A Level Chemistry: Colour, Stereoisomerism and Stability Constants — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Work through these before looking at the answers. Write full answers — the marks are for the reasoning, not the conclusion.

Related: Transition Elements: Colour, Stereoisomerism and Stability Constants revision notes

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Section A — short answer

1. Explain why [Sc(H₂O)₆]³⁺ is colourless. [2]

2. State the number of d electrons in Ti³⁺, and hence explain whether it can be coloured. [2]

3. State how the five d orbitals split in an octahedral complex, giving the number of orbitals in each energy level. [2]

4. Define stability constant, Kstab. [2]

5. State whether the trans isomer of [Pt(NH₃)₂Cl₂] is polar or non-polar, and explain why. [2]

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Section B — structured

6. Aqueous cobalt(II) chloride is pink and contains the octahedral complex [Co(H₂O)₆]²⁺. Adding excess concentrated hydrochloric acid turns the solution blue, forming [CoCl₄]²⁻.

(a) State the coordination number and shape of [CoCl₄]²⁻. [2]

(b) Explain, in terms of ΔE, why this change causes the colour to change. [3]

(c) Explain why a solution containing only Zn²⁺ complexes would stay colourless even under the same conditions. [2]

7. [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ contains three bidentate 1,2-diaminoethane (“en”) ligands.

(a) State the coordination number of nickel in this complex. [1]

(b) Explain why this complex shows optical isomerism. [2]

(c) A related complex, [Ni(en)₂(H₂O)₂]²⁺, can show both geometrical and optical isomerism. State which of its two geometrical isomers, cis or trans, is optically active, and explain why the other is not. [3]

8. Aqueous silver ions react with excess aqueous ammonia: Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq).

(a) Write the expression for Kstab for this reaction. [1]

(b) Kstab for this reaction is 1.7 × 10⁷ mol⁻² dm⁶. At equilibrium, [[Ag(NH₃)₂]⁺] = 0.0400 mol dm⁻³ and [NH₃] = 0.200 mol dm⁻³. Calculate [Ag⁺]. [3]

(c) State and explain what a very large value of Kstab indicates about the position of this equilibrium. [2]

Section C

9. [Pt(NH₃)₂Cl₂] (the platinum complex used in the anti-cancer drug cisplatin) exists as cis and trans geometrical isomers. Deduce which isomer is polar and which is non-polar, explaining your reasoning in terms of the symmetry of the bond dipoles around the square planar platinum centre. [3] (The drug-action mechanism itself is background beyond 9701’s outcome 28.4, which requires only deducing overall polarity from geometry — it is not examined here.)

10. Adding excess aqueous ammonia to a solution of [Cu(H₂O)₆]²⁺ causes a ligand exchange reaction, forming [Cu(NH₃)₄(H₂O)₂]²⁺. Explain, in terms of Kstab, why this ligand exchange occurs. [3]

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Answers

1. Sc³⁺ is d⁰ — it has no d electrons to promote between the split d orbitals [1], so no d–d electron transition can occur and no visible light is absorbed [1].

2. Ti³⁺ is d¹ [1]. Since it has one d electron that can be promoted (and a vacancy to promote into), it can be coloured [1].

3. Two higher-energy orbitals and three lower-energy orbitals [2]. The reverse split (three higher, two lower) applies to tetrahedral complexes, not octahedral — a very common mix-up.

4. The equilibrium constant [1] for the formation of a complex ion in solution from its constituent ions/molecules [1].

5. Non-polar [1]. The two Cl (and two NH₃) ligands are opposite each other, so the complex is symmetric and the bond dipoles cancel [1].

6. (a) Coordination number 4 [1]; shape tetrahedral [1].

(b) The ligand and geometry change together alter ΔE (the splitting energy between the two sets of d orbitals) [1]; tetrahedral splitting is inherently smaller than octahedral splitting for comparable ligands [1], so a different frequency of light is absorbed, giving a different (complementary) colour [1].

(c) Zn²⁺ is d¹⁰ — the d subshell is full, so there is no vacancy to promote an electron into [1]; with no d–d electron transition possible, no visible light is absorbed regardless of ligand or geometry [1].

7. (a) 6 [1].

(b) The complex has no internal mirror plane [1], so it and its mirror image are non-superimposable — a pair of optical isomers [1].

(c) The cis isomer is optically active [1]. The trans isomer has a mirror plane (the two water ligands sit directly opposite each other, giving the complex internal symmetry) [1], so it is not chiral and shows no optical isomerism [1].

8. (a) Kstab = [[Ag(NH₃)₂]⁺] / ([Ag⁺][NH₃]²) [1].

(b) [NH₃]² = 0.200² = 0.0400 [1]. [Ag⁺] = 0.0400 ÷ (1.7 × 10⁷ × 0.0400) = 0.0400 ÷ (6.8 × 10⁵) [1] = 5.9 × 10⁻⁸ mol dm⁻³ [1].

(c) A large Kstab means the equilibrium lies strongly towards the complex [1] — the ammine complex is much more thermodynamically stable than the free Ag⁺ ion, so almost all the silver is present as [Ag(NH₃)₂]⁺ at equilibrium [1].

9. The cis isomer is polar [1]: its two Cl and two NH₃ ligands are not positioned symmetrically opposite each other around the square-planar platinum centre, so the Pt–Cl and Pt–N bond dipoles do not cancel [1]. The trans isomer is non-polar: its two Cl ligands are opposite each other, and its two NH₃ ligands are opposite each other, so by symmetry the bond dipoles cancel exactly [1]. (Cisplatin’s actual anti-cancer action is background beyond 9701’s syllabus, which requires only this polarity deduction from the complex’s geometry.)

10. Kstab for forming the ammine complex, [Cu(NH₃)₄(H₂O)₂]²⁺, from [Cu(H₂O)₆]²⁺ is large [1], meaning the ammine complex is much more thermodynamically stable than the starting hexaaqua ion [1] (by convention, water is omitted from the expression and [Cu(H₂O)₆]²⁺ is the reference species being converted, not a species with its own separate stability constant); because NH₃ binds more strongly than H₂O, once enough NH₃ is available the equilibrium shifts strongly towards the ammine complex, displacing water from the coordination sphere [1].

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Where marks are usually lost

• Reversing the octahedral d-orbital split (writing three higher, two lower — that’s the tetrahedral pattern).
• Explaining colour by “colour absorbed” instead of the complementary colour being observed.
• Missing the electron-count requirement: d⁰ has nothing to promote from, d¹⁰ has nowhere to promote to — both are colourless regardless of ligand.
• Assuming any complex with bidentate ligands is automatically chiral — check whether a mirror plane exists (the trans isomer above does, and is not chiral).
• Forgetting to square [NH₃] in the Kstab expression before substituting.