A2 Unit F325: Equilibria, energetics and elements

Module 6: Transition elements
Question 1 / Total marks: 15
(a)Iron is a typical transition element.
(i)What is meant by the term transition element?
(ii)Write the electron configuration for a Fe2+ ion.
(iii)Using iron as your example, state the typical properties of a transition element.
(iv)Describe what you would see when Fe3+(aq) ions react with aqueous sodium hydroxide and write the ionic equation.
Marks available: (i) 1 (ii) 1 (iii) 4
(iv) 2
Student answer:
(a)(i)An element which can form an ion with an incomplete d sub-shell.
(ii)1s22s22p63s23p63d6
(iii)Using iron:
Form coloured compounds, e.g. Fe2+ ions are green
Form complex ions, e.g. Fe(H2O)62+
Have ions with different oxidation states, e.g. Fe2+ is +2 and Fe3+ is +3
It shows good catalytic properties, e.g. Fe in the Haber process.
(iv) An orange/brown precipitate would form.
Fe3+(aq) + 3OH–(aq)  Fe(OH)3(s)
Examiner comments:
(a)(i)The student has correctly stated an ion with an incomplete d sub-shell (would also allow half-filled d-orbital).
(ii)A common error is to remove two 3d electrons and leave the two 4s electrons, instead of removing the two 4s electrons.
(iii)Iron is a good example to use as it is relatively easy to show how it exhibits all four of the typical properties of a transition element.
(iv)Although not asked for here, you should include statesymbols when writing ionic equations.
(b)What is meant by the term bidentate ligand?
Marks available: 2
Student answer:
(b)A ligand is a species with a lone pair of electrons with which it can form a coordinate bond to a transition metal ion.
Bidentate means the ligand species has two lone pairs of electrons with which it can form a coordinate bond to a transition metal ion.
Examiner comments:
(b)This is an example of where there are two words in italics –you need to address both terms.
(c)Draw a diagram to show and explain the isomerism exhibited by [Ni(NH2CH2CH2NH2)3]2+.
Marks available: 2
Student answer:
(c)

Examiner comments:
(c)The diagrams have correctly been drawn to represent 3D drawings.
(d)State the shape of the following complex ions.
(i)CuCl42–
(ii)Ni(NH3)2Cl2
(iii)Fe(H2O)62+
Marks available: 3
Student answer:
(d)(i)Tetrahedral
(ii)Square planar
(iii)Octahedral
Examiner comments:
(d)The three shapes given here are the three most common, so are likely to feature widely.
Module 6: Transition elements
Question 2 / Total marks: 15
(a)A student added concentrated hydrochloric acid to a solution of aqueous cobalt chloride.
(i)Describe what the student would observe.
(ii)Write the equation for the reaction which occurred and state what type of process happened during the reaction.
Marks available: (i) 1 (ii) 3
Student answer:
(a)(i)The pink solution turned blue.
(ii) Co(H2O)62+(aq) + 4Cl–(aq)  CoCl42–(aq) + 6H2O(l)
The process is called ligand exchange.
Examiner comments:
(a)(i)It’s a good idea to include the initial colour as well as the final colour.
(ii)Although the state symbols are not required, you do need to know the formulae of the complexes, including the charges. The equation must also be balanced.
(b)Many aqueous transition element ions undergo reactions with cyanide CN–(aq) ions, according to the equation shown below:
Fe(H2O)62+(aq) + 6CN–(aq)  Fe(CN)6y(aq) + 6H2O(l)
(i)Calculate the value of y.
(ii)Write the equation for the stability constant Kstab for this reaction, giving its units.
Marks available: (i) 1 (ii) 2
Student answer:
(b)(i)y = 4-.
(ii)Kstab = [Fe(CN)64–](aq) units dm18 mol–6
[Fe(H2O)62+(aq)] + [CN–(aq)]5
Examiner comments:
(b)(i)The student has realised that the Fe2+ contributes +2 in terms of oxidation number and the 6CN– ions contributes 6 × -1 in terms of oxidation number, giving an overall charge of –4 on the left-hand side of the equation. So, y must be 4– to balance the equation in terms of charge.
(ii)Note how the student has calculated the units. Each [ ] means mol dm–3 so:
  • on the bottom line the units are {mol dm–3 + (6 × mol dm–3)} giving mol7 dm–21 in total
  • on the top line the units are mol dm–3 in total
  • top units divided by bottom units gives dm18 mol–6.

(c)A student carried out a series of steps to determine the percentage of copper in brass.
Step 10.724 g of brass was added to excess nitric acid.
Step 2Excess aqueous potassium iodide was added to this mixture.
Step 3The excess iodine formed was titrated against 0.400 mol dm–3 aqueous sodium thiosulfate and an end-point of 24.35 cm3 was recorded as the colour of iodine disappeared to leave a white precipitate.
(i)Write the equation for the reaction that occurred in Step 2.
(ii)Write the equation for the reaction that occurred in Step 3.
(iii)Calculate the percentage of copper in this sample of brass.
Marks available: (i) 1 (ii) 1 (iii) 4
Student answer:
(c)(i)2Cu2+(aq) + 4I– (aq)  2CuI(s) + I2(aq)
(ii)I2(aq) + 2S2O32–(aq)  2I– (aq) + S4O62–(aq)
(iii)Moles of S2O32–(aq) in equation (ii) = 0.400 × 24.35 = 9.74 × 10–3 mol
1000
Moles of Cu2+(aq) in equation (i) = 9.74 × 10–3 mol
Mass of Cu metal = 9.74 × 10–3 mol × 63.5 = 0.618 g (3 sf) (0.61849)
% Cu present in sample = 0.618 × 100 = 85.4%
0.724
Examiner comments:
(c)(i)The student has correctly identified that copper(I) iodide is formed and has balanced the equation accordingly.
(ii)The S4O62–(aq) ion is unusual but the equation has been balanced correctly.
(iii)Although all three sub-answers are shown rounded to three significant figures (3 sf), it is acceptable to leave your sub-answers in the calculator’s memory and round off your final answer.
(d)Copper(I) sulfate is a white solid. When added to water, the white solid dissolves and a blue solution with some brown particles in the bottom remains. Explain the redox processes which occurred during the washing process.
Marks available: 2
Student answer:
(d)The copper has changed from a +1 oxidation state in copper(I) sulfate to Cu2+ ions (which have a +2 oxidation state) and solid Cu metal (which has a 0 oxidation state).
This process is called disproportionation.
Examiner comments:
(d)Remember to write in terms of oxidation numbers and not ionic charge.