OCR A-Level Chemistry Unified chemistry (03), June 2023: Question 6

23 marks · Hard difficulty · Structured Questions

Predict bond angles in nitric acid, write reaction equations including redox and acid-base reactions, calculate Kc for sulfur trioxide equilibrium, and determine formulae of compounds in reactions involving sulfuric acid.

Practise this question

Question

A structured multi-part question about nitric acid, hydrochloric acid, and sulfuric acid. Part (a) shows the structure of nitric acid and asks for bond angles and reasoning. Part (b) asks for an equation reacting dilute nitric acid with aluminium oxide and a dot-and-cross diagram for the nitrate ion. Part (c) covers the reaction of aqua regia with gold, balancing a half-equation and determining reduction products. Part (d) provides a chemical equilibrium calculation involving sulfur dioxide, oxygen, and sulfur trioxide. Part (e) describes three reactions involving sulfuric acid, requiring identification of compounds D, E, and F and construction of equations with organic structures.
Question text

6 This question is about nitric acid, hydrochloric acid and sulfuric acid.

(a) Nitric acid has 2 single covalent bonds, 1 double covalent bond and 1 dative covalent bond

as shown below.

H O

N O

O

Nitric acid

Predict the H–O–N and O–N–O bond angles in nitric acid.

Explain your reasoning.

… [4]

(b) Dilute nitric acid reacts with aluminium oxide to form a solution of aluminium nitrate.

(i) Write an equation for this reaction.

… [2]

(ii) The solution contains nitrate ions, NO –.

Draw a ‘dot-and-cross’ diagram for the NO – ion.

Use a different symbol for the extra electron.

[2]

(c) A mixture of concentrated nitric and hydrochloric acid is called ‘aqua regia’. Aqua regia can

dissolve gold.

The reaction of aqua regia with gold is a redox reaction which forms chlorauric acid, HAuCl4.

(i) Balance the half-equation for the oxidation process in this reaction.

Au + … HCl … H+ + AuCl – + … e–

[1]

(ii) In the reduction process in this reaction, HNO and H+ react together to form 2 oxides:

X (Mr = 30) and Z (Mr = 18).

Determine the formulae of X and Z and write the half-equation for this reduction.

X = …

Z = …

half-equation … [3]

(d) In the UK, most sulfuric acid, H2SO4, is manufactured by the Contact process.

One stage in the Contact process involves the equilibrium between sulfur dioxide, oxygen

and sulfur trioxide.

2SO2(g) + O2(g) 2SO3(g)

This equilibrium is investigated:

Step 1 5.82 × 10–2 mol of SO is mixed with 7.40 × 10–2 mol of O in a 2.00 dm3 container.

Step 2 The container is sealed and allowed to reach equilibrium at constant temperature.

Step 3 At equilibrium, 5.20 × 10–2 mol of SO is formed.

Determine the equilibrium concentrations and calculate Kc, including units.

Kc = … units … [5]

(e)* Three reactions involving sulfuric acid are shown below.

Reaction 1

Dilute sulfuric acid is reacted with nickel(II) hydroxide to form a green solution.

The solvent is allowed to evaporate leaving hydrated crystals of compound D, with the

percentage composition by mass: Ni, 22.33%; S, 12.20%; O, 60.87%; H, 4.60%.

Reaction 2

Concentrated sulfuric acid is reacted with hydrogen bromide, HBr, to form three products:

• an element which exists as diatomic molecules

• a gaseous compound E

• a liquid.

At RTP, 1.00 dm3 of compound E has a mass of 2.67 g.

Reaction 3

Concentrated sulfuric acid acts as a catalyst when 2-hydroxypropanoic acid reacts to form

compound F (Mr = 144).

In this reaction, 2 mol of 2-hydroxypropanoic acid forms 1 mol of compound F and 2 mol of

water.

Identify compounds D, E and F and construct equations for the reactions.

Show structures for any organic compounds. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme The official mark scheme providing answers, acceptable alternative responses, mark allocations, and guidance notes for each part of question 6, including expected bond angles, equations, dot-and-cross electron diagrams, redox half-equations, Kc equilibrium calculations with common error checks, and level-of-response criteria for the organic synthesis identification question.

