Edexcel A-Level Chemistry Paper 1, June 2018: Question 8
12 marks · Hard difficulty · Extended Writing
Identify the orbitals, write equations, and explain successive ionisation energies of calcium, calculate electron affinity from a Born–Haber cycle, and discuss differences between theoretical and experimental lattice energies.
Practise this questionQuestion
Question text
8 This question is about ions and ionic compounds.
(a) The first three ionisation energies of calcium are shown in the table.
First ionisation Second ionisation Third ionisation
Ionisation energy
−1 590 1145 4912
/ kJmol
Orbital
(i) Complete the table by identifying the specific orbital from which each
electron is removed.
(2)
(ii) Write the equation for the third ionisation energy of calcium.
Include state symbols.
(1)
(iii) Explain why the difference between the second and third ionisation energies
of calcium is much larger than the difference between the first and second
ionisation energies.
(2)
(b) The diagram, which is not drawn to scale, shows the Born-Haber cycle for
lithium fluoride. The energy changes are given in kJ mol−1.
Li+(g) + F(g) + e−
+79
Y
Li+(g) + ½F (g) + e−
+520 Li+(g) + F−(g)
Li(g)*P52302RA01628*+½F2(g)
+159
Li(s) + ½F2(g)
−1031
−616
LiF(s)
What is the value for Y, in kJ mol−1?
(1)
A −273
B −343
C −432
D −889
*(c) The table shows the theoretical and experimental lattice energy values of two compounds.
Theoretical lattice energy Experimental lattice energy
Compound −1 −1
/ kJmol / kJmol
lithium chloride, LiCl −845 −848
magnesium iodide, MgI2 −1944 −2327
Comment on the theoretical and experimental lattice energy values, giving the
reasons for any differences and similarities.
(6)
… 17
… *P52302RA01728*
… 18
… *P52302RA01828*
(Total for Question 8 = 12 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional Guidance Mark
Number
8(a)(i) Example of table (2)
any 2 correct (1)
all 3 correct (2)
Accept 3p / 3p / 3p for 3rd IE
x y z
Ignore any superscript numbers by 4s and 3p
Allow (1) for just ‘s, s, p’ or ‘s, s, p’ with one
or more incorrect numbers in front
Question
Acceptable Answer Additional Guidance Mark
Number
8(a)(ii) Examples of equations (1)
correct equation Ca2+(g) → Ca3+(g) + e(―)
or
Ca2+(g) − e(―) → Ca3+(g)
Correct state symbols are required
Ignore any state symbol for the electron
Question
Acceptable Answer Additional Guidance Mark
Number
8(a)(iii) An explanation that makes reference to the following (2)
points:
(there is a much larger difference between the 2nd and Ignore electron is lost from a full (sub-)shell
3rd ionisation energies because the) / a full (sub-)shell is more stable
3rd electron is lost from a shell / energy level / sub-
shell / (3p) orbital closer to the nucleus Ignore just ‘3rd electron lost is more strongly
or attracted to the nucleus’
the 3rd electron is lost from a shell / energy level /
sub-shell / (3p) orbital of lower energy (1)
(there is a smaller difference between the 1st and 2nd
ionisation energies because the) 1st and 2nd electrons
removed from the same shell / energy level / sub-
level / orbital
or
the first two electrons experience similar shielding Allow the same amount of shielding
(from the inner electrons)
Allow the 3rd electron (to be lost)
experiences less shielding (from inner
or electrons)
there is only a small change in electron-electron
repulsion as the first two electrons are removed (1)
Question Answer
Mark
Number
8(b) The only correct answer is B (1)
A is incorrect because (−1031) + (79 + 520 + 159) is incorrect
C is incorrect because (−1031) + (79 + 520) is incorrect
D is incorrect because (−1031) + 79 +520 +159 – 616 is incorrect
Question
Acceptable Answers Additional Guidance Mark
Number
8(c)* This question assesses a student’s ability to show a Guidance on how the mark scheme should (6)
coherent and logically structured answer with linkages be applied:
and fully-sustained reasoning. The mark for indicative content should be
added to the mark for lines of reasoning. For
Marks are awarded for indicative content and for how the 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 scores 4 marks (3 marks for indicative
awarded 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
same five indicative marking points would
yield an overall score of 3 marks (3 marks
for indicative content and no marks for
linkages).
The following table shows how the marks should be
awarded for structure and lines of reasoning.
In general it would be expected that 5 or 6
indicative points would get 2 reasoning
marks, and 3 or 4 indicative points would get
1 mark for reasoning, and 0, 1 or 2
indicative points would score zero marks for
reasoning.
