Edexcel International A Level Chemistry: Transition metals: electronic structure, complexes and colour (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers outcomes 17.1 to 17.17 of Topic 17: Transition Metals and their Chemistry, in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It is IA2 content, taught in Unit 5: Transition Metals and Organic Nitrogen Chemistry. Work through the study guide and revision notes first if the topic is new. Record weak outcomes on the chemistry checklist, find more on the Edexcel course hub, or take a free diagnostic.

Data for this set: Ar values H = 1.0, N = 14.0, Cl = 35.5, Pt = 195.1; molar gas volume = 24.0 dm³ mol⁻¹.

Questions

1. This question is about electronic structure.

(a) Write the electronic configurations of an iron atom (Z = 26) and of the Fe³⁺ ion, using [Ar] shorthand. [2] (b) Scandium is a d-block element. Its only stable ion is Sc³⁺. Explain why scandium is not classed as a transition metal. [2]

2. Give the electronic configuration, in [Ar] shorthand, of each species: Cr (Z = 24), Cr³⁺, Cu (Z = 29), Ni²⁺ (Z = 28) and V²⁺ (Z = 23). [5]

3. Chromium forms CrO₃ and CrCl₂.

(a) Deduce the oxidation number of chromium in each compound. [2] (b) Explain why transition metals can show several oxidation numbers. [2]

4. Nickel(II) ions form the complex ion [Ni(NH₃)₆]²⁺.

(a) State what is meant by the term ligand. [1] (b) Describe the bonding between each NH₃ and the nickel ion. [2] (c) State the coordination number and the shape of this complex. [2]

5. Deduce the oxidation number of the metal and the coordination number in each complex.

(a) [Fe(H₂O)₅(OH)]²⁺ [2] (b) [NiCl₄]²⁻ [2] (c) [Co(NH₂CH₂CH₂NH₂)₃]³⁺ [2]

6. Aqueous nickel(II) sulfate is green; aqueous zinc sulfate is colourless.

(a) Explain why [Ni(H₂O)₆]²⁺ is coloured. [3] (b) Explain why [Zn(H₂O)₆]²⁺ is colourless. [2]

7. A student makes three observations. For each, identify the factor or factors from outcome 17.11 that cause the colour change.

(a) A pale green solution of iron(II) sulfate left open to the air slowly turns yellow-brown. [1] (b) Excess aqueous ammonia turns pale blue aqueous copper(II) sulfate deep blue, forming [Cu(NH₃)₄(H₂O)₂]²⁺. [1] (c) Concentrated hydrochloric acid turns a pink solution of [Co(H₂O)₆]²⁺ blue, forming [CoCl₄]²⁻. [2] (d) Explain, in terms of the d-orbitals, why changing a ligand changes the colour. [1]

8. This question is about the shapes of complexes.

(a) Explain why [Cr(H₂O)₆]³⁺ is octahedral, and give its bond angle. [3] (b) Suggest why chloride ions form [CoCl₄]²⁻ rather than a six-coordinate complex with Co²⁺. [1] (c) State the Cl–Co–Cl bond angle in [CoCl₄]²⁻. [1]

9. Cis-platin, [Pt(NH₃)₂Cl₂], is a square planar complex used in cancer treatment.

(a) Deduce the oxidation number of platinum. [1] (b) Describe how the cis and trans isomers differ, giving the Cl–Pt–Cl angle in each. [2] (c) Suggest why the drug is supplied as a single isomer. [1] (d) A dose contains 75.0 mg of cis-platin. Calculate the amount of cis-platin in moles and the mass of platinum, in mg, in the dose. [2]

10. NH₂CH₂CH₂NH₂ and EDTA⁴⁻ are both ligands.

(a) Explain why NH₂CH₂CH₂NH₂ is described as bidentate. [2] (b) State the number and type of donor atoms in EDTA⁴⁻. [1] (c) Fe²⁺ bonds to one EDTA⁴⁻ ion. Give the formula, with charge, of the complex and its coordination number. [2]

