Edexcel A-Level Chemistry AS Paper 1, June 2017: Question 6
16 marks · Hard difficulty · Open Response
Write the equation for the reaction of boron with chlorine, draw a dot-and-cross diagram and explain the shape of BCl3, calculate the molecular formula of aluminium chloride using analytical and ideal gas data, and explain the difference in sublimation temperatures between aluminium fluoride and aluminium chloride.
Practise this questionQuestion
Question text
6 Boron and aluminium are in the same group of the Periodic Table. Both form
compounds with chlorine and with fluorine.
(a) Boron reacts directly with chlorine to produce a covalently bonded compound, BCl3.
(i) Write the equation for this reaction. State symbols are not required.
(1)
(ii) Draw a dot-and-cross diagram for BCl3 showing only the outer shell electrons
of the atoms.
(1)
(iii) Use your diagram to explain why BCl3 has a trigonal planar shape with
bond angles of 120q.
(2)
(b) Aluminium also reacts directly with chlorine to form a compound,
aluminium chloride, containing only aluminium and chlorine.
A 0.500 g sample of aluminium chloride was analysed and found to contain
0.101 g of aluminium.
Another 0.500 g sample was heated to 473 K. The gas produced occupied a
volume of 73.6 cm3 at a pressure of 1.00 × 102 kPa.
Determine the molecular formula of the gas.
You will need to use the equation pV = nRT and R = 8.31 J mol–1 K–1
*P49858A01424* (6)
*(c) Aluminium fluoride and aluminium chloride are both crystalline solids at room
temperature. Aluminium fluoride sublimes to form a gas at 1291qC (1564 K),
whilst aluminium chloride sublimes at 178qC (451 K).
Use the Pauling electronegativity values in the Data Booklet to explain these
differences in sublimation temperature.
(6)
… *P49858A01524*
(Total for Question 6 = 16 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional Guidance Mark
Number
6 (a) (i) 2B + 3Cl2 → 2BCl3 Allow multiples (1)
Ignore state symbols even if incorrect
Question
Acceptable Answer Additional Guidance Mark
Number
6 (a) (ii) Ignore inner shell electrons and circles (1)
ALLOW
All dots or all crosses
Question
Acceptable Answer Additional Guidance Mark
Number
6 (a)(iii) An explanation that makes reference to the following points: (2)
3 bonding pairs of electrons (bonding environments) (and Accept 3 pairs of electrons
no non-bonding / lone pairs of electrons in the outer shell
of boron) (1)
(the bonding pairs of electrons) move apart to minimise Do not award 3 bonding pairs repel
repulsion (1) each other equally
Accept move as far apart as possible
/ maximise separation
Question
Acceptable Answer Additional Guidance Mark
Number
6 (b) Determine empirical formula Example of calculation (6)
finds mass of Cl 0.500 - 0.101 = 0.399(g)
AND AND
finds moles of aluminium and chlorine (1) 0.101/27.0 = 0.00374074 / 3.74...
x 10-3
AND
0.399/35.5 = 0.01123944 / 1.12...
x 10-2
determines ratio and hence empirical formula is AlCl3 (1)
0.01123944 = 3.005
0.00374074
Could use (0.101/0.5) x 100 =
20.2%
20.2/27.0 = 0.74814815
AND
79.8/35.5 = 2.2478873
2.2478873
0.74814815 = 3.005
Determine molecular mass
converts p into Pa / N m-2 and V into m3 (1)
p = 1.00 x 102 x 103 =100 000 / 1 x
AND
rearrange pV = nRT and finds number of moles (1)
V = 73.6 / 1 000 000 or 7.36 x 10-
finds molecular mass (1)
n = 100000 x (73.6/1000000) =
0.001872 or
finds molecular formula (1) 8.31 x 473
1.872473 x 10-3 (mol)
Mr = 0.500 = 267.03
1.872473 x 10-3
267.03 = 2 so Al2Cl6
27.0 + (35.5 x 3)
COMMENT MP 3-5 and identity of
Al2Cl6 without incorrect working
scores 6 marks
Question
Acceptable Answer Additional Guidance Mark
Number
*6 (c) This question assesses a student’s ability to show a Guidance on how the mark scheme should be (6)
coherent and logically structured answer with applied:
linkages and fully-sustained reasoning.
The mark for indicative content should be
Marks are awarded for indicative content and for added to the mark for lines of reasoning.
how the answer is structured and shows lines of
reasoning. For example, an answer with five indicative
marking points, which is partially structured
The following table shows how the marks should be with some linkages and lines of reasoning,
awarded for indicative content. scores 4 marks (3 marks for indicative
content and 1 mark for partial structure and
Number of indicative Number of marks awarded some linkages and lines of reasoning).
marking points seen in for indicative marking
answer points If there are no linkages between points, the
64 same five indicative marking points would
5–4 3 yield an overall score of 3 marks (3 marks for
3–2 2 indicative content and no marks for linkages).
Question
Acceptable Answer Additional Guidance Mark
Number
*6 (c) The following table shows how the marks should be (6)
contd awarded for structure and lines of reasoning.
Number of marks awarded
for structure of answer and In general it would be expected
sustained line of reasoning that 5 or 6 indicative points
Answer shows a coherent and 2 would get 2 reasoning marks,
logical structure with linkages and and 3 or 4 indicative points
fully sustained lines of reasoning would get 1 mark
demonstrated throughout. for reasoning, and 0, 1 or 2
Answer is partially structured with 1 indicative points would score
some linkages and lines of zero marks for reasoning.
reasoning.
