OCR A-Level Chemistry Unified chemistry (03), June 2024: Question 4

16 marks · Hard difficulty · Structured Questions

Answer questions about the chemistry of compounds containing phosphorus, including acid-base equilibria, redox reactions with phosphine, systematic naming, bond angles and structures, percentage composition, and evidence for molecular structure.

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Question text

4 This question is about the chemistry of compounds containing phosphorus.

(a) Phosphorus forms several acids including H3PO4 and H3PO3.

H3PO4 is a tribasic acid. The equilibria for the dissociations are shown below.

1 H PO H+ + H PO –

34 2 4

2 H PO – H+ + HPO 2–

24 4

3 HPO 2– H+ + PO 3–

(i) During the equilibria, H PO – behaves both as an acid and as a base.

Explain this statement, using the equilibria 1, 2 and 3, as required.

… [2]

(ii) In a H3PO3 molecule, the O atoms are covalently bonded to the P atom. The H atoms are

bonded to the O atoms.

Draw the structure of a H3PO3 molecule, showing all the bonds.

On your diagram, add the values for the O–P–O and P–O–H bond angles.

[3]

(iii) The systematic name of H3PO4 is phosphoric(V) acid.

What is the systematic name of H3PO3?

… [1]

(b) Phosphine, PH3, is a poisonous gas.

(i) Phosphine reacts with oxygen gas to form phosphorus(V) oxide and water.

Write the equation for this reaction.

… [1]

(ii) Aqueous silver nitrate, AgNO3, is reduced by PH3.

The unbalanced equation is shown below.

Balance the equation and use oxidation numbers to explain why this is a redox reaction.

… AgNO3 + … PH3 + … H2O … Ag + … H3PO3 + … HNO3

Explanation …

… [3]

compound P3N3Cl6 is formed, together with HCl gas.

P3N3Cl6 has a cyclic structure, like the Kekulé structure of benzene.

(i) Write an equation for the reaction of PCl5 and ammonium chloride to form P3N3Cl6.

… [1]

(ii) Calculate the percentage by mass of P in P3N3Cl6.

Give your answer to 2 decimal places.

percentage by mass of P = … % [2]

(iii) Suggest one example of evidence that could show that P3N3Cl6 has a Kekulé structure rather

than a delocalised structure.

… [1]

(iv) In a molecule of P3N3Cl6 all the N and Cl atoms are bonded to P atoms.

Suggest a possible structure for a molecule of P3N3Cl6.

[2]

Mark scheme

Show the mark scheme

MnO – using the correct mean titre from the candidate’s

titres Mean titre and n(MnO –)

AND (21.65 + 21.55)

Mean titre = = 21.6(0) (cm3)

Obtains correct value of x as 2 2

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

logically structured. 21.6(0)

The information presented is relevant and substantiated. n(MnO –) = 0.0200 = 4.32 10–4 (mol)

4 1000

Level 2 (3–4 marks)

– Amount of FeC2O4 in mol

Analyses titration results to determine an amount of MnO4 3 –

n(FeC2O4) in 25.0 cm = 5/3 n(MnO4 )

from a mean titre of the candidate’s titres –4

AND = 7.2(0) 10 (mol)

n(FeC O ) in 250 cm3 = 7.2(0) 10–3 (mol)

amount of FeC O in 25.0 cm3 OR 250 cm3 2 4

uses a mass of FeC2O4 to obtain a value of x with few

errors Value of x (final answer)

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some 1.295

Molar mass FeC2O4•xH2O = –3

evidence. 7.2(0) 10

= 179.9

Level 1 (1–2 marks) Molar mass of xH2O = 179.9 – 143.8 = 36.(….)

Analyses results to determine an amount of MnO – from

the candidate’s titres x = 36/18 = 2

Analyses the information to obtain values of n(MnO –) and Credit other correct methods,

n(FeC2O4) with some errors. e.g. For value of x

There is an attempt at a logical structure with a line of reasoning. Mass of FeC O = 7.2(0) 10–3 143.8 = 1.03536 g

The information is in the most part relevant. Mass of H O = 1.295 – 1.035 = 0.25964 g

0.25964

0 marks – No response or no response worthy of credit. n(H2O) = 18 = 0.0144 mol

0.0144

x = –3 = 2

7.2 10

Question Answer Marks Guidance

Responses using 25.0 cm3 rather than the titres are

limited to Level 1

For communication, a typical ‘logical structure’ would

label most calculation steps in response

e.g.