11. Haemoglobin carries oxygen in the blood.

(a) State the oxidation number of iron in haemoglobin. [1] (b) Haemoglobin contains a polydentate ligand. State what polydentate means. [1] (c) Explain how carbon monoxide reduces the oxygen carried by blood. [3] (d) A patient’s blood contains 9.0 g of haemoglobin per 100 cm³ and the total blood volume is 5.0 dm³. Each haemoglobin molecule (Mr = 64 500) has four iron(II) sites, each able to bind one O₂. 20% of the sites hold CO. Calculate the largest volume of O₂, in dm³, the blood can carry. [4]

12. Element X is in Period 4. Its ion X³⁺ has three 3d electrons.

(a) Identify X and give the electronic configuration of its atom. [2] (b) Give the electronic configuration of X²⁺. [1] (c) [X(NH₃)₆]³⁺ and [X(H₂O)₆]³⁺ have different colours. Explain why each complex is coloured, and why the colours differ. [3] (d) State the shape of these two complexes. [1]

Answers

1. (a) Fe: [Ar] 3d⁶ 4s² [1]. Fe³⁺: [Ar] 3d⁵ [1]. (b) Sc³⁺ is [Ar], with no 3d electrons [1]. A transition metal must form a stable ion with an incompletely filled d sub-shell, and scandium’s only ion has an empty one [1]. Examiner insight: For (b), name the ion and its configuration; a bare “it has no d electrons” is about the atom, which does have one, and is unlikely to earn both marks.

2. Cr [Ar] 3d⁵ 4s¹ [1]; Cr³⁺ [Ar] 3d³ [1]; Cu [Ar] 3d¹⁰ 4s¹ [1]; Ni²⁺ [Ar] 3d⁸ [1]; V²⁺ [Ar] 3d³ [1]. Examiner insight: Cr and Cu atoms are the exceptions; their ions follow the normal rule of losing 4s first, so do not carry the exception into Cr³⁺.

3. (a) CrO₃: x + 3(−2) = 0, so +6 [1]. CrCl₂: x + 2(−1) = 0, so +2 [1]. (b) The 4s and 3d sub-shells are close in energy [1], so different numbers of electrons can be lost or used in bonding without a very large jump in ionisation energy [1]. Examiner insight: “Similar energy” alone is only half the answer; link it to electrons being removed from both sub-shells.

4. (a) A species (molecule or anion) with a lone pair it can donate to the metal ion at the centre of a complex [1]. (b) Dative (coordinate) covalent bond [1]; the nitrogen of NH₃ donates its lone pair to the Ni²⁺ ion [1]. (c) 6 [1]; octahedral [1]. Examiner insight: Name the donor atom (N); “ammonia shares electrons” does not show which species supplies both electrons.

5. (a) x + 5(0) + (−1) = +2, so Fe is +3 [1]; coordination number 6 [1]. (b) x + 4(−1) = −2, so Ni is +2 [1]; coordination number 4 [1]. (c) The ligand is neutral, so Co is +3 [1]; three bidentate ligands give 3 × 2 = 6 [1]. Examiner insight: In (c), “3” is the number of ligands, not the coordination number; count dative bonds.

6. (a) The water ligands split the 3d orbitals into two energy levels [1]. A d electron absorbs visible light of energy ΔE (ΔE = hf) and is promoted to the higher level [1]. The light not absorbed is transmitted, and this is the green colour seen [1]. (b) Zn²⁺ is [Ar] 3d¹⁰, a full d sub-shell [1]. No electron can move to the upper d level, so no visible light is absorbed [1]. Examiner insight: Answers that describe electrons “falling back and giving out light” describe emission, which is the wrong model for colour in complexes.

7. (a) Oxidation number of iron changes, from +2 to +3 [1]. (b) The ligand changes, from H₂O to NH₃ [1]. (c) The ligand changes, H₂O to Cl⁻ [1], and the coordination number changes, from 6 to 4 [1]. (d) A different ligand splits the d-orbitals by a different amount, so ΔE and the frequency of light absorbed change [1]. Examiner insight: Where two factors change, as in (c), naming only one leaves a mark behind.

