AQA A-Level Chemistry Paper 1, June 2025: Question 3

23 marks · Hard difficulty · State/Explain/Numerical

Analyze the chemistry of copper, including electron configuration, ligand substitution, color theory, redox titration calculations to identify an unknown metal, and drawing stereoisomers of a complex ion.

Practise this question

Question

Question text

03 This question is about the chemistry of copper.

03.1 Deduce the number of unpaired electrons in a Cu2+ ion.

[1 mark]

03.2 Describe how water acts as a ligand in the formation of [Cu(H O) ]2+

[1 mark]

03.3 A solution of sodium carbonate is added to a blue solution containing

[Cu(H O) ]2+ ions.

State what is observed.

Give an equation for the reaction.

[2 marks]

Observation

Equation

03.4 When a solution containing [Cu(H O) ]2+ ions reacts with an excess of

concentrated hydrochloric acid, a ligand exchange reaction occurs.

[Cu(H O) ]2+(aq) + 4 Cl–(aq) → [CuCl ]2–(aq) + 6 H O(l)

26 4 2

Blue Yellow

The shape and the colour of the [CuCl ]2– ion are different from the shape and the

colour of the [Cu(H O) ]2+ ion.

Explain

• the shape of each complex ion and why these shapes are different

• how colour arises in complex ions

• why the [CuCl ]2– ion has a yellow colour, different from the blue colour of the

[Cu(H O) ]2+ ion.

[6 marks]

03.5 An excess of water, followed by an excess of concentrated aqueous ammonia, is

added to a solution containing [CuCl ]2– ions.

State what is observed.

Give an equation for the overall change that occurs.

[2 marks]

Observation

Equation

A solution contains M2[Cu(C2O4)2(H2O)2] where M is a metal.

A student does a titration to determine the identity of M.

• 2.52 g of solid M2[Cu(C2O4)2(H2O)2] are dissolved in dilute sulfuric acid.

• This solution is transferred to a volumetric flask and made up to 250 cm3 with

dilute sulfuric acid.

*09* • A 25.0 cm3 portion is pipetted into a conical flask.

• The mixture in the conical flask is heated to 60 °C and then titrated with

0.0200 mol dm–3 KMnO solution.

• Several titrations are done and the student’s results are shown in Table 3.

In this titration, only the C O 2– ions react with the MnO – ions.

24 4

2 mol of MnO – ions react with 5 mol of C O 2– ions.

42 4

Table 3

Rough 1 2 3

Final reading / cm3 29.5 28.60 30.35 34.05

Initial reading / cm3 0.0 0.00 1.55 5.55

Titre / cm3 29.5 28.60 28.80 28.50

03.6 Give an equation for the reaction between C O 2– ions and

MnO – ions in acidic solution.

[1 mark]

03.7 Calculate the mean titre from the student’s results.

[1 mark]

cm3

03.8 Calculate the Mr of M2[Cu(C2O4)2(H2O)2]

Identify M.

[7 marks]

Mr of M2[Cu(C2O4)2(H2O)2]

03.9 Draw the structure of two stereoisomers of the complex ion in M2[Cu(C2O4)2(H2O)2]

and include the charge on each structure.

[2 marks]

Identity of M

Structure 1 Structure 2

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

1 or one 1

03.1

(1 x AO3)

(Water) has a lone pair (of electrons on the O) that it can donate (to Allow lone pair donor/electron pair donor/both

copper) electrons

03.2 Or

(1 x AO1)

(Water) has a lone pair (of electrons on the O) that forms a

co-ordinate/dative covalent bond (to copper)

M1 Blue-green/green-blue/green precipitate/solid M1 Do not accept blue

M2 [Cu(H O) ]2+ + CO 2– → CuCO + 6 H O

03.3 2 6 3 3 2 M2 [Cu(H O) ]2++Na CO →CuCO +6 H O +2Na+

26 2 3 3 2 (1 x AO1,

Ignore state symbols 1 x AO2)

Allow multiples

– A-LEVEL CHEMISTRY – –

Stage 1

This question is marked using levels of response. Refer to the 2+

1a Octahedral [Cu(H2O6)] (with 6 ligands) and

Mark Scheme Instructions for Examiners for guidance on how to 2–

tetrahedral [CuCl4] (with 4 ligands) or diagram of

mark this question.

