Transition elements Banker

25.0 cm³ of an acidified solution of iron(II) sulfate is titrated with 0.0200 mol dm⁻³ potassium manganate(VII). The mean titre is 22.40 cm³.

1. (a) Write the ionic equation for the reaction. [1]
2. (b) State the colour change at the end-point and explain why no indicator is needed. [2]
3. (c) Calculate the concentration of Fe²⁺ in the solution. [2]

One iron tablet (containing iron(II) sulfate) is crushed and dissolved in dilute sulfuric acid and the solution is made up to 250 cm³. 25.0 cm³ portions are titrated with 0.00200 mol dm⁻³ KMnO₄; the mean titre is 11.70 cm³.MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

1. (a) Calculate the mass of iron, in mg, in one tablet. (Ar Fe = 55.8) [3]
2. (b) Explain why the tablet is dissolved in dilute sulfuric acid rather than in water or hydrochloric acid. [2]

A 0.200 g sample of iron ore from a mine in Chiniot is dissolved in acid and all the iron is reduced to Fe²⁺. The solution needs 24.30 cm³ of 0.0167 mol dm⁻³ K₂Cr₂O₇ for complete oxidation.

1. (a) Write the ionic equation for the reaction between dichromate(VI) and iron(II) ions in acid. [1]
2. (b) Calculate the percentage by mass of iron in the ore. (Ar Fe = 55.8) [3]
3. (c) State the colour change of the dichromate ion as it reacts. [1]

A student standardises a KMnO₄ solution. 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate, Na₂C₂O₄, is acidified with dilute H₂SO₄ and warmed to 60 °C, then titrated with KMnO₄ from a burette.2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂Otitration | rough | 1 | 2 | 3final reading / cm³ | 23.80 | 23.15 | 23.60 | 23.25initial reading / cm³ | 0.00 | 0.00 | 0.20 | 0.00titre / cm³ | 23.80 | 23.15 | 23.40 | 23.25

1. (a) Select the concordant titres and calculate the mean titre. [2]
2. (b) Calculate the concentration of the KMnO₄ solution, to 3 significant figures. [3]
3. (c) Explain why the flask is warmed, and why the first few drops of KMnO₄ take a long time to lose their colour but later drops decolourise quickly. [2]

1. (a) Calculate the mass of potassium dichromate(VI) needed to make 250 cm³ of a 0.0200 mol dm⁻³ solution. (Mr K₂Cr₂O₇ = 294.2) [2]
2. (b) Write the half-equation for the reduction of dichromate(VI) ions in acid. [1]
3. (c) Calculate the amount of Fe²⁺ that reacts with 20.0 cm³ of this solution. [1]
4. (d) Unlike KMnO₄, dichromate titrations need a redox indicator. Explain why. [1]

10.0 cm³ of a hair bleach containing hydrogen peroxide is diluted to 250 cm³. 25.0 cm³ of the diluted solution is acidified and titrated with 0.0200 mol dm⁻³ KMnO₄; 21.80 cm³ is needed.

1. (a) Write the equation for the reaction, using the half-equation H₂O₂ → O₂ + 2H⁺ + 2e⁻. [1]
2. (b) Calculate the concentration of H₂O₂ in the original hair bleach in mol dm⁻³. [3]
3. (c) Convert your answer into g dm⁻³. (Mr H₂O₂ = 34.0) [1]

Spinach leaves contain ethanedioic (oxalic) acid. An extract from 25.0 g of spinach is made up to 100 cm³. 25.0 cm³ of this is acidified, warmed to 60 °C and titrated with 0.0100 mol dm⁻³ KMnO₄; the titre is 14.80 cm³.2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

1. (a) Give the oxidation number of carbon in C₂O₄²⁻ and in CO₂. [1]
2. (b) Calculate the mass of ethanedioic acid, H₂C₂O₄ (Mr 90.0), in mg per 100 g of spinach. Give your answer to 3 significant figures. [4]

A 0.400 g sample of brass from a Peshawar metal workshop is dissolved in nitric acid; the solution is neutralised, and excess KI(aq) is added.2Cu²⁺ + 4I⁻ → 2CuI + I₂I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻The iodine needs 20.20 cm³ of 0.200 mol dm⁻³ Na₂S₂O₃.

