Edexcel A-Level Chemistry Paper 3, June 2025: Question 5

12 marks · Medium difficulty · Synoptic Questions

Determine the empirical formula and 3D structure of a cobalt complex, define dynamic equilibrium, and explain complex stability using equilibrium and entropy concepts.

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Question

Question text

5 This question is about cobalt complexes.

(a) A sample of cobalt(II) chloride was dissolved in aqueous ammonia, NH3(aq).

During the process, a stream of oxygen was gently bubbled through the solution

to oxidise the cobalt(II) ion.

A coloured solid, R, was extracted from the solution. A pure sample of this solid

was found to contain the masses of the elements shown and had a molar mass

of 233.4 g mol−1.

Element Mass / g

Co 1.26

N 1.20

H 0.257

Cl 2.28

(i) Show that the empirical formula and the molecular formula of R is CoN4H12Cl3.

You must show your working.

(2)

(ii) 1.00g of R was dissolved in water forming a coloured solution of

the complex ion, Q.

The solution was reacted with excess silver nitrate solution, forming

0.614g of silver chloride, AgCl(s).

Calculate the number of moles of chloride ions released when 1 mol of

R dissolves.

(2)

(iii) Draw the three-dimensional shape of the complex ion, Q, using your answer

to (a)(ii) to help deduce the formula and overall charge.

(3)

14 2+

(b) An aqueous solution of cobalt(II) chloride contains the complex ion [Co(H*P77834A01436*2O)6] .

The dynamic equilibria shown occur when NH (aq) and EDTA4−(aq) are each

added to separate solutions of this complex ion.

[Co(H O) ]2+(aq) + 6NH (aq) [Co(NH )6]2+(aq) + 6H O(l) log K = 4.39

26 3 3 2 c

[Co(H O) ]2+(aq) + EDTA4−(aq) [Co(EDTA)]2−(aq) + 6H O(l) log K = 16.5

26 2 c

(i) State what is meant by the term ‘dynamic equilibrium’.

(2)

(ii) Explain, in terms of equilibria and entropy, which of the three complex ions is

the most stable.

(3)

… *P77834A01536*

(Total for Question 5 = 12 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

5(a)(i) An answer that makes reference to the following (2)

points: Co N H Cl

mol 1.26 ÷ 1.20 ÷ 0.257 ÷ 2.28 ÷

• calculate moles of each element (1) 58.9 14.0 1.0 35.5

= 0.0214 = 0.0857 = 0.257 = 0.0642

• show ratio by division by smallest number (1) ratio 0.0214 / 0.0857 / 0.257 / 0.0642 /

and 0.0214 0.0214 0.0214 0.0214

show RFM of CoN4H12Cl3 is 233.4 = 1 = 4 = 12 = 3

(58.9 + (4 × 14) + 12 + (3 × 35.5) = 233.4 (so

molecular formula = CoN4H12Cl3 )

M1 % of each element by mass in sample of 4.997 g

M2 % of each element by comparing ΣAr to Mr

Question

Answer Additional Guidance Mark

Number

5(a)(ii) EITHER Example of calculation (2)

− (1) moles of AgCl = 0.614 ÷ 143.4 = 4.2817 × 10–3 mol

• calculation of moles of AgCl / Cl

(1) moles of R = 1.00 ÷ 233.4 = 4.2845 × 10–3 mol

• calculation of moles of R

and 3 3

4.2845 × 10– mol : 4.2817 × 10– mol ≈ 1:1, (so 1 mol of R

show ratio is 1:1 (so one mole of R releases

1 mol of chloride ions) releases 1 mol of chloride ions)

Do not award 3Cl− ions

• calculation of moles of R moles of R = 1.00 ÷ 233.4 = 4.2845 × 10–3 mol

(1)

• calculation of moles of AgCl /Cl− moles of AgCl = 0.614 ÷ 143.4 = 4.2817 × 10–3 mol

(1) –3 –3

and 4.2845 × 10 mol : 4.2817 × 10 mol ≈ 1:1, (so 1 mol of R

show ratio is 1:1 (so one mole of R releases releases 1 mol of chloride ions)

1 mol of chloride ions)

Do not award 3Cl−ions

Ignore SF except 1 SF

Question

Answer Additional Guidance Mark

Number

5(a)(iii) An answer that makes reference to the following points: (3)

• octahedral shape (around cobalt) using wedges (1)

• correct number and type of ligand (1)

• correct overall charge (1)

Allow dashed line for dashed wedges

Allow ‘+’ on cobalt

Allow correct calculation of charge from incorrect

ligands

Ignore position of Cl ligands relative to NH3 ligands

Ignore connectivity

Ignore omission of square brackets

Ignore additional chloride ion (to form R)

Arrows if drawn must be pointing towards the cobalt

Question

Answer Additional Guidance Mark

Number

5(b)(i) An answer that makes reference to the following points: (2)