both complexes 15

Level 3

1b Cl– ligands are bigger (so only 4 fit around the

5–6 marks 2+

Cu ) or reverse argument

All stages are covered and the description of each stage is

generally correct and virtually complete. Stage 2

Answer is communicated coherently and shows a logical 2a d-orbitals split (into ground state and excited

progression from stage 1 to stage 2 and stage 3. state)

Level 2

2b (some wavelengths/frequencies) of white/visible

3–4 marks light are absorbed (by the complex)

All stages are covered but the description of each stage may be

03.4 incomplete or may contain inaccuracies OR two stages are 2c electrons are excited/promoted (from ground (2 x AO1,

covered and the explanations are generally correct and virtually state to excited state) OR d–d transitions occur 2 x AO2,

complete. 2 x AO3)

Stage 3

Answer is mainly coherent and shows progression from stage 1 to 3a Different ligand/co-ordination number changes

stage 2 and/or stage 3. the energy gap/splitting OR octahedral and

Level 1 tetrahedral complexes have different energy

1–2 marks gap/splitting (in the d orbitals)

Two stages are covered but the description of each stage may be 3b The energy gap is related to the

incomplete or may contain inaccuracies, OR only one stage is wavelength/frequency OR ∆E = h𝜈 OR ∆E = hf OR

covered but the explanation is generally correct and virtually ∆E = hc ÷ 𝜆

complete.

Answer includes isolated statements and these are presented in a 3c Wavelengths/frequencies/colour not absorbed

logical order. are transmitted/reflected (not emitted) OR light

reflected is complementary to light absorbed.

– A-LEVEL CHEMISTRY – –

Level 0

03.4 0 marks

(cont)

Insufficient correct chemistry to gain a mark.

M1 (yellow solution turns to a) deep/dark blue solution 2

03.5 M2 [CuCl ]2− + 4 NH + 2 H O → [Cu(NH ) (H O) ]2+ + 4 Cl− (1 x AO1,

43 2 3 4 2 2

1 x AO2)

2 MnO – + 5 C O 2– + 16 H+ → 2 Mn2+ + 10 CO + 8 H O Allow multiples 1

42 4 2 2

03.6

Ignore state symbols (1 x AO2)

28.55 cm3 1

03.7

(1 x AO2)

– A-LEVEL CHEMISTRY – –

M1 n MnO – = 28.55 × 10–3 × 0.0200 = 5.71 ×10–4 mol M1 n MnO – = 28.55 × 10–3 × 0.0200 = 5.710 × 10–4 mol

2– 5 2– –4 5 –3

M2 n C2O4 = M1 × 2 M2 n C2O4 = 5.71 × 10 × 2 = 1.4275 × 10 mol

M3 n C O 2– in 250 cm3 = M2 × 10 M3 n C O 2– in 250 cm3 = 1.4275 × 10–3 × 10 = 1.4275 ×10–2

24 2 4

M3 mol

M4 n complex = 2 –2

1.4275 × 10 –3

M4 n complex = 2 = 7.1375 × 10 mol

2.52

M5 Mr complex = M4 and answer on line 7

2.52

03.8 M5 Mr complex = –3 = 353 and answer on line

7.1375× 10 (5 x AO2,

2 x AO3)

If answer on line is not correct but final answer is correct lose

M6 Mr of 2M = M5 – 275.5 M6 Mr of 2M =353 – 275.5 = 77.5

M7 Mr of M = M6 ÷ 2 and metal with +1 oxidation state M7 Mr of M = 38.8 so potassium / K on answer line

and answer on line M7 Allow K+

– A-LEVEL CHEMISTRY – –

any 2 from these 3 structures Ligands must be bonded via correct atoms

1 mark for structure

1 mark for 2- charge outside both brackets

03.9

(2 x AO3)

allow one mark for 2 structures with no charge shown

How to answer it

Mastering Transition Metal Chemistry & Titrations

What this question tests

Electronic Configuration: Deducing unpaired d-electrons in transition metal ions ( Cu²⁺ ).
Ligand Bonding: Explaining coordinate (dative covalent) bonding in hexaaqua complexes.
Inorganic Reactions: Precipitation reactions with carbonates and ligand substitution reactions with ammonia.
Physical Chemistry of Complexes: Explaining shapes, d-orbital splitting, light absorption, and the origin of color in 6-mark structured responses.
Quantitative Analysis: Performing multi-step redox titration calculations to determine molar mass ( M_r ) and identify an unknown metal.
Stereoisomerism: Drawing 3D representations of cis/trans isomers containing bidentate ligands.

