Edexcel A-Level Chemistry Paper 1, June 2017: Question 6

11 marks · Medium difficulty · Calculations

Draw a dot-and-cross diagram for magnesium bromide, complete a Born-Haber cycle to calculate the first electron affinity of bromine, explain the difference in first ionisation energy between magnesium and sodium, and write an equation for the third ionisation energy of magnesium.

Practise this question

Question

Question text

6 Magnesium bromide, MgBr2, is an ionic compound.

(a) (i) Draw a dot-and-cross diagram to show the bonding in magnesium bromide.

Only outer shell electrons are required.

(1)

(ii) State all the conditions under which magnesium bromide conducts electricity.

(1)

(b) The table shows the enthalpy changes needed to calculate the first electron affinity

of bromine.

Value

Enthalpy change –1

/ kJ mol

enthalpy change of atomisation of magnesium, ¨atH [Mg(s)] +148

1st ionisation energy of magnesium, 1st IE[Mg(g)] +738

2nd ionisation energy of magnesium, 2nd IE[Mg+(g)] +1 451

enthalpy change of atomisation of bromine, ¨atH [½Br2(l)] +112

lattice energy of magnesium bromide, LE[MgBr2(s)] –2 440

enthalpy change of formation of magnesium bromide, ¨fH [MgBr2(s)] –524

(i) Complete the Born-Haber cycle for magnesium bromide with formulae,

electrons and labelled arrows. The cycle is not drawn to scale.

(3)

*P48058A01328*

Mg(g) + Br2(l)

¨atH [Mg(s)]

Mg(s) + Br2(l)

¨fH [MgBr2(s)]

MgBr2(s)

(ii) Calculate the first electron affinity of bromine, in kJ mol*P48058A01428*–1.

(2)

Explain why the first ionisation energy of magnesium is higher than that of sodium.

(3)

(ii) Write the equation, including state symbols, to show the

third ionisation energy of magnesium.

(1)

(Total for Question 6 = 11 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

6(a)(i) Example of diagram (1)

dot-and-cross diagram

and

charges

Circles are not needed

Allow no electrons or 8 electrons on outer

shell of Mg

Allow dots or crosses for all electrons

Allow diagrams without square brackets,

provided charges are shown

Allow alternative ways of showing that

there are 2 bromide ions

Ignore inner shell electrons

Question

Answer Additional Guidance Mark

Number

6(a)(ii) (1)

(conducts electricity when) molten / liquid Both needed for the mark

and

dissolved in water / (in) aqueous (solution) Ignore gaseous

Allow ‘in solution / dissolved’

Question

Answer and Additional Guidance Mark

Number

6(b)(i) (3)

Mg2+(g) + 2Br(g) + 2e(―)

Box 3

2∆ atH[½Br2(l)]/2 x(+)112/(+)224

2 x EA[Br(g)]

Mg2+(g) + Br (l) + 2e(―)

Box 2 2

2nd IE[Mg+(g)]/(+)1451

Mg2+(g) + 2Br―(g) Box 4

Box 1 Mg+(g) + Br (l) + e(―)

st LE[MgBr2(s)]/

1 IE[Mg(g)]/(+)738

―2440

Mg(g) + Br2(l)

∆atH[Mg(g)]/(+)148

Mg(s) + Br2(l)

∆fH[MgBr2(s)]/−524

MgBr2(s)

Correct arrows with 1st and 2nd IE of Mg labelled and correct Mg symbols with state symbols in

boxes 1 and 2 or 2 and 3 (1)

o 1

2 x ∆atH [2Br2(l)] / 2 x(+)112 / (+)224 and 2Br(g) in box 3 or 1 and 2 x EA[Br(g)] labelled and

2Br―(g) in box 4 and correct arrows (1)

LE[MgBr2(s)] / −2440 labelled and arrow in correct direction (1)

Allow any unambiguous labels for the arrows with words and/or numbers – state symbols not required

Accept enthalpy change of atomisation of bromine before IEs of magnesium

Ignore missing electrons / 2e(―) in boxes 1, 2 and 3

Allow 1 state symbol missing but penalise 2 missing, or an incorrect state symbol in boxes once only

Question

Answer Additional Guidance Mark

Number

6(b)(ii) Example of calculation (2)

correct expression for 2 x EA(Br) in numbers or 2 x EA(Br) = − (2 x +112) − (+1451) − (+738)

symbols (1) − (+148) + (−524) − (−2440)

calculation of EA(Br) (1) EA(Br) = −645 = −322.5 /−323 (kJ mol―1)

Correct answer with no working scores (2)

Allow for 1 mark:

(+)322.5 / (+)323 (wrong sign)

−266.5 /−267 (2 missing from ∆atH (Br))

−645 (2 missing from EA)

−533 (both 2s missing for Br)

Ignore units

No TE on incorrect arrows in (b)(i)

Question

Answer Additional Guidance Mark

Number

6(c)(i) An explanation that makes reference to the following points: Penalise reference to ion once only (3)

Ignore reference to atomic radius

Nuclear charge

magnesium (atom) / Mg has more protons than sodium

(atom) / Na

magnesium / Mg has a greater (effective) nuclear charge

(than sodium / Na) (1)

Shielding

(outer) electron in magnesium (atom) / Mg in the same Allow correct E.C of both atoms

(quantum) shell / energy level / sub-shell / orbital as in a

sodium atom / Na Allow same number of (quantum) shells

or / energy levels in Mg and Na

shielding in magnesium atom / Mg similar to / same as

that in sodium atom / Na (1)

