}}$ . The reproducible $4.00$ point lies below the linear prediction of $0.600$ , so it shows a negative deviation from Beer-Lambert behaviour rather than a random outlier and must be excluded from the linear calibration.	Check proportionality before fitting every standard. From zero to $3.00\,\mathrm{mmol\,dm^{-3}}$ , absorbance rises by exactly $0.150$ per concentration unit. The linear prediction at $4.00\,\mathrm{mmol\,dm^{-3}}$ is $0.600$ , but the reproducible measured value is only $0.560$ , so the high-concentration standard deviates negatively. The unknown absorbance lies safely within the retained range and gives $2.50\,\mathrm{mmol\,dm^{-3}}$ by interpolation.	5
04.1	Comparing U with V isolates ligand identity because oxidation state and co-ordination number are both unchanged. Any colour difference between U and V can therefore be attributed to the water-to-ammonia ligand change altering the d-level gap. Comparing V with X isolates oxidation state because ligand and co-ordination number are unchanged. No pair in the set isolates co-ordination number while keeping both ligand and oxidation state constant. U and W differ in both ligand identity and co-ordination number, so either change could account for a different absorbed wavelength.	Treat this as a controlled-comparison problem. U and V differ in only one variable, ligand identity. V and X differ in only cobalt oxidation state. W is four-coordinate, but its chloride ligands differ from every six-coordinate comparison partner, so no observed colour difference involving W can be assigned uniquely to co-ordination number.	5
05.1	$\Delta E_A=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(480\times10^{-9})=4.14\times10^{-19}\,\mathrm{J}$ . $\Delta E_B=3.21\times10^{-19}\,\mathrm{J}$ . Complex A has the larger d-electron energy gap because it absorbs the shorter wavelength. Using unrounded energies, $4.14375\times10^{-19}-3.20806\times10^{-19}=9.36\times10^{-20}\,\mathrm{J}$ per photon (carry at least 4 significant figures in the intermediate energies). The difference from the unrounded energy gap is $56.3\,\mathrm{kJ\,mol^{-1}}$ (accept $56.0$ – $56.3\,\mathrm{kJ\,mol^{-1}}$ if rounded energies were carried).	Convert both wavelengths to metres and apply $\Delta E=hc/\lambda$ without rounding the individual energies prematurely. Subtract the smaller gap from the larger to obtain the per-photon difference, then multiply by the Avogadro constant and divide by $1000$ to express the molar result in kilojoules per mole. The inverse wavelength relationship fixes the ordering independently of the arithmetic.	5

3.2.5.5 · Variable oxidation states (A-level only)

Tier 1 · Easy

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01.1	Vanadium is in oxidation state $+4$ and the ion is blue.	Oxygen contributes $-2$ and the ion has charge $+2$ , so $x-2=+2$ and $x=+4$ . The specified vanadium(IV) oxo ion is blue.	2
02.1	Oxidation state $+3$ ; $\mathrm{V^{3+}}$ .	The specified sequence is yellow vanadium(V), blue vanadium(IV), green vanadium(III), then violet vanadium(II). Green therefore identifies $\mathrm{V^{3+}}$ .	2

Tier 2 · Standard

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01.1	$\mathrm{2MnO_4^-+16H^++5C_2O_4^{2-}\rightarrow 2Mn^{2+}+8H_2O+10CO_2}$ .	Two manganate(VII) ions gain ten electrons in total to form two $\mathrm{Mn^{2+}}$ . Five ethanedioate ions lose ten electrons while forming ten $\mathrm{CO_2}$ . Add $16\mathrm{H^+}$ and eight waters to balance hydrogen and oxygen; the charges are $-2+16-10=+4$ on the left and $+4$ on the right.	4
02.1	$\mathrm{VO_2^+(aq)+2H^+(aq)+e^-\rightarrow VO^{2+}(aq)+H_2O(l)}$ : yellow $\mathrm{VO_2^+}$ forms blue $\mathrm{VO^{2+}}$ . $\mathrm{VO^{2+}(aq)+2H^+(aq)+e^-\rightarrow V^{3+}(aq)+H_2O(l)}$ : blue $\mathrm{VO^{2+}}$ forms green $\mathrm{V^{3+}}$ .	Balance oxygen with water, hydrogen with $\mathrm{H^+}$ and charge with one electron in each one-step reduction. The first equation has total charge $+2$ on both sides and the second has total charge $+3$ on both sides.	6
03.1	The blue-to-green change demonstrates variable oxidation states because M changes from $+3$ to $+2$ by reduction. The green-to-yellow change is ligand substitution because the oxidation state remains $+2$ while L replaces a ligand. Colour change alone is insufficient because both oxidation-state changes and ligand changes alter the d-electron energy gap.	Track oxidation state separately from colour. Electron gain changes $\mathrm{M^{3+}}$ to $\mathrm{M^{2+}}$ , directly showing two oxidation states. In the second step M stays at $+2$ and only its ligand environment changes. Either factor can change the wavelengths absorbed, so formulas or redox evidence are needed in addition to colour.	4