Part 03.1: Unpaired Electrons in Cu²⁺

Correct Answer

1 Mark (1 x AO3)

Key Knowledge

Copper atom configuration: [Ar] 3d¹⁰ 4s¹

When forming a Cu²⁺ ion, electrons are lost from the 4s subshell first, then the 3d subshell:

Cu²⁺ = [Ar] 3d⁹

In the five d-orbitals, four pairs are fully filled, leaving exactly one unpaired electron.

Part 03.2: Water as a Ligand

Correct Answer

Water has a lone pair of electrons on the oxygen atom that it donates to the copper ion to form a co-ordinate (dative covalent) bond.

1 Mark (1 x AO1)

Exam Technique

To secure this mark, you must explicitly state:

Where the lone pair is located (specifically on the oxygen atom).
The action (donating the lone pair to form a co-ordinate/dative bond).

Part 03.3: Reaction with Sodium Carbonate

Correct Answer

Observation: Blue-green (or green) precipitate / solid.

Equation:
[Cu(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → CuCO₃(s) + 6H₂O(l)
(Alternatively: [Cu(H₂O)₆]²⁺ + Na₂CO₃ → CuCO₃ + 6H₂O + 2Na⁺ )

2 Marks (1 x AO1, 1 x AO2)

Common Pitfalls

❌ Writing "blue precipitate"

The precipitate is specifically blue-green or green. Pure blue is not accepted here.

❌ Writing bubbles/effervescence

Unlike 3+ metal ions (which are highly polarizing and release CO₂ gas), 2+ metal ions like Cu²⁺ undergo simple precipitation with carbonate ions. No gas is produced!

Part 03.4: 6-Mark Level of Response (Ligand Substitution & Color)

Level 3: 5-6 Marks

To score full marks, your answer must be structured logically into three distinct stages:

Stage 1: Shapes & Coordination

[Cu(H₂O)₆]²⁺ is octahedral (coordination number 6).
[CuCl₄]²⁻ is tetrahedral (coordination number 4).
Why they differ: Chloride ( Cl⁻ ) ligands are larger than water ( H₂O ) ligands, so only 4 chloride ligands can fit around the central copper ion due to steric hindrance.

Stage 2: Origin of Color

In the presence of ligands, the d-orbitals split into two non-degenerate energy levels (ground state and excited state).
Electrons absorb specific frequencies/wavelengths of visible light to become excited (promoted) from the lower to the higher d-orbital level (d-d transitions).
The remaining frequencies of light are transmitted/reflected to show the complementary color.

Stage 3: Why Colors Differ

The different ligands ( Cl⁻ vs H₂O ) and different coordination numbers cause a different energy gap ( ΔE ) between the split d-orbitals.
Since ΔE = hν (or ΔE = hc/λ ), a different frequency/wavelength of light is absorbed.
This results in a different complementary color being transmitted/reflected.

Part 03.5: Reaction of [CuCl₄]²⁻ with Excess Water & Ammonia

Correct Answer

Observation: Yellow solution turns into a deep blue / dark blue solution.

Equation:
[CuCl₄]²⁻ + 4NH₃ + 2H₂O → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4Cl⁻

2 Marks (1 x AO1, 1 x AO2)

Exam Technique

When excess ammonia is added to any aqueous copper complex, the stable product is always the octahedral tetraamminediplastic complex: [Cu(NH₃)₄(H₂O)₂]²⁺ . Make sure to include the two remaining water ligands in your formula!

Parts 03.6 & 03.7: Titration Preparation

03.6 Redox Equation

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

1 Mark (1 x AO2)

03.7 Mean Titre Calculation

Concordant titres are within 0.10 cm³ of each other.

