OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 16

18 marks · Hard difficulty · Structured Questions

Calculate the enthalpy change of hydration of bromide ions, total ions in magnesium bromide, and explain physical properties and lattice enthalpy.

Practise this question

Question

An A-Level Chemistry multi-part exam question about magnesium, bromine, and magnesium bromide. Part (a) asks to explain 'weighted mean mass' (1 mark). Part (b) has two sub-parts: (i) draw a dot-and-cross diagram for MgBr2 showing outer shells (2 marks), (ii) calculate the total number of ions in 1.74 g of MgBr2 to 3 sig figs (3 marks). Part (c) gives Table 16.1 showing melting points and electrical conductivities for magnesium, bromine, and magnesium bromide, and asks to explain these physical properties using structure and bonding (6 marks). Part (d) provides Table 16.2 of enthalpy changes, asks to complete an energy cycle (2 marks), calculate the enthalpy change of hydration of bromide ions (2 marks), and write an equation and calculate the lattice enthalpy of magnesium bromide (3 marks).
Question text

16 This question is about magnesium, bromine and magnesium bromide.

(a) Relative atomic mass is defined as ‘the weighted mean mass compared with 1/12th mass of

carbon-12’.

Explain what is meant by the term weighted mean mass.

… [1]

(b) (i) Draw a ‘dot-and-cross’ diagram for MgBr2.

Show outer electron shells only.

[2]

(ii) Calculate the total number of ions in 1.74 g of magnesium bromide, MgBr2.

Give your answer to 3 significant figures.

number of ions = … [3]

(c)* Table 16.1 shows some physical properties of magnesium, bromine and magnesium bromide.

Electrical conductivity

Substance Melting point / °C

Solid Liquid

Magnesium 711 Good Good

Bromine −7 Poor Poor

Magnesium bromide 650 Poor Good

Table 16.1

Explain the physical properties shown in Table 16.1 using your knowledge of structure and

bonding. [6]

Additional answer space if required

(d) The enthalpy change of hydration of bromide ions can be determined using the enthalpy

changes in Table 16.2.

Enthalpy change Energy / kJ mol−1

1st ionisation energy of magnesium +736

2nd ionisation energy of magnesium +1450

atomisation of bromine +112

atomisation of magnesium +148

electron affinity of bromine −325

formation of magnesium bromide −525

hydration of bromide ion to be calculated

hydration of magnesium ion −1926

solution of magnesium bromide −186

Table 16.2

(i) An incomplete energy cycle based on Table 16.2 is shown below.

On the dotted lines, add the species present, including state symbols.

Mg2+(g) + 2Br–(g)

Mg+(g) + 2Br(g) + e–

Mg(g)+2Br(g)

Mg(s)+2Br(g)

… 2+ –

Mg (aq)+2Br (g)

MgBr2(s)

Mg2+(aq) + 2Br–(aq)

[2]

(ii) Using your completed energy cycle in 16(d)(i), calculate the enthalpy change of hydration

of bromide ions.

enthalpy change of hydration = … kJ mol−1 [2]

(iii) Write the equation for the lattice enthalpy of magnesium bromide and calculate the lattice

enthalpy of magnesium bromide.

Equation …

Calculation

lattice enthalpy = … kJ mol−1 [3]

Mark scheme

Show the mark scheme Official mark scheme providing detailed answers and marking guidance for all parts of question 16, including acceptable definitions, dot-and-cross diagrams, mole calculations, level-based descriptors for structure and bonding, completed Born-Haber / energy cycle steps, and lattice enthalpy calculations.

AO

Question Answer Marks Guidance

element

16 (a) (The mean/average mass) taking into account the 1 1.1 ALLOW

relative abundancies of the isotopes sum of (isotopic mass × %abundance)

sum of (isotopic mass × abundance) / total

abundance

DO NOT ALLOW average mass of the isotopes

(i) 2+ – 2 ALLOW 8 electrons in Mg2+ BUT ‘extra’ electron

Mg x Br in Br– must match symbol for electrons in Mg2+

– IGNORE inner shells and circles

x Br

ALLOW 1 mark if both electron arrangements

Mg with no (or 8) outer electrons and charges are correct but only one Br is

AND drawn.

2 × Br with ‘dot-and-cross’ outer octet 1.2

ALLOW 2[Br–], 2[Br]– (brackets not required)

Correct charges 2.5

(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 3 2.2×3 ALLOW ECF

If answer = 1.71 × 1022 award 3 marks

-------------------------------------------------------------------------

1.74 Calculator answer = 9.451385117 × 10–3

n(MgBr2) = = 0.00945….. mol

184.1

ALLOW ECF from incorrect moles of ions.

