OCR A-Level Chemistry Unified chemistry (03), June 2022: Question 1
16 marks · Hard difficulty · Structured Questions
Answer questions on nitrogen compounds, brass isotope mass spectrometry, proton NMR of a diketone, and amino acid reactions and transition metal complexes.
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Question text
1 These questions are from different areas of chemistry.
(a) Ammonia, NH3, and ammonium nitrate, NH4NO3, are compounds of nitrogen.
(i) The boiling point of NH3 is –33 °C.
The boiling point of NH4NO3 is 210 °C.
Explain why there is a large difference in boiling points.
… [2]
(ii) Two students discuss the oxidation numbers in ammonium nitrate, NH4NO3.
One student claims that the two nitrogen atoms have the same oxidation number. The
other student disagrees and claims that the nitrogen atoms have different oxidation
numbers.
Explain with reasons which student is correct.
… [1]
(b) Brass is an alloy of copper and zinc.
The mass spectrum of a sample of brass is shown below.
45.64%
Relative
abundance 20.36%
16.82%
9.53%
6.27%
1.38%
60 61 62 63 64 65 66 67 68 69 70
m/z
The peaks at m/z = 63 and m/z = 65 are from the 63Cu and 65Cu isotopes of copper.
The remaining four peaks are from isotopes of zinc.
(i) What are the percentage compositions of copper and zinc in the brass sample?
Cu = … % Zn = … % [1]
(ii) Calculate the relative atomic mass of zinc in the sample of brass.
Give your answer to 2 decimal places.
relative atomic mass = … [2]
(c) The structure of an organic compound is shown below.
The protons are in four different environments, which are labelled 1–4.
CH2 CH2 O
H3C C C
O H
(i) Fill in the table to predict the splitting patterns in the proton NMR spectrum of the
organic compound.
Proton environment Splitting pattern
[2]
(ii) The table shows the chemical shifts for the peaks in the proton NMR spectrum at
proton environments 2 and 3.
Proton environment 2 3
Chemical shift, δ 2.5 ppm 3.6 ppm
Suggest why the peaks for proton environments 2 and 3 have the chemical shifts which
are shown in the table.
… [2]
(d) Glycine, H2NCH2COOH, is an α-amino acid.
(i) Glycine reacts with NaOH to form the salt H2NCH2COONa.
Glycine reacts with HCl to form the salt HOOCCH2NH3Cl.
The salts have different H–N–H bond angles.
State the different H–N–H bond angles and explain why they are different.
H2NCH2COONa H–N–H bond angle = … °
HOOCCH2NH3Cl H–N–H bond angle = … °
explanation …
… [3]
(ii) Glycine reacts with aqueous copper(II) ethanoate to form copper(II) glycinate,
Cu(H2NCH2COO)2, and ethanoic acid. Copper(II) glycinate is a complex which exists
as two square planar isomers.
Write an equation for this reaction and draw the structures of the two square planar
isomers of the complex Cu(H2NCH2COO)2.
equation
structures
[3]
Mark scheme
Show the mark scheme
Question Answer Marks AO Guidance
element
1 (a) (i) Structure and bonding 2 AO1.1 For intermolecular bonds/forces
NH3 is (simple) molecular/simple covalent/ ×2 ALLOW hydrogen bonds
/has intermolecular forces OR London Forces/induced dipole
AND forces/permanent dipole forces
NH4NO3 is ionic OR van der Waals’ forces
ALLOW NH4NO3 has molecular ions
NH + and NO – are molecular ions
Comparison of strength
Ionic bonds are stronger than intermolecular bonds / ORA
forces between molecules
OR ALLOW:
Ionic bonds need more energy to break than Intermolecular bonds are weak
intermolecular bonds AND ionic bonds are strong
(ii) (NH +) nitrogen has oxidation number of –3 1 AO1.2 Statement that one student is correct is
AND NOT required.