• forward and backward reactions proceed at the same (1) Do not award rates of forward and backward

rate reactions are constant

• concentration of reactants and products remain constant (1) Accept resulting in no observable change in the

/ remain the same system

Question

Answer Additional Guidance Mark

Number

5(b)(ii) An explanation that makes reference to the following points: (3)

• (The most stable is) [Co(EDTA)]2–(aq)

as it has the larger (log) Kc / eqm lies furthest to RHS (1) Ignore calculated Kc values

• 2 moles (of ions) forms 7 moles (of product in EDTA (1) Allow more moles of product than reactant

equation) Allow species

Do not award ‘molecules’

• (greater) increase in entropy (change of the system) / (1) Do not award increase in entropy if clear that

∆Ssys is positive it is referring to the EDTA (complex) and not

the reaction

(Total for Question 5 = 12 marks)

How to answer it

Cobalt Coordination Complexes & Ligand Exchange Equilibria

WHAT THIS QUESTION TESTS

This question assesses fundamental analytical and inorganic chemistry competencies across transition metals:

Empirical & Molecular Formula: Determining mole ratios from quantitative mass data and confirming molecular formulas with molar mass.
Precipitation Stoichiometry: Deducing inner-sphere vs. outer-sphere ions using gravimetric analysis with silver nitrate (Ag⁺ + Cl⁻ → AgCl).
3D Complex Geometry: Representing 6-coordinate octahedral transition metal complexes with stereochemical wedge/dash conventions, coordinating ligands, and correct formal complex ions charges.
Equilibrium & The Chelate Effect: Defining dynamic equilibrium precisely and using log K_c values alongside entropy changes (ΔS_system) to explain complex stability.

Part (a)(i)

Empirical and Molecular Formula of Solid R

📐 Step-by-Step Calculation

Calculate moles of each element:
Co: 1.26 g ÷ 58.9 g mol⁻¹ = 0.02139 mol
N: 1.20 g ÷ 14.0 g mol⁻¹ = 0.08571 mol
H: 0.257 g ÷ 1.0 g mol⁻¹ = 0.2570 mol
Cl: 2.28 g ÷ 35.5 g mol⁻¹ = 0.06423 mol
Divide by the smallest value (0.02139):
Co = 0.02139 ÷ 0.02139 = 1
N = 0.08571 ÷ 0.02139 = 4.01 ≈ 4
H = 0.2570 ÷ 0.02139 = 12.01 ≈ 12
Cl = 0.06423 ÷ 0.02139 = 3.00 ≈ 3
⇒ Empirical formula = CoN₄H₁₂Cl₃
Compare empirical mass to molar mass:
M_r(CoN₄H₁₂Cl₃) = 58.9 + (4 × 14.0) + (12 × 1.0) + (3 × 35.5) = 233.4 g mol⁻¹
Since empirical formula mass = given molar mass (233.4 g mol⁻¹), the molecular formula is also CoN₄H₁₂Cl₃.

🧠 Exam Technique

The question explicitly says "Show that the empirical formula and the molecular formula...". Many students lose the second mark simply by stopping after finding the ratio 1 : 4 : 12 : 3 without summing the relative atomic masses to show it matches 233.4 g mol⁻¹.

❌ Common Errors

Dividing by molecular masses (e.g. N₂ = 28, Cl₂ = 71, H₂ = 2) instead of atomic masses (A_r).
Premature rounding during the initial division steps, distorting the integer ratio.

Mark Scheme Breakdown:
• Mark 1: Calculate moles of all four elements correctly.
• Mark 2: Divide by smallest mole value to show 1 : 4 : 12 : 3 AND show calculated M_r = 233.4 g mol⁻¹ to conclude molecular formula is CoN₄H₁₂Cl₃.

Part (a)(ii)

Determining Free Chloride Ions Precipitated

📐 Step-by-Step Calculation

Calculate moles of solid R:
Moles of R = mass ÷ M_r = 1.00 g ÷ 233.4 g mol⁻¹ = 4.285 × 10⁻³ mol
Calculate moles of AgCl precipitate formed:
M_r(AgCl) = 107.9 + 35.5 = 143.4 g mol⁻¹
Moles of AgCl = 0.614 g ÷ 143.4 g mol⁻¹ = 4.282 × 10⁻³ mol
Deduce reacting ratio:
Moles of Cl⁻ precipitated = moles of AgCl = 4.282 × 10⁻³ mol
Ratio of moles of R : moles of Cl⁻ released = (4.285 × 10⁻³) : (4.282 × 10⁻³) ≈ 1 : 1
Conclusion: 1 mol of Cl⁻ ions is released per mole of R.

✅ Correct Conclusion

1 mole of chloride ions is released when 1 mole of R dissolves.

