OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 21

17 marks · Medium difficulty · Structured Questions

Complete electron configurations, describe precipitation tests, determine an empirical formula and coordination number, draw a stereoisomer, and calculate percentage mass from a redox titration.

Practise this question

Question

Question 21 covers iron and copper chemistry across parts (a) to (d). Part (a) asks for full electron configurations of Fe and Cu2+, and observations when aqueous sodium hydroxide is added to FeSO4 and CuSO4 solutions. Part (b) shows the skeletal structure of pyridine, asks for the definition of a monodentate ligand, and gives percentage compositions by mass for an iron(II) complex with pyridine and chloride ligands to determine its empirical formula, ligand count, and coordination number. Part (c) depicts a copper(II) salicylate complex ion with a 2- charge and asks to draw its other stereoisomer and name the type of stereoisomerism. Part (d) describes an experimental procedure where a 9.51 g mixture of CuSO4.5H2O and FeSO4.7H2O is dissolved to 250.0 cm3, and 25.0 cm3 aliquots are reacted with excess KI and titrated against 0.0800 mol dm-3 Na2S2O3 (mean titre 31.15 cm3). Students must calculate the percentage by mass of CuSO4.5H2O to 3 significant figures, and explain why the titre increases if the solution is left exposed to air causing Fe(II) to oxidise to Fe(III).

Mark scheme

Show the mark scheme Mark scheme for Question 21: (a)(i) Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2; Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9 [2 marks]. (a)(ii) FeSO4: Green precipitate; CuSO4: (Pale) blue precipitate [2 marks]. (b)(i) Donates one electron pair (to a metal ion) AND forms a dative covalent / coordinate bond [1 mark]. (b)(ii) Ratio of Fe:C:H:N:Cl is calculated as 0.226:4.525:4.5:0.900:0.451 giving 1:20:20:4:2, empirical formula FeC20H20N4Cl2, 4 pyridine ligands, coordination number 6 [4 marks]. (c) Drawing of cis-isomer with carboxylate oxygens adjacent and phenolate oxygens adjacent; type of stereoisomerism is cis/trans or geometric [2 marks]. (d)(i) Moles of S2O3 2- = 2.492 x 10^-3 mol; moles Cu2+ in 250 cm3 = 2.492 x 10^-2 mol; mass of CuSO4.5H2O = 6.22 g; percentage = 65.4% [5 marks]. (d)(ii) Fe(III) reacts with I- to form I2 / oxidises I- [1 mark].

How to answer it

Reactions of Iron and Copper Compounds

What this question tests

  • Electronic Configurations: Sub-shell notation for transition elements and d-block ions (loss of 4s before 3d).
  • Precipitation Reactions: Identifying precipitates formed by Fe²⁺ and Cu²⁺ with aqueous hydroxide ions.
  • Ligands & Coordination: Defining monodentate ligands, calculating empirical formulae from percentage composition, and identifying coordination numbers.
  • Stereoisomerism: Recognising and drawing cis-trans isomerism in complex ions with bidentate ligands.
  • Redox Titrations: Performing multi-step quantitative calculations involving Cu²⁺, I⁻, and S₂O₃²⁻, including scaling factors and percentage mass.
Part (a) — Electronic Structure & Qualitative Analysis

Electronic Configurations and Precipitation with NaOH

Question parts (a)(i) and (a)(ii) [4 Marks Total]

✅ Correct Answers

(a)(i) Electron configurations:

  • Fe atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² (or 3d⁶ 4s² )
  • Cu²⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ (or 3d⁹ )

(a)(ii) Observations with NaOH(aq):

  • With FeSO₄(aq): Green precipitate
  • With CuSO₄(aq): (Pale) blue precipitate

💡 Key Knowledge

  • Transition metal atoms lose their 4s electrons before 3d when forming positive ions. A neutral Cu atom is [Ar] 3d¹⁰ 4s¹ ; removing 2 electrons gives [Ar] 3d⁹ .
  • With OH⁻(aq), [Fe(H₂O)₆]²⁺ forms insoluble [Fe(H₂O)₄(OH)₂](s), which is a green ppt (oxidises to brown on standing).
  • [Cu(H₂O)₆]²⁺ forms [Cu(H₂O)₄(OH)₂](s), a distinct pale blue ppt.

🧠 Exam Technique

Notice that the question stem starts with 1s² . You cannot use the noble gas shorthand [Ar] here because the paper specifically requires you to complete the full sequence starting from 1s².

❌ Common Errors

  • Writing 4s² 3d⁷ for Cu²⁺ (forgetting Cu is an exception with a 4s¹ 3d¹⁰ ground state).
  • Writing "dark blue precipitate" for CuSO₄. Dark blue solutions only form with excess aqueous ammonia via ligand substitution, not with NaOH!
  • Omitting the word "precipitate" or "solid" (e.g. stating just "blue solution").
Mark allocation: 1 mark for Fe config, 1 mark for Cu²⁺ config; 1 mark for green ppt, 1 mark for blue ppt.
Part (b) — Ligands & Empirical Formula of Complexes

Pyridine Complex of Iron(II)

Question parts (b)(i) and (b)(ii) [5 Marks Total]

✅ Correct Answers

(b)(i) Monodentate ligand definition:

Donates one electron pair to a metal ion AND forms one dative covalent / coordinate bond.