8. (a) There are six bonding pairs around the chromium ion [1]. They repel each other equally and get as far apart as possible [1], giving an octahedral shape with bond angles of 90° [1]. (b) A chloride ion is bigger than an H₂O molecule, so there is room for just four round Co²⁺ [1]. (c) 109.5° [1]. Examiner insight: Refer to bonding pairs repelling; “six ligands make it octahedral” restates the question without explaining it.

9. (a) x + 2(0) + 2(−1) = 0, so +2 [1]. (b) Cis: the two Cl ligands are adjacent, at 90°; trans: they are opposite, at 180° [1]. Both are flat, with Pt and the four donor atoms in one plane [1]. (c) The two isomers are different compounds with different effects in the body, and only the cis form is the intended drug [1]. (d) Mr = 195.1 + 2(17.0) + 2(35.5) = 300.1; n = 0.0750 ÷ 300.1 = 2.50 × 10⁻⁴ mol [1]. Mass of Pt = 2.50 × 10⁻⁴ × 195.1 = 0.0488 g = 48.8 mg [1]. Examiner insight: Convert mg to g before dividing by Mr, and convert back at the end; unit slips are the usual source of a factor of 1000 error.

10. (a) Each molecule has two donor atoms, the nitrogen atoms [1], so it forms two dative bonds to the same metal ion [1]. (b) Six: two nitrogen and four oxygen atoms [1]. (c) [Fe(EDTA)]²⁻ (+2 − 4 = −2) [1]; coordination number 6 [1]. Examiner insight: “Bidentate means it has two lone pairs” is not enough; the two pairs must bond to the same metal ion.

11. (a) +2 [1]. (b) A ligand that forms several dative bonds to the same metal ion [1]. (c) CO replaces O₂ on the iron(II) by ligand exchange [1]. CO bonds more strongly than O₂ [1], so the change is not easily reversed and those sites cannot carry oxygen to tissues [1]. (d) Mass of haemoglobin = 9.0 × (5000 ÷ 100) = 450 g [1]. Amount = 450 ÷ 64 500 = 6.98 × 10⁻³ mol [1]. Sites free for O₂ = 4 × 0.80 × 6.98 × 10⁻³ = 2.23 × 10⁻² mol [1]. Volume = 2.23 × 10⁻² × 24.0 = 0.54 dm³ [1]. Examiner insight: Show each stage of a multi-step calculation; a wrong final value with clear working can still show the method, while a bare wrong number shows nothing.

12. (a) X is chromium [1]; atom [Ar] 3d⁵ 4s¹ [1]. (b) [Ar] 3d⁴ [1]. (c) In each, ligands split the 3d orbitals and d electrons absorb visible light to move to the higher level [1]. NH₃ and H₂O split the d-orbitals by different amounts, so ΔE differs [1]. Different frequencies are absorbed, so the light transmitted, and the colour seen, differs [1]. (d) Both octahedral [1]. Examiner insight: Working back from the d-electron count to the element needs the configuration of the ion, then the atom; check the Cr exception at the last step.

Where marks are usually lost

• Taking electrons out of 3d ahead of 4s in an ion’s configuration.
• Writing Cr as [Ar] 3d⁴ 4s² or Cu as [Ar] 3d⁹ 4s².
• Leaving out “ion” or “incompletely filled” from the transition metal definition.
• Not stating that the ligand donates a lone pair in a dative bond.
• Counting ligands, not dative bonds, when a ligand is bidentate or hexadentate.
• Saying the colour seen is the colour absorbed.
• Missing one of two factors when the ligand and coordination number both change.
• Giving 90° for tetrahedral complexes, or confusing cis (90°) and trans (180°).
• Describing CO poisoning without ligand exchange or stronger bonding.

Next steps

• Recap with the revision notes
• Re-read the study guide
• Edexcel chemistry course hub
• Printable chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 17: Transition Metals and their Chemistry, outcomes 17.1 to 17.17.