OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 22
9 marks · Medium difficulty · Structured Questions
Determine the electron configuration, equation, lattice type and molar mass/molecular formula for compounds of bromine.
Practise this questionQuestion
Question text
22 This question is about compounds of bromine.
(a) Bromine reacts with phosphorus, P4, to form phosphorus tribromide, PBr3.
(i) Complete the electron configuration of a bromine atom.
1s2 … [1]
(ii) Write the equation for the reaction of phosphorus with bromine.
… [1]
(b) A compound of bromine is a solid at room temperature. The electrical conductivity of the
compound at different physical states is shown in the table.
Physical state Electrical conductivity
solid poor
liquid good
Name the type of lattice in the compound at room temperature and explain the different
electrical conductivities.
Name of lattice …
Explanation for different conductivities …
[2]
(c) Bromine reacts with fluorine to form compound A.
Compound A is a liquid at room temperature and pressure but can easily be vaporised.
When vaporised, 0.428 g of A produces 76.0 cm3 of gas at 1.00 × 105 Pa and 100 °C.
Determine the molar mass and molecular formula of compound A.
molar mass of A = … g mol−1
molecular formula of A = …
[5]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
22 (a) (i) (1s2)2s22p63s23p63d104s24p5 1 1.2 ALLOW 3d after 4s2,
e.g. 1s22s22p63s23p64s23d104p5
Look carefully at 1s22s22p63s23p6 ALLOW upper case D, etc and subscripts,
– there may be a mistake e.g … 4S23D1
DO NOT ALLOW [Ar] as shorthand for
1s22s22p63s23p6
IGNORE 1s2 repeated
(a) (ii) P4 + 6Br2 → 4PBr3 1 2.6 ALLOW multiples
(b) Giant ionic 2 1.1 ‘Giant’ is essential
In solid state/lattice, Mark independently of 1st structure mark
ions are fixed (in position) OR cannot move
AND IGNORE comments about electrons for solid
In liquid state,
ions are mobile OR can move 1.2 IGNORE ‘free’ ions
AO
7 element
(c) FIRST CHECK ANSWER LINES 5 ALLOW ECF throughout
If molecular formula = BrF5 AND 174.6/175 AND
working showing use of ideal gas equation
Award 5 marks for calculation
----------------------------------------------------------------------
Rearranging ideal gas equation pV
IF n = is omitted, ALLOW when values are
pV RT
n =
RT 2.2×4 substituted into rearranged ideal gas equation.
pV
Unit conversion AND substitution into n = :
RT
• R = 8.314 OR 8.31 ALLOW conversion of V into dm3 AND p in kPa
• V = 76(.0) × 10–6 (m3) Gives same answer in powers of 10
• T in K: 373 K
1.00 × 105 × 76.0 × 10–6 Calculator value:
e.g. from 8.314 = 2.450725899 × 10–3
8.314 × 373 –3
Calculation of n using p, V, R AND T from 8.31 = 2.45190555 × 10
n = 2.45 × 10–3 (mol) IGNORE figures after 5 in 2.45
Calculation of M
ALLOW ECF f rom a value of n that has been
0.428
M = = 174.6 derived from pV = nRT
2.45 × 10–3
e.g. 0.174.6 OR 0.175 from 2.45
Molecular formula
BrF5 OR F5Br 3.2 ALLOW ECF matching ECF M f rom pV = nRT
Use of Final 2 marks possible for use of 76.0 cm3 OR 0.760 dm3 by ECF
24 dm3 76.0
e.g. n = = 3.17 × 10–3 No mark (calculation much simpler)
24000
0.428
M = –3 = 135 ECF
3.17 × 10
BrF3 ECF
How to answer it
Compounds of Bromine Study Guide
What this question tests
This question evaluates your foundational knowledge of atomic structure (electron configurations), bonding and physical properties (giant ionic lattices and electrical conductivity), and quantitative chemistry (rearranging the ideal gas equation, unit conversions, molar mass, and determining molecular formulae).
Question 22 (a) — Electron Configuration & Equation Writing
✅ Correct Answers
- (i) 2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵
- (ii) P₄ + 6Br₂ → 4PBr₃ (or correct multiples)
💡 Key Knowledge
- Bromine has an atomic number of 35. Make sure to fill orbitals following the aufbau principle, remembering that the 4s orbital fills before the 3d orbital.
- Phosphorus reacts directly with halogens to form phosphorus(III) halides like PBr₃ .
❌ Common Errors
- Using noble gas shorthand like [Ar] when the question explicitly begins the line with 1s² .
- Forgetting that bromine exists as a diatomic molecule ( Br₂ ) and phosphorus as P₄ when writing equations.
Question 22 (b) — Bonding and Electrical Conductivity
✅ Correct Answers
- Name of lattice: Giant ionic
- Explanation: In the solid state, ions are fixed in position / cannot move. In the liquid (molten) state, ions are mobile / can move.
🧠 Exam Technique
- The word "Giant" is essential for the lattice mark—do not just write "ionic".
- You must contrast both states clearly: explain why it doesn't conduct as a solid, and why it conducts as a liquid. Focus on the mobility of ions, not electrons.
Question 22 (c) — Ideal Gas Calculations & Molecular Formula
📐 Step-by-Step Calculation
- Rearrange the ideal gas equation: n = pV / RT
- Convert units carefully:
- Pressure ( p ) = 1.00 × 10⁵ Pa
- Volume ( V ) = 76.0 cm³ = 76.0 × 10⁻⁶ m³
- Temperature ( T ) = 100 + 273.15 = 373 K
- Gas constant ( R ) = 8.314 J mol⁻¹ K⁻¹
- Calculate moles ( n ):
n = (1.00 × 10⁵ × 76.0 × 10⁻⁶) / (8.314 × 373) = 2.45 × 10⁻³ mol - Calculate molar mass ( M ):
M = mass / n = 0.428 / (2.45 × 10⁻³ ) = 174.6 g mol⁻¹ - Determine molecular formula:
Bromine ( Br ) = 79.9 , Fluorine ( F ) = 19.0 .
Br + 5F = 79.9 + 5(19.0) = 174.9 ≈ 174.6 g mol⁻¹ . Therefore, formula is BrF₅ .
❌ Common Calculation Traps
- Forgetting to convert volume from cm³ to m³ ( × 10⁻⁶ ) or dm³ ( ÷ 1000 ).
- Failing to convert Celsius to Kelvin by adding 273.
- Rounding intermediate values too early, which drifts the final molar mass away from the correct integer ratio for atomic masses.
Topics
Module 2: Foundations in chemistry · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.