General points to note
Comment: If there is any incorrect chemistry, deduct
Look for the indicative marking points first, then mark(s) from the reasoning. If no reasoning
consider the mark for structure of answer and mark(s) awarded do not deduct mark(s).
sustained line of reasoning e.g.
penalise any reference to ‘molecule’ once
only
or
penalise ‘ion’ not mentioned in word or
formula at least once in answer, once only
Allow reverse arguments for IP3 to IP6 Ignore
mention of stoichiometry Ignore references to
electronegativity
Indicative content
IP1 - Ionic
lithium chloride / LiCl (has very similar theoretical Allow very small amount of / no covalent
and experimental lattice energy values so) is character in LiCl
(almost 100%) ionic Allow assumption that ions act as point
charges / are spherical is true for LiCl
IP2 - Covalency
magnesium iodide / MgI2 (has different theoretical Allow MgI2 more covalent character than LiCl
and experimental lattice energy values so) has
(some) covalent character
IP3 - Charge on cations
magnesium is Mg2+ and lithium is Li+ Allow magnesium has 2+ charge and lithium
has 1+ charge / magnesium ion has a larger
charge than a lithium ion
Allow charge density for charge
IP4 - Polarising – what does the polarising
magnesium ion/Mg2+ is (more) polarising / has a
large(r) polarising power (than lithium ion)
IP5 - Size of anion −
− − Allow iodine ion / I is a large atom / has a
iodide ion / I is larger (than chloride ion / Cl )
large atomic radius
Ignore size of cation
Do not award iodide has a larger charge
density
IP6 – Polarisable – what is polarised
− Allow this shown in a diagram
iodide ion / I is (more easily) polarised / distorted
Ignore just ‘greater attraction to cation’
(Total for Question 8 = 12 marks)
How to answer it
Atomic Structure, Born-Haber Cycles & Lattice Energies
This question assesses fundamental physical chemistry concepts spanning thermodynamics and atomic structure:
- Electronic Configurations & Orbitals: Identifying specific subshells/orbitals from which successive electrons are removed.
- Ionisation Energy Equations: Constructing accurate thermochemical equations with state symbols.
- Trends & Shielding: Explaining large jumps in successive ionisation energies using nuclear distance and shielding arguments.
- Born-Haber Cycles: Applying Hess's Law to calculate electron affinity from enthalpy data.
- Theoretical vs Experimental Lattice Energies: Explaining polarisation, ion deformability, and the covalent character in ionic compounds (Fajans' Rules).
Part (a)(i): Identifying Specific Orbitals
Calcium Successive Ionisation Energies [2 Marks]
✅ Correct Answer
| Ionisation | 1st IE (590 kJ mol⁻¹) | 2nd IE (1145 kJ mol⁻¹) | 3rd IE (4912 kJ mol⁻¹) |
|---|---|---|---|
| Orbital | 4s | 4s | 3p (or 3px / 3py / 3pz) |
• Any 2 correct = [1 mark]
• All 3 correct = [2 marks]
💡 Key Knowledge
- Calcium (Z = 20) electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² .
- Electrons are removed from the highest occupied energy level first.
- Removing 1st electron: from the outermost 4s subshell.
- Removing 2nd electron: from the remaining 4s subshell.
- Removing 3rd electron: now breaks into the full 3p subshell.
❌ Common Errors & Examiner Tips
- Omitting principal quantum number: Writing just "s, s, p" loses a mark. You must specify the shell number ( 4s, 4s, 3p ).
- Writing 3d: Calcium does not occupy 3d in its ground state or ions ( Ca²⁺ is [Ar] ).
Part (a)(ii): Third Ionisation Energy Equation
Writing the Equation with State Symbols [1 Mark]
✅ Correct Answer
Ca²⁺(g) → Ca³⁺(g) + e⁻
Alternative acceptable format: Ca²⁺(g) - e⁻ → Ca³⁺(g)
🧠 Exam Technique
- nth Ionisation Energy Rule: The nth ionisation starts with the ion of charge +(n - 1) and forms a charge of +n .
- Third IE must start with Ca²⁺(g) , never Ca(s) or Ca(g) .
- Always check state symbols immediately: (g) is mandatory on both reactant and product ions!
❌ Common Errors
- Writing Ca(g) → Ca³⁺(g) + 3e⁻ : That is the sum of the first three ionisation energies, not the third ionisation energy.
- Omitting the state symbols (g) .
Part (a)(iii): Explanation of Energy Differences
Comparing 1st/2nd vs 2nd/3rd Ionisation Energy Jumps [2 Marks]
✅ Model Response
Point 1 (The 3rd electron removal):
The third electron is removed from an inner quantum shell (3p) that is closer to the nucleus / of lower energy (and experiences less shielding from inner electrons) [1 mark].
Point 2 (The 1st and 2nd electron removal):
The first two electrons are removed from the same shell / subshell (4s), so they experience similar shielding (and there is only a small change in electron-electron repulsion) [1 mark].
