A-LevelChemistryChemical bonding9701/42 · Feb/Mar 2025 · Q317 Marks

Fe2+ and Fe3+ ions are able to form a variety of complexes with different species.

(a)(i)Define complex.[1]
(a)(ii)Table 3.1 gives some details of different complexes of Fe2+ and of Fe3+. Complete Table 3.1. Table 3.1 complex ion ligand coordination number formula and charge of complex E Fe2+ NH3 6 F [FeC14]2- G en [Fe(en)3]3+[3]
(a)(iii)Complete Fig. 3.1 to show the splitting of the d-orbitals in a tetrahedral complex. [Figure 3.1] energy[1]
(b)(i)Table 3.2 gives details of some complexes of Fe3+. Table 3.2 complex colour value of Kstab [Fe(H2O)6]3+ violet 1 [Fe(H2O)5SCN]2+ red 1.40 x 102 [Fe(H2O)5F]2+ colourless 2.40 x 105 Explain the reason for the difference in colour of the two complexes [Fe(H2O)6]3+ and [Fe(H2O)5SCN]2+.[2]
(b)(ii)Write an expression for Kstab of [Fe(H2O)5SCN]2+. Kstab = .......................................................................................................................................[1]
(b)(iii)Use information in Table 3.2 to calculate the value of the equilibrium constant, Kc, for the following reaction. [Fe(H2O)5SCN]2+ + F¯ ⇌ [Fe(H2O)5F]2+ + SCN- Kc = ..............................................[1]
(b)(iv)A few drops of KF(aq) are added to a solution of [Fe(H2O)6]3+(aq), followed by a few drops of KSCN(aq). Use information in Table 3.2 to describe any observations after each addition. Explain your answer.[2]
(c)(i)Hydrated compound J, K3Fe(C2O4)3•xH2O, contains the green complex ion [Fe(C2O4)3]³¯. The value of x can be determined by titration of a sample of J with acidified MnO4¯ions. MnO4 ions oxidise C2O42- ions in acidic conditions. 2MnO4¯ + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O Write half equations for the oxidation of C2O42- ions and for the reduction of MnO4¯ ions. • oxidation of C2O42- • reduction of MnO4¯[2]
(c)(ii)A student prepares a solution containing 0.100 g of J. The student titrates this solution with 0.0200 moldm⁻³ acidified KMnO4(aq). The titre obtained is 12.20 cm³. Assume all of the C2O42– ions are oxidised. Calculate the value of x in K3Fe(C2O4)3•XH2O. Give your answer to the nearest whole number. Show your working. [Mr: K3Fe(C2O4)3, 437.1] x = ..............................................[4]

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 (Feb/Mar 2025). The mark or mark type is shown in brackets.

[1] (central) metal / transition element atom / ion AND surrounded by / bonded to ligands
[3] complex E: [Fe(NH3)6]2+
complex F: Fe2+, Cl-, 4
complex G: Fe3+, 6
any two [1] any four [2] all six [3]
[1] three(3) higher AND two(2) lower boxes / energy levels
[1] d-d energy gap / ΔE is different
[1] different frequency / wavelength (of light) absorbed / transmitted / reflected
[1] Kstab = [[Fe(H₂O)₅SCN]²⁺] / ([[Fe(H₂O)₆]³⁺] [SCN⁻]])
[1] Kc = 2.40 x 105 / 1.40 × 102 = 1714 min 2sf
[2] M1: solution turns (violet / purple to) colourless AND solution then remains / stays colourless / no change
M2: Kstab of [Fe(H2O)5F]2+ > [Fe(H2O)5SCN]2+ OR [Fe(H2O)5F]2+ is most stable (more stable than [Fe(H2O)5SCN]2+)
[2] C2O42- → 2CO2 + 2e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
[4] M1 moles of manganate = 12.20/1000 × 0.0200 = 2.44 x 10-4 mol AND moles of ethanedioate = 5/2 x 2.44 × 10-4 = 6.10 x 10-4 mol
M2 moles of K3Fe(C2O4)3 = 6.10 × 10-4 ÷ 3 = 2.03 × 10-4 mol AND mass of K3Fe(C2O4)3 = 437.1 x 2.03 × 10-4 = 0.0889 g
M3 mass of water = 0.100 -0.0889 = 0.0111 g AND moles of water = 0.0111/18.0 = 6.18 x 10-4 mol
M4 x = molar ratio = 6.18 x 10-4/2.03 × 10-4 = 3