Question Answer Marks AO Guidance

element

6 (a) 4 Throughout,

• IGNORE names of shapes

(even if wrong)

• IGNORE ‘electrons repel’

• DO NOT ALLOW ‘atoms repel’

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

H–O–N

104.5º AO1.2 ALLOW 104–105º

2 bonded pairs/regions AND 2 lone pairs (around O) ALLOW lp for lone pair (of electrons)

AND lone pairs repel more AO2.1 bp for bonding pair (of electrons)

Independent of bond angle ‘bond’ for ‘bonded pair’

IGNORE electron density

O–N–O

120º AO1.2 ALLOW 115–125º

3 bonded regions/pairs (around N) AO2.1 ALLOW 3 bonded areas/environments

Independent of bond angle 3 regions/areas of electron density

3 bonded groups

ALLOW 2 bonded pairs and 1 double bond

OR 2 bonded pairs and 1 bonded region

(b) (i) Al2O3 + 6HNO3 → 2Al(NO3)3 + 3H2O 2 ALLOW multiples

Any THREE species correct AO2.5 IGNORE state symbols (even if wrong)

Correct balanced equation AO2.6

ALLOW ionic equation

DO NOT ALLOW more than 4 species in equation Al O + 6H+ → 2Al3+ + 3H O

23 2

Mark using same criteria

element

(b) (ii) 🐀 2 AO2.1 NOT REQUIRED

Always 5 • Charge (‘–‘)

around N AO2.5 • Brackets

🐀 🐀 • Circles

unbonded paired in O–N • N and O symbols

O O IGNORE inner shells

N O N O ALLOW rotated diagram

OR

O O In N=O bond, ALLOW sequence × × • •

ALLOW non-bonding electrons unpaired

❌ ❌ 🐀

32 + 1 ALLOW dot and cross labels swapped:

around N around N i.e. • for O electrons and for N electrons

🐀❌🐀

= N electron

= O electron

= extra electron

1st mark: 8 Electrons around N as above

1 single covalent bond,

1 dative covalent bond

1 double bond

2nd mark: 8 electrons around each O

AND 6 O electrons around each O

Only award 2nd mark if 1st mark awarded

NO ECF

element

(c) (i) Au + 4 HCl → 4 H+ + AuCl – + 3 e– ✓ 1 AO1.2

(c) (ii) Formulae 3 AO3.1 If X and Z in wrong order award 1 out of 2

3 formula marks

X = NO ✓ i.e. X = H2O and Z = NO 1 mark

Z = H2O ✓

Equation Independent from ID of X and Z ALLOW multiples

HNO + 3 H+ + 3 e– → NO + 2 H O

OR

NO – + 4 H+ + 3 e– → NO + 2 H O ✓

CHECK BELOW ANSWER SPACE FOR RESPONSES

element

(d) FIRST CHECK THE ANSWERS ON ANSWER LINE 5 Use of fractions is fine but final answer

If Kc value = 2931 OR 2930 award 4 calc marks MUST be shown using normal numbers

If units = dm3 mol–1 OR mol–1 dm3 award 1 unit mark ---------------------------------------------------------

--------------------------------------------------------------------------------- COMMON ERRORS

SO2 and O2 equilibrium moles Kc = 1,465 (2,930/2) → 3 calc marks

n(SO ) = 6.20 10–3 ( 5.82 10–2 – 5.20 x 10–2 ) AO2.6 Moles not converted to concentration (No ÷2)

5.20 10–2 3 (5.20 10–2)2

AND n(O ) = 4.80 10–2 ✓ (7.4 10–2 – )

22 (6.2 10–3)2 (4.80 10–2)

Equilibrium concentrations (moles 2) --------------------------------------------

6.20 10–3 Kc = 21.6 → 3 calc marks

[SO ] = 3.10 10–3 (mol dm–3) Original values used,

(2.60 10–2)2

4.80 10–2

AND [O ] = 2.40 10–2 (mol dm–3) (2.91 10–2)2 (3.70 10–2)

–2 --------------------------------------------

–2 5.20 10 –3

AND [SO3] = 2.60 10 ✓ (mol dm ) Kc = 10.8 → 2 calc marks

2 Original values used and no ÷2,

Kc calculation (5.20 10–2)2

(2.60 10–2)2

(5.82 10–3)2 (7.40 10–2)

Kc = –3 2 –2

(3.10 10 ) (2.40 10 ) --------------------------------------------

Kc = 732.74 → 3 calc marks

= 2,930 OR 2,931 At least 3 SF required AO1.2 2 instead of ÷ 2 for concentration

2 (0.104)2

Calc value from unrounded values: 2,930.974679 (0.0124)2 (0.096)

Units --------------------------------------------

dm3 mol–1 DO NOT ALLOW dm3 mol–

Kc = 112729.8 → 3 calc marks

2.60 10–2 not squared

(2.60 10–2)

For units, ALLOW ECF using incorrect Kc expression (3.10 10–3)2 (2.40 10–2)

Units must match Kc expression used --------------------------------------------

K = 3.41… 10–4 Calculator 3.41183432 10–4

c

Inverted Kc → 3 calc marks

(3.10 10–3)2 (2.40 10–2)

Units mol dm–3

(2.60 10–2)2

27 element

(e)* Please refer to the marking instructions on page 6 of this mark 6 AO3.1 Indicative scientific points may include:

scheme for guidance on how to mark this question. 3

Identify of D, E and F

Level 3 (5–6 marks) AO3.2 • D: NiSO •6H O

• Reaches a comprehensive conclusion to determine all three 3

OR NiSO4(H2O)6 OR NiSO10H12

correct formulae of D, E AND F

• AND constructs most equations with few errors

• E: SO2

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

structured.