(NO –) nitrogen has oxidation number of +5 Implicit in answer
i.e. nitrogens are –3 AND +5 gets the mark ALLOW 3– AND 5+
BOTH signs essential
Question Answer 9 Marks AO Guidance
element
(b) (i) Cu: 66% AND Zn 34% 1 AO2.6
(ii) FIRST CHECK ANSWER ON THE ANSWER LINE 2 AO1.2 Refer to answer to 1b(i) for ECF from
If answer = 65.42 (to 2 DP) award 2 marks ×2 incorrect % composition of Zn and Cu
--------------------------------------------------------------------------------
Numerator from Zn isotopes
(64 × 16.82) + (66 × 9.53) + (67 × 1.38) + (68 × 6.27)
OR
2224.28
Relative atomic mass
Numerator ÷ 34 AND answer to 2 DP ECF ÷ by Zn % in b(i)
Mark ECF from numerator
----------------------------------------------------
(64 × 16.82) + (66 × 9.53) + (67 × 1.38) + (68 × 6.27) Common errors
22.24
= 65.42 (to 2 DP) ÷100 and answer to 2 DP
→ 1 mark for numerator
64.23
All 6 isotopes used → No marks
188.91
All 6 isotopes used
→ 6423 for numerator
÷34 and 2 DP → 1 mark by ECF
10 element
(c) (i) 2 AO1.2
Proton ×2 For quartet,
Splitting pattern ALLOW Quad….
environment
e.g. quadruplet, quadlet, quadret, etc
1 Triplet Triplet
AND For doublet, ALLOW duplet
2 Quartet quartet
ALLOW diagrams to show splitting pattern
3 Doublet Doublet e.g.
AND
4 Triplet triplet for triplet for quartet
ALLOW splitting patterns shown as
numbers
i.e. ‘3’ for triplet, ‘4’ for quartet
(c) (ii) Environment 2: 2 AO3.1
(Protons) adjacent to (one) C=O ×2 ALLOW HC–C=O
DO NOT ALLOW H–C=O
Environment 3:
(Protons) adjacent/between/surrounded by DO NOT ALLOW HC–O
2 C=O / a ketone AND aldehyde Simply reading δ = 3.6 ppm
OR from data sheet)
C=O on both sides
IGNORE ‘next to 2 Os’
11 element
(d) (i) Bond angles 3 AO1.2
H2NCH2COONa, bond angle = 107º ×3 ALLOW 107 ± 0.5
AND
HOOCCH2NH3Cl, bond angle = 109.5º ALLOW 109 OR 110º
Number of electron pairs
Mark independently of angles
In NaOH/107º, (NH2 has) 3 bonded pairs / 3 bonds ALLOW NH2 has 4 pairs, one of which is a
AND lone pair
1 lone pair
For bonded pairs/bonds
ALLOW bonded groups, atoms,
elements, regions
In HCl/109.5º, (NH + has) 4 bonded pairs / 4 bonds Bonded essential
IGNORE electron region OR electron
density
IGNORE NH3 has no lone pairs
IGNORE lone pairs repel more
(than bonded pairs)
IGNORE shapes, even if wrong
ALLOW bp for bonded pair
and lp for lone pair
(ii) Equation: 3
2 H2NCH2COOH + Cu(CH3COO)2 ALLOW molecular formulae or mixture,
→ Cu(H2NCH2COO)2 + 2 CH3COOH AO2.6 e.g. 2C2H5NO2 + CuC4H6O4
→ CuC4H8N2O4 + 2C2H4O2
IGNORE charges
element
i.e. IGNORE wrong or missing charges in
ionic compounds if formula is correct/ e.g.
ALLOW Cu(CH COO–) , Cu+(CH COO–)
32 3 2
Structures ALLOW any combination of skeletal OR
AO2.5 structural OR displayed formula as long as
×2 unambiguous
IGNORE charges
ALLOW arc to represent –CH2– between:
C of C=O and NH2
ALLOW 1 mark for 2 ‘correct’ structures
OR shown as tetrahedral e.g.
IGNORE missing Hs on C, e.g.
Ligands must shown as bidentate rings
IGNORE connectivity for NH2
BUT connectivity must be to O of COO
How to answer it
Multi-Topic Chemistry Assessment Study Guide
This comprehensive multi-topic OCR A-Level paper tests your core knowledge across Physical, Inorganic, Organic, and Transition Element chemistry. Key competencies assessed include explaining boiling point trends based on bonding types, determining oxidation numbers in ionic lattices, interpreting mass spectra for isotopic abundance calculations, predicting proton NMR splitting patterns and chemical shifts, applying VSEPR theory to amine/ammonium bond angles, and drawing complex stereoisomers (square planar cis/trans complexes).
Part (a): Nitrogen Compounds, Bonding & Oxidation Numbers
✅ Correct Answers
- (a)(i): NH₃ has simple molecular structure with intermolecular forces (hydrogen bonding), whereas NH₄NO₃ is an ionic lattice. Ionic bonds are much stronger and require significantly more energy to break than intermolecular forces.
