OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 18
18 marks · Hard difficulty · Structured Questions
Analyze reactions of transition metal ions including chromium complexes, vanadium redox potentials, and titration calculations involving iron(II) gluconate.
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Question text
18 This question is about reactions of ions and compounds of transition elements.
(a) A student carries out two experiments on a solution containing [Cr(H O) ]3+(aq).
Experiment 1
The student adds an excess of aqueous ammonia to a solution containing [Cr(H O) ]3+(aq)
until a purple solution is formed.
Experiment 2
The student carries out the following reaction sequence.
Step 1 NaOH(aq) is added slowly to a solution containing [Cr(H O) ]3+(aq) in a boiling
tube.
A grey–green precipitate forms.
Step 2 An excess of NaOH(aq) is added to the boiling tube.
The precipitate dissolves and a green solution forms containing a 6 coordinate
complex ion.
Step 3 H2O2 is added to the mixture and the boiling tube is heated.
A yellow solution forms.
Step 4 The solution in the boiling tube is acidified.
The solution now contains Cr O 2−(aq).
(i) What is the formula of the complex ion in the purple solution that forms in Experiment 1?
… [1]
(ii) Suggest an equation for the reaction in Experiment 2, Step 1.
Include state symbols.
… [1]
(iii) Draw a 3-D diagram for the shape of the complex ion that forms in Experiment 2, Step 2.
Include the charge of the ion.
[2]
(iv) What is the formula of the ion that causes the yellow colour in Experiment 2, Step 3?
… [1]
(v) State the colour of the solution that forms in Experiment 2, Step 4.
… 15 [1]
(b) Vanadium ions have four common oxidation states. Table 18.1 shows the colours of the ions
in aqueous solution.
Oxidation state of
Vanadium ion Colour
vanadium
+5 VO +(aq) yellow
+4 VO2+(aq) blue
+3 V3+(aq) green
+2 V2+(aq) violet
Table 18.1
(i) Complete the electron configuration of a V3+ ion.
1s2 … [1]
(ii) The student adds excess iron to a solution containing VO2+(aq) ions, and observes that
the colour of the solution changes from blue to green and then to violet.
Use the relevant standard electrode potentials shown in Table 18.2 to explain these
observations.
Redox system Eө / V
1 V2+(aq) + 2e− V(s) −1.18
2 Fe2+(aq) + 2e− Fe(s) −0.44
3 V3+(aq) + e− V2+(aq) −0.26
4 VO2+(aq) + 2H+ + e− V3+(aq) + H O(l) +0.34
5 Fe3+(aq) + e− Fe2+(aq) +0.77
6 VO +(aq) + 2H+ + e− VO2+(aq) + H O(l) +1.00
Table 18.2
… [3]
(iii) Construct an equation for the first colour change from blue to green.
… [1]
(c) Iron(II) gluconate, C12H22FeO14, is the active ingredient in some brands of iron supplements.
A student carries out an experiment to determine the mass of iron(II) gluconate in one tablet
of an iron supplement, using the method below.
Stage 1 The student crushes two tablets and dissolves the powdered tablets in dilute
sulfuric acid.
Stage 2 The student makes up the solution from Stage 1 to 250.0 cm3 in a volumetric flask.
Stage 3 The student then titrates 25.0 cm3 portions of the solution obtained in Stage 2 with
0.00200 mol dm−3 potassium manganate(VII).
The student obtains a mean titre of 13.50 cm3.
In this titration, 1 mol of manganate(VII) ions reacts with 5 mol of iron(II) ions.
(i) Explain why the student used 0.00200 mol dm−3 potassium manganate(VII) solution
for this titration, rather than the more usual concentration of 0.0200 mol dm−3 used in
manganate(VII) titrations.
… [1]
(ii) Use the student’s results to determine the mass, in mg, of iron(II) gluconate in one tablet.
Give your answer to 3 significant figures.
mass of iron(II) gluconate in one tablet = … mg [5]
(iii) Some iron supplements contain iron(II) sulfate or iron(II) fumarate.
The information in Table 18.3 is taken from the labels of two iron supplements, A and B.
Mass of iron compound in
Iron supplement Iron compound
one tablet / mg
A iron(II) sulfate, FeSO4 180
B iron(II) fumarate, C4H2FeO4 210
Table 18.3
Choose which iron supplement, A or B, would provide the greater mass of iron per tablet.
