OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 22

15 marks · Hard difficulty · Structured Questions

Identify unknown transition metal compounds B to H and write ionic equations, determine the formula and oxidation number of a complex nickel compound containing SCN- and NH3 ligands, and calculate the concentration of sulfite in burger meat using a manganate(VII) titration.

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Question text

22 (a)* B and C are compounds of two different transition elements.

A student carries out test tube reactions on aqueous solutions of B and C.

The observations of the student’s tests are shown below.

Test B(aq) C(aq)

NH3(aq) added dropwise green precipitate D grey-green precipitate E

excess NH3(aq) added no further change purple solution F

HNO3(aq) no change no change

followed by Ba(NO3)2(aq) white precipitate G no change

HNO3(aq) no change no change

followed by AgNO3(aq) no change white precipitate H

Analyse the results to identify B to H, and construct ionic equations for the formation of

products D to H. [6]

Additional answer space if required

(b) A compound of nickel, J, has the formula (NH ) [Ni(SCN) (NH ) ] and contains SCN− and

42 x 3 y

NH3 ligands.

The percentage by mass of three of the elements in compound J is shown below:

Ni, 16.26%; S, 35.56%; N, 31.00%.

(i) Calculate the values of x and y in the formula of compound J.

x = …

y = …

[3]

(ii) Determine the oxidation number of nickel in compound J.

oxidation number: … [1]

Food safety legislation allows a maximum of 850 mg Na2SO3 per kg of burger meat.

A chemist determines the amount of Na2SO3 in a sample of burger meat using a

manganate(VII) titration.

Step 1 The Na2SO3 from 525 g of burger meat is extracted to form a solution containing

SO 2−(aq) ions.

Step 2 The solution from step 1 is made up to 250.0 cm3 in a volumetric flask with water.

25.0 cm3 of this diluted solution is pipetted into a conical flask.

Step 3 The pipetted solution from step 2 is acidified with dilute sulfuric acid and then

titrated with 0.0100 mol dm−3 potassium manganate(VII), KMnO .

2MnO −(aq) + 6H+(aq) + 5SO 2−(aq) 2Mn2+(aq) + 3H O(l) + 5SO 2−(aq)

43 2 4

12.60 cm3 of KMnO (aq) is required to reach the endpoint.

Analyse the results to determine whether the burger meat complies with food safety

legislation.

[5]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

22 (a)* (i) Refer to marking instructions on page 5 of mark scheme for 6 3.3×3 Indicative scientific points may include:

guidance on marking this question. 3.4×3

Identification of unknowns

Level 3 (5–6 marks) Can be identified within labelled equation.

All three tests are covered in detail, with at least six of B B is FeSO4 OR Iron(II) sulfate

to H identified correctly and equations mostly correct. • Test 1: Fe2+ present

• Test 2: SO42– present

There is a well-developed line of reasoning which is

clear and logically structured. The information presented D is Fe(OH)2OR [Fe(H2O)4(OH)2] OR iron(II)

is relevant and substantiated. hydroxide

G is BaSO4 OR barium sulfate

Level 2 (3–4 marks)

All three tests are covered with at least four of B to H C is CrCl3OR chromium(III) chloride

identified correctly. Some attempt at writing equations, 3+

• Test 1: Cr present

but with several omissions or incorrect formulae. –

• Test 3: Cl present

There is a line of reasoning presented with some

structure. The information presented is relevant and E is Cr(OH)3 OR [Cr(H2O)3(OH)3]OR

supported by some evidence. chromium(III) hydroxide

F is [Cr(NH3)6]3+

Level 1 (1–2 marks) H is silver chloride OR AgCl

Only two tests covered with at least two of B to H

identified correctly, and little attempt at writing Equations

equations. D: [Fe(H2O)6]2+ +2OH– → Fe(OH)2 + 6H2O OR

Fe2+ + 2OH– → Fe(OH)2 OR

There is an attempt at a logical structure with a line of [Fe(H2O)6]2++ 2OH- → [Fe(H2O)4(OH)2] + 2H2O OR

reasoning. The information is in the most part relevant. [Fe(H2O)6]2++ 2NH3 → [Fe(H2O)4(OH)2] +2NH4+

0 marks [Fe(H2O)6]2+ + 2NH3 → Fe(OH)2 + 4H2O + 2NH4+

No response or no response worthy of credit.

