Edexcel A-Level Chemistry Paper 1, November 2020: Question 3

18 marks · Hard difficulty · Open Response

Describe chromium bonding and high melting point, chromium(III) complex shape and colour, entropy changes for a ligand substitution equilibrium involving EDTA4-, and calculate the final oxidation state of vanadium in a redox reaction with chlorine.

Practise this question

Question

A multi-part exam question about transition metals and chromium complexes. Part (a) asks to describe bonding in chromium and high melting point. Part (b) covers the shape and colour of [Cr(H2O)6]3+. Part (c) gives the structure of EDTA4- and a ligand substitution equilibrium with [Cr(H2O)6]3+, asking to comment on the equilibrium constant using entropy. Part (d) provides titration data for the oxidation of VCl2 by Cl2 gas, asking to calculate the final oxidation state and state colour changes.
Question text

3 This question is about transition metals and transition metal complexes.

(a) Describe the bonding in the element chromium and use your answer to justify

why it has such a high melting temperature.

You may find it helpful to draw a labelled diagram.

(4)

(b) When chromium(III) sulfate dissolves in water, a green solution containing the

[Cr(H O) ]3+ ion forms.

(i) Give the shape of this complex ion.

(1)

(ii) Explain why the chromium complex ion is coloured.

(3)

(c) The ligand ethylenediaminetetraacetate, EDTA*P62668A0724*4−, has the structure shown.

O

−O O

O−

N

N

−O

O O−

O

When a solution of EDTA4− is added to a solution of [Cr(H O) ]3+ ions, a new

complex ion is formed.

[Cr(H O) ]3+ + EDTA4− ⇌ [Cr(EDTA)]− + 6H O

26 2

The equilibrium constant for this equilibrium is 2.51 × 1023 dm3 mol−1.

By considering the equilibrium for this reaction and changes in entropy, comment

on the value of the equilibrium constant. No calculations are required.

(3)

(d) Aqueous vanadium(II) chloride, VCl2(aq), can be oxidised by bubbling

gaseous chlorine, Cl2(g), through the solution in the absence of air.

40.0 cm3 of 0.100 mol dm−3 VCl solution was oxidised by 144 cm3 of chlorine gas,

at room temperature and pressure (r.t.p.).

The chlorine was reduced to chloride ions, according to the half-equation

Cl (g) + 2e− → 2Cl−(aq)

[Molar volume of a gas at r.t.p. = 24.0 dm3 mol−1]

(i) Use these data to calculate the final oxidation state of vanadium.

You must show your working.

8 (5)

*P62668A0824*

(ii) State the initial and final colours you would see as the chlorine bubbles

through the aqueous vanadium(II) chloride, VCl2(aq).

(2)

(Total for Question 3 = 18 marks)

Mark scheme

Show the mark scheme The mark scheme providing detailed marking points for all parts of question 3. Part (a) awards marks for lattice of positive ions, delocalised electrons, strong forces of attraction, and energy required. Part (b)(i) accepts octahedral. Part (b)(ii) covers d-orbital splitting, light absorption, and transmitted complementary colour. Part (c) awards marks for a large equilibrium constant, positive entropy change from increased moles. Part (d)(i) outlines calculations for moles of VCl2 and Cl2, mole ratio, electrons lost, and final oxidation state (+5). Part (d)(ii) lists initial purple/lilac and final yellow colours.

How to answer it

Transition Metals and Transition Metal Complexes Study Guide

What this question tests

This question assesses core understanding of transition metals and their compounds. Key areas tested include metallic bonding and physical properties of chromium, shapes and optical/electronic properties (colour) of complex ions, entropy changes during ligand substitution reactions involving multidentate ligands (the chelate effect), and quantitative redox titration calculations determining final oxidation states and associated colour changes.