Answer has no linkages between 0 Reasoning marks may be
points and is unstructured.
reduced for extra incorrect
chemistry
Indicative content:
aluminium and chlorine electronegativity difference 1.5 AND Allow all 3 electronegativity
aluminium and fluorine electronegativity difference 2.5 values / difference between F
and Cl is 1 / difference between
differences is 1/ F is 4, CL is 3
and this is a significant
difference
aluminium chloride (mostly) covalent / (small) molecule
aluminium fluoride (bonds) more polar Allow mostly/more ionic
aluminium chloride molecular so weak(er) intermolecular forces Allow weak(er) dipole-dipole
/ London forces interactions
Do not award any suggestion of
breaking covalent bonds
aluminium fluoride is a giant structure/ strong electrostatic Allow stronger dipole-dipole
forces of attraction between the ions attractions
more energy needed to break the stronger bonds to cause Allow (dative) covalent bonds
sublimation in aluminium fluoride breaking (to form small
molecule / AlF3)
(Total for Question 6 = 16 marks)
How to answer it
Boron and Aluminium Halides Study Guide
What this question tests
This question assesses your understanding of Group 3 chemistry, chemical equations, electron pair repulsion theory (VSEPR), ideal gas calculations (using pV = nRT), empirical and molecular formulae determination, and bonding trends linked to electronegativity differences.
Writing the Synthesis Equation
✅ Correct Answer
2B + 3Cl₂ → 2BCl₃
Multiples are accepted. State symbols are not required by the question.
❌ Common Errors
- Writing chlorine as Cl instead of the diatomic molecule Cl₂ .
- Incorrect balancing leading to improper stoichiometry.
Drawing BCl₃
✅ Correct Answer
Boron shares all 3 of its outer shell electrons with 3 chlorine atoms, forming 3 single covalent bonds. Boron will have 6 electrons in its outer shell (an incomplete octet). Chlorine atoms have 3 non-bonding pairs and 1 bonding pair each.
💡 Key Knowledge
Boron is an electron-deficient element. It does not expand or fill a full octet in simple halides like BCl₃, stopping at 6 electrons in its outer shell.
Explaining Trigonal Planar Geometry
✅ Correct Answer
- There are 3 bonding pairs of electrons in the outer shell of boron (and no lone pairs).
- These electron pairs move as far apart as possible to minimise mutual repulsion, resulting in a trigonal planar shape with a bond angle of 120°.
🧠 Exam Technique
You must explicitly state: (1) the number of bonding pairs vs lone pairs, and (2) that electron pairs repel and try to get as far apart as possible. Avoid phrasing like "atoms repel" — it must be electron pairs.
Determining the Formula of Aluminium Chloride
📐 Step-by-Step Calculation
Step 1: Find the mass of chlorine in the sample
Mass of Cl = 0.500 g - 0.101 g = 0.399 g
Step 2: Find moles of Al and Cl
Moles of Al = 0.101 / 27.0 = 0.00374 mol
Moles of Cl = 0.399 / 35.5 = 0.01124 mol
Step 3: Determine the empirical formula ratio
Ratio Al : Cl = 0.00374 / 0.00374 : 0.01124 / 0.00374 = 1 : 3005 ≈ 1 : 3
Empirical formula = AlCl₃
Step 4: Convert units for gas equation (pV = nRT)
p = 1.00 × 10² kPa = 1.00 × 10⁵ Pa
V = 73.6 cm³ = 73.6 × 10⁻⁶ m³ (or 73.6 / 1000000)
T = 473 K
Step 5: Calculate moles of gas (n)
n = pV / (RT) = (1.00 × 10⁵ × 73.6 × 10⁻⁶) / (8.31 × 473) = 0.00187 mol
Step 6: Find relative molecular mass (Mr) and molecular formula
Mr = mass / moles = 0.500 / 0.00187 = 267.03
Empirical mass of AlCl₃ = 27.0 + (35.5 × 3) = 133.5
267.03 / 133.5 = 2, so molecular formula = Al₂Cl₆
❌ Common Calculation Traps
- Forgetting to convert cm³ into m³ by dividing by 10⁶.
- Failing to convert kPa into Pa by multiplying by 10³.
- Stopping at the empirical formula instead of calculating Mr using the ideal gas law to find the dimer Al₂Cl₆ .
Explaining Sublimation Temperature Differences
✅ Correct Answer & Indicative Content
- Electronegativity: Aluminium and chlorine have a smaller electronegativity difference (~1.5) compared to aluminium and fluorine (~2.5).
- Bonding Type: Aluminium chloride is mostly covalent (simple molecular), whereas aluminium fluoride is much more ionic.
- Intermolecular vs Ionic Forces: AlCl₃ has weak intermolecular forces (London / van der Waals forces) between molecules, requiring little energy to break (sublimes at 178 °C).
- Giant Structures: AlF₃ is a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions, requiring significantly more energy to break (sublimes at 1291 °C).
🧠 Top-Level Exam Strategy
This is a starred (*c) question where quality of written communication and logical structure count towards your marks. Link your points explicitly: state the electronegativity difference, deduce the bonding type, identify the type of structure/forces holding the particles together, and relate this directly to the energy required.
Topics
Physical Chemistry · Inorganic Chemistry · Topic 2: Bonding and Structure · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.