E: [Cr(H2O)6]3+ +3OH– → Cr(OH)3 + 6H2O OR

Cr3+ + 3OH– → Cr(OH)3 OR

[Cr(H2O)6]3++ 3OH- → [Cr(H2O)3(OH)3] + 3H2O OR

[Cr(H2O)6]3++ 3NH3 → [Cr(H2O)3(OH)3] +3NH4+ OR

22 element

[Cr(H2O)6]3++ 3NH3 → Cr(OH)3 + 3H2O + 3NH4+

F: [Cr(H2O)6]3+ + 6NH3 → [Cr(NH3)6]3+ + 6H2O

Cr(OH)3 + 6NH3 → [Cr(NH3)6]3+ + 3OH– OR

[Cr(H2O)3(OH)3]

+6NH3 → [Cr(NH3)6]3++3H2O+3OH-

G: Ba2+ + SO42– → BaSO4

H: Ag+ + Cl – → AgCl

element

(b) (i) 16.26 35.36 31.0 3 3.1×1 ALLOW any correct method

Ni : S : N = : : OR 0.277 : 1.10 : 2.21

58.7 32.1 14

OR 1 : 4 : 8 ALLOW NiS4N8 for ratio

x = 4 3.2×2

2 + x + y = 8 y = 2 ALLOW ECF for y from incorrect x

(ii) +2 1 2.1 + required

ALLOW 2+

n(MnO4–) in titration ALLOW ECF throughout

12.6 –4

= 0.01 × = 1.26 × 10

1000 1.2×1

n(SO32–) in 25.0 cm3

= 1.26 × 10–4 × 2.5 = 3.15 × 10–4 (mol) 2.8×3

n(SO32–) in 250 cm3

= 10 × 3.15 × 10–3 = 3.15 × 10–3 (mol)

mass Na2SO3 in 525 g meat

= 3.15 × 10–3 × 126.1 = 0.397 (g) 3.2×1

Calculator = 0.397215 g

mass Na2SO3in 1 kg of meat

1000

= 0.397215 × = 0.7566 g OR 756.6 mg ALLOW within range: 756 to 757 mg

AND less than the maximum permitted level OR

AW ALLOW 0.397 g<0.446 g per 525 g meat.

Total 15

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How to answer it

Transition Metals, Formula Determination & Redox Titrations

OCR A-Level Chemistry • Comprehensive Study Guide

What this question tests

This multi-step synoptic question tests your knowledge of transition metal ion identification using qualitative tests (ligand exchange and precipitation), ionic equations, empirical formula determination from percentage composition, oxidation number rules, and complex back-titration calculations involving redox stoichiometry.

Part (a): Qualitative Analysis & Identification of B to H

Qualitative testing of transition metal and halide/sulfate ions

✅ Correct Identifications

B: FeSO₄ or Iron(II) sulfate
C: CrCl₃ or Chromium(III) chloride
D: Fe(OH)₂ (green precipitate)
E: Cr(OH)₃ (grey-green precipitate)
F: [Cr(NH₃)₆]³⁺ (purple solution)
G: BaSO₄ (white precipitate)
H: AgCl (white precipitate)

💡 Key Knowledge

Test 1 (NH₃): Iron(II) gives a green ppt of Fe(OH)₂ which is insoluble in excess. Chromium(III) gives a grey-green ppt of Cr(OH)₃ which dissolves in excess NH₃ to form a purple/violet solution due to ligand substitution.
Test 2 (Ba²⁺): Confirms presence of sulfate ( SO₄²⁻ ) forming an insoluble white precipitate of BaSO₄ .
Test 3 (Ag⁺): Confirms presence of chloride ( Cl⁻ ) forming a white precipitate of AgCl .

🧠 Exam Technique & Level Response

This is a Level of Response question (6 marks max). To achieve Level 3 (5–6 marks), you must identify at least 6 out of B to H correctly AND provide accurate corresponding ionic equations. Structure your answer systematically by test or by substance.

❌ Common Errors

Omitting state symbols or getting charges wrong in complex ionic equations.
Confusing iron(II) green precipitates with iron(III) orange-brown precipitates.
Forgetting to add acid ( HNO₃ ) before testing with barium or silver nitrate.