Question Part (a)

Metallic Bonding and High Melting Temperature of Chromium

✅ Correct Answer Requirements (4 Marks)

  • Lattice of positive ions / regular arrangement of positive ions (1)
  • Sea of delocalised electrons (1)
  • Strong forces of attraction between positive ions and delocalised electrons (1)
  • Lots of heat energy needed to break the strong metallic bonds/attractions (1)

🧠 Exam Technique & Diagram Tip

Drawing a labelled diagram can effortlessly secure your points. Sketch a grid of positive ions (marked with 2+ or 3+) surrounded by interspersing dots or minus signs labelled "delocalised electrons". Ensure you explicitly mention the attraction between them in text or labels.

Question Part (b)

Shape and Colour of Chromium(III) Complex Ions

(i) Shape of [Cr(H₂O)₆]³⁺

✅ Correct Answer (1 Mark)

Octahedral (Allow "octahedral" or "octagonal" as a spelling slip, but avoid writing non-specific descriptions).

(ii) Explanation of Colour

💡 Key Knowledge (3 Marks)

  • Ligands (water molecules) cause the d orbitals to split into two energy levels (1)
  • Light/energy in the visible region is absorbed to promote d electrons to higher energy d orbitals (1)
  • The remaining/unabsorbed light (complementary colour / green light) is transmitted or reflected (1)

❌ Common Errors

Students often lose marks by stating that light is "emitted" when electrons return to the ground state. Make sure you state that light is absorbed to promote electrons, and unabsorbed light is transmitted.

Question Part (c)

Ligand Substitution, Entropy, and Equilibrium Constants

✅ Correct Answer (3 Marks)

  • The equilibrium constant is very large, so equilibrium lies far to the right / [Cr(EDTA)]⁻ is very stable (1)
  • Delta-S system is positive / increases entropy of the system (1)
  • Because 2 moles of reactants form 7 moles of products (an increase in number of particles/moles) (1)

🧠 Examiner Insights

This tests the classic chelate effect driven by entropy. To score the third mark, you must explicitly link the balanced equation coefficients: counting 1 complex + 1 ligand = 2 moles of reactants on the left, going to 1 complex + 6 water molecules = 7 moles of products on the right.

Question Part (d)

Redox Titration Calculations and Colour Changes

(i) Calculating Final Oxidation State of Vanadium

📐 Step-by-Step Calculation (5 Marks)

  1. Calculate moles of VCl₂:
    (40.0 / 1000) × 0.100 = 0.00400 mol (1)
  2. Calculate moles of Cl₂ gas:
    144 / 24000 = 0.00600 mol (1)
  3. Deduce reacting molar ratio (V²⁺ : Cl₂):
    0.00400 : 0.00600 = 2 : 3 (meaning 2V²⁺ : 3Cl₂ ) (1)
  4. Determine electrons lost per vanadium ion:
    Each Cl₂ gains 2 electrons (total 6 electrons for 3 Cl₂). These 6 electrons are lost by 2 V²⁺ ions, meaning 3 electrons lost per V ion (1)
  5. Deduce final oxidation state:
    Starting oxidation state = +2. Losing 3 electrons increases the oxidation state by 3: (+2) + 3 = +5 (1)

❌ Calculation Traps & Error Warning

Watch out for volume units! Always convert cm³ to dm³ by dividing by 1000 for solutions, and by 24000 for gases at r.t.p. Make sure your final answer is clearly stated as a sign-preceded integer (+5).

(ii) Initial and Final Colours

✅ Correct Answer (2 Marks)

  • Initial colour: Purple / lilac / violet (for V²⁺) (1)
  • Final colour: Yellow (for V⁵⁺ as VO₂⁺) (1)

🧠 Consequential Marking

Mark (ii) is fully marked consequentially based on your calculated final oxidation state in part (i). If your oxidation state was incorrect, ensure your stated colours match your derived oxidation state according to standard vanadium redox charts!

Topics

Physical Chemistry · Inorganic Chemistry · Topic 2: Bonding and Structure · Topic 5: Formulae, Equations and Amounts of Substance · Topic 13: Energetics II · Topic 15: Transition Metals

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.