OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 22

16 marks · Hard difficulty · Structured Questions

Determine the formulae, structures, ionic equations, and explanations related to transition metal complexes of iron and copper, including isomerism, redox potentials, and stoichiometry.

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Question text

22 This question is about reactions of transition metal compounds.

(a) Ethane-1,2-diamine, H2NCH2CH2NH2, is a bidentate ligand.

The structure of ethane-1,2-diamine is shown below.

H2N NH2

(i) Explain why ethane-1,2-diamine can act as a bidentate ligand.

… [1]

(ii) The iron(III) ion, Fe3+, forms a complex ion A with two ethane-1,2-diamine ligands and

two chloride ligands.

Complex ion A has cis and trans stereoisomers.

One of these stereoisomers exists as optical isomers.

Determine the empirical formula, with charge, of complex ion A and draw the 3-D

structures of the three stereoisomers.

Empirical formula with charge …

Structures

[4]

(b) Aqueous sodium hydroxide is added to an aqueous solution of iron(II) sulfate.

A pale green precipitate forms which turns brown when left to stand in air.

(i) Write an ionic equation for the formation of the pale green precipitate.

… [1]

(ii) Use the information below to explain why the pale green precipitate turns brown when

left to stand in air and construct an equation for the reaction which occurs.

Redox System Equation E ө / V

1 Fe(OH) (s) + e– Fe(OH) (s) + OH–(aq) –0.56 V

2 O (g) + 2H O(l) + 4e– 4OH–(aq) +0.40 V

… [4]

(c)* This question is about copper and copper compounds.

Experiment 1

Hydrochloric acid, HCl (aq), is added to an aqueous solution containing [Cu(H O) ]2+

complex ions.

A yellow-green solution forms containing complex ion B.

Experiment 2

A piece of copper metal is heated with concentrated sulfuric acid.

A reaction takes place forming a pale blue solution C and 45 cm3 of a gas D, measured at

RTP. The mass of gas D is 0.12 g.

Experiment 3

An excess of copper(II) oxide is heated with dilute nitric acid. The resulting mixture is

filtered. The filtrate is a blue solution E.

Aqueous potassium iodide, KI(aq), is added to the blue solution E.

A white precipitate F and a brown solution G form.

Determine the formulae of B–G.

Construct equations for the reactions taking place, include any changes in oxidation number,

and show your working where appropriate. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

22 (a) (i) (N) donates two electron pairs (to a metal ion/metal/Fe(3+)) 1 AO1.2 ALLOW lone pairs for electron pairs

AND

forms two coordinate / dative (covalent) bonds TWO is only needed once if bonds are plural

e.g. donates 2 electron pairs to form co-

ordinate bonds

OR donates electron pairs to form 2

coordinate bonds.

28 AO

element

(ii) Empirical formula 4 DO NOT ALLOW Fe(NH2CH2CH2NH2)Cl2 for

FeC4H16N4Cl2 (any order) empirical formula

AND charge = (1)+ AO1.2

×1 ALLOW any order

Structures ------------------------------------------------------

TAKE CARE: structures may be in different

i.e. Optical isomers (cis) orientations and in different order

IGNORE charges (anywhere)

IF connectivity between Fe AND N of NH2 is

AO3.1 incorrect then penalise first time ONLY

×3

i.e. trans isomer

29 AO

element

For NH2CH2CH2NH2, ALLOW skeletal, structural, displayed Each structure to contain

2 ‘out wedges’, 2 ‘in wedges’ and 2 lines in

plane of paper OR 4 lines, 1 ‘out wedge’ and 1

formula AND C-C without Hs and NH2 NH2 ‘in wedge’:

Bond into paper can be shown as:

IF NH2 shown with incorrect number of H, eg. N N, penalise

first time ONLY ALLOW

IF ALL 3 isomers are ‘correct’, but 2 x Cl AND no Ns, e.g.