The information presented is relevant and substantiated. • F: Cyclic diester

O

Level 2 (3–4 marks)

• Reaches a comprehensive conclusion to determine two O

correct formulae of D, E AND F

• AND constructs some equations with some errors O

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some O

evidence. OR unsaturated ester/acid

O CH3

Level 1 (1–2 marks)

• Determines a correct formula for one of D, E AND F O COOH

• AND provides some evidence to support the formula

There is an attempt at a logical structure with a line of reasoning. The OR unsaturated acid anhydride

information is in the most part relevant. O O

0 marks No response or no response worthy of credit. O

EQUATIONS SHOULD BE USED TO INFORM THE OH

COMMUNICATION STRAND OR cyclic acid anhydride

See next page for details O

O

CHECK TOP OF QUESTION FOR RESPONSES

O

IGNORE CONNECTIVITY FOR F O

element

SUMMARY Equations

H2SO4 + Ni(OH)2 → NiSO4 + 2H2O

Setting the level OR

For Level 3 (5–6 marks), H2SO4 + Ni(OH)2 + 4H2O → NiSO4•6H2O

• All 3 identified: D, E and F

• Most equations For equation

For Level 2 (3–4 marks), ALLOW NiSO4•6H2O OR NiSO4(H2O)6

• 2 identified from D, E and F -----------------------------------------------------

H2SO4 + 2HBr → Br2 + SO2 + 2H2O

• 2 equations

For Level 1 (1–2 marks), O

• 1 identified from D, E and F

• Evidence O

---------------------------------------------------------- 2CH3CH(OH)COOH + 2H2O

Evidence to support a formula for Level 1 O

Molar ratios of D O

Ni : S : O : H OR

22.33 12.20 60.87 4.60 O CH3

58.7 : 32.1 : 16.0 : 1.0

2CH3CH(OH)COOH

0.38 : 0.38 : 3.80 : 4.60 + 2H2O

O COOH

1 : 1 10 12 OR NiSO10H12 OR

--------------------------------------------------------- O O

H432/03 Molar mass of E Mark Scheme June 2023

Molar mass = 2.67 24 = 64(.08) g mol–1 2CH CH(OH)COOH + 2H2O

O

OH

If structure of F is shown, ALLOW equation

using molecular formulae,

e.g. 2 C3H6O3 → C6H8O4 + 2H2O

BLANK PAGE

How to answer it

Inorganic Chemistry & Calculations Masterclass

What this question tests

This comprehensive multi-topic question assesses your mastery of VSEPR theory (bond angles and shapes), dot-and-cross bonding diagrams, redox half-equations and balancing, equilibrium calculations involving Kc expressions and unit manipulation, and complex organic-inorganic reaction synthesis involving stoichiometry and structural elucidation.

Part (a) • 4 Marks

Nitric Acid Bond Angles & Shapes

✅ Correct Answer

  • H-O-N bond angle: 104.5° (Accept 104°–105°)
  • O-N-O bond angle: 120° (Accept 115°–125°)

💡 Key Knowledge (VSEPR)

  • Around O atom: 2 bonded pairs and 2 lone pairs give a bent/non-linear geometry, compressing the tetrahedral angle to ~104.5°.
  • Around N atom: 3 regions of electron density (1 single bond, 1 double bond, 1 dative covalent bond) with no lone pairs give trigonal planar geometry at 120°.

🧠 Exam Technique

  • Always state the angle and explicitly count the number of bonding pairs and lone pairs/regions of electron density around the central atom.
  • Mention that electron pairs repel each other to positions of maximum separation.

❌ Common Errors

  • Confusing the oxygen coordination with nitrogen and writing 104.5° for both.
  • Treating the double bond or dative covalent bond as multiple regions affecting the planar geometry.
Part (b) • 4 Marks Total

Reaction with Aluminium Oxide & Nitrate Ion Dot-and-Cross

✅ Correct Answers

(i) Equation: Al₂O₃ + 6HNO₃ → 2Al(NO₃)₃ + 3H₂O

(ii) Nitrate Ion (NO₃⁻): Dot-and-cross diagram showing total 8 electrons around N (1 single, 1 dative, 1 double bond) and 8 electrons around each O atom, including the extra electron symbol.

