Edexcel A-Level Chemistry Paper 3, June 2017: Question 7
15 marks · Hard difficulty · Calculations
Explain why two d-block elements in Period 4 are not transition metals, calculate the final oxidation state of manganese from a redox titration, and determine the identity of a metal X in an electrochemical cell.
Practise this questionQuestion
Question text
7 This question is about the chemistry of elements in the d-block of the Periodic Table.
*(a) Many of the d-block elements are also classified as transition metals.
Explain why two of the d-block elements within Period 4 (scandium to zinc) are
not classified as transition metals.
You should include full electronic configurations where relevant.
(6)
(b) Under certain conditions, dichromate(VI) ions, Cr*P48954A02536*2O72−, can oxidise manganese(II) ions, Mn2+.
In this reaction, dichromate(VI) ions are reduced to chromium(III) ions, in acidic
conditions, according to the half-equation
Cr O2−(aq) + 14H+(aq) + 6e− U 2Cr3+(aq) + 7H O(l)
27 2
In an experiment it was found that 20.0 cm3 of 0.100 mol dm−3 potassium dichromate(VI)
was required to oxidise 30.0 cm3 of 0.200 mol dm−3 manganese(II) sulfate solution.
Use these data to calculate the final oxidation state of the manganese.
(5)
(c) A student constructed an electrochemical cell as follows:
• a half-cell was made from a strip of chromium metal and a solution of
aqueous chromium(III) sulfate
• a second half-cell was made from a piece of metal, X, and a solution of its sulfate, XSO4(aq)
• the two half-cells were connected and a current allowed to pass for some time.
Results
• the chromium electrode increased in mass by 1.456 g
• the electrode made of metal X decreased in mass by 1.021 g.
Use these data to determine the identity of the metal, X.
(4)
*P48954A02636*
(Total for Question 7 = 15 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answers Additional Guidance Mark
Number
*7(a) This question assesses a student’s ability to show a coherent Guidance on how the mark scheme should (6)
and logically structured answer with linkages and fully- be applied:
sustained reasoning. The mark for indicative content should be
added to the mark for lines of reasoning.
Marks are awarded for indicative content and for how the For example, an answer with five indicative
answer is structured and shows lines of reasoning. marking points that is partially structured
with some linkages and lines of reasoning
The following table shows how the marks should be awarded scores 4 marks (3 marks for indicative
for indicative content. content and 1 mark for partial structure
and some linkages and lines of reasoning).
If there are no linkages between points, the
Number of Number of same five indicative marking points would
indicative marks yield an overall score of 3 marks (3 marks
marking awarded for for indicative content and no marks for
points seen indicative linkages).
in answer marking
points In general it would be expected that 5 or 6
64 indicative points would score 2 reasoning
5–4 3 marks, and 3 or 4 indicative points would
3–2 2 score 1 reasoning mark. A total of 2, 1 or 0
11 indicative points would score 0 marks for
00 reasoning.
Reasoning marks may be subtracted for
The following table shows how the marks should be awarded extra incorrect chemistry.
for structure and lines of reasoning.
Number of marks
awarded for
structure of
answer and
sustained line of
reasoning
Answer shows a coherent and 2
logical structure with linkages
and fully sustained lines of
reasoning demonstrated
throughout.
Answer is partially structured 1
with some linkages and lines of
reasoning.
Answer has no linkages between 0
points and is unstructured.
Indicative content (IPs) Allow ‘partially-filled’ for incomplete
IP1: Allow d-orbital(s)
(transition metal) forms an ion with an incomplete d Do not award “d-shell”
sub-shell Allow “D” for “d” throughout
IP2: Allow if only Sc and Zn are used to
illustrate d-block elements that are not
scandium and zinc are not transition metals transition metals
IP3: Allow 4s0 and/or 3d0
Penalise use of [Ar] once only
Sc3+ and 1s2 2s2 2p6 3s2 3p6
IP4:
Zn2+ and 1s2 2s2 2p6 3s2 3p6 3d10
IP5: Allow “Sc3+ has no d sub-shell”
Sc3+ and d sub-shell empty / d-orbitals empty
IP6:
Allow ‘d orbital is full’ if clarified by 3d10
Zn2+ and d sub-shell full / ALL d-orbitals are full
Question
Acceptable Answers Additional Guidance Mark
Number
7(b) Example of calculation (5)
calculation of moles of Cr O 2−
(1) moles of Cr O 2− = 0.100 x 20.0
1000
= 2(.00) x 10−3 (mol)
calculation of moles of Mn2+
(1) moles of Mn2+ = 0.200 x 30.0
1000
= 6(.00) x 10−3 (mol)
deduction of whole number mole ratio of Cr O 2− : Mn2+
(1) mole ratio Cr O 2− : Mn2+
= 1 : 3
deduction of total number of electrons lost by 3 mol of
Mn2+
(1) 3 mol Mn2+ lose a total of 6e−
deduction of final oxidation state of manganese
(1) each Mn2+ loses 2e−, so final
oxidation state of Mn is (+)4 / IV
/Mn4+
MP3 and MP4 may be awarded via
alternative methods e.g. use of oxidation
numbers / moles of electrons
correct final oxidation state with no working
scores M5 only
Question
Acceptable Answers Additional Guidance Mark
Number
7(c) Example of calculation (4)
calculation of moles of Cr
(1) Moles Cr = 1.456 = 0.028(0)
52(.0)
MP2, 3 & 4 are only available for answers using a 3:2
mole ratio
deduction of mole ratio of X to Cr3+
(1) 3 mol X : 2 mol Cr3+ / Cr
Allow
2Cr3+ + 3X → 3X2+ + 2Cr
calculation of moles of X
(1) Moles X = 0.028(0) x 1.5
= 0.042(0)
Correctly multiplying by 1.5 for MP3 implies
MP2
calculation of molar mass / Ar of X
and Mr = 1.021
identification of X accordingly 0.042(0)
(1) = 24.3 (g mol−1)
and
(so) X is magnesium/Mg
COMMENT:
If transpose 3:2 ratio,
X has M = 54.7 (g mol−1) and X = Mn
r
so scores M1, then M3 and M4 by TE
(i.e. (3) marks overall)
(Total for Question 7 = 15 marks)
Question
Acceptable Answers Additional Guidance Mark
Number
How to answer it
Chemistry of the d-block Elements & Electrochemical Analysis
What this question tests
This question assesses your mastery of transition metal definitions, full electronic configurations of ions, stoichiometry in redox titrations involving dichromate(VI), oxidation state determinations, and quantitative analysis of electrode mass changes in electrochemical cells.