iron supplement: … [1]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
18 (a) (i) [Cr(NH ) ]3+(aq) 1 1.1 IGNORE state symbols
(ii) CrCl3(aq) + 3NaOH(aq) → Cr(OH)3(s) + 3NaCl(aq) 1 2.8 IGNORE square brackets around precipitate formulae
or
Cr3+(aq) + 3OH–(aq) → Cr(OH) (s) ALLOW
3 3+ –
state symbols required [Cr(H2O)6] (aq) + 3OH (aq) Cr(OH)3(H2O)3(s)+3H2O(l)
ALLOW ‘hybrid’ equations,
Eg 3+ +
Cr (aq) + 3NaOH(aq) → Cr(OH)3(s) + 3Na (aq)
3+ –
[Cr(H2O)6] (aq) + 3OH (aq) → Cr(OH)3(s) + 6H2O(l)
3+
[Cr(H2O)]6 (aq)+ 3NaOH(aq)
+
Cr(OH)3(s) + 6H2O(l) +3Na (aq)
(iii) 2 Must contain 2 ‘out wedges’, 2 ‘in wedges’ and 2 lines
in plane of paper OR 4 lines, 1 ‘out wedge’ and 1 ‘in
wedge’:
3-D diagram with all bonds through O in OH 1.1
3– charge 2.3 ALLOW dotted line OR unfilled wedge as alternatives
for dotted wedge
IGNORE charges inside brackets
(iv) CrO 2– 1 3.1 IGNORE compounds e.g. Na CrO
42 4
(v) orange 1 1.1
(b) (i) (1s2)2s22p63s23p63d2 1 1.1 ALLOW upper case D, etc. and subscripts, e.g. 3D
If included, ALLOW 4s0
18 b (ii) Explanation of colours 3 3.1 ×2
VO2+ goes to V3+ (green) AND then V3+ goes to V2+
H432/01 Mark Schemes June 2019
12 AO
element
(violet)
Explanation using Eo values IGNORE ‘lower/higher’
(Eo of) system 4 (VO2+/V3+) is more positive / less ALLOW reverse argument System 2 more negative
3.2 ×1
negative than system 2 (Fe2+/Fe ) OR than system 4 etc
(Eo of) system 3 (V3+/V2+,) is more positive / less E = (+)0.78 V for system 4 + system 2 reaction
negative than system 2 (Fe2+/Fe ) OR
E = (+)0.18 V for system 3 + system 2 reaction
Equilibrium shift related to Eo values For shifts right’
More positive/less negative system 4 ALLOW (VO2+) is reduced OR gains electrons
(VO2+/V3+) shifts right (maybe seen as an equation)
AND AND
More positive/less negative system 3 ‘For shifts right’
(V3+/V2+) shifts right ALLOW (V3+) is reduced OR gains electrons
(maybe seen as an equation)
IGNORE Fe oxidised
(iii) + 2+ 2+ 3+ 2.8 IGNORE state symbols
Fe + 4H + 2VO → Fe + 2H2O + 2V
1 ALLOW multiples
ALLOW ‘⇌’
(c) (i) (0.00200 mol dm–3 solution gives) a large titre 1 3.4 ALLOW (0.0200 mol dm–3 solution gives) a small titre
which leads to a small (percentage) error / which leads to a large (percentage) error / uncertainty
uncertainty
Assume ‘it’ means dilute solution
ALLOW 13.50 cm3 gives a lower percentage error
than 1.35 cm3
18 c (ii) FIRST CHECK THE ANSWER ON ANSWER LINE 5 2.8 ×5 ALLOW ECF throughout
If answer = 301 mg award 5 marks
ALLOW working to 3SF minimum throughout
H432/01 Mark Schemes June 2019
AO
element
– 13.50 –5
n(MnO4 ) = 1000 × 0.00200 = 2.7(0) × 10 (mol) Common errors
602 (mg) (not dividing by 2) = 4 marks
n(Fe2+) (in.25.0 cm3) = 2.7(0) × 10-5 × 5 = 1.35 × 37.7 (using 55.8 instead of 445.8) = 4 marks
10–4 (mol)
n(Fe2+) (in 250 cm3) = 1.35 × 10-4 × 10 = 1.35 ×
10–3
Mass C12H22FeO14 in 2 tablets
= 1.35 × 10–3 × 445.8 = 0.6018 (g)
Mass C12H22FeO14 in 1 tablet = 301 (mg)
AND to 3 SF Last mark involves dividing by two and converting g to
mg. These steps may be seen earlier
(iii) 180 x 55.8 1 3.1 ×1 ALLOW correct working if iron supplement is not
A: Mass Fe = = 66 mg named
151.8
210 x 55.8
B: Mass Fe = = 69 mg
169.8
Iron supplement:
B provides more Fe per tablet ALLOW iron(II) fumarate or C4H2FeO4
How to answer it
Reactions of Ions and Compounds of Transition Elements
This comprehensive OCR A-Level Chemistry question tests transition metal chemistry, including ligand substitution, precipitation reactions, stereochemistry (3D drawing of octahedral complexes), electron configurations of ions, standard electrode potentials (E°) applied to redox color changes, titration error minimization, multi-step redox titration calculations, and mass comparisons of active pharmaceutical ingredients.
Part (a): Chromium Chemistry & Reactions
(i) Purple Complex Ion in Experiment 1
✅ Correct Answer
[Cr(NH₃)₆]³⁺(aq)
💡 Key Knowledge
Adding excess aqueous ammonia to hexaaquachromium(III) undergoes ligand substitution where neutral NH₃ molecules replace water ligands entirely, forming a purple hexaammine complex.
