An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V2) Cambridge International AS Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all 40 multiple-choice questions. For each question, choose the correct answer from A, B, C, or D using soft pencil on the separate answer sheet.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
A sample containing \(0.120\text{ g}\) of a volatile liquid hydrocarbon, \(\text{Y}\), was vaporised at \(373\text{ K}\) and \(1.00 \times 10^5\text{ Pa}\). The vapour occupied a volume of \(44.5\text{ cm}^3\).
What is the relative molecular mass, \(M_r\), of \(\text{Y}\)? [The molar gas constant, \(R = 8.31\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1}\)]
A.27.8
B.83.6
C.112
D.836
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Worked solution
Using the ideal gas equation: \[pV = nRT = \frac{m}{M_r}RT\]
Rearranging for \(M_r\): \[M_r = \frac{mRT}{pV}\]
Convert units: - \(m = 0.120\text{ g}\,\) (if using \(p\) in \(\text{Pa}\) and \(V\) in \(\text{m}^3\), \(n\) is in \(\text{mol}\), so \(m\) in \(\text{g}\) gives \(M_r\) in \(\text{g}\,\text{mol}^{-1}\)) - \(p = 1.00 \times 10^5\text{ Pa}\,\) - \(V = 44.5 \times 10^{-6}\text{ m}^3\) - \(T = 373\text{ K}\) - \(R = 8.31\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1}\)
What is the standard enthalpy change of formation, \(\Delta H^\ominus_f\), of propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH(l)}\), in \(\text{kJ}\,\text{mol}^{-1}\)?
A.-510.9
B.+510.9
C.-847.7
D.+847.7
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Worked solution
The equation for the standard enthalpy change of formation of propanoic acid is: \[3\text{C(s)} + 3\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(l)}\]
Using enthalpy changes of combustion: \[\Delta H^\ominus_f = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})\]
Award 1 mark for option A. [1] M1: Correct use of Hess's Law cycle with combustion data: \(\Delta H^\ominus_f = 3(-393.5) + 3(-285.8) - (-1527.0) = -510.9\text{ kJ}\,\text{mol}^{-1}\).
Question 3 · multiple_choice
1 marks
Two solid Period 3 chlorides, \(\text{E}\) and \(\text{F}\), are each added separately to excess water.
- When \(\text{E}\) is added to water, it dissolves without reaction to give a solution with a pH of 7. - When \(\text{F}\) is added to water, vigorous hydrolytic fuming occurs, giving an acidic solution with a pH of 2.
Which row correctly identifies \(\text{E}\) and \(\text{F}\)?
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Worked solution
Period 3 chlorides behave differently with water: - \(\text{NaCl}\) undergoes simple hydration (dissolves) forming neutral \(\text{Na}^+\text{(aq)}\) and \(\text{Cl}^-\text{(aq)}\), pH = 7. - \(\text{MgCl}_2\) dissolves with slight hydrolysis to give a slightly acidic solution (pH \(\approx\) 6.5). - \(\text{AlCl}_3\), \(\text{SiCl}_4\), and \(\text{PCl}_5\) undergo vigorous hydrolysis releasing \(\text{HCl}\) gas (white fumes) and forming strongly acidic solutions (pH 1–2).
Among the choices: - \(\text{E} = \text{NaCl}\) (pH = 7, simple dissolution) - \(\text{F} = \text{SiCl}_4\) (vigorous hydrolysis giving \(\text{SiO}_2\cdot x\text{H}_2\text{O}\) / \(\text{Si(OH)}_4\) and \(\text{HCl}\) fumes, pH \(\approx 1-2\)).
Therefore, row C is correct.
Marking scheme
Award 1 mark for option C. [1] M1: Identifies NaCl as the neutral dissolving chloride and SiCl4 as the vigorously hydrolysing covalent chloride.
Question 4 · multiple_choice
1 marks
Solid sodium iodide is treated with concentrated sulfuric acid.
Which products are formed by the redox reactions that take place?
A.\(\text{HI}\) and \(\text{NaHSO}_4\) only
B.\(\text{I}_2\) and \(\text{SO}_2\) only
C.\(\text{I}_2\), \(\text{HI}\), and \(\text{SO}_2\) only
D.\(\text{I}_2\), \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\)
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Worked solution
Iodide ions, \(\text{I}^-\), are very strong reducing agents and reduce concentrated \(\text{H}_2\text{SO}_4\) (where sulfur is in oxidation state +6) to: - \(\text{SO}_2\) (sulfur in