Transition Metal Criteria & Electronic Configurations
💡 Key Knowledge
- A transition metal is defined as a d-block element that forms at least one stable ion with an incompletely filled d-sub-shell.
- Scandium forms only the Sc³⁺ ion, which has an empty d-sub-shell.
- Zinc forms only the Zn²⁺ ion, which has a completely full 3d¹⁰ sub-shell.
✅ Required Answer Points
- IP1: Transition metal forms an ion with an incomplete d-sub-shell.
- IP2: Identify that Scandium and Zinc are not transition metals.
- IP3 & IP5: Sc³⁺ has full configuration 1s² 2s² 2p⁶ 3s² 3p⁶ (empty 3d sub-shell / no electrons in d-sub-shell).
- IP4 & IP6: Zn²⁺ has full configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ (full d-sub-shell).
🧠 Exam Technique (Quality of Written Communication)
This is a 6-mark extended-response question assessed using both indicative content (up to 4 marks) and structure/lines of reasoning (up to 2 marks). Ensure you explicitly link the electronic configurations of Sc³⁺ and Zn²⁺ to the definition of a transition metal in a logical sequence.
❌ Common Errors
- Writing condensed electron configurations using [Ar] when the question explicitly demands full electronic configurations.
- Stating that Sc and Zn don't have d-electrons in their atoms, rather than focusing on their ions.
Dichromate(VI) Titration & Final Oxidation State
📐 Step-by-Step Calculation
- Calculate moles of Cr₂O₇²⁻ used:
20.0 × 10⁻³ dm³ × 0.100 mol dm⁻³ = 2.00 × 10⁻³ mol (1 mark) - Calculate moles of Mn²⁺ reacted:
30.0 × 10⁻³ dm³ × 0.200 mol dm⁻³ = 6.00 × 10⁻³ mol (1 mark) - Determine whole-number mole ratio (Cr₂O₇²⁻ : Mn²⁺):
2.00 × 10⁻³ : 6.00 × 10⁻³ = 1 : 3 (1 mark) - Calculate total electrons lost by 3 moles of Mn²⁺:
From the half-equation provided, 1 mole of Cr₂O₇²⁻ gains 6 electrons. Therefore, 3 moles of Mn²⁺ must lose a total of 6 electrons to balance the redox process. (1 mark) - Deduce final oxidation state of manganese:
Since 3 moles of Mn²⁺ lose 6 electrons, each mole of Mn loses 2 electrons. Moving from +2 by losing 2 more electrons gives a final oxidation state of +4 (or Roman numeral IV / Mn⁴⁺ ). (1 mark)
🧠 Exam Technique
Always start titration/stoichiometry questions by calculating the number of moles for every species where volume and concentration are known. Work methodically through mole ratios before attempting to deduce oxidation state changes.
❌ Common Errors
Failing to convert cm³ into dm³ by multiplying by 10⁻³ leads to magnitude errors that break subsequent mole ratios.
Determining the Identity of Metal X
📐 Step-by-Step Calculation
- Calculate moles of Chromium (Cr) deposited:
Mass increase = 1.456 g . Relative atomic mass ( Aᵣ ) of Cr = 52.0 .
Moles = 1.456 / 52.0 = 0.0280 mol (1 mark) - Deduce the mole ratio of X to Cr³⁺:
Chromium ions gain 3 electrons to form solid Cr ( Cr³⁺ + 3e⁻ → Cr ). Metal X loses electrons to form X²⁺ ( X → X²⁺ + 2e⁻ ). To balance electrons transferred ( 6e⁻ ), the reacting ratio is 3 mol X : 2 mol Cr (or a 3:2 mole ratio). (1 mark) - Calculate moles of metal X reacted:
Moles of X = Moles of Cr × (3 / 2) = 0.0280 × 1.5 = 0.0420 mol (1 mark) - Calculate Molar Mass / Aᵣ of X and identify the metal:
Mass of X lost = 1.021 g .
Aᵣ = Mass / Moles = 1.021 / 0.0420 = 24.3 g mol⁻¹ .
Looking up the periodic table, metal X is Magnesium (Mg). (1 mark)
🧠 Exam Technique & Error Carried Forward (ECF)
Examiners apply strict rules here: MP2, MP3, and MP4 are only available if a 3:2 mole ratio is used. However, if a student incorrectly transposes the ratio (e.g. 2:3 ratio yielding Mᵣ = 54.7 , identifying Mn), Error Carried Forward allows them to pick up subsequent method marks.
❌ Common Errors
Confusing electrode mass increases with decreases—remember reduction happens at the cathode (increasing mass of chromium electrode) and oxidation happens at the anode (decreasing mass of metal X).
Topics
Inorganic Chemistry · Physical Chemistry · Topic 15: Transition Metals · Topic 5: Formulae, Equations and Amounts of Substance · Topic 14: Redox II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.