Tier 3 · Hard

Mark scheme for 3.2.5.5 Tier 3 · Hard
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01.1	$23.3\%$ iron by mass	Moles of $\mathrm{MnO_4^-}$ are $(0.00220)(17.30\times10^{-3})=3.806\times10^{-5}$ . The aliquot contains $5(3.806\times10^{-5})=1.903\times10^{-4}\,\mathrm{mol}$ of $\mathrm{Fe^{2+}}$ . The full solution contains ten times this amount, $1.903\times10^{-3}\,\mathrm{mol}$ , with iron mass $(1.903\times10^{-3})(55.8)=0.1062\,\mathrm{g}$ . Hence the percentage is $(0.1062/0.455)\times100=23.3\%$ .	6
02.1	A silver mirror, or a grey-black precipitate of silver, forms. $\mathrm{RCHO+2[Ag(NH_3)_2]^++3OH^-\rightarrow RCOO^-+2Ag+4NH_3+2H_2O}$ . The aldehyde is oxidised to a carboxylate ion and silver(I) is reduced to silver metal.	Each silver(I) ion gains one electron, so two complex ions produce two silver atoms. The aldehyde loses two electrons as it becomes carboxylate. The equation balances ligand atoms, hydrogen, oxygen and total charge at $-1$ on each side.	7
03.1	Before L is added, $\mathrm{M^{3+}}$ is reduced and $\mathrm{N^{2+}}$ is oxidised: $\mathrm{M^{3+}+N^{2+}\rightarrow M^{2+}+N^{3+}}$ . $E^\circ_{\mathrm{cell}}=+0.62-(+0.45)=+0.17\,\mathrm{V}$ . After L binds, $\mathrm{N^{3+}}$ is reduced and $\mathrm{[ML]^{2+}}$ is oxidised: $\mathrm{N^{3+}+[ML]^{2+}\rightarrow N^{2+}+[ML]^{3+}}$ . $E^\circ_{\mathrm{cell}}=+0.45-(+0.20)=+0.25\,\mathrm{V}$ . Ligand binding changes the relative stability of M's oxidation states and lowers the M(III)/M(II) reduction potential enough to reverse the spontaneous direction.	Use the more positive reduction potential as the reduction half-cell and reverse the other half-equation for oxidation. Without L, M(III) is the stronger oxidising agent and gives a positive $0.17\,\mathrm{V}$ cell. With L, the M couple is only $+0.20\,\mathrm{V}$ , so N(III) is now reduced and the reverse pairing gives $+0.25\,\mathrm{V}$ . No Nernst calculation is involved; the supplied ligand-dependent potentials describe the two conditions.	6
04.1	$\mathrm{2MnO_4^-+5H_2C_2O_4+6H^+\rightarrow2Mn^{2+}+10CO_2+8H_2O}$ . $n(\mathrm{MnO_4^-})=(0.0200)(0.01980)=3.96\times10^{-4}\,\mathrm{mol}$ . The aliquot contains $(5/2)(3.96\times10^{-4})=9.90\times10^{-4}\,\mathrm{mol}$ of ethanedioic acid. The full solution contains $9.90\times10^{-3}\,\mathrm{mol}$ of the hydrate. The hydrate has $M_r=1.250/(9.90\times10^{-3})=126.3$ . $x=(126.3-90.0)/18.0=2.02$ , so $x=2$ .	Balance manganate(VII) reduction against oxidation of ethanedioic acid to carbon dioxide to obtain the $2{:}5$ ratio. Convert the titre to manganate amount, scale by $5/2$ for ethanedioic acid, then scale the aliquot by ten. Dividing sample mass by hydrate amount gives its molar mass. The excess above anhydrous ethanedioic acid corresponds to approximately two water molecules.	6
05.1	Vanadium changes from $+5$ to $+2$ , so each vanadium atom gains three electrons. The total electron amount accepted is $3(0.0180)=0.0540\,\mathrm{mol}$ . Each zinc atom supplies two electrons, so $n(\mathrm{Zn})=0.0540/2=0.0270\,\mathrm{mol}$ . The minimum zinc mass is $(0.0270)(65.4)=1.77\,\mathrm{g}$ . The first intermediate is blue $\mathrm{VO^{2+}}$ , followed by green $\mathrm{V^{3+}}$ . The final species is violet $\mathrm{V^{2+}}$ .	Track the oxidation-state fall from vanadium(V) to vanadium(II): each vanadium requires three electrons. Zinc oxidation to $\mathrm{Zn^{2+}}$ supplies two electrons per zinc atom, so divide the total electron demand by two before converting to mass. The specified reduction sequence is yellow vanadium(V), blue vanadium(IV), green vanadium(III), then violet vanadium(II).	6