1. (a) Deduce the mole ratio Cu²⁺ : S₂O₃²⁻. [1]
2. (b) Calculate the percentage by mass of copper in the brass. (Ar Cu = 63.5) [3]
3. (c) Starch is the indicator. State when it is added and the colour change at the end-point. [2]

6.95 g of hydrated iron(II) sulfate, FeSO₄·xH₂O, is dissolved in dilute sulfuric acid and made up to 250 cm³. 25.0 cm³ portions need 25.00 cm³ of 0.0200 mol dm⁻³ KMnO₄.MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

1. (a) Calculate the amount of Fe²⁺ in the 250 cm³ solution. [2]
2. (b) Calculate the Mr of the crystals and hence the value of x. (Mr FeSO₄ = 151.9, Mr H₂O = 18.0) [2]
3. (c) Old crystals of this salt often have a brown coating. Suggest what this is and how it would affect the calculated value of x. [2]

A standard solution is made by dissolving 9.80 g of ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂·6H₂O, in dilute sulfuric acid and making it up to 250 cm³. 25.0 cm³ portions are pipetted into a flask, a redox indicator is added, and the solution is titrated with potassium dichromate(VI) of unknown concentration.Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂Otitration | rough | 1 | 2 | 3titre / cm³ | 26.00 | 25.55 | 25.65 | 26.10

1. (a) Calculate the mean titre using concordant results only. [1]
2. (b) Calculate the concentration of the K₂Cr₂O₇ solution, to 3 significant figures. (Mr of ammonium iron(II) sulfate = 392.0) [3]
3. (c) Each burette reading has an uncertainty of ±0.05 cm³. Calculate the percentage error in the mean titre. [1]
4. (d) Run 3 is not used. Suggest one practical mistake that could cause a titre that is too large. [1]

Two students titrate 25.0 cm³ portions of the same iron(II) solution with the same KMnO₄ solution. Student A acidifies with dilute sulfuric acid and gets a mean titre of 18.20 cm³. Student B uses dilute hydrochloric acid instead and gets a mean titre of 19.35 cm³.

1. (a) Calculate the percentage difference between student B's titre and student A's titre. [1]
2. (b) Explain why student B's titre is larger. [2]
3. (c) Explain why the top of the meniscus is read for KMnO₄ in a burette. [1]
4. (d) Student A obtained titres of 18.20, 18.25 and 18.95 cm³. Identify the anomalous titre and suggest a possible cause. [2]

Pyrolusite ore contains manganese(IV) oxide. A 0.400 g sample is warmed with 50.0 cm³ of 0.100 mol dm⁻³ sodium ethanedioate in acid (an excess):MnO₂ + C₂O₄²⁻ + 4H⁺ → Mn²⁺ + 2CO₂ + 2H₂OThe excess ethanedioate is then titrated with 0.0200 mol dm⁻³ KMnO₄, needing 26.40 cm³:2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

1. (a) Calculate the amount of ethanedioate that remained after reaction with the MnO₂. [2]
2. (b) Calculate the percentage by mass of MnO₂ in the ore. (Mr MnO₂ = 86.9) [3]
3. (c) Give the change in oxidation number of manganese in each of the two reactions. [1]

25.0 cm³ of a 0.0400 mol dm⁻³ solution of violet V²⁺ ions is titrated with acidified 0.0200 mol dm⁻³ KMnO₄. 30.00 cm³ is needed. MnO₄⁻ is reduced to Mn²⁺.

