Sketch the shape of a 3dxy orbital.

i) Some transition elements and their compounds behave as catalysts. Explain why transition elements behave as catalysts.
[2]
ii) Catalysis can be classified as heterogeneous or homogeneous. Complete Table 1.1 by placing one tick () in each row to indicate the type of catalysis in each reaction.
Table 1.1
| type of catalysis | |
reaction | heterogeneous | homogeneous |
Fe in the Haber process |
|
|
Fe2+ in the I– / S2O82– reaction |
|
|
NO2 in the oxidation of SO2 |
|
|
[1]
A solution containing a mixture of Sn2+ (aq) and Sn4+ (aq) is added to a solution containing a mixture of Fe2+ (aq) and Fe3+ (aq).
Table 2.2 lists electrode potentials for some electrode reactions of these ions.
Table 2.2
electrode reaction | Eɵ / V |
|---|---|
Fe2+ + 2e– ⇌ Fe | –0.44 |
Fe3+ + 3e– ⇌ Fe | –0.04 |
Fe3+ + e– ⇌ Fe2+ | +0.77 |
Sn2+ + 2e– ⇌ Sn | –0.14 |
Sn4+ + 2e– ⇌ Sn2+ | +0.15 |
Eɵ data from the table can be used to predict the reaction that takes place when the two solutions are mixed.
i) Construct an equation for this reaction.
[1]
ii) Calculate Eɵcell for this reaction.
[1]
Hexaaquairon(III) ions are pale violet. They form a colourless complex with fluoride ions, F–, as shown in equilibrium 1, and a deep-red complex with thiocyanate ions, SCN–, as shown in equilibrium 2.
equilibrium 1:
[Fe(H2O)6]3+ + F– ⇌ [Fe(H2O)5F]2+ + H2O
violet ⇌ colourless
Kstab = 2.0 × 105 mol–1 dm3
equilibrium 2:
[Fe(H2O)6]3+ + SCN– ⇌ [Fe(H2O)5SCN]2+ + H2O
violet ⇌ deep-red
Kstab = 1.0 × 103 mol–1 dm3
The following two experiments are carried out.
Experiment 1: A few drops of KSCN (aq) are added to 5 cm3 of Fe3+ (aq), followed by a few drops of KF (aq).
Experiment 2: A few drops of KF (aq) are added to 5 cm3 of Fe3+ (aq), followed by a few drops of KSCN (aq).
i) State and explain the sequence of colour changes you would observe in each of Experiment 1 and Experiment 2.
[4]
ii) State the type of reaction occurring during the experiments in (d)(i).
[1]
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. Show your working.
pH = ..........................................................
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