OCR A-Level Chemistry Unified chemistry (03), June 2024: Question 3
17 marks · Hard difficulty · Structured Questions
Analyze optical isomerism of an iron(III) ethanedioate complex, write oxidation half-equations, complete a titration table, calculate percentage uncertainty, and determine the formula of hydrated iron(II) ethanedioate using titration data.
Practise this questionQuestion
Question text
3 This question is about compounds and ions of iron(II) and iron(III) that contain ethanedioate ions,
C O 2–.
(a) The C O 2– ion, shown below, is an example of a bidentate ligand.
O O
C C
–O O–
(i) Explain what is meant by the term bidentate ligand.
… [2]
(ii) A complex ion E contains three C O 2– ions bonded to an iron(III) ion in an octahedral shape.
Complex ion E exists as a mixture of two optical isomers.
Draw 3D diagrams to show the structures of the optical isomers of E.
Include any overall charge.
[3]
(b) A student plans an investigation to find the number of waters of crystallisation, x, in a sample of
hydrated iron(II) ethanedioate, FeC2O4•xH2O.
The student decides to carry out a redox titration between solutions of iron(II) ethanedioate and
potassium manganate(VII) in acidic conditions.
(i) In the titration, both iron(II) ions and ethanedioate, C O 2–, ions are oxidised.
Construct half-equations for the oxidation of iron(II) and ethanedioate ions.
Oxidation of iron(II) ions
Oxidation of ethanedioate ions
[2]
(ii) The student prepares a 250.0 cm3 solution of iron(II) ethanedioate by dissolving 1.295 g of
FeC2O4•xH2O, in dilute sulfuric acid.
The student titrates 25.0 cm3 samples of this solution with 0.0200 mol dm−3 KMnO in the burette.
The student carries out a trial, followed by three further titrations.
The diagrams show the initial burette readings and the final burette readings for the student’s
three further titrations.
Titration 1 Titration 2 Titration 3
Initial reading Final reading Initial reading Final reading Initial reading Final reading
0 23 23 45 0 21
1 24 24 46 1 22
2 25 25 47 2 23
All burette readings are measured to the nearest 0.05 cm3.
Complete the titration table.
12 3
Final reading / cm3
Initial reading / cm3
Titre / cm3
[3]
(iii) The uncertainty in each burette reading is ±0.05 cm3.
Calculate the percentage uncertainty for the titre in Titration 1.
percentage uncertainty = … % [1]
(iv)* In the titration, 5 mol of iron(II) ethanedioate reacts with 3 mol of manganate(VII) ions.
Analyse the student’s results to find the number of waters of crystallisation, x, in the hydrated
iron(II) ethanedioate, FeC2O4•xH2O. [6]
Extra answer space if required.
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
3 (a) (i) species with two lone pairs (of electrons) 2 ALLOW species with lone pairs that form two
dative/coordinate bonds
forming dative (covalent)/co-ordinate bond(s)
OR ALLOW non-bonding pair for lone pair
donates electrons to a (central) metal atom/ion IGNORE LP for lone pair
IGNORE donates two pairs of electrons alone
(a) (ii) 3 IGNORE charges or dipoles on atoms within diagrams
(even if wrong)
Square brackets NOT required
Charge ALLOW unambiguous structures
Overall 3– charge shown (outside brackets) on at -------------------------------------------------------
least ONE optical isomer ✓ ALLOW –3 for 3–
3– must apply to the overall charge of structures
--------------------------- ------------------------
3D structures 3D: Must contain 2 ‘out wedges’, 2 ‘in wedges’ and 2
O O lines in plane of paper
3– 3–
O O OR 4 lines, 1 ‘out wedge’ and 1 ‘in wedge’:
O O
O O O O O O For bond into paper, ALLOW:
Fe Fe
O O
O O
O O O O ALLOW following geometry throughout:
O O
O O DO NOT ALLOW 3D structures
Fe simplified loop for oxalate, e.g.
1 mark for each isomer ✓✓
• Bonds MUST go to O– of (COO–) ligands
2 O O
DO NOT ALLOW impossible 3D diagrams, e.g. For incorrect element in centre, e.g. Cu,
AWARD 2 marks max
(b) (i) 2 For both half-equations,
Fe2+ → Fe3+ + e– ✓ ALLOW multiples
OR Fe2+ – e –→ Fe3+ ALLOW e for e–
IGNORE state symbols.
C O 2– → 2CO + 2e– ✓ ALLOW C O 2– → C O – + e–
24 2 2 4 2 4
OR C O 2– – 2e –→ 2CO C O 2– → C O + 2e–
24 2 2 4 2 4
2H O + C O 2– → 2CO 2– + 4H+ + 2e–
22 4 3
ALLOW 2C O 2– → C O 2– + 2e–
24 4 8
(b) (ii) 1 2 3 3
Final reading/cm3 23.55 45.40 22.75
Initial reading/cm3 1.90 23.55 1.20
Titre/cm3 21.65 21.85 21.55
Readings recorded to accuracy of burette
All readings recorded to two decimal places with
the last figure either 0 or 5
AND
Final and initial readings in correct rows ✓
Correct titres
All 3 titres correct to 2 DP: ✓ ✓
2 titres correct to 2 DP: ✓
(b) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE 1 Check Titres from 3b(ii) at top of response
IF % error = 0.46, award 1 mark ----------------------------------------------------------------------
---------------------------------------------------------------------- ALLOW % error from ANY of the 3 titres from 3b(ii)
20.05 OR from the mean titre
21.65 100 = 0.46 (%) 2 DP minimum
Calculator value: 0.46189… DO NOT ALLOW 0.50%
Question Answer 16 Marks Guidance
(b)* (iv) Please refer to the marking instructions on page 4 of this mark 6 *For mean titre,
scheme for guidance on how to mark this question. Check Titres from 3b(ii) at top of response*
Level 3 (5–6 marks)
How to answer it
Transition Metals & Redox Titrations Study Guide
What this question tests
This multi-step synoptic question assesses your understanding of transition metal complexes, bidentate ligands, optical isomerism in octahedral complexes, constructing redox half-equations, reading burette values, calculating percentage uncertainty, and performing complex multi-stage stoichiometry calculations to find water of crystallisation.