1. (a) Calculate the amount of electrons removed from each vanadium ion and deduce the final oxidation state of vanadium. [3]
2. (b) The vanadium product is VO₂⁺. Write the balanced ionic equation for the reaction. [2]

A solution contains a mixture of Fe²⁺ and Fe³⁺ ions.Titration 1: 25.0 cm³ of the solution, acidified, needs 15.00 cm³ of 0.0200 mol dm⁻³ KMnO₄.Titration 2: another 25.0 cm³ is warmed with excess zinc to reduce all the Fe³⁺ to Fe²⁺, the zinc is filtered off, and the solution needs 24.00 cm³ of the same KMnO₄.MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

1. (a) Calculate the concentrations of Fe²⁺ and of Fe³⁺ in the solution. [4]
2. (b) Explain why all the excess zinc must be removed before the second titration. [1]
3. (c) Write the equation for the reduction of Fe³⁺ by zinc. [1]

A student plans to determine the concentration of a KMnO₄ solution using ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂·6H₂O, as the primary standard.

1. (a) Describe how to prepare 250 cm³ of a standard solution containing about 2.45 g of the solid. Include the apparatus and the steps that make the concentration accurate. [4]
2. (b) The balance reads to ±0.005 g. Calculate the percentage error in the mass of solid used. [2]
3. (c) Suggest why dilute sulfuric acid, rather than distilled water alone, is used to dissolve the solid. [1]

Step 1: 25.0 cm³ of 0.0150 mol dm⁻³ K₂Cr₂O₇ is acidified and excess KI is added. The iodine released needs 22.75 cm³ of sodium thiosulfate solution.Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O; I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻Step 2: 10.0 cm³ of household bleach is diluted to 250 cm³. 25.0 cm³ of this is acidified and excess KI is added; the iodine needs 18.30 cm³ of the same thiosulfate.ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O

1. (a) Deduce the mole ratio Cr₂O₇²⁻ : S₂O₃²⁻ and calculate the concentration of the thiosulfate, to 3 significant figures. [3]
2. (b) Calculate the concentration of ClO⁻ in the original bleach. [3]

1. (a) Define the term transition element. [1]
2. (b) Use electronic configurations to explain why zinc and scandium are d-block elements but not transition elements. [2]
3. (c) Give three characteristic properties of transition elements or their compounds, other than variable oxidation states. [2]

1. (a) Give the oxidation number of manganese in each: KMnO₄, K₂MnO₄, MnO₂, MnSO₄. [2]
2. (b) Explain why transition elements can show several different oxidation states. [1]
3. (c) State the maximum oxidation state of chromium and explain it in terms of its electronic configuration. [1]

1. (a) Give two reasons why transition elements and their compounds are often good catalysts. [2]
2. (b) Classify each catalyst as homogeneous or heterogeneous:(i) iron in the Haber process(ii) Fe²⁺(aq) in the S₂O₈²⁻/I⁻ reaction(iii) nickel in the hydrogenation of vegetable oils to make vanaspati ghee(iv) V₂O₅ in the Contact process [2]

Peroxodisulfate ions oxidise iodide ions: S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂E°: S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻ +2.01 V; Fe³⁺ + e⁻ ⇌ Fe²⁺ +0.77 V; I₂ + 2e⁻ ⇌ 2I⁻ +0.54 V

1. (a) Suggest why the uncatalysed reaction is slow even though it is energetically feasible. [1]
2. (b) Write two equations to show how Fe²⁺ ions catalyse the reaction, and calculate E°cell for each step. [4]
3. (c) Explain why Fe³⁺ ions would catalyse the reaction equally well. [1]
4. (d) State whether this is homogeneous or heterogeneous catalysis. [1]

Data for calcium and iron (both in Period 4):property | Ca | Femelting point / °C | 842 | 1538density / g cm⁻³ | 1.55 | 7.87atomic radius / nm | 0.197 | 0.126

1. (a) Explain why iron has a much higher melting point than calcium. [2]
2. (b) Explain why iron is much denser than calcium. [1]
3. (c) Compare the colour of aqueous Ca²⁺ ions with that of aqueous Fe²⁺ ions and explain the difference. [2]

