AQA A-Level Chemistry Paper 1, 2018: Question 8
17 marks · Medium difficulty · Long Answer
Explain differences in melting point between NaBr, Na, and NaI; calculate gas volume and ion concentration from sodium reacting with water; and determine the shape and bond angle of the NH₂⁻ ion.
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Question text
08 This question is about sodium and some of its compounds.
08.1 Use your knowledge of structure and bonding to explain why sodium bromide has a
melting point that is higher than that of sodium, and higher than that of sodium iodide.
[6 marks]
When 250 mg of sodium were added to 500 cm3 of water at 25 °C a gas was
08.2
produced.
Give an equation for the reaction that occurs.
Calculate the volume, in cm3, of the gas formed at 101 kPa
The gas constant, R = 8.31 J K–1 mol–1
[6 marks]
Equation
Volume cm3
Calculate the concentration, in mol dm–3, of sodium ions in the solution produced in
08.3
the reaction in Question 08.2.
[1 mark]
23Concentration mol dm–3
08.4 Sodium reacts with ammonia to form the compound NaNH2 that contains the
–
NH2 ion.
*22* –
Draw the shape of the NH2 ion.
Include any lone pairs of electrons that influence the shape.
Predict the bond angle.
Justify your prediction.
[4 marks]
Shape
Bond angle
Justification
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark
This question is marked using Levels of Response. Indicative chemistry content. Contradictions (eg
Examiners should apply a ‘best-fit’ approach to the marking. molecules, IMFs, covalent bonding,) negate 6
statements.
Level 3 All stages are covered and the explanation of each
stage is generally correct and virtually complete. Stage 1- Na
5-6 marks
Answer is communicated coherently and shows a 1a) Na has metallic bonding
logical progression from stage 1 to stage 2 and
then stage 3. 1b) there is attraction/ bonding between the positive
nucleus/ ion and the delocalised electrons in Na
Coherent communication requires that there is a
comparison between the types of bonding and that the 1c) Na has a giant/lattice structure
bonding is correct for each substance.
Stage 2 – NaBr or NaI
Level 2 All stages are covered but the explanation of each stage
may be incomplete or may contain inaccuracies 2a) Ionic bonding in NaBr and/or NaI
08.1 3-4 marks
OR two stages are covered and the explanations are 2b) There is attraction/ bonding between the + and –
generally correct and virtually complete. ions in NaBr and/or NaI
Answer is mainly coherent and shows some 2c) NaBr and/or NaI have a giant/lattice structure
progression from stage 1 to stage 2 and then stage 3.
Stage 3 - comparison of bonding
Level 1 Two stages are covered but the explanation of each
stage may be incomplete or may contain inaccuracies 3a) The ionic bonds are stronger (or wtte) than the
1-2 marks metallic bonds
OR only one stage is covered but the explanation is
generally correct and virtually complete. 3b) there is stronger attraction (or wtte) between the
+ and – ions in NaBr than in NaI
Answer shows some progression between two
stages 3c) since the Br– – ion is smaller than the I–– ion–
Level 0 Insufficient correct chemistry to gain a mark.
08.1
cont.
0 marks
M1 Na + H2O NaOH + ½ H2 Allow multiples 1
M2 (Mass Na = 0.250 g so moles Na = 0.250/23.0) = 0.0109 CE: If not divided by 23, max 3/5 calculation marks – 1
M3, M4 and M5
AE: If not divided by 1000 and final answer is 1.33 x
105 cm3 4/5
M3 moles H = 5.43 x 10-3 to 5.45 x 10-3 M3 = M2 /2 1
08.2 CE: If incorrect ratio used max 3/5 calculation marks
– M2, M4 and M5
M4 T = 298 (K) and P = 101000 (Pa) 1
M5 V = nRT/P or (5.435 x 10-3 x 8.31 x 298)/101000 or 1.33 x 10-4 (m3 ) 1
M6 V = 133 – 134 cm3 Allow to 2 significant figures or more 1
– – –
08.3 Conc = 0.0109/ 500 x 10-3 = 0.0217-0.022 (mol dm-3 ) Allow M2 from question 08.2 / 0.5 1
_ Ignore charge and brackets
M1 1
H H
N
08.4
o 1
M2 104.5 Allow 104-106
M3 (4) electron pairs repel to be as far apart as possible 1
M4 lp/lp repulsion> lp/bp repulsion (> bp/bp repulsion) For M4 allow lone pairs repel more than bonding
pairs
Mark independently
Total 16
How to answer it
Sodium & Its Compounds: Structure, Gas Stoichiometry & Shapes
- Structure & Bonding Comparisons: Distinguishing giant metallic lattices from giant ionic lattices, explaining electrostatic forces, and linking ionic radius to lattice enthalpy/melting point.
