A-LevelChemistryChemical bonding9701/42 · May/June 2017 · Q517 Marks

1,2-diaminoethane, en, H2NCH2CH2NH2, is a bidentate ligand.

(a)(i)What is meant by the terms bidentate and ligand? bidentate ligand[2]
(a)(ii)There are three isomeric complex ions with the formula [Cr(en)2Cl2]+. Complete the three-dimensional diagrams of the isomers in the boxes. You may use N N to represent en. Cr Cr Cr[3]
(b)(i)Write the expressions for the stability constants, Kstab1 and Kstab2, for equilibria 1 and 2. Include units in your answers. Kstab1 = ... units = ... Kstab2 = ... units = ...[3]
(b)(ii)Write an expression for the equilibrium constant, Keq3, in terms of Kstab1 and Kstab2. Keq3 = ...[1]
(b)(iii)Calculate the value of Keq3 stating its units. Keq3 = ... unit = ...[2]
(c)(i)Explain why ΔS°eq2 is so different from ΔS°eq1-[1]
(c)(ii)Calculate ΔG°eq2 at 298 K. ΔG°eq2 = ... kJ mol-1[2]
(c)(iii)What conclusion can be made about the relative feasibility of equilibria 1 and 2? Explain your answer.[1]
(c)(iv)Using data from the table, suggest a value of ΔH° for equilibrium 3.[1]
(c)(v)State the type of reaction that is occurring in equilibrium 2.[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

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

[1] bidentate: (a species that) forms two dative bonds / donates two lone pairs
[1] ligand: a species that uses a lone pair to form a dative bond to a metal atom / metal ion
[3] Structures for three isomeric complex ions [Cr(en)2Cl2]+ are provided in the mark scheme.
[1] Kstab1 = [Cu(NH3)4²⁺]/[Cu²⁺][NH3]⁴
[1] Kstab2 = [Cu(en)2²⁺]/[Cu²⁺][en]²
[1] mol⁻⁴ dm¹² AND mol⁻² dm⁶
[1] Keq3 = Kstab2 / Kstab1
[1] Keq3 = Kstab2 / Kstab1 = 4.4(2) × 10⁶
[1] mol⁻² dm⁻⁶
[1] (ΔS°eq1 is negative as) more / 5 moles of reactants are forming (one mole of) the complex OR (ΔS°eq2 is positive as) fewer / 3 moles of reactants are forming (one mole of) the complex
[1] ΔG°eq2 = -100 – 298 × 40 / 1000 OR ∆G =∆H – T∆S
[1] = -112 or -111.9 (kJ mol⁻¹) correct answer [2]
[1] Since (ΔGeq2) is more negative (than ΔGeq1) AND equilibrium 2 is more feasible
[1] ΔH(3) = −8 (kJ mol⁻¹)
[1] ligand exchange / replacement / substitution / displacement

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