Definition of a Bidentate Ligand
✅ Correct Answer
A species that contains two lone pairs of electrons, both of which can form coordinate (dative covalent) bonds to a single central metal ion/atom.
❌ Common Errors
- Saying it "donates two pairs of electrons alone" without mentioning coordinate/dative bonds.
- Confusing bidentate with multidentate or polydentate definitions.
3D Optical Isomers of Octahedral Complex E
✅ Correct Answers & Mark Scheme
- An overall 3- charge shown clearly outside square brackets for at least one optical isomer.
- Correct 3D octahedral geometry featuring 2 wedges, 2 dashes, and 2 in-plane bonds representing three bidentate ethanedioate ( C₂O₄²⁻ ) ions.
- Bonds must connect directly to the oxygen atoms ( O⁻ ) of the ligands, not carbon atoms.
🧠 Exam Technique
When drawing optical isomers of octahedral complexes with three bidentate ligands, draw the mirror image carefully. Ensure bonds connect to O . Avoid simplified loop-style diagrams as they are explicitly disallowed by OCR examiners.
Redox Half-Equations
✅ Correct Answers
Oxidation of iron(II):
Fe²⁺ → Fe³⁺ + e⁻ (or Fe²⁺ - e⁻ → Fe³⁺ )
Oxidation of ethanedioate:
C₂O₄²⁻ → 2CO₂ + 2e⁻ (or C₂O₄²⁻ - 2e⁻ → 2CO₂ )
💡 Key Knowledge
Oxidation is the loss of electrons. Ensure charges and balancing coefficients (especially the 2CO₂ ) are correct before moving into titration calculations.
Completing the Titration Table
✅ Correct Table Values
| Measurement | Titration 1 | Titration 2 | Titration 3 |
|---|---|---|---|
| Final reading / cm³ | 23.55 | 45.40 | 22.75 |
| Initial reading / cm³ | 1.90 | 23.55 | 1.20 |
| Titre / cm³ | 21.65 | 21.85 | 21.55 |
🧠 Exam Technique
All burette readings must be recorded to two decimal places, ending in either .00 or .05 . Watch out for Titration 2 where the initial reading matches Titration 1's final reading.
Percentage Uncertainty Calculation
📐 Step-by-Step Calculation
- Identify the equipment error: Each burette reading has an uncertainty of ±0.05 cm³. Because a titre involves two readings (initial and final), total uncertainty = 0.05 × 2 = 0.10 cm³ .
- Use Titration 1 titre value: 21.65 cm³ .
- Calculate percentage uncertainty:
( 0.10 / 21.65 ) × 100 = 0.46189...% - Round to required significant figures/decimal places: 0.46% (minimum 2 decimal places required).
❌ Common Errors
Forgetting to multiply the burette uncertainty by 2 (accounting for both initial and final volume measurements). Do not round to 0.50%.
Determining Water of Crystallisation (x)
📐 Step-by-Step Calculation Guide
- Calculate Mean Titre: Use concordant titres from 3b(ii). Titrations 1, 2, and 3 give a mean value of ( 21.65 + 21.85 + 21.55 ) / 3 = 21.68 cm³ (or use selected concordant pair giving 21.60 cm³ depending on strict concordance checks; examiner accepts valid mean choices from student tables).
- Moles of MnO₄⁻ used:
Moles = ( concentration × volume ) / 1000
= ( 0.0200 × mean titre ) / 1000 - Moles of FeC₂O₄ in the 25.0 cm³ aliquot:
Use the stoichiometry given in the stem: 5 mol Fe²⁺/C₂O₄²⁻ react with 3 mol MnO₄⁻ . Scale moles of MnO₄⁻ accordingly. - Moles in whole 250 cm³ volumetric flask:
Multiply aliquot moles by 10 (since 250 / 25 = 10). This gives total moles of FeC₂O₄·xH₂O in the original 1.295 g sample. - Calculate Molar Mass of FeC₂O₄·xH₂O :
Molar Mass = Mass / Moles = 1.295 / total moles . - Find x :
Subtract the mass of anhydrous FeC₂O₄ ( 55.8 + (2 × 12.0) + (4 × 16.0) = 143.8 g mol⁻¹ ) from the total molar mass, then divide by the molar mass of water ( 18.0 g mol⁻¹ ) to find x .
Expected value: x = 2 ( FeC₂O₄·2H₂O ).
🧠 Examiner's Top Tips for Level 3 (5–6 Marks)
- Set out your working clearly with annotated step headers.
- Clearly state volume scale-up factors (×10 for 25 cm³ to 250 cm³).
- Double-check the 5:3 stoichiometric ratio provided in the question stem rather than assuming standard 1:1 or 5:1 ratios.
Topics
Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Practical Activity Groups · 5.3 Transition elements · 2.1 Atoms and reactions · PAG 2: Acid-base titration
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.