3.2.5.6 · Catalysts (A-level only)

Tier 1 · Easy

Mark scheme for 3.2.5.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Adsorption, reaction at active sites and desorption; a poison occupies active sites so fewer reactant particles can adsorb and react.	Follow a reactant through the catalytic cycle: it first binds to the surface, reacts by the lower-activation-energy route, then leaves so the site can be reused. A poison blocks these sites and reduces the available catalytic surface.	4
02.1	Heterogeneous catalysis; the reaction occurs at active sites on the iron surface.	The catalyst and reactants are in different phases, which defines heterogeneous catalysis. Gas particles adsorb and react at surface active sites.	2

Tier 2 · Standard

Mark scheme for 3.2.5.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{V_2O_5+SO_2\rightarrow V_2O_4+SO_3}$ and $\mathrm{V_2O_4+\tfrac{1}{2}O_2\rightarrow V_2O_5}$ .	$\mathrm{SO_2}$ first removes an oxygen atom from $\mathrm{V_2O_5}$ , producing $\mathrm{SO_3}$ and $\mathrm{V_2O_4}$ . Oxygen then reoxidises $\mathrm{V_2O_4}$ to the original $\mathrm{V_2O_5}$ . Adding the equations cancels both vanadium oxides and gives $\mathrm{SO_2+\tfrac{1}{2}O_2\rightarrow SO_3}$ .	4
02.1	The $\mathrm{2Mn^{2+}}$ consumed in the first step is formed again in the second step, so it cancels when the equations are added. The overall equation is $\mathrm{S_2O_8^{2-}+2Fe^{2+}\rightarrow2SO_4^{2-}+2Fe^{3+}}$ .	Add the left- and right-hand sides. Cancel both $\mathrm{2Mn^{3+}}$ and $\mathrm{2Mn^{2+}}$ because each appears on opposite sides. Regeneration and cancellation show that $\mathrm{Mn^{2+}}$ changes the route but is not consumed overall; the remaining equation balances atoms and has total charge $+2$ on each side.	4
03.1	The support disperses the metal and exposes a larger surface area. More active sites are therefore available for adsorption and reaction, increasing the rate from $1.0$ to $8.5$ . The poison binds to or blocks active sites. Fewer reactant particles can adsorb and react, so the poisoned supported catalyst is slower even than the unsupported lump.	Keep the amount of active metal fixed and compare accessible sites. Dispersal prevents most of the metal being buried inside a lump, so many more surface sites can participate. Poisoning then makes those exposed sites unavailable, accounting for the large loss of activity.	4