Moles of ions = 0.00945…. × 3 = 0.0283…. mol e.g. 0.00945

Common error

23 22 5.69 × 1021 no × 3 2 marks

Number of ions = 0.0283... × 6.02 × 10 = 1.71 × 10

3SF required

AO

element

(c)* Refer to marking instructions on page 5 of mark scheme for 6 1.1×3 Indicative scientific points may include:

guidance on marking this question. 2.1×3

Structure and bonding

Level 3 (5–6 marks) Magnesium

Explains all three melting point values and conductivities in • Structure: giant lattice

terms of structure, bonding, particles and relative strengths of • Metallic bonding

the f orces. • Delocalised electrons

There is a well-developed line of reasoning which is clear and Bromine

logically structured. The information presented is relevant and • Structure: simple molecular

substantiated.

• induced dipole dipole forces (London forces)

Level 2 (3–4 marks) • (Between) molecules

Attempts to explain all three melting point values and DO NOT ALLOW (between) atoms

conductivities in terms of the structure, bonding, particles of all

three substances, but explanations may be incomplete or may Magnesium bromide

contain only some correct statements or comparisons. • Structure: giant lattice

OR • Ionic bonding

Correctly explains two of the melting point values and

conductivities in terms of the structure, bonding, particles. • (Between) oppositely charged ions

Comparison of bond strengths

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some • Metallic and ionic bonds are stronger than

evidence. London forces

OR Metallic and Ionic bonds need more energy to

Level 1 (1–2 marks) break than London forces

Identif ies only some of the structures, forces and particles

AND Conductivity

Attempts to explain the melting point values OR conductivities • Magnesium: conducts due to delocalised

in terms of the structure, bonding, particles electrons can move/mobile.

IGNORE ‘Carry’ charge for movement

There is an attempt at a logical structure with a line of

reasoning. The information is in the most part relevant. • Magnesium bromide: In solid IONS cannot move;

in solution IONS can move.

0 marks DO NOT ALLOW electrons.

No response or no response worthy of credit. • Bromine: Does not conduct as no mobile charge

carriers.

7 AO

element

(d) (i) Mg2+(g) + 2Br(g) + 2e– 2 1.2× 2 State symbols required.

Mg(s) + Br2(l) CARE: Liquid state symbol for Br2

(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 2 2.2×2 ALLOW -347 (kJ mol–1) for 2 marks.

If answer = –346.5 award 2 marks

-----------------------------------------------------------------

2∆H hyd = ALLOW f or 1 mark ONE error with sign OR use of 2:

– 525 – 186 – (2 x 112) – 148 – 736 – 1450 + (2 x -325) –693 (not divided by 2 at the end)

346.5 (wrong sign on answer)

+ 1926

OR Common errors for 1 mark

– 525 – 186 – 224 – 148 – 736 – 1450 + 650 + 1926 –2272.5 (–1926 instead of 1926)

OR -1386 (2 x -693 instead of -693)

= – 693 –996.5 (–650 instead of 650)

–509 (2 × 325 not used)

∆H hyd = –346.5 (kJ mol–1) –290.5 (2 × 112 not used)

–198.5 (148 instead of –148)

–160.5 (186 instead of –186)

–122.5 (224 instead of –224)

178.5 (525 instead of –525)

389.5 (736 instead of –736)

1103.5 (1450 instead of –1450)

For other answers, check for a single transcription

error or calculation error which could merit 1 mark

DO NOT ALLOW any answer which involves two

errors

e.g. –453 (2 × 325 not used AND 2 x 112 not used)

AO

Question Answer 8 Marks Guidance

element

(iii) Equation: Mg2+(g) + 2Br –(g) → MgBr2(s) 3 1.2 State symbols required

For other answers, check for a single transcription

CHECK THE ANSWER ON ANSWER LINE error or calculation error which could merit 1 mark

If answer = –2433 award 2 marks 2.2 x 2

DO NOT ALLOW any answer which involves two

----------------------------------------------------------------- errors

Lattice enthalpy =

∆hyH(Mg2+) + 2 × ∆hyH(Br–) – ∆solH(MgBr2) OR ALLOW ECF f rom incorrect answer to d(ii)

-1926 + (2 x -346.5) – (-186)

OR

∆fH(MgBr2) – 2∆atH(Br) – ∆atH(Mg)

– 1st IE(Mg) – 2nd IE(Mg) – 2∆eaH(Br) OR

-525 – (2 x 112) – 148 – 736 – 1450 – (2 x -325)

Lattice enthalpy = –2433 kJ mol–1

Total 18

How to answer it

Magnesium, Bromine and Magnesium Bromide

OCR A-Level Chemistry • Comprehensive Study Guide

What this question tests

This multi-topic question assesses foundational and advanced physical chemistry concepts: relative isotopic mass definitions, ionic dot-and-cross structures, stoichiometry and molar calculations, structure and bonding links to physical properties, Born-Haber / enthalpy cycles, and lattice enthalpy calculations. It tests your precision with state symbols, significant figures, and multi-step Hess's Law cycles.

Part (a) — Definition of Weighted Mean Mass

✅ Correct Answer

The mean/average mass taking into account the relative abundances of the isotopes.

❌ Common Errors

Students often lose the mark by stating "average mass of the isotopes" without mentioning relative abundances or weighted/mean mass.

