OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 22
12 marks · Hard difficulty · Structured Questions
Deduce transition metal electron configurations, formulas of cobalt complexes and precipitates, determine a complex ion formula from molar mass, and evaluate blood pH using ligand substitution and acid-base buffer calculations.
Practise this questionQuestion
Question text
22 This question is about transition elements.
(a) Iron is in the d block of the periodic table and contains s, p and d orbitals.
• Draw diagrams to show the shapes of an s orbital and a p orbital.
• Complete the electron configurations of an iron atom and an iron(II) ion.
Shapes
s orbital p orbital
Electron configurations
Iron atom: 1s2 …
Iron(II) ion 1s2 …
[2]
(b) The flowchart shows some reactions of cobalt(II) chloride, CoCl2.
dissolve in water Pale pink solution containing a
CoCl2(s)
six-coordinate complex ion A
aqueous NaOH concentrated HCl
Blue solution containing a
Blue-green solid B
four-coordinate complex ion C
In A, B and C, cobalt has an oxidation number of +2.
(i) Suggest the formulae of A, B and C.
Complex ion A: …
Solid B: …
Complex ion C: …
[3]
(ii) Cobalt (III) forms an octahedral complex ion D, which contains both ammonia and chloride
ligands.
Complex ion D has a molar mass of 197.9 g mol–1.
Determine the formula and charge of complex ion D.
… [2]
(c) Red blood cells contain haemoglobin which transports oxygen around the body.
For efficient transportation of oxygen, healthy human blood must be maintained at a pH value
between 7.35 and 7.45.
Human blood acts as a buffer due to the presence of carbonic acid, H2CO3, and
hydrogencarbonate, HCO –, ions as shown below.
H CO (aq) H+(aq) + HCO –(aq) K = 4.27 × 10–7 mol dm–3
23 3 a
• Explain, using ligand substitution, how haemoglobin transports oxygen around the body.
• Determine whether a sample of blood with a [HCO –] : [H CO ] ratio of 8.5:1 is healthy.
32 3
… [5]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
22 (a) 2
s orbital p orbital
IGNORE shading
IGNORE axes directions x, y, z
DO NOT ALLOW multiple p orbitals
Fe = (1s2)2s22p63s23p64s23d6 For electron configuration,
ALLOW 4s2 after 3d6
AND
Fe2+ = (1s2)2s22p63s23p63d6
i.e. 1s22s22p63s23p63d64s2
ALLOW upper case D, etc and subscripts,
e.g … 4S23D1
ALLOW 4s0
IGNORE [Ar]3d6 4s2
(b) (i) (A =) [Co(H O) ]2+ 3 IGNORE state symbols even if incorrect
[ ] essential
(B =) Co(OH)2 ALLOW [Co(OH)2(H2O)4] OR Co(OH)2(H2O)4
2– 2– ALLOW -2 for 2-
(C =) [CoCl4] OR CoCl4 -2
i.e. [CoCl4]
Question Answer 31 Marks Guidance
(ii) Complex : [Co(NH ) Cl ] 2 IGNORE Any charges for 1st mark
34 2
Charge +1 / + / 1+ ALLOW [CoCl2(NH3)4]
ALLOW [Co(Cl)2(NH3)4]
DO NOT ALLOW [Co(Cl2)(NH3)4]
DO NOT ALLOW if charges shown in formula within
brackets for 2nd mark
(c) Oxygen (O lone pair) forms a coordinate/dative bond to 5 ALLOW word equations using → and ⇌
Fe(II)/Fe/Iron/Fe2+
IGNORE number of coordinate bonds
replaced by H2O or CO2
OR
O2 bonds reversibly (with metal ion) ALLOW ORA
Check for alternative methods on mark scheme.
FIRST CHECK ANSWER ON ANSWER LINE ALLOW ECF throughout
If 7.3(0) AND not healthy / below 7.35 award three
calculation marks ALLOW [A-] for [HCO -] AND/OR [HA] for [H CO ]
32 3
------------------------------------------------------------------------ (asked for in 19 a) ii))
+ [H₂CO₃]
[H ] = Ka [HCO₃⁻]
[HCO₃⁻] +
ALLOW [H ] = Ka
OR [H₂CO₃]
[HCO₃⁻] Kₐ
= [H₂CO₃] [H⁺]
[H₂CO₃] [H⁺]
ALLOW = Kₐ
[HCO₃⁻]
[H+] = 5.02 10–8 [H+] value subsumes MP3
ALLOW [H+] = 5.02 10–8 up to the calculator value
(5.023529412 x 10–8)
DO NOT ALLOW a weak acid approach for marking
points 3 and 5. i.e. [H+] can be awarded.
pH = –log(5.02 10–8) = 7.3(0) ALLOW 7.3 up to calculator value (pH =7.298991951)
AND not healthy / below 7.35
Alternative method 1:
ALLOW [H+] = 3.98 10-8 from average pH 7.40
pH of healthy blood is between 7.35 and 7.45 used.