- Stoichiometry & Ideal Gas Equation: Writing balanced redox equations of alkali metals with water and calculating gas volume using pV = nRT with standard unit conversions (mg to g, kPa to Pa, °C to K, m³ to cm³).
- Solution Concentration: Determining molarity ( mol dm⁻³ ) in aqueous solutions.
- VSEPR Theory: Predicting 3D shapes, electron pair repulsions, lone pair distortions, and bond angles for simple molecular ions ( NH₂⁻ ).
Question 08.1
Structure & Bonding: Melting Points of Na, NaBr, and NaI [6 Marks]
💡 Key Knowledge (3-Stage Framework)
This is a 6-mark banded question requiring all three logical stages:
- Stage 1: Sodium (Na)
• Giant metallic lattice.
• Electrostatic attraction between positive ions (Na⁺) and delocalised electrons. - Stage 2: Sodium Halides (NaBr & NaI)
• Giant ionic lattice.
• Electrostatic attraction between oppositely charged ions (Na⁺ and Br⁻ / I⁻). - Stage 3: Comparison
• NaBr vs Na: Ionic bonds in NaBr are stronger than the metallic bonds in Na.
• NaBr vs NaI: The Br⁻ ion is smaller than the I⁻ ion (higher charge density), resulting in stronger electrostatic attraction between Na⁺ and Br⁻ than between Na⁺ and I⁻.
✅ Model Answer (Level 3: 5–6 Marks)
"Sodium has a giant metallic structure with electrostatic attraction between Na⁺ ions and delocalised electrons."
"Sodium bromide and sodium iodide both have giant ionic lattice structures with strong electrostatic attraction between oppositely charged ions (Na⁺ and halide ions)."
"Sodium bromide has a higher melting point than sodium because its ionic bonds are stronger and require more energy to break than the metallic bonds in sodium."
"Sodium bromide has a higher melting point than sodium iodide because the Br⁻ ion has a smaller ionic radius than the I⁻ ion. This means the Na⁺ and Br⁻ ions can pack closer together, resulting in stronger electrostatic attraction between the ions in NaBr than in NaI."
🧠 Exam Technique & Examiner Guidance
- Top-Band Requirement: You must cover all 3 stages with virtually complete explanations and logical progression to access 5–6 marks.
- Always use the word "ion": Saying "bromine is smaller than iodine" loses marks. You must state that the bromide ion (Br⁻) is smaller than the iodide ion (I⁻).
- Delocalised is mandatory: When describing metallic bonding, simply stating "electrons" is not enough; they must be named as delocalised electrons.
❌ Contradiction Traps (Immediate Mark Negation)
- Mentioning Intermolecular Forces: Mentioning "van der Waals", "dipole-dipole", or "hydrogen bonds" for Na, NaBr, or NaI is a direct scientific contradiction and caps/negates your marks.
- Calling them "molecules": Referring to NaBr or NaI as "molecules" or referring to "covalent bonds" being broken.
- Vague comparisons: Stating "NaBr is more reactive" instead of comparing bond strengths and electrostatic attraction.
• Level 3 (5–6 marks): All 3 stages covered; virtually complete explanations; coherent comparison of bond types.
• Level 2 (3–4 marks): 2 stages complete OR all 3 stages covered with minor omissions/inaccuracies.
• Level 1 (1–2 marks): Only 1 stage complete OR 2 stages incomplete.
Question 08.2
Reaction of Sodium with Water & Ideal Gas Calculation [6 Marks]
✅ Part 1: Chemical Equation (1 Mark)
Na + H₂O → NaOH + ½ H₂
Alternative standard balanced form accepted: 2Na + 2H₂O → 2NaOH + H₂ (state symbols not required).