Communication strand met

Communication strand not met

4 (a) (i) In (Equilibrium) 1, 2 ALLOW description for 1 or 2 as long as

H PO –/It acts as a base unambiguous, e.g. Equation 1, etc

AND

accepts/gains H+/a proton IGNORE missing charge on H PO – throughout

OR H PO – forms H PO

24 3 4

IGNORE reference to HPO 2– acting as an

In (Equilibrium) 2 acid/base OR Equilibrium 3

H PO –/It acts as an acid, Question is about H PO –

24 2 4

AND

donates/loses H+/a proton ALLOW ‘dissociates into H+ and HPO 2–‘

OR H PO – forms HPO 2– IGNORE ‘partially’

24 4

(a) (ii) Diagram showing all bonds correctly 3 IGNORE geometry

ALLOW dot and cross diagram showing 2

shared electrons for each bond

… and IGNORE any lone pairs

e.g.

• 3 bonds only around each P

• 2 bonds only around each O

• Each O bonded to an H

Unambiguous bond angles may be shown on

dot and cross diagram

Bond angles

O–P–O = 107º ALLOW 106–108º

P–O–H = 104.5º ALLOW 104–105º

(a) (iii) phosphoric(III) acid 1 DO NOT ALLOW phosphoric acid (III)

Oxidation number MUST be in correct place

DO NOT ALLOW phosphorous acid

(b) (i) 4PH3 + 8O2 → P4O10 + 6H2O 1 ALLOW multiples

ALLOW 2PH3 + 4O2 → P2O5 + 3H2O

IGNORE state symbols, even if wrong

(b) (ii) 6AgNO3 + (1)PH3 + 3H2O → 6Ag + (1)H3PO3 + 6HNO3 3 ALLOW equation with ‘1’ omitted, i.e.

6AgNO3 + PH3 + 3H2O

→ 6Ag + H3PO3 + 6HNO3

BUT DO NOT ALLOW ‘0’

Ag is reduced from +1 to 0 ALLOW 1 mark for BOTH correct oxidation

number changes with ‘reduced’ and ‘oxidised’

P is oxidised from –3 to +3 omitted

OR ‘oxidised and reduced the wrong way round

IGNORE oxidation numbers written around equation

Treat as rough working + signs required for +1 and +3

IGNORE reference to electrons For oxidation numbers,

Question states oxidation numbers ALLOW 1+, 3– and 3+

IGNORE state symbols, even if wrong

IF % by mass = 26.72, award 2 marks ALLOW 1 mark total for 26.7

IF % by mass = 26.7, award 1 mark Question asks for 2 DP

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

Mr of P3N3Cl6 = 348(.0)

31.0 3 ALLOW ECF from incorrect Mr

% by mass of P = 348 100 = 26.72

2 DP required ALLOW 1 mark for 8.91 (omission of 3):

31.0

348 100 = 8.91

(P–N) bond lengths are different IGNORE C–C bond lengths are different

enthalpy change of hydrogenation is more exothermic (than IGNORE hydration

delocalised structure)

reacts with bromine/electrophiles/by addition ALLOW decolourises bromine (without a

catalyst/halogen carrier)

IGNORE more reactive without example

IGNORE alternating single and double bonds

Structure shown with molecular formula P3N3Cl6 1st mark

1st mark Meets criteria for 1st mark

• Each P bonded to 2 Cl atoms

• Each P bonded to N AND Cl

• Each N has at least 2 bonds

• Each Cl has 1 bond

2nd mark (dependent on 1st mark) ZERO marks

• Each N has 3 bonds N bonded to Cl

• Each P has 3 OR 5 bonds

IGNORE charges

Examples for 2 marks

Cl Cl Cl Cl

P N atom(s) with 1 bond only

N N N N

P P P P

Cl Cl Cl Cl

N N

Cl Cl Cl Cl

How to answer it

Chemistry of Phosphorus Compounds

What this question tests

This exam question evaluates your understanding of Brønsted-Lowry acid-base equilibria, molecular structure and bond angles, systematic nomenclature, balancing redox equations alongside oxidation numbers, stoichiometric mass calculations, and inorganic cyclic structures compared with benzene.

Question 4(a)

Phosphorus Acids, Structures, and Nomenclature

Part (i): Brønsted-Lowry Behaviour of H₂PO₄⁻

✅ Correct Answer

In equilibrium 1, H₂PO₄⁻ acts as a base because it accepts/gains an H⁺ ion to form H₃PO₄. In equilibrium 2, H₂PO₄⁻ acts as an acid because it donates/loses an H⁺ ion to form HPO₄²⁻.

❌ Common Errors

Students often mix up the equilibrium numbers or incorrectly reference equilibrium 3, which involves HPO₄²⁻ instead of H₂PO₄⁻.