🧠 Exam Technique & Guidance

  • For part (b)(i), state symbols are ignored unless incorrect, but balanced stoichiometry is essential. Ionic equations are also accepted ( Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O ).
  • For part (b)(ii), use a distinct symbol (e.g., a shaded circle or square) for the extra electron. Do not worry about inner shells—focus strictly on valence shells.
Part (c) • 4 Marks Total

Aqua Regia Redox Chemistry

✅ Correct Answers

(i) Oxidation Half-Equation:
Au + 4HCl → H⁺ + AuCl₄⁻ + 3e⁻ (or starting with Au + 4Cl⁻ → AuCl₄⁻ + 3e⁻ )

(ii) Formulae & Reduction Half-Equation:
X = NO , Z = H₂O
Half-equation: HNO₃ + 3H⁺ + 3e⁻ → NO + 2H₂O (or nitrate ionic equivalent)

❌ Common Errors

  • Swapping the identities of X and Z in part (ii). If X and Z formulae are inverted, you only secure 1 out of 2 formula marks.
  • Failing to balance charges with electrons ( 3e⁻ ) in reduction and oxidation steps.
Part (d) • 5 Marks

Contact Process Equilibrium & Kc Calculation

📐 Step-by-Step Calculation Guide

  1. Find moles of SO₃ at equilibrium: Given as 5.20 × 10⁻² mol .
  2. Determine moles of SO₂ reacted and remaining: Initial SO₂ = 5.82 × 10⁻² mol . From stoichiometry (2:2 ratio), moles of SO₂ reacted = moles of SO₃ formed = 5.20 × 10⁻² mol .
    Equilibrium n(SO₂) = 5.82 × 10⁻² - 5.20 × 10⁻² = 6.20 × 10⁻³ mol .
  3. Determine moles of O₂ reacted and remaining: Initial O₂ = 7.40 × 10⁻² mol . From stoichiometry (2SO₂ : 1O₂ ratio), moles of O₂ reacted = 5.20 × 10⁻² / 2 = 2.60 × 10⁻² mol .
    Equilibrium n(O₂) = 7.40 × 10⁻² - 2.60 × 10⁻² = 4.80 × 10⁻² mol .
  4. Calculate equilibrium concentrations (Container volume = 2.00 dm³):
    • [SO₂] = (6.20 × 10⁻³) / 2.00 = 3.10 × 10⁻³ mol dm⁻³
    • [O₂] = (4.80 × 10⁻³) / 2.00 = 2.40 × 10⁻³ mol dm⁻³
    • [SO₃] = (5.20 × 10⁻³) / 2.00 = 2.60 × 10⁻³ mol dm⁻³
  5. Substitute into Kc expression and evaluate:
    Kc = [SO₃]² / ([SO₂]²[O₂])
    Kc = (2.60 × 10⁻³)² / ((3.10 × 10⁻³)² × (2.40 × 10⁻³)) = 2930 dm³ mol⁻¹ (to 3 Significant Figures).

❌ Calculation Traps

  • Forgetting to divide equilibrium moles by the container volume ( 2.00 dm³ ) to get concentrations.
  • Incorrectly handling the stoichiometric factor of 2 for O₂ when calculating moles reacted.
  • Incorrect unit derivation: dm³ mol⁻¹ . Do not write dm³mol⁻¹ with missing spaces if penalized by specific board formatting, though standard notation is accepted.

🧠 Significant Figures Note

Your final calculated value must be rounded to 3 significant figures ( 2930 or 2.93 × 10³ ) to secure the accuracy mark.

Part (e)* • 6 Marks

Organic & Inorganic Synthesis Elucidation (Compounds D, E, F)

✅ Correct Identifications

  • Compound D: NiSO₄·6H₂O (Hydrated nickel(II) sulfate)
  • Compound E: SO₂ (Sulfur dioxide gas, derived from mass calculation 2.67 g at RTP)
  • Compound F: Cyclic diester or unsaturated acid/anhydride structures (e.g., lactide dimer structure from 2-hydroxypropanoic acid) with Mr = 144 .

💡 Analytical Breakdown

  • Reaction 1: Nickel(II) hydroxide reacts with dilute sulfuric acid to form green nickel sulfate crystals NiSO₄·6H₂O . Percentage mass analysis confirms water of crystallization stoichiometry.
  • Reaction 2: Concentrated sulfuric acid oxidises HBr to form diatomic bromine ( Br₂ ), sulfur dioxide gas ( E ), and water. Molar mass check of E confirms SO₂ .
  • Reaction 3: Condensation of 2 mol of 2-hydroxypropanoic acid (lactic acid) eliminates 2 mol of water to form 1 mol of cyclic ester F ( C₆H₈O₄ ).

🧠 Level of Response Strategy

To achieve Level 3 (5–6 marks), you must correctly deduce all three formulae (D, E, and F) and provide clear, balanced equations for the accompanying reactions with correct organic connectivity shown.

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 3.2 Physical chemistry · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.