(ii) Equation for Experiment 2, Step 1
✅ Correct Answer
CrCl₃(aq) + 3NaOH(aq) → Cr(OH)₃(s) + 3NaCl(aq)
Alternative ionic form: Cr³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(s)
❌ Common Errors
Forgetting state symbols on ionic equations will lose the mark. Ensure formula for chromium(III) hydroxide is neutral: Cr(OH)₃ .
(iii) 3D Diagram of Experiment 2, Step 2 Complex
✅ Correct Answer
An octahedral complex: [Cr(OH)₆]³⁻
- 2 in-plane bonds, 2 wedges (out of plane), 2 dashes (into plane).
- All coordinate bonds must connect through the O atom in OH.
- Overall charge: 3− outside square brackets.
🧠 Exam Technique
When drawing octahedral complexes, remember the convention: 2 straight lines in the plane, 2 wedged bonds pointing forward, and 2 dashed/unfilled wedges pointing backward. Always link the bond directly to the oxygen atom, not hydrogen.
(iv) & (v) Subsequent Oxidation States
✅ Correct Answers
(iv) CrO₄²⁻ (Chromate ion giving yellow colour)
(v) Orange (Colour of dichromate solution, Cr₂O₇²⁻ )
Part (b): Vanadium Redox Systems & Electrode Potentials
(i) Electron Configuration of V³⁺
✅ Correct Answer
1s² 2s² 2p⁶ 3s² 3p⁶ 3d² (or [Ar] 3d² / 4s⁰ 3d² )
❌ Common Errors
Students often forget that 4s electrons are lost *first* when transition metal ions form. Vanadium atom is 1s²... 4s² 3d³ , so losing 3 electrons removes both 4s electrons and one 3d electron.
(ii) Explanation of Colour Changes using E° Values
💡 Key Knowledge & Explanation
- Observation 1 (Blue to Green): VO²⁺ is reduced to V³⁺ .
- Observation 2 (Green to Violet): V³⁺ is further reduced to V²⁺ .
- Electrode Potential Rationale: The E° value for system 4 ( VO²⁺/V³⁺ at +0.34 V) is more positive than system 2 ( Fe²/Fe or zinc systems), meaning VO²⁺ is reduced. Similarly, system 3 ( V³⁺/V²⁺ at −0.26 V) drives further reduction to V²⁺ because the more positive/less negative system shifts right.
(iii) Equation for the First Colour Change
✅ Correct Answer
Fe + 4H⁺ + 2VO²⁺ → Fe²⁺ + 2H₂O + 2V³⁺
Part (c): Iron(II) Gluconate Titration Calculations
(i) Titrant Concentration Rationale
✅ Correct Answer
A dilute 0.00200 mol dm⁻³ solution gives a larger titre, which results in a smaller percentage error / uncertainty.
🧠 Exam Technique
Always tie titration concentration choices back to volumetric glassware uncertainties. Larger titres reduce the relative percentage error of the burette reading.
(ii) Step-by-Step Titration Calculation
📐 Step-by-Step Calculation (Target: 301 mg)
- Moles of MnO₄⁻ titrated:
(13.50 / 1000) × 0.00200 = 2.70 × 10⁻⁵ mol - Moles of Fe²⁺ in 25.0 cm³ portion:
Ratio is 1 mol MnO₄⁻ : 5 mol Fe²⁺
2.70 × 10⁻⁵ × 5 = 1.35 × 10⁻⁴ mol - Moles of Fe²⁺ in full 250 cm³ flask:
1.35 × 10⁻⁴ × (250 / 25.0) = 1.35 × 10⁻³ mol - Mass of iron(II) gluconate ( C₁₂H₂₂FeO₁₄ , Mr = 445.8) in 2 tablets:
Since 2 tablets were used:
1.35 × 10⁻³ mol × 445.8 = 0.6018 g - Mass in ONE tablet (converted to mg, 3 SF):
(0.6018 / 2) × 1000 = 300.9 mg → 301 mg
❌ Common Calculation Traps
- Forgetting to divide by 2: Calculating the mass for two tablets instead of one yields 602 mg (loses final mark).
- Using incorrect Mr: Using atomic mass of Fe alone (55.8) instead of the full ligand complex Mr (445.8) leads to 37.7 mg.
- Significant Figures: Ensure final answer rounds correctly to 3 SF.
(iii) Comparing Iron Supplements
✅ Correct Answer
Iron supplement: B (Iron(II) fumarate)
Working:
A (sulfate): (180 × 55.8) / 151.8 = 66 mg of Fe per tablet.
B (fumarate): (210 × 55.8) / 169.8 = 69 mg of Fe per tablet.
Topics
Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Practical Activity Groups · 5.3 Transition elements · 5.2 Energy · PAG 2: Acid-base titration · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.