Marks available: 2 (1 mark for identifying base behaviour with equilibrium 1; 1 mark for identifying acid behaviour with equilibrium 2).

Part (ii): Structure and Bond Angles of H₃PO₃

💡 Key Knowledge

H₃PO₃ is a diprotic acid (phosphonic acid). Its structure features a central P bonded to one H directly, two −OH groups, and a double-bonded oxygen (=O).

3 single covalent bonds around P (plus one P=O double bond).
2 bonds around each O atom.
Each H atom is bonded to an O atom (except the H directly bonded to P).

🧠 Exam Technique & Angles

Remember that lone pairs and regions of electron density dictate shape.
• O−P−O angle: 107° (Acceptable range: 104°–108°)
• P−O−H angle: 104.5° (Acceptable range: 104°–105°)

Marks available: 3 (1 mark for correct bonding connectivity, 1 mark for O−P−O angle, 1 mark for P−O−H angle).

Part (iii): Systematic Nomenclature

✅ Correct Answer

phosphoric(III) acid (Roman numerals must be included in parentheses).

❌ Common Errors

Writing phosphorous(III) acid or omitting the Roman numeral will lose the mark. Stock nomenclature requires oxidation states in brackets.

Marks available: 1

Question 4(b)

Phosphine Reactions and Redox Chemistry

Part (i): Combustion of Phosphine

✅ Correct Answer

4PH₃ + 8O₂ → P₄O₁₀ + 6H₂O
(Multiples such as 2PH₃ + 4O₂ → P₂O₅ + 3H₂O are also fully accepted).

🧠 Exam Technique

Always balance elements in order of complexity: balance phosphorus and hydrogen first using PH₃ and H₂O, then balance oxygen last.

Marks available: 1

Part (ii): Balancing Redox Equations & Oxidation Numbers

✅ Correct Answer

Balanced Equation:
6AgNO₃ + 1PH₃ + 3H₂O → 6Ag + 1H₃PO₃ + 6HNO₃

Explanation:
• Ag is reduced from +1 to 0.
• P is oxidised from −3 to +3.

❌ Common Errors

Forgetting to state both oxidation number changes or failing to explicitly link the changes to the terms 'reduced' and 'oxidised'.

Marks available: 3 (1 mark for correct balancing coefficients, 1 mark for silver oxidation state change, 1 mark for phosphorus oxidation state change).

Question 4(c)

Cyclic Phosphorus-Nitrogen Compounds (P₃N₃Cl₆)

Part (i): Formation Equation

✅ Correct Answer

3PCl₅ + 3NH₄Cl → P₃N₃Cl₆ + 12HCl

💡 Key Knowledge

This synthesis mimics inorganic condensation reactions forming ring systems analogous to organic nitrogen-phosphorus polymers.

Marks available: 1

Part (ii): Percentage by Mass Calculation

📐 Step-by-Step Calculation

Find the molar mass of P₃N₃Cl₆:
(3 × 30.97) + (3 × 14.01) + (6 × 35.45) = 92.91 + 42.03 + 212.70 = 347.64 g mol⁻¹ (using atomic masses: P=31.0, N=14.0, Cl=35.5 gives 348.0 g mol⁻¹).
Calculate percentage by mass of Phosphorus (P):
(3 × 31.0 / 348.0) × 100 = (93.0 / 348.0) × 100 = 26.72%

❌ Common Calculation Traps

• Forgetting to multiply the atomic mass of P by 3 (since there are 3 atoms in the formula).
• Failing to round to the requested 2 decimal places.

Marks available: 2 (1 mark for correct molar mass / working, 1 mark for final percentage rounded to 2 decimal places).

Part (iii) & (iv): Benzene Analogy and Structure

💡 Key Knowledge (Part iii)

To prove it has a Kekulé structure (localized alternating single and double bonds) rather than a fully delocalised ring like benzene, you can point out that:
• P−N bond lengths are different (whereas benzene has equal C−C bond lengths).
• Enthalpy change of hydrogenation is more exothermic than expected for delocalised systems.
• It reacts readily with electrophiles/bromine via addition.

🧠 Structure Rules (Part iv)

To draw the alternative structure of P₃N₃Cl₆ :
• Alternate P and N atoms in a 6-membered ring.
• Each P is bonded to 2 Cl atoms.
• Each N has at least 2 bonds, and each Cl has 1 bond.

Marks available: 3 total (1 mark for evidence of Kekulé structure in part iii; 2 marks for valid cyclic molecular structure satisfying valency in part iv).

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 3.1 The periodic table · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.