Attraction

so the force of attraction between the nucleus and the Allow the (outer) electron in Mg is held

(outer) electron is greater in magnesium (atom) / Mg more tightly to the nucleus (than in Na)

(than in sodium atom / Na) (1)

Note

An answer that describes the trend

across a period, without one reference to

either sodium or magnesium, scores

maximum (2) marks

Question

Answer Additional Guidance Mark

Number

6(c)(ii) Examples of equations (1)

correct equation with state symbols Mg2+(g) → Mg3+(g) + e(−)

Mg2+(g) − e(−) → Mg3+(g)

Ignore state symbol for the electron

Do not allow ⇌

(Total for Question 6 = 11 marks)

How to answer it

Magnesium Bromide, Born-Haber Cycles & Ionisation Energy

What this question tests

This assessment evaluates your understanding of ionic bonding representations, electrical conductivity conditions, thermodynamic cycles (Born-Haber calculations), periodic trends in ionisation energies (comparing magnesium and sodium), and writing precise ionisation energy equations including state symbols.

Question 6 (a) — Ionic Bonding & Conductivity

Structure and Physical Properties

✅ Correct Answer: Part (a)(i)

A dot-and-cross diagram showing:

Two bromide ions [Br]⁻ (each with 8 electrons in the outer shell and a 1- charge) and one magnesium ion [Mg]²⁺ (with 0 or 8 electrons and a 2+ charge).
Square brackets around all ions with charges clearly displayed outside.

✅ Correct Answer: Part (a)(ii)

Magnesium bromide conducts electricity only when molten (liquid) AND when dissolved in water (aqueous).

💡 Key Knowledge

Ionic compounds have giant ionic lattices held together by strong electrostatic forces.
Solid ionic compounds do not conduct because ions are fixed in position.
Mobile charge carriers (delocalised ions) are required for electrolysis/conductivity.

❌ Common Errors

Forgetting to include square brackets or charges in the dot-and-cross diagram.
Only stating one condition (e.g., just "molten") in part (a)(ii) when both are required for the mark.

🎯 Marks: (a)(i) = 1 mark, (a)(ii) = 1 mark

Question 6 (b) — Born-Haber Cycle & Electron Affinity

Enthalpy Calculations

✅ Correct Answer: Part (b)(i) - Cycle Completion

Box 1: Mg⁺(g) + Br₂(l) + e⁻
Box 2: Mg²⁺(g) + Br₂(l) + 2e⁻
Box 3: Mg²⁺(g) + 2Br(g) + 2e⁻
Box 4: Mg²⁺(g) + 2Br⁻(g)
Arrows must correctly point upwards for atomisation/ionisation and downwards for lattice energy.

📐 Calculation: Part (b)(ii) - First Electron Affinity

Step 1: Set up the Hess's Law equation / cycle balance.
Δ_fH = Δ_atH[Mg] + 1st IE + 2nd IE + 2(Δ_atH[½Br₂]) + 2(EA) + LE
Step 2: Rearrange to solve for 2 × EA(Br).
2 × EA(Br) = Δ_fH - (Δ_atH[Mg] + 1st IE + 2nd IE + 2(Δ_atH[½Br₂]) + LE)
2 × EA(Br) = (-524) - (+148 + 738 + 1451 + (2 × +112) + (-2440))
Step 3: Calculate final value for one Br atom.
2 × EA(Br) = -645 kJ mol⁻¹
EA(Br) = -645 / 2 = -322.5 kJ mol⁻¹ (or -323 kJ mol⁻¹)

🧠 Exam Technique

Always multiply the atomisation of bromine and electron affinity by 2 because one mole of MgBr₂ contains two moles of bromide ions.
Watch your signs carefully when inputting exothermic (negative) lattice energies and endothermic (positive) ionisation energies.

❌ Common Errors

Failing to divide the final answer by 2, calculating the electron affinity for two moles instead of one.
Omitting state symbols or electron terms ( e⁻ ) in the Born-Haber cycle boxes.

🎯 Marks: (b)(i) = 3 marks, (b)(ii) = 2 marks

Question 6 (c) — Ionisation Energy Trends & Equations

Periodic Trends: Magnesium vs. Sodium

✅ Correct Answer: Part (c)(i)

To explain why Mg first IE > Na first IE, you must make three mandatory comparison points:

Nuclear Charge: Magnesium has a greater nuclear charge / more protons (12 compared to 11 in sodium).
Shielding: The outer electron is in the same principal energy level / shell (3s), offering similar shielding.
Attraction: Therefore, there is a stronger electrostatic attraction between the nucleus and the outer electron in magnesium, requiring more energy to remove.

✅ Correct Answer: Part (c)(ii)

Equation for the third ionisation energy of magnesium:

Mg²⁺(g) → Mg³⁺(g) + e⁻

(Alternative accepted format: Mg²⁺(g) - e⁻ → Mg³⁺(g) )

🧠 Exam Technique

When comparing properties across a period, explicitly mention both elements to avoid losing comparative marks.
For ionisation energy equations, always include state symbols (g) for gaseous species—it is a strict marking point.

❌ Common Errors

Stating that magnesium has more shells (both Na and Mg have their outer electrons in the 3rd shell).
Writing a 2nd or 1st ionisation equation instead of the 3rd in part (c)(ii).

🎯 Marks: (c)(i) = 3 marks, (c)(ii) = 1 mark

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 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.