pH 7.35 pH 7.45
[H+] = 4.47 10-8 OR [H+] = 3.55 10-8
[HCO₃⁻] Kₐ [HCO₃⁻] Kₐ
= =
[H₂CO₃] [H⁺] [H₂CO₃] [H⁺]
[HCO₃⁻] [HCO₃⁻]
= =
[H₂CO₃] [H₂CO₃]
4.27 x 10-7 4.27 x 10-7
4.47 x 10-8 3.55 x 10-8
[HCO₃⁻] [HCO₃⁻]
= 9.55:1 = 12:03:1
[H₂CO₃] [H₂CO₃]
8.5:1 does not lie in the range of 9.55:1 to 12.03:1 AND
unhealthy
Alternative method 2:
[HCO₃⁻]
pH = pKa + log
[H₂CO₃]
pKa = 6.37
(8.5)
6.37 + log (1)
7.3(0) AND not healthy / below 7.35
How to answer it
Transition Elements, Complexes & Blood Buffers Study Guide
What this question tests
This comprehensive OCR A-Level Chemistry question assesses your knowledge of atomic structure (orbital shapes and electron configurations), transition metal complex formulas, ligand substitution mechanisms in biological systems (haemoglobin and oxygen transport), and advanced buffer calculations using acid dissociation constants (Ka) and pH.
Orbitals & Electron Configurations
✅ Correct Answers
- Shapes: s-orbital = sphere; p-orbital = figure-of-eight (dumbbell shape).
- Iron atom (Fe): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
- Iron(II) ion (Fe²⁺): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ (Note: electrons are lost from the 4s orbital first!)
💡 Key Knowledge
- Remember that 4s electrons fill before 3d and are lost before 3d when forming transition metal ions.
- Acceptable configurations include both chronological order and 4s written after 3d.
❌ Common Errors
- Removing electrons from the 3d subshell instead of the 4s subshell when writing the Fe²⁺ ion configuration.
- Drawing multiple p-orbitals instead of a single p-orbital shape as requested.
Cobalt Complex Formulas
✅ Correct Formulas
- Complex A: [Co(H₂O)₆]²⁺ (Pale pink, 6-coordinate)
- Solid B: Co(OH)₂ (Blue-green solid precipitate)
- Complex C: [CoCl₄]²⁻ (Blue solution, 4-coordinate tetrahedral)
🧠 Exam Technique
- Square brackets [] and overall ionic charges are essential for complex ions.
- State symbols are ignored here, but charge placement outside the bracket is critical.
Complex D Molar Mass & Formula Determination
✅ Correct Answer
Formula and Charge: [Co(NH₃)₄Cl₂]⁺
📐 Step-by-Step Calculation
- Identify components: Cobalt(III) has oxidation state +3. The complex contains neutral ammonia ( NH₃ ) ligands and chloride ( Cl⁻ ) ligands in an octahedral (6-coordinate) geometry.
- Calculate ligand combinations matching molar mass (197.9 g mol⁻¹):
Co (58.9) + 4 × NH₃ (4 × 17.0 = 68.0) + 2 × Cl (2 × 35.5 = 71.0) = 58.9 + 68.0 + 71.0 = 197.9 g mol⁻¹. - Determine overall charge: Co(III) = +3. Four neutral NH₃ = 0. Two chloride ions = 2 × (-1) = -2. Overall charge = (+3) + 0 + (-2) = +1.
Haemoglobin, Ligand Substitution & Blood Buffering
✅ Correct Answers & Explanation
- Haemoglobin mechanism: Oxygen lone pair forms a coordinate/dative bond to Fe(II) in haemoglobin; oxygen is bonded reversibly so it can be released to tissues.
- Buffer calculation conclusion: The blood sample is not healthy. (Calculated pH = 7.30, which falls below the healthy range of 7.35 - 7.45).
📐 Step-by-Step Calculation (Method 1)
- Set up Ka expression:
Ka = ([H⁺][HCO₃⁻]) / [H₂CO₃] = 4.27 × 10⁻⁷ - Rearrange for [H⁺]:
[H⁺] = Ka × ([H₂CO₃] / [HCO₃⁻]) - Substitute the given ratio:
Since [HCO₃⁻] : [H₂CO₃] = 8.5 : 1, the ratio ([H₂CO₃] / [HCO₃⁻]) = 1 / 8.5.
[H⁺] = (4.27 × 10⁻⁷) × (1 / 8.5) = 5.0235 × 10⁻⁸ mol dm⁻³. - Calculate pH:
pH = -log[H⁺] = -log(5.02 × 10⁻⁸) = 7.30. - Final evaluation: 7.30 is below 7.35, therefore the blood is acidic/unhealthy.
❌ Common Calculation Traps
- Inverting the buffer ratio: Always check whether the question gives [HCO₃⁻] : [H₂CO₃] or the reverse before substituting into Ka expressions!
- Forgetting to state whether the blood is healthy or not alongside the calculated numerical value. Both parts are required for full marks.
Topics
Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 5.3 Transition elements · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 5.1 Rates, equilibrium and pH
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.