Tier 3 · Hard

Mark scheme for 3.2.5.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{S_2O_8^{2-}+2Fe^{2+}\rightarrow 2SO_4^{2-}+2Fe^{3+}}$ and $\mathrm{2Fe^{3+}+2I^-\rightarrow 2Fe^{2+}+I_2}$ ; adding cancels both iron species.	Peroxodisulfate first oxidises two $\mathrm{Fe^{2+}}$ ions to $\mathrm{Fe^{3+}}$ while it is reduced to sulfate. The resulting $\mathrm{Fe^{3+}}$ then oxidises iodide to iodine and is reduced back to $\mathrm{Fe^{2+}}$ . Add the steps and cancel $\mathrm{2Fe^{2+}}$ and $\mathrm{2Fe^{3+}}$ from opposite sides, leaving exactly the stated overall reaction.	6
02.1	Vary the initial concentration of added $\mathrm{Mn^{2+}}$ and measure the time for the purple manganate(VII) colour to disappear, or use a colorimeter to follow absorbance. Include a control with no added $\mathrm{Mn^{2+}}$ . Keep temperature, total volume, acid concentration and initial reactant concentrations constant. Trials with more added $\mathrm{Mn^{2+}}$ decolourise sooner; the control is slow initially but accelerates as $\mathrm{Mn^{2+}}$ forms.	Change only the initial catalyst concentration and use the same objective colour endpoint or absorbance threshold. The no-catalyst control demonstrates autocatalytic acceleration, while the fixed conditions prevent concentration or temperature from explaining the rate difference.	8
03.1	Claim 1 is false: an intermediate is formed in one step and consumed in another, so it cancels from the overall equation. Claim 2 is false: a catalyst does not change the equilibrium constant or equilibrium position at a fixed temperature. Claim 3 is false: the alternative route lowers the activation energy for both forward and reverse reactions, so equilibrium is reached sooner. Claim 4 is false: adsorption at active sites describes heterogeneous catalysis; a homogeneous catalyst is in the same phase as the reactants and forms intermediates. Claim 5 is true: a homogeneous catalyst forms an intermediate species with a reactant. Claim 6 is true: a support increases the surface area available per unit mass of metal.	Judge all six claims before correcting the four false ones. In a homogeneous catalytic cycle the catalyst forms an intermediate with a reactant; adding the steps makes the intermediates and regenerated catalyst cancel from the overall equation. A catalyst changes the pathway rather than the thermodynamic energies of reactants and products, so it accelerates both directions without changing $K$ . Adsorption is characteristic of a heterogeneous catalyst, whose metal can be dispersed on a support to expose more surface area per unit mass.	6
04.1	$\mathrm{S_2O_8^{2-}+2Co^{2+}\rightarrow2SO_4^{2-}+2Co^{3+}}$ is spontaneous. Its $E^\circ_{\mathrm{cell}}=+2.01-(+1.82)=+0.19\,\mathrm{V}$ . $\mathrm{2Co^{3+}+2I^-\rightarrow2Co^{2+}+I_2}$ is spontaneous. Its $E^\circ_{\mathrm{cell}}=+1.82-(+0.54)=+1.28\,\mathrm{V}$ . Adding the steps gives the stated overall reaction and cancels both cobalt species, so the $\mathrm{Co^{2+}/Co^{3+}}$ couple can provide a catalytic cycle. The vanadium(III)/vanadium(II) potential is below the iodine/iodide potential, so oxidation of iodide by $\mathrm{V^{3+}}$ would have $E^\circ_{\mathrm{cell}}=-0.26-(+0.54)=-0.80\,\mathrm{V}$ and is not spontaneous.	A mediator must lie between the two substrate reduction potentials. First pair peroxodisulfate reduction with oxidation of cobalt(II), then pair reduction of cobalt(III) with iodide oxidation. Write each resulting reaction in its spontaneous direction and only then subtract the relevant reduction potentials. Both cobalt steps have positive cell potentials and sum to the uncatalysed equation. The vanadium couple fails the iodide-oxidation step because its potential lies below $+0.54\,\mathrm{V}$ .	6
05.1	The unsupported catalyst activity is $18.0/2.40=7.50\,\mathrm{mol\,min^{-1}\,kg^{-1}}$ of metal. The unpoisoned supported catalyst activity is $9.00/0.300=30.0\,\mathrm{mol\,min^{-1}\,kg^{-1}}$ of metal. The supported catalyst is $30.0/7.50=4.00$ times as active per unit mass of metal because dispersal exposes more active surface sites. Poisoning lowers the supported catalyst activity from $30.0$ to $2.25/0.300=7.50\,\mathrm{mol\,min^{-1}\,kg^{-1}}$ , a $75.0\%$ loss, because poison blocks active sites. The poisoned bed causes lost production and must be regenerated or replaced, adding treatment or new-catalyst cost and shutdown time; the operator should compare regeneration cost with replacement cost and the value of restored output.	Divide each production rate by the mass of active metal before comparing catalyst performance. Supporting raises activity per unit metal fourfold by exposing more surface sites. After poisoning, the supported bed returns only one quarter of its original activity, so activity loss is $75.0\%$ . Blocked sites reduce output even though the metal remains present, creating both a direct regeneration-or-replacement expense and an indirect downtime or lost-production cost.	5