📝 Required Ionic Equations

D: Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂IS(s) (or with hexaaqua complexes)
E: Cr³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(s)
F: [Cr(H₂O)₆]³⁺ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 6H₂O
G: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
H: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Part (b)(i): Calculating Values of x and y in Compound J

Empirical formula and complex stoichiometry determination

✅ Correct Answers

x = 4

y = 2

Awarded 3 marks for correct ratio and derivation of x and y.

📐 Step-by-Step Calculation

Find % of NH₃: Total percentages must equal 100%. % N = 31.00%, Ni = 16.26%, S = 35.56%. Remainder for NH₃ or calculate moles directly. Given N = 31.00% total.
Calculate moles / atomic ratios for Ni : S : N (from SCN⁻ and NH₃):
Ni: 16.26 / 58.7 = 0.277
S: 35.56 / 32.1 = 1.10
N (total): 31.00 / 14 = 2.21
Ratio Ni : S : N = 0.277 : 1.10 : 2.21 = 1 : 4 : 8
Use formula structure (NH₄)₂[Ni(SCN)ₓ(NH₃)₂] :
From formula, moles of SCN⁻ = x, moles of NH₃ = y.
S comes from SCN⁻, so x = 4 .
Total nitrogens in outside ammonium + ligand: 2(from NH₄⁺) + x(from SCN⁻) + y(from NH₃) ... wait, let's look at the stoichiometry check: 2 + x + y = 8 . Since x = 4 , 2 + 4 + y = 8 ⇒ y = 2 .

❌ Common Calculation Traps

Forgetting that the ammonium ion (NH₄⁺) outside the square brackets also contains nitrogen, which affects total nitrogen stoichiometry.
Rounding division numbers too early leading to distorted whole-number ratios.

Part (b)(ii): Oxidation Number of Nickel

Assigning oxidation states in transition metal complexes

✅ Correct Answer

Oxidation number: +2 (or 2 )

1 mark

💡 Key Knowledge

Compound J is (NH₄)₂[Ni(SCN)₄(NH₃)₂] .

Two NH₄⁺ ions contribute +2 overall charge.
Four SCN⁻ ligands each contribute -1 (total -4 ).
Two NH₃ ligands are neutral ( 0 ).
Therefore, Nickel must be +2 to balance the complex anion charge [Ni(SCN)₄(NH₃)₂]²⁻ .

Part (c): Redox Titration & Food Safety Legislation Analysis

Back-calculation, dilution factors, and ppm limit checks

✅ Correct Answer & Conclusion

Mass of Na₂SO₃ in 525 g of meat = 0.397 g (or 397 mg).

Equivalent per kg of meat = 757 mg kg⁻¹ .

Conclusion: 757 mg is less than the maximum permitted level ( 850 mg ), so the burger meat complies with food safety legislation.

5 marks

📐 Step-by-Step Calculation

Moles of MnO₄⁻ used in titration:
n(MnO₄⁻) = (12.60 / 1000) × 0.0100 = 1.26 × 10⁻⁴ mol
Moles of SO₃²⁻ in 25.0 cm³ pipette sample:
Using equation ratio ( 2 MnO₄⁻ : 5 SO₃²⁻ ):
n(SO₃²⁻) = 1.26 × 10⁻⁴ × (5 / 2) = 3.15 × 10⁻⁴ mol
Moles of SO₃²⁻ in volumetric flask (250 cm³):diluted solution:
n = 3.15 × 10⁻⁴ × (250 / 25.0) = 3.15 × 10⁻³ mol
Mass of Na₂SO₃ in 525 g burger meat:
Molar mass of Na₂SO₃ = (2 × 23.0) + 32.1 + (3 × 16.0) = 126.1 g mol⁻¹
Mass = 3.15 × 10⁻³ × 126.1 = 0.397 g (397 mg)
<Scale up to 1 kg (1000 g) of meat:
(0.3972 g / 525 g) × 1000 g = 0.7566 g = 756.6 mg kg⁻¹

❌ Common Calculation Traps

Forgetting to multiply by the volumetric flask scale factor ( 250 / 25 ).
Inverting the stoichiometry ratio from the balanced equation ( 5/2 vs 2/5 ).
Failing to convert final mg units correctly back to kg comparison limits.

Topics

Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Practical Activity Groups · 5.3 Transition elements · PAG 2: Acid-base titration · PAG 4: Qualitative analysis of ions · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.