💡 Coordination Chemistry Context

Only chloride ions residing outside the coordination sphere (counter-ions) dissociate in aqueous solution to react with Ag⁺(aq). Chlorides bound directly to the cobalt as ligands do not precipitate as AgCl.

❌ Common Errors

Stating "3Cl⁻" because the formula contains 3 chlorides, or giving the answer as "3" without calculating. The mark scheme strictly instructs: "Do not award 3 Cl⁻ ions".

Mark Scheme Breakdown:
• Mark 1: Moles of AgCl = 4.28 × 10⁻³ mol (or moles of R = 4.28 × 10⁻³ mol).
• Mark 2: Showing ratio is 1 : 1, giving final answer of 1 mol of chloride ions.

Part (a)(iii)

Deducing Formula, Charge, and 3D Shape of Complex Ion Q

💡 Deducing the Structure

From (a)(i), R contains: 1 Co, 4 N, 12 H (i.e. 4 × NH₃ ligands), and 3 Cl.
From (a)(ii), only 1 Cl⁻ is an uncoordinated counter-ion in solution.
Therefore, the remaining 2 Cl atoms must act as ligands inside the coordination sphere!
Ligands coordinated to cobalt: 4 × NH₃ and 2 × Cl⁻ → coordination number = 6 (octahedral).
Formula of complex ion Q: [Co(NH₃)₄Cl₂]⁺
Charge deduction: Co was oxidized from Co(II) to Co(III) by O₂ bubbling: Co³⁺ + 2(Cl⁻) + 4(NH₃°) = +1 overall charge.

✅ How to Draw the 3D Octahedral Shape

Examiner Drawing Description

• Central Co atom.
• Two vertical bonds in the plane of the paper (one straight up, one straight down).
• Four bonds in the equatorial plane: two wedged bonds (pointing out/forward) and two dashed/hatched bonds (pointing back/away).
• Place four NH₃ ligands (bonded through N) and two Cl ligands (bonded through Cl). (cis or trans arrangement are both credited).
• Large square brackets around the whole complex with an overall charge of + (or +1) outside the top right.

🧠 Ligand Connectivity Tip

Always bond the nitrogen directly to the cobalt (e.g. write H₃N—Co on the left side, not NH₃—Co where H appears bonded). Although connectivity was ignored here, it is strictly penalised on other papers!

Mark Scheme Breakdown:
• Mark 1: Octahedral shape drawn around cobalt using 3D wedge conventions.
• Mark 2: Correct ligands shown: 4 × NH₃ and 2 × Cl.
• Mark 3: Correct overall charge of + (or +1).

Part (b)(i)

Definition of 'Dynamic Equilibrium'

✅ Essential Marking Points

The rate of the forward reaction equals the rate of the backward (reverse) reaction.
The concentrations of reactants and products remain constant (in a closed system).

❌ Pitfalls to Avoid

DO NOT say: "Concentrations of reactants and products are equal." (Concentrations are constant, not necessarily equal).
DO NOT say: "Rates are constant." (Rates must be equal / the same). Saying "rates of forward and reverse reactions are constant" gets 0 marks!

Mark Scheme Breakdown:
• Mark 1: Forward and backward reactions proceed at the same rate.
• Mark 2: Concentration of reactants and products remain constant / remain the same.

Part (b)(ii)

Equilibria, Entropy, and Complex Ion Stability (The Chelate Effect)

✅ Model Answer (Scoring 3/3)

Identify most stable complex: [Co(EDTA)]²⁻(aq) is the most stable complex because it has the largest log K_c value (16.5), meaning its position of equilibrium lies furthest to the right.
Particle number comparison: In the reaction forming [Co(EDTA)]²⁻, 2 moles of ions/particles react to form 7 moles of products/ions (1 [Co(EDTA)]²⁻ + 6 H₂O).
Entropy change: The increase in the total number of particles leads to a significant increase in system entropy (ΔS_system is positive). This makes ΔG much more negative, driving the reaction strongly forward (the chelate effect).

🧠 Exam Structure Guide

Notice the question demands an explanation in terms of both equilibria AND entropy. You must address both:

Equilibria aspect: Mention log K_c = 16.5 and equilibrium position furthest to RHS.
Entropy aspect: Count species on left vs. right (2 → 7) and explicitly state ΔS_system increases.

❌ Examiner Warnings

Do not refer to the species as "molecules" because EDTA⁴⁻ and the complexes are ions. Use the terms moles, particles, or ions.

Mark Scheme Breakdown:
• Mark 1: [Co(EDTA)]²⁻ is the most stable as it has the largest log K_c / equilibrium lies furthest to RHS.
• Mark 2: 2 moles of reactant particles/ions form 7 moles of product particles/ions in the EDTA reaction.
• Mark 3: Resulting in a greater increase in entropy of the system (ΔS_sys > 0).

Topics

Physical Chemistry · Inorganic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 10: Equilibrium I · Topic 13: Energetics II · Topic 15: Transition Metals

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.