(b)(ii) Complex Deduction:

  • Empirical formula: FeC₂₀H₂₀N₄Cl₂ or Fe(C₅H₅N)₄Cl₂
  • Number of C₅H₅N ligands: 4
  • Coordination number of iron(II) ion: 6

📐 Step-by-Step Empirical Formula

1
Divide % by atomic mass:
Fe: 12.6 / 55.8 = 0.2258 mol
C: 54.3 / 12.0 = 4.525 mol
H: 4.5 / 1.0 = 4.500 mol
N: 12.6 / 14.0 = 0.900 mol
Cl: 16.0 / 35.5 = 0.4507 mol
2
Divide all by the smallest value (0.2258):
Fe = 1.00 : C = 20.0 : H = 19.9 : N = 3.99 : Cl = 2.00
Ratio = 1 : 20 : 20 : 4 : 2
3
Molar mass check:
55.8 + (20 × 12) + (20 × 1) + (4 × 14) + (2 × 35.5) = 442.8 g mol⁻¹.
Matches given molar mass! Each pyridine is C₅H₅N, so 4 × (C₅H₅N) accounts for all C, H, and N.

🧠 Exam Technique: Coordination Number

Coordination number is the total number of coordinate bonds formed to the central metal ion. Here, the complex has 4 monodentate pyridine ligands AND 2 chloride ligands (which also act as monodentate ligands here inside the coordination sphere): 4 + 2 = 6.

❌ Common Errors

  • Writing coordination number as 4 (forgetting the two Cl⁻ ions act as ligands to complete an octahedral complex).
  • Forgetting both halves of the ligand definition: you must state donating one lone pair and forming one coordinate bond.
Mark allocation: (b)(i) 1 mark; (b)(ii) 2 marks for empirical formula, 1 mark for 4 ligands, 1 mark for coordination number = 6.
Part (c) — Stereoisomerism in Complexes

Copper(II) Salicylate Complex

Question part (c) [2 Marks Total]

✅ Correct Answers

Type of stereoisomerism: Cis-trans (allow geometric or E/Z)

Description of the other isomer (cis-isomer):
In the given structure (trans), both phenolic oxygen atoms are opposite (180°) and both carboxylate oxygen atoms are opposite (180°).

To draw the cis-isomer:
  • Central Cu atom bonded to 4 oxygen atoms in square planar arrangement.
  • Place the two phenolic O atoms adjacent (at 90°) to each other.
  • Place the two carboxylate C=O / C–O groups adjacent (at 90°) to each other.
  • Retain the benzene rings completing the chelating rings, with overall charge 2−.

💡 Key Knowledge

  • Salicylate acts as a bidentate ligand, coordinating via the deprotonated phenolic oxygen and one carboxylate oxygen.
  • Square planar or octahedral complexes containing two non-symmetrical bidentate ligands exhibit cis-trans isomerism based on whether identical donor atoms are adjacent (90°, cis) or opposite (180°, trans).
Mark allocation: 1 mark for correct cis-structure (brackets/charges can be ignored); 1 mark for cis-trans / geometric.
Part (d) — Quantitative Redox Titration

Determination of CuSO₄·5H₂O in a Mixture

Question parts (d)(i) and (d)(ii) [6 Marks Total]

📐 Calculation Breakdown: Part (d)(i) [5 Marks]

1
Calculate moles of thiosulfate used in titration:
n(S₂O₃²⁻) = (31.15 cm³ × 0.0800 mol dm⁻³) / 1000 = 2.492 × 10⁻³ mol
2
Relate S₂O₃²⁻ to Cu²⁺ using stoichiometric ratios:
From Step 3 equation: 1 mol I₂ ≡ 2 mol S₂O₃²⁻ → n(I₂) = 2.492 × 10⁻³ / 2 = 1.246 × 10⁻³ mol.
From Step 2 equation: 2 mol Cu²⁺ ≡ 1 mol I₂.
Therefore, 2 mol Cu²⁺ ≡ 2 mol S₂O₃²⁻ (a 1:1 overall ratio).
n(Cu²⁺ in 25.0 cm³) = 2.492 × 10⁻³ mol
3
Scale up to the full 250.0 cm³ volumetric flask:
Scaling factor = 250.0 / 25.0 = 10
n(Cu²⁺ in 250.0 cm³) = 2.492 × 10⁻³ × 10 = 2.492 × 10⁻² mol
4
Calculate mass of hydrated copper(II) sulfate (CuSO₄·5H₂O):
M(CuSO₄·5H₂O) = 63.5 + 32.1 + (4 × 16.0) + (5 × 18.0) = 249.6 g mol⁻¹
Mass = 2.492 × 10⁻² mol × 249.6 g mol⁻¹ = 6.220 g
5
Calculate percentage by mass (to 3 significant figures):
% mass = (6.220 g / 9.51 g) × 100 = 65.4%

✅ Part (d)(ii) Explanation [1 Mark]

Why the titre is greater than 31.15 cm³ when exposed to air:

  • Fe²⁺ oxidises to Fe³⁺ upon exposure to air.
  • Fe³⁺ / iron(III) oxidises iodide ions (I⁻) to iodine (I₂):
    2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
  • This produces extra I₂, which requires a larger volume of Na₂S₂O₃ to reach the endpoint.

❌ Common Calculation Traps in (d)(i)

  • Molar mass error: Using 159.6 g mol⁻¹ (anhydrous CuSO₄) instead of 249.6 g mol⁻¹ for hydrated crystals → gives 41.8%.
  • Missing the scaling factor: Forgetting to multiply by 10 for the volumetric flask dilution → gives 6.54%.
  • Incorrect Cu²⁺:S₂O₃²⁻ ratio: Dividing or multiplying by 2 unnecessarily → gives 32.7%.
  • Rounding too early: Rounding intermediate values to 2 SF leading to 65.3% or 65.5%. Always keep full calculator values until the final step.
Mark allocation: (d)(i) 5 marks total (1 mark per calculation step shown above); (d)(ii) 1 mark for Fe³⁺ reacts with I⁻ / produces extra I₂.

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 1.1 Practical skills assessed in a written examination · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 5.3 Transition elements

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.