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Transition metals test answers and mark scheme for A Level chemistry (free PDF download)

The questions are on the transition metals test. Take it under real conditions before reading the mark scheme.

Quick answer: Transition metals are d-block elements that form ions with a partially filled d subshell. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on transition metals, from electron configurations to cisplatin, with the equivalent answers to accept.

Transition metals test answers and mark scheme for A Level chemistry (free PDF download)

Compare your answers with the model answers and the marking points. Credit is given for correct working even if a later number is wrong; use the gaps to decide which topic to revise.

Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.2.5

Question 1 Electron configurations (2 marks)

  1. [Ar] 3d⁵ 4s¹.
  2. Cu⁺ loses the 4s electron: [Ar] 3d¹⁰.

Marking: 1 mark each.

Question 2 Definition (3 marks)

  1. A d-block element that forms at least one ion with a partially filled d subshell.
  2. Zinc forms only Zn²⁺, which is 3d¹⁰ and so has a full d subshell; it has no ion with a partially filled d subshell.

Marking: 1 for definition, 1 for Zn²⁺ 3d¹⁰, 1 for full subshell link.

Question 3 Read the complex (2 marks)

  1. Coordination number 6, octahedral.
  2. NH₃ is neutral, so cobalt is +3.

Marking: 1 for number and shape together, 1 for +3.

Question 4 Colour of a complex (3 marks)

  1. Ligands split the 3d orbitals into two energy sets. An electron absorbs a photon of energy ΔE (red-orange light) to jump to the higher set. The remaining blue light is transmitted.

Marking: 1 for orbital splitting, 1 for absorbing light of energy ΔE, 1 for transmitted colour.

Question 5 Ligand substitution (2 marks)

  1. [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O; deep blue.

Marking: 1 for the balanced equation, 1 for the colour.

Question 6 Titration with manganate(VII) (4 marks)

  1. n = 0.02000 × 0.0200 = 4.00 × 10⁻⁴ mol.
  2. n(Fe²⁺) = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol.
  3. c = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 mol dm⁻³.

Marking: 1 each for (a) and (b); 1 for correct ratio use and 1 for concentration in (c).

Question 7 Catalysts (2 marks)

  1. They can change oxidation state and so form intermediates, giving an alternative route with lower activation energy; solid catalysts also adsorb reactants on the surface.

Marking: 1 for variable oxidation states or surface, 1 for lower activation energy.

Question 8 Cisplatin (2 marks)

  1. Square planar.
  2. cis-trans (geometric) isomerism.

Marking: 1 mark each.

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