AWARD 1 mark

(b) (i) 1 AO2.6

Fe2+ + 2OH– → Fe(OH) IGNORE state symbols, even if wrong

ALLOW

[Fe(H O) ]2+ + 2OH– → Fe(OH) (H O) + 2H O

26 2 2 4 2

[Fe(H O) ]2+ + 2OH– → Fe(OH) + 6H O

26 2 2

30 AO

element

(ii) Explanation of the brown precipitate 4 AO3.1 ORA

The brown ppt is Fe(OH)3 ×4

Fe(OH) loses electrons/ Fe(OH) oxidised ALLOW Fe2+ is oxidised to Fe3+

Comparison of E values

(E of) Fe/Redox system 1 is more negative/less positive ALLOW Fe

(than E of O2/redox system 2) ALLOW Ecell is (+) 0.96V

OR IGNORE ‘lower/higher’

(E of) O2/Redox system 2 is more positive/less negative

(than E of Fe/redox system 1)

Equilibrium shift

More negative/less positive OR Fe system OR Redox system For equilibrium shift

1 shifts left ALLOW Ecell is +ve therefore the reaction is

OR feasible.

More Positive/less negative OR O2 system OR Redox system OR

2 shifts right Direction of half equation correctly written.

Equation

4Fe(OH)2(s) + O2(g) + 2H2O(l) → 4Fe(OH)3(s) ALLOW multiples

ALLOW equilibrium

IGNORE state symbols, even if wrong

DO NOT ALLOW uncancelled species

31 element

scheme for guidance on how to mark this question. ×3

Formula

Level 3 (5–6 marks) AO3.2 B CuCl 2–

Reaches a comprehensive conclusion to determine the correct ×3 OR

formulae of almost all of B, C, D, E, F and G. [CuCl ]2–

AND C [Cu(H O) ]2+

most correct equations and identifies some changes in oxidation OR

number CuSO4

AND D SO2

Calculation of Mr of the gas E Cu(NO3)2

There is a well-developed line of reasoning which is clear and [Cu(H O) ]2+

logically structured. The information presented is relevant and F CuI

substantiated. G

Level 2 (3–4 marks)

Experiment 1

Reaches a conclusion to determine the correct formulae of at least

Equation

half of B, C, D, E, F and G. 2+ – 2–

[Cu(H2O)6] + 4Cl → [CuCl4] + 6H2O

AND EITHER 2+ 2–

[Cu(H2O)6] + 4HCl → [CuCl4] + 6H2O +

some correct equations +

Any one correct equation and a relevant change in oxidation

Experiment 2

number

Evidence

45 –3

any one correct equation and a correct calculation of the Mr n(D) = = 1.875 × 10

24000

0.12

Molar mass (D) = –3 = 64

1.875 × 10

There is a line of reasoning presented with some structure. The Equation

information presented is relevant and supported by some evidence. Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

Oxidation numbers

Cu 0 → Cu +2; S +6 → S +4

32 AO

element

Level 1 (1–2 marks) Experiment 3

Reaches a simple conclusion to determine the correct formulae of Equation

some of B, C, D, E, F and G CuO + 2HNO3 → Cu(NO3)2 + H2O

The correct formulae for 1 of B, C, D, E, F and G with correct 2Cu2+ + 4I – → 2CuI + I

equation or calculation. OR

2Cu(NO3)2 + 4KI → 2CuI + I 2 +4KNO3

There is an attempt at a logical structure with a line of reasoning.

The information is in the most part relevant. Oxidation numbers

Cu +2 → Cu +1; I –1 to 0

0 marks

No response or no response worthy of credit.

How to answer it

Reactions of Transition Metal Compounds Study Guide

OCR A-Level Chemistry • Comprehensive Exam Breakdown

What this question tests

This synoptic question evaluates your deep understanding of transition metal chemistry. Key areas assessed include: defining and identifying bidentate ligands, drawing 3D stereoisomers (cis, trans, and optical) of octahedral complexes, applying electrode potentials (E°) to explain redox changes in metal hydroxides, balancing complex ionic equations, calculating molar mass from gas volumes at RTP, and tracking oxidation number changes across multiple reaction pathways involving copper compounds.

Part (a): Ligands, Complex Formulae, and Stereoisomerism

💡 Key Knowledge

Bidentate Ligands: Species that donate two lone pairs of electrons to a metal ion, forming two coordinate (dative covalent) bonds per ligand molecule.
Octahedral Stereoisomerism: Complexes with coordination number 6 can exhibit cis/trans geometric isomerism and optical isomerism when bidentate ligands are unsymmetrical or mixed.

✅ Correct Answers

(a)(i): N donates two electron pairs AND forms two coordinate bonds.
(a)(ii) Empirical Formula: FeC₄H₁₆N₄Cl₂ (or Fe(NH₂CH₂CH₂NH₂)₂Cl₂ ) with charge (1)+ .
Structures: Three distinct 3D diagrams required showing two en bidentate ligands and two Cl ligands: two cis isomers (one pair of optical enantiomers) and one trans isomer.