🧠 Examiner Commentary & Trap Warning
- DO NOT SAY: "The 3rd electron is removed from a stable full shell" or "a full subshell is stable." The mark scheme explicitly states: "Ignore electron is lost from a full shell / a full shell is more stable".
- DO NOT JUST SAY: "The 3rd electron is more strongly attracted." You must give the physical cause: closer to the nucleus or less shielding.
- Always address both comparisons: explain why 3rd is much higher and why 1st vs 2nd difference is small.
Part (b): Born-Haber Cycle Calculation
Determining the First Electron Affinity of Fluorine (Y) [1 Mark]
✅ Correct Option: B (-343 kJ mol⁻¹)
📐 Step-by-Step Calculation
Using Hess's Law, following the clockwise and anticlockwise routes around the Born-Haber cycle:
- Identify the Formation route:
ΔHբ[LiF(s)] = -616 kJ mol⁻¹ - Identify all alternative steps from elements to ionic solid:
• Sublimation/Atomisation of Li: +159 kJ mol⁻¹
• First Ionisation Energy of Li: +520 kJ mol⁻¹
• Atomisation of Fluorine ( ½F₂(g) → F(g) ): +79 kJ mol⁻¹
• Electron Affinity of Fluorine: Y
• Lattice Energy of LiF: -1031 kJ mol⁻¹ - Set up the cycle equation:
ΔHբ = ΔHₐₜ(Li) + 1st IE(Li) + ΔHₐₜ(F) + Y + ΔHₗₐₜₜ
-616 = (+159) + (+520) + (+79) + Y + (-1031) - Simplify numerical values:
-616 = 758 + Y - 1031
-616 = -273 + Y - Solve for Y:
Y = -616 - (-273) = -616 + 273 = -343 kJ mol⁻¹
❌ Distractor Analysis
- A (-273): Incomplete calculation ( -1031 + 758 = -273 ), completely omitting ΔHբ (-616) .
- C (-432): Forgot the atomisation of fluorine (+79).
- D (-889): Subtracted 616 instead of adding (sign inversion error: -273 - 616 = -889 ).
Part (c)*: Theoretical vs Experimental Lattice Energy
6-Mark Extended Response: Bonding & Polarisation
💡 The Theoretical Ionic Model vs Real Bonds
Theoretical Model assumes: Purely ionic bonding with perfectly spherical point-charge ions and no electron distortion.
Discrepancy means: Covalent character arising from polarisation (distortion of the anion's electron cloud by a small, highly charged cation).
✅ 6 Indicative Content Points (IP1 to IP6)
1. LiCl is purely ionic:
Theoretical (-845) and experimental (-848) values are almost identical, showing LiCl is almost 100% ionic with spherical ions behaving as point charges.
2. MgI₂ has covalent character:
Large difference between theoretical (-1944) and experimental (-2327) values indicates significant covalent character / polarisation.
3. Cation charge difference:
Magnesium ion has a 2+ charge ( Mg²⁺ ) whereas lithium ion has a 1+ charge ( Li⁺ ).
4. Cation polarising power:
Mg²⁺ has a higher charge density and is therefore more polarising (has greater polarising power) than Li⁺ .
5. Anion size difference:
The iodide ion ( I⁻ ) has a larger ionic radius than the chloride ion ( Cl⁻ ).
6. Anion polarisability:
The larger I⁻ electron cloud is more easily polarised / distorted than the smaller Cl⁻ ion.
🧠 Structuring for Level 3 (5–6 Marks)
This is a starred (*) question assessing logical structuring and sustained lines of reasoning:
- Paragraph 1 (The Comparison): State clearly what the numbers tell you about the bonding type of each compound (IP1 & IP2).
- Paragraph 2 (Cations): Compare Mg²⁺ vs Li⁺ in terms of charge, radius/density, and polarising power (IP3 & IP4).
- Paragraph 3 (Anions): Compare I⁻ vs Cl⁻ in terms of ionic radius and polarisability (IP5 & IP6).
- Conclusion: Link distortion back to extra covalent bonding, explaining why experimental is more exothermic than theoretical for MgI₂.
❌ Critical Penalties to Avoid
- Calling them "molecules": Penalised immediately! These are giant ionic lattices with ions, NOT molecules.
- Confusing the terms: Cations polarise; anions are polarised (polarisable). Cations do NOT get polarised here!
- Omitting "ion": Always say "magnesium ion" or " Mg²⁺ ", never just "magnesium". The mark scheme penalises omitting the word/formula "ion".
- Electronegativity arguments: Explaining purely through electronegativity differences does NOT answer lattice energy discrepancy questions. Focus on polarising power and polarisability.
Topics
Physical Chemistry · Topic 1: Atomic Structure and the Periodic Table · Topic 2: Bonding and Structure · Topic 13: Energetics II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.