📐 Step-by-Step Gas Calculation (5 Marks)
1 Calculate moles of Na reacted:
Convert mass to grams: 250 mg = 0.250 g
n(Na) = mass / Aᵣ = 0.250 / 23.0 = 0.01087 mol (M2)
2 Determine moles of gas (H₂) produced:
From equation ratio: 2 Na : 1 H₂ (or 1 Na : ½ H₂ )
n(H₂) = 0.01087 / 2 = 5.435 × 10⁻³ mol (allow 5.43 × 10⁻³ to 5.45 × 10⁻³) (M3)
3 Convert values to SI units:
• Temperature: T = 25 °C + 273 = 298 K
• Pressure: P = 101 kPa = 101 000 Pa (M4)
4 Rearrange pV = nRT and solve for V in m³:
V = nRT / P = (5.435 × 10⁻³ × 8.31 × 298) / 101 000
V = 1.3468 × 10⁻⁴ / 1.01 = 1.333 × 10⁻⁴ m³ (M5)
5 Convert volume to cm³:
V(cm³) = 1.333 × 10⁻⁴ × 10⁶ = 133 to 134 cm³ (e.g. 133 cm³) (M6)
❌ Common Calculation Traps
- Missing unit conversion on mass: Leaving 250 mg as 250 g produces 1.33 × 10⁵ cm³ (loses M2, capping marks to max 4/5).
- Forgetting 2:1 stoichiometric ratio: Using n(H₂) = n(Na) is a Chemical Error (CE) and caps calculation marks to a maximum of 3/5.
- Unit conversion from m³ to cm³: Remember 1 m³ = 1 000 000 cm³ ( × 10⁶ ), NOT × 10³ .
🧠 Exam Technique Tip
Always write down your rearranged formula V = nRT / P before substituting numbers. If you make an arithmetic slip, writing down the substituted numbers explicitly secures partial method marks (M4 & M5).
Question 08.3
Concentration of Sodium Ions [1 Mark]
📐 Step-by-Step Calculation
1 Identify moles of Na⁺ ions:
Every mole of Na produces 1 mole of Na⁺ ions in solution:
n(Na⁺) = 0.01087 mol (from 08.2 M2)
2 Convert volume of water to dm³:
V = 500 cm³ = 500 / 1000 = 0.500 dm³
3 Calculate concentration:
Concentration = n / V = 0.01087 / 0.500 = 0.0217 to 0.022 mol dm⁻³
✅ Final Answer
0.0217 mol dm⁻³ (or 0.022 mol dm⁻³)
Note: Error Carried Forward (ECF) applies directly from Question 08.2 step M2 divided by 0.5 dm³.
Question 08.4
Shape and Bonding in the Amide Ion (NH₂⁻) [4 Marks]
✅ Shape & Structure Drawing (1 Mark)
Shape Name: Bent / V-shaped / Non-linear
[ N ]⁻
[ / \ ]
[ H H ]
How to draw: Show central N atom bonded to two H atoms at an angle, with 2 distinct lone pairs drawn either as lobes with electron dots or as simple dot pairs / lines on the nitrogen atom. (Square brackets and charge are not strictly required for the mark).
✅ Angle & Justification (3 Marks)
- Predicted Bond Angle: 104.5° (allow 104°–106°) [1 Mark]
- Electron Pair Repulsion: Central N has 4 electron pairs (2 bonding pairs, 2 lone pairs) which repel as far apart as possible. [1 Mark]
- Lone Pair vs Bonding Pair Repulsion: Lone pair–lone pair repulsion is greater than lone pair–bonding pair repulsion (which is greater than bonding pair–bonding pair repulsion) / "lone pairs repel more than bonding pairs". [1 Mark]
💡 Why 104.5°?
Nitrogen is in Group 5 (5 valence electrons) + 1 extra electron from the negative charge + 2 electrons shared from 2 H atoms = 8 valence electrons = 4 pairs.
A tetrahedral base angle is 109.5°. Each lone pair reduces the bond angle by approximately 2.5°:
109.5° - (2 × 2.5°) = 104.5° (identical to H₂O).
❌ Common Errors in Shape Justification
- Forgetting the negative charge: Treating N as having only 1 lone pair like ammonia ( NH₃ , pyramidal, 107°).
- Incomplete repulsion statement: Saying "lone pairs repel" without explicitly stating that lone pairs repel MORE than bond pairs.
- Omitting total repulsion: Forgetting to state that all 4 electron pairs repel to achieve maximum separation / minimum repulsion.
Topics
Physical Chemistry · 3.1.2 Amount of Substance · 3.1.3 Bonding
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.