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

Tier 1 · Easy

Mark scheme for 3.2.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{Cu^{2+}}$ gives a blue precipitate; $\mathrm{Fe^{3+}}$ gives a brown precipitate.	Hydroxide deprotonates water ligands and forms insoluble metal hydroxides. Copper(II) hydroxide is blue, whereas iron(III) hydroxide is brown.	2
02.1	$\mathrm{Al^{3+}}$ gives a white precipitate that is insoluble in excess ammonia. $\mathrm{Cu^{2+}}$ gives a blue precipitate that dissolves in excess ammonia to form a deep-blue solution.	A small amount of ammonia acts as a base and forms the two hydroxide precipitates. Excess ammonia does not dissolve aluminium hydroxide, but it replaces water ligands around copper(II), forming the soluble deep-blue ammine complex.	4

Tier 2 · Standard

Mark scheme for 3.2.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The precipitate dissolves to give a deep-blue solution of $\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}$ . $\mathrm{[Cu(H_2O)_6]^{2+}+4NH_3\rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}+4H_2O}$ .	A small amount of ammonia acts as a base and forms the blue hydroxide precipitate. With excess ammonia, ligand substitution becomes dominant: four water ligands are replaced by ammonia, the solid dissolves, and the soluble deep-blue tetraammine complex forms with charge still $2+$ .	5
02.1	$\mathrm{Fe^{2+}}$ gives a green precipitate: $\mathrm{Fe^{2+}(aq)+2OH^-(aq)\rightarrow Fe(OH)_2(s)}$ . $\mathrm{Fe^{3+}}$ gives a brown precipitate: $\mathrm{Fe^{3+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)}$ .	Add hydroxide to fresh portions and distinguish the precipitates by colour. Each neutral hydroxide needs enough $\mathrm{OH^-}$ ions to cancel the metal-ion charge, so both atoms and charge balance.	6
03.1	$\mathrm{Fe^{2+}}$ requires $10.0\,\mathrm{cm^3}$ and forms a green precipitate. $\mathrm{Fe^{3+}}$ requires $15.0\,\mathrm{cm^3}$ and forms a brown precipitate.	Each sample contains $(0.0500)(0.0100)=5.00\times10^{-4}\,\mathrm{mol}$ of metal ions. Forming $\mathrm{Fe(OH)_2}$ needs $1.00\times10^{-3}\,\mathrm{mol}$ of hydroxide, which occupies $0.0100\,\mathrm{dm^3}$ . Forming $\mathrm{Fe(OH)_3}$ needs $1.50\times10^{-3}\,\mathrm{mol}$ , which occupies $0.0150\,\mathrm{dm^3}$ . The precipitates are green and brown respectively.	4