- (a)(ii): The disagreeing student is correct. The two nitrogen atoms have different oxidation numbers: the nitrogen in the ammonium ion (NH₄⁺) is -3 , and in the nitrate ion (NO₃⁻) it is +5 .
💡 Key Knowledge
- Always differentiate clearly between the type of bonding/forces present (intermolecular vs. ionic) rather than just stating "ionic bonds are stronger than covalent bonds" (remember, the covalent bonds inside the ions do not break during boiling).
- Oxidation state rules dictate that oxygen is usually -2 and hydrogen is +1, allowing you to easily solve for nitrogen in complex ions. Both signs (+ and -) are essential for full marks.
Part (b): Mass Spectrometry & Relative Atomic Mass of Brass
✅ Correct Answers
- (b)(i): Cu = 66% , Zn = 34% (calculated from summed percentage abundances: Cu peaks at 63 and 65 sum to 45.64 + 20.36 = 66%).
- (b)(ii): Relative atomic mass = 65.42 (to 2 decimal places).
📐 Step-by-Step Calculation
- Identify the 4 zinc isotope peaks from the spectrum: 64 (16.82%), 66 (9.53%), 67 (1.38%), and 68 (6.27%).
- Calculate the numerator (sum of [isotopic mass × percentage abundance]):
(64 × 16.82) + (66 × 9.53) + (67 × 1.38) + (68 × 6.27) = 2224.28 - Divide by the total percentage abundance of zinc (34%):
2224.28 / 34 = 65.420588... - Round to 2 decimal places to get 65.42 .
❌ Common Errors & Traps
- Dividing by 100 instead of dividing by the total percentage of zinc (34%) is a major trap.
- Including copper isotopes (m/z 63 and 65) in the zinc calculation will completely ruin your numerator and denominator ratios.
Part (c): Proton NMR Spectroscopy
✅ Correct Answers
- (c)(i) Splitting Patterns:
Environment 1: Triplet
Environment 2: Quartet
Environment 3: Doublet
Environment 4: Triplet - (c)(ii) Chemical Shift Explanations:
Environment 2 is adjacent to one C=O group .
Environment 3 is adjacent to two C=O groups (or sandwiched between two carbonyls / an aldehyde-like deshielding environment), which causes greater deshielding and a higher delta value (3.6 ppm vs 2.5 ppm).
🧠 Exam Technique
- Use the n + 1 rule for splitting: count the number of equivalent protons on adjacent carbon atoms and add 1.
- For chemical shifts, explicitly mention electronegative environments or electron-withdrawing groups (like anisotropic C=O carbonyl carbons) causing deshielding , which shifts peaks to higher ppm values.
Part (d): Amino Acids, VSEPR Theory & Transition Metal Isomerism
✅ Correct Answers
- (d)(i) Bond Angles:
H₂NCH₂COONa bond angle = 107° (Pyramidal, due to 3 bond pairs and 1 lone pair on nitrogen).
HOOCCH₂NH₃Cl bond angle = 109.5° (Tetrahedral, due to 4 bond pairs and 0 lone pairs on the protonated ammonium nitrogen NH₃⁺ ). - (d)(ii) Equation:
2 H₂NCH₂COOH + Cu(CH₃COO)₂ → Cu(H₂NCH₂COO)₂ + 2 CH₃COOH - (d)(ii) Structures (Stereoisomerism):
Draw two square planar cis/trans isomers of Cu(H₂NCH₂COO)₂ . Ensure that the bidentate amino acid ligands coordinate via the nitrogen atom of the amino group and an oxygen atom of the carboxylate group, forming 5-membered chelate rings with copper. Show cis (similar donor atoms at 90°) and trans (similar donor atoms at 180°) arrangements clearly.
💡 Key Knowledge & VSEPR
- Lone pairs exert greater repulsive force than bonding pairs, compressing bond angles from the tetrahedral 109.5° down to roughly 107° in neutral amines ( -NH₂ ).
- When protonated to form -NH₃⁺ , the lone pair is used to form a dative covalent bond with H⁺, leaving 4 bonding pairs and restoring the tetrahedral angle of 109.5°.
- In transition metal complex isomers, verify that connectivity is maintained through the correct donor atoms (N bonded to Cu, O bonded to Cu) to avoid connectivity errors penalized by examiners.
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 5.3 Transition elements · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.