1. (a) Vanadium(V) oxide is the catalyst in the Contact process. In the first step V₂O₅ oxidises SO₂ to SO₃ and is itself reduced to V₂O₄; in the second step the catalyst is regenerated by oxygen. Write an equation for each step. [2]
2. (b) Give the oxidation number of vanadium in V₂O₅ and in V₂O₄. [1]
3. (c) Explain which property of transition elements makes vanadium(V) oxide an effective catalyst here. [1]
4. (d) State the type of catalysis. [1]

The reaction S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ can be catalysed by a transition-metal ion that cycles between two oxidation states.E°: S₂O₈²⁻/SO₄²⁻ +2.01 V; I₂/I⁻ +0.54 VCandidate couples: Fe³⁺/Fe²⁺ +0.77 V; VO₂⁺/VO²⁺ +1.00 V; Cr³⁺/Cr²⁺ −0.41 V; Cu²⁺/Cu⁺ +0.15 V

1. (a) State the two conditions that the E° of a couple must meet for it to catalyse the reaction. [2]
2. (b) Use the data to identify which of the four couples could catalyse the reaction. Explain your answer for one couple that cannot. [3]

Acidified KMnO₄ is mixed with excess ethanedioate ions at 25 °C and the absorbance of MnO₄⁻ (proportional to its concentration) is recorded:time / s | 0 | 30 | 60 | 90 | 120 | 150 | 180absorbance | 1.00 | 0.97 | 0.90 | 0.70 | 0.38 | 0.15 | 0.05

1. (a) Calculate the mean rate of decrease of absorbance between 0 and 30 s and between 60 and 120 s, in s⁻¹. Calculate how many times faster the second rate is. [2]
2. (b) Explain the shape of the results. [3]
3. (c) Describe an experiment to confirm that Mn²⁺ is the catalyst. [1]

1. (a) Give the full electronic configuration of: Fe, Fe²⁺, Fe³⁺, Cr, Cu⁺ and Mn²⁺. [4]
2. (b) Explain why iron loses its 4s electrons before its 3d electrons when it forms ions. [1]

1. (a) State how many d orbitals there are in the 3d subshell and the maximum number of electrons it can hold. [1]
2. (b) Describe in words the shapes of the 3dxy and 3dz² orbitals, including how they are arranged relative to the x, y and z axes. [2]
3. (c) Name the two 3d orbitals whose lobes point along the axes. [1]

1. (a) Give the number of unpaired electrons in each gaseous ion: Ti³⁺, V³⁺, Cr³⁺, Mn²⁺, Fe²⁺, Ni²⁺, Cu²⁺, Zn²⁺. [3]
2. (b) Identify the ion in the list that has the same 3d configuration as Fe³⁺. [1]
3. (c) Paramagnetism increases with the number of unpaired electrons. Identify the most paramagnetic and the least paramagnetic ion in the list. [1]

Third ionisation energies / kJ mol⁻¹: Mn 3248; Fe 2957

1. (a) Give the electronic configurations of Mn²⁺ and Fe²⁺ in the form [Ar]3dⁿ. [1]
2. (b) Explain why the third ionisation energy of manganese is greater than that of iron. [3]
3. (c) Use the data to explain why aqueous Fe²⁺ is easily oxidised to Fe³⁺ in air, while Mn²⁺ is much harder to oxidise to Mn³⁺. [1]

1. (a) An ion M²⁺ has 25 electrons. Identify M, give the configuration of M²⁺ and the number of unpaired electrons. [3]
2. (b) An ion Y³⁺ has the same number of electrons as V²⁺. Identify Y and give the configuration of Y³⁺. [2]
3. (c) Copper(I) compounds such as CuCl are white, whereas copper(II) compounds are usually blue or green. Explain this in terms of electronic configuration. [2]

The first five ionisation energies of a Period 4 d-block element Z are (kJ mol⁻¹): 659, 1310, 2652, 4175, 9581.

1. (a) Calculate the ratio IE5 ÷ IE4 and use it to deduce the group of Z. [2]
2. (b) Identify Z and give the electronic configuration of the Z⁴⁺ ion. [2]
3. (c) Ti⁴⁺ compounds such as TiO₂ are white. Explain why. [1]