Marks: 1 mark

Part (b) — Bonding, Ions and Stoichiometry

(b)(i) Dot-and-Cross Diagram for MgBr₂

💡 Key Knowledge

  • Magnesium loses 2 electrons to form Mg²⁺ with no outer electrons shown (or an empty outer shell with a visible inner shell).
  • Each Bromine gains 1 electron to form Br⁻ with a full outer octet (8 electrons).
  • Include square brackets around each ion with correct charges outside. Show two separate Br⁻ ions or use a coefficient.

🧠 Exam Technique

When drawing ionic lattices or separate ionic species, make sure charges are clearly superscripted outside square brackets. Examiners penalise missing brackets or incorrect electron counts.

(b)(ii) Calculating Total Number of Ions

📐 Step-by-Step Calculation

  1. Find molar mass of MgBr₂:
    24.3 + (2 × 79.9) = 184.1 g mol⁻¹
  2. Calculate moles of MgBr₂:
    1.74 g / 184.1 g mol⁻¹ = 0.009451... mol
  3. Calculate total moles of ions:
    Each formula unit splits into 3 ions ( Mg²⁺ and 2 Br⁻ ).
    0.009451... × 3 = 0.02835... mol of ions
  4. Multiply by Avogadro constant (Nₐ):
    0.02835... × 6.02 × 10²³ = 1.71 × 10²² ions

❌ Common Calculation Traps

  • Forgetting to multiply by 3: Failing to account for the fact that MgBr₂ dissociates into three ions.
  • Significant Figures: The question asks for 3 significant figures. Final answer must be 1.71 × 10²² .
Marks: 5 marks total ((i) = 2 marks, (ii) = 3 marks)

Part (c) — Structure, Bonding and Physical Properties

💡 Key Knowledge (Level 3 Response Guide)

To score Level 3 (5–6 marks), you must systematically link particles, bonding type, and relative strength of forces for all three substances:

  • Magnesium: Giant metallic lattice. Positive ions surrounded by a sea of delocalized electrons. Strong metallic bonds require a lot of energy to break (high melting point). Delocalized electrons can move, allowing electrical conductivity in both solid and liquid states.
  • Bromine: Simple molecular structure. Held together by weak induced dipole-dipole forces (London forces) between molecules. Little energy is required to overcome these forces (low melting point). No mobile charge carriers, so it does not conduct electricity in solid or liquid states.
  • Magnesium Bromide: Giant ionic lattice. Strong electrostatic forces of attraction between oppositely charged ions ( Mg²⁺ and Br⁻ ). High melting point. Does not conduct electricity in the solid state because ions are fixed in position; conducts in the liquid (molten) state because ions are mobile.

❌ Common Errors

  • Describing bromine as having covalent bonds between molecules rather than weak intermolecular forces.
  • Stating that solid ionic compounds can conduct electricity (ions are fixed in a lattice, not free to move).
Marks: 6 marks (Level-based assessment)

Part (d) — Enthalpy Cycles and Hydration

(d)(i) Completing the Energy Cycle

✅ Correct Answer / Species to Add

  • Top box: Mg²⁺(g) + 2Br(g) + 2e⁻
  • Bottom-middle box: Mg(s) + Br₂(l)

Crucial: State symbols must be exact ( (l) for bromine, (s) for magnesium).

(d)(ii) Calculating Enthalpy of Hydration of Bromide Ions

📐 Step-by-Step Calculation

Using Hess's Law / energy cycle routes to find ΔH(hydration of Br⁻):

  1. Set up the energy balance equation:
    ΔH(solution) = ΔH(lattice, solution) + ΔH(hydration)
    Alternatively, equate the two pathways from elements to aqueous ions:
  2. Substitute values from Table 16.2:
    -525 - 186 - (2 × 112) - 148 - 736 - 1450 - (2 × ΔH_hyd(Br⁻)) + (-1926) = 0
  3. Rearrange and solve for 2ΔH_hyd(Br⁻) , then divide by 2:
    ΔH_hyd = -346.5 kJ mol⁻¹ (Accept -347 kJ mol⁻¹ )

❌ Common Calculation Traps

  • Forgetting to multiply by 2: Bromine is Br₂ , meaning there are 2 bromide ions per formula unit. The hydration of two moles of Br⁻ must be accounted for.
  • Sign errors when transposing values from the enthalpy table.

(d)(iii) Lattice Enthalpy Equation and Calculation

✅ Equation

Mg²⁺(g) + 2Br⁻(g) → MgBr₂(s)

📐 Calculation

Lattice enthalpy is the enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions:

Lattice Enthalpy = ΔH(formation) - [ΔH(atomisation of Mg) + 1st IE + 2nd IE + (2 × atomisation of Br) + (2 × electron affinity of Br)]

Or using cycle values: -525 - 148 - 736 - 1450 - (2 × 112) - (2 × -325) = -2433 kJ mol⁻¹

Marks: 7 marks total ((i) = 2 marks, (ii) = 2 marks, (iii) = 3 marks)

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 3.2 Physical chemistry · 5.2 Energy

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.