Looking at Table 3:

Titre 1 = 28.60 cm³
Titre 2 = 28.80 cm³
Titre 3 = 28.50 cm³

Concordant titres are Titre 1 and Titre 3.

Mean = (28.60 + 28.50) / 2 = 28.55 cm³

1 Mark (1 x AO2)

Part 03.8: 7-Mark Titration Calculation

Difficulty: Hard

Step-by-Step Solution

1 Calculate moles of manganate(VII) used:
n(MnO₄⁻) = Mean Titre × Concentration
n(MnO₄⁻) = 28.55 × 10⁻³ dm³ × 0.0200 mol dm⁻³ = 5.710 × 10⁻⁴ mol

2 Calculate moles of oxalate ions in the 25.0 cm³ sample:
Using the reacting ratio from 03.6 ( 2 MnO₄⁻ : 5 C₂O₄²⁻ ):
n(C₂O₄²⁻) = 5.710 × 10⁻⁴ mol × (5 / 2) = 1.4275 × 10⁻³ mol

3 Scale up to the 250 cm³ volumetric flask:
n(C₂O₄²⁻) in 250 cm³ = 1.4275 × 10⁻³ mol × 10 = 1.4275 × 10⁻² mol

4 Calculate moles of the complex:
The formula of the complex is M₂[Cu(C₂O₄)₂(H₂O)₂] . Each mole of complex contains 2 moles of oxalate ( C₂O₄²⁻ ) ligands.
n(complex) = n(C₂O₄²⁻) / 2 = 1.4275 × 10⁻² mol / 2 = 7.1375 × 10⁻³ mol

5 Calculate the molar mass (M_r) of the complex:
M_r = mass / moles = 2.52 g / 7.1375 × 10⁻³ mol = 353.06 g mol⁻¹ (rounds to 353)

6 Calculate the mass of 2M:
Subtract the known parts of the complex from the total M_r :
M_r([Cu(C₂O₄)₂(H₂O)₂]²⁻) = 63.5 (Cu) + 2 × 88.0 (C₂O₄) + 2 × 18.0 (H₂O) = 275.5
M_r of 2M = 353 - 275.5 = 77.5

7 Identify Metal M:
A_r of M = 77.5 / 2 = 38.75
Looking at the periodic table, the metal with an A_r closest to 38.8 is Potassium (K) (A_r = 39.1).
Since the complex ion has a 2- charge, two K⁺ ions are needed to balance it, confirming the identity.

7 Marks (5 x AO2, 2 x AO3)

Calculation Traps

The 2:1 Complex Ratio: Many students forget that there are 2 oxalate ligands per complex molecule. Failing to divide by 2 in Step 4 leads to an incorrect M_r of 176.5.
The 2M Trap: Remember that the formula contains M₂ . You must divide the remaining mass (77.5) by 2 to find the atomic mass of a single atom of M.

Part 03.9: Stereoisomers of [Cu(C₂O₄)₂(H₂O)₂]²⁻

How to Draw the Isomers

You must draw the cis and trans isomers of this octahedral complex:

Isomer 1 (Cis-isomer):

Draw a central Cu atom with 6 octahedral bonds (4 in a square plane, 1 up, 1 down).
Place the two monodentate H₂O ligands adjacent to each other (e.g., one on the top position, one on an adjacent equatorial position).
Draw the two bidentate oxalate ( C₂O₄²⁻ ) ligands connecting the remaining adjacent positions, forming rings.

Isomer 2 (Trans-isomer):

Place the two monodentate H₂O ligands directly opposite each other (one on the top vertical bond, one on the bottom vertical bond).
Draw the two bidentate oxalate ( C₂O₄²⁻ ) ligands wrapping around the equatorial positions.

2 Marks (2 x AO3)

Examiner's Drawing Rules

Correct Bonding Atoms: The coordinate bonds must point directly to the Oxygen (O) atoms of both the water and oxalate ligands. Do not point them at H or C!
Overall Charge: Put square brackets around both structures and write the 2- charge outside the top right of both brackets.

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance · 3.1.7 Oxidation, Reduction and Redox Equations · 3.2.5 Transition Metals · 3.2.6 Reactions of Ions in Aqueous Solution

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.