🧠 Exam Technique

When drawing 3D octahedral complexes, clearly show bonds going into the page (dotted lines) and coming out of the page (wedges). For optical isomers, draw a non-superimposable mirror image. Ensure connectivity is correct: nitrogen bonds directly to iron, not through the hydrogen atoms.

❌ Common Errors

Stating that the ligand itself has two lone pairs rather than noting nitrogen donates two pairs in total across the molecule.
Incorrect bond connectivity, such as drawing the metal attached to the hydrogen atoms of the amine group instead of the nitrogen.

Mark Allocation: (a)(i) = 1 mark; (a)(ii) = 4 marks (1 for formula/charge, 3 for 3D structures).

Part (b): Iron Hydroxide Precipitation and Redox Reactions

💡 Key Knowledge

Precipitation: Aqueous Fe²⁺ reacts with hydroxide ions to form a pale green precipitate of Fe(OH)₂ .
Redox in Air: Transition metal hydroxides can be oxidised by atmospheric oxygen. Compare standard electrode potentials ( E° ) to determine feasibility: a more negative E° system gets oxidised (loses electrons) by a more positive E° system.

✅ Correct Answers

(b)(i) Ionic Equation:
Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂s)
(Allow complex ion formulations).
(b)(ii) Explanation & Equation:
Pale green Fe(OH)₂ is oxidised to brown Fe(OH)₃ because system 1 has a more negative E° (-0.56 V) than system 2 (+0.40 V), shifting equilibrium 1 to the left.
Equation: 4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

🧠 Exam Technique

To secure full marks in electrode potential explanation questions, explicitly state three points: 1) which species is oxidised, 2) compare the numerical values of the two E° systems, and 3) state the resulting direction of the equilibrium shift.

❌ Common Errors

Forgetting balancing coefficients when constructing the overall redox equation from two half-equations, especially balancing the electrons and oxygen molecules.

Mark Allocation: (b)(i) = 1 mark; (b)(ii) = 4 marks.

Part (c): Copper Chemistry and Calculation of Gas M_r

💡 Key Knowledge

Ligand Substitution: Addition of concentrated HCl replaces H₂O ligands with Cl⁻ ions, causing a colour change from pale blue to yellow-green ( [CuCl₄]²⁻ ).
Redox Titrations & Precipitation: Cu²⁺ reacts with excess iodide ions ( I⁻ ) to form a white precipitate of copper(I) iodide ( CuI ) and brown aqueous iodine ( I₂ ).

✅ Correct Answers (Formulae B to G)

B: [CuCl₄]²⁻ (or [CuCl₄(H₂O)₂]²⁻ )
C: [Cu(H₂O)₆]²⁺ or CuSO₄
D: SO₂
E: Cu(NO₃)₂
F: CuI
G: I₂

📐 Step-by-Step Calculation: Determining M_r of Gas D

Step 1: Calculate moles of gas D at RTP.
Using molar volume at RTP (24000 cm³ mol⁻¹):
n(D) = 45 / 24000 = 1.875 × 10⁻³ mol
Step 2: Use mass to find Molar Mass (M_r).
Mass of gas D = 0.12 g.
M_r = mass / moles = 0.12 / (1.875 × 10⁻³) = 64 g mol⁻¹ ( SO₂ ).

🧠 Exam Technique

Level 3 responses require systematic tracking across all three experiments. Clearly link observations (colour changes, precipitates, gases) to specific species. Write balanced equations with correct state symbols and highlight changes in oxidation numbers (e.g., Cu from 0 to +2, S from +6 to +4).

❌ Common Errors

Using the wrong molar volume constant (e.g., 22.4 instead of 24.0 dm³ mol⁻¹ or 24000 cm³ mol⁻¹).
Failing to identify the white precipitate CuI correctly alongside the brown I₂ solution in Experiment 3.

Mark Allocation: Total 6 marks (Assessed via Level of Response criteria: Level 3 requires formulae for almost all of B–D, E, F, G, correct equations, oxidation number changes, and M_r calculation).

Topics

Module 5: Physical chemistry and transition elements · Module 3: Periodic table and energy · 5.3 Transition elements · 3.1 The periodic table

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.