Tier 3 · Hard

Mark scheme for 3.2.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{Fe^{2+}}$ gives a green $\mathrm{FeCO_3}$ precipitate with no gas: $\mathrm{[Fe(H_2O)_6]^{2+}+CO_3^{2-}\rightarrow FeCO_3(s)+6H_2O}$ . $\mathrm{Fe^{3+}}$ gives a brown hydroxide precipitate and effervescence: $\mathrm{2[Fe(H_2O)_6]^{3+}+3CO_3^{2-}\rightarrow 2[Fe(H_2O)_3(OH)_3](s)+3CO_2+3H_2O}$ . The $3+$ aqua ion is more acidic and transfers protons to carbonate, forming $\mathrm{CO_2}$ .	For the $2+$ ion, carbonate acts as a precipitating ligand and forms green iron(II) carbonate. The higher charge-to-size ratio of $\mathrm{Fe^{3+}}$ strongly polarises its water ligands, so carbonate instead accepts protons; hydrolysis gives the brown hydroxide and carbonic acid decomposes to carbon dioxide and water. Atom and charge counts in the two equations confirm the different pathways.	8
02.1	Use equal fresh portions and add aqueous sodium hydroxide dropwise. A white precipitate indicates $\mathrm{Al^{3+}}$ , green indicates $\mathrm{Fe^{2+}}$ , blue indicates $\mathrm{Cu^{2+}}$ and brown indicates $\mathrm{Fe^{3+}}$ . Add excess hydroxide only to the white precipitate; it dissolves to give a colourless solution. Compare with known-ion controls under the same volumes and concentrations. $\mathrm{Al^{3+}(aq)+3OH^-(aq)\rightarrow Al(OH)_3(s)}$ and $\mathrm{Al(OH)_3(s)+OH^-(aq)\rightarrow[Al(OH)_4]^-(aq)}$ .	The four initial precipitate colours give a one-reagent split, and the excess step confirms the only white result by amphoteric dissolution. Known-ion controls protect against assigning an ion from a poor-quality or contaminated reagent.	9
03.1	The $3+$ metal ion has a greater charge and is smaller than the corresponding $2+$ ion, so it has greater charge density. It attracts electron density from the oxygen atoms of its water ligands more strongly and polarises their O-H bonds more strongly. A proton is therefore transferred to water more readily, so the $3+$ aqua ion produces a higher concentration of $\mathrm{H_3O^+}$ . $\mathrm{[M(H_2O)_6]^{3+}(aq)+H_2O(l)\rightleftharpoons[M(H_2O)_5(OH)]^{2+}(aq)+H_3O^+(aq)}$ . $\mathrm{[M(H_2O)_6]^{2+}(aq)+H_2O(l)\rightleftharpoons[M(H_2O)_5(OH)]^+(aq)+H_3O^+(aq)}$ .	Compare charge density before considering hydrolysis. The extra positive charge and smaller radius of $\mathrm{M^{3+}}$ strengthen its attraction for ligand oxygen and draw electron density away from each O-H bond. This makes proton transfer to a neighbouring water molecule more favourable. Both ions can hydrolyse as shown, but equilibrium lies further to the right for the $3+$ aqua ion.	6
04.1	Adding sodium hydroxide initially forms a white aluminium hydroxide precipitate and a blue copper(II) hydroxide precipitate. Excess sodium hydroxide dissolves the white precipitate as colourless $\mathrm{[Al(OH)_4]^-}$ while blue $\mathrm{Cu(OH)_2}$ remains. Filter the mixture to recover the blue copper(II) hydroxide solid. $\mathrm{Cu^{2+}(aq)+2OH^-(aq)\rightarrow Cu(OH)_2(s)}$ . $\mathrm{Al(OH)_3(s)+OH^-(aq)\rightarrow[Al(OH)_4]^-(aq)}$ . Add dilute acid carefully to the filtrate to re-form white aluminium hydroxide: $\mathrm{[Al(OH)_4]^-(aq)+H^+(aq)\rightarrow Al(OH)_3(s)+H_2O(l)}$ .	Use amphoteric behaviour to create a solubility difference. Hydroxide first precipitates both ions, but excess hydroxide converts only aluminium hydroxide into soluble tetrahydroxoaluminate. Filtration then separates the blue copper solid from the colourless aluminium-containing filtrate. Controlled acid addition removes one hydroxide equivalent from the aluminate ion and recovers aluminium hydroxide; too much acid would dissolve it again.	6
05.1	If the aluminium and copper amounts are $x$ and $y$ , then $x+y=0.0100$ . Hydroxide stoichiometry gives $3x+2y=0.0260$ . Solving gives $x=0.00600\,\mathrm{mol}$ of $\mathrm{Al^{3+}}$ . The copper amount is $y=0.00400\,\mathrm{mol}$ . The mixture is $60.0\%$ aluminium ions and $40.0\%$ copper ions by amount. Excess hydroxide dissolves the white $\mathrm{Al(OH)_3}$ , leaving a blue $\mathrm{Cu(OH)_2}$ precipitate in a colourless solution.	Let the two metal-ion amounts sum to the stated total. Complete precipitation needs three hydroxide ions per aluminium ion and two per copper ion, producing the second simultaneous equation from the $0.0260\,\mathrm{mol}$ hydroxide amount. Solve for both amounts and divide by the total for the percentages. In excess hydroxide, amphoteric aluminium hydroxide forms soluble aluminate but copper(II) hydroxide remains insoluble.	6