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A-LevelChemistryEquilibria9701/41 · Oct/Nov 2017 · Q713 Marks

Ethanedioate ions, ⁻O₂CCO₂⁻, are bidentate ligands. The abbreviation ed²⁻ can be used to represent ethanedioate ions.

  1. (a)Complete the following electronic structures. • the iron atom, Fe 1s²2s²2p⁶ .............................. • the iron(III) ion, Fe³⁺ 1s²2s²2p⁶ ..............................[1]
  2. (b)Solutions of iron(III) salts are acidic due to the equilibrium shown. [Fe(H₂O)₆]³⁺(aq) ⇌ [Fe(H₂O)₅(OH)]²⁺(aq) + H⁺(aq) Kₐ = 8.9 × 10⁻⁴ mol dm⁻³ Calculate the pH of a 0.25 mol dm⁻³ FeCl₃ solution. pH = ..............................[2]
  3. (c)(i)What is meant by the term stability constant, Kstab?[1]
  4. (c)(ii)Use the data in the table to predict the formula of the complex formed in the greatest amount when • a solution containing equal concentrations of both F⁻ and SCN⁻ ions is added to Fe³⁺(aq), • a solution containing equal concentrations of both Fe³⁺ and Hg²⁺ ions is added to Cl⁻(aq).[1]
  5. (d)The complex [Fe(ed)₂Cl₂]³⁻ can be formed according to the equation shown. [Fe(H₂O)₄Cl₂]⁺(aq) + 2ed²⁻(aq) ⇌ [Fe(ed)₂Cl₂]³⁻(aq) + 4H₂O(l) Write the expression for the equilibrium constant, Kstab, and state its units. Kstab = .............................. units ..............................[2]
  6. (e)(i)Complete the three-dimensional diagrams to show the three stereoisomers of [Fe(ed)₂Cl₂]³⁻. You may use [symbol for ethanedioate] to represent ed²⁻.[3]
  7. (e)(ii)Give the letters of two isomers of [Fe(ed)₂Cl₂]³⁻ which are geometrical isomers of each other.[1]
  8. (e)(iii)Give the letters of the two isomers of [Fe(ed)₂Cl₂]³⁻ which show optical isomerism.[1]
  9. (e)(iv)Give the letter of the isomer which has no dipole moment.[1]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

  1. [1] (a) Fe atom= (1s²2s²2p⁶)3s²3p⁶3d⁶4s²
  2. [1] (a) Fe³⁺ ion= (1s²2s²2p⁶)3s²3p⁶3d⁵
  3. [1] (b) ([H⁺]² = = 8.9 × 10⁻⁴ × 0.25 or 2.225 × 10⁻⁴)
  4. [1] (b) [H⁺] = 0.0149
  5. [1] (b) pH = -log(0.0149) = 1.83
  6. [1] (c)(i) (Kstab is) the equilibrium constant for the formation of a complex (ion) (in a solvent from its constituent ions/molecules)
  7. [1] (c)(ii) [Fe(H₂O)₅F]²⁺ and [Hg(H₂O)₅Cl]⁺
  8. [1] (d) Kstab = [Fe(ed)₂Cl₂³⁻] / ([Fe(H₂O)₄Cl₂⁺][ed]²)
  9. [1] (d) mol⁻² dm⁶
  10. [1] (e)(i) [Diagram for isomer A (cis with Cl-Cl trans to each other, bidentate ed ligands are cis to each other)]
  11. [1] (e)(i) [Diagram for isomer B (cis with Cl-Cl cis to each other, bidentate ed ligands are cis to each other, enantiomer of A)]
  12. [1] (e)(i) [Diagram for isomer C (trans with Cl-Cl trans to each other, bidentate ed ligands are cis to each other)]
  13. [1] (e)(ii) any cis isomer and the trans isomer identified
  14. [1] (e)(iii) both correct cis isomers identified
  15. [1] (e)(iv) trans isomer identified

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About this question

Topic
Equilibria, A-Level Chemistry (syllabus 9701). Structured question, 13 marks.
Source
A-Level Chemistry, Oct/Nov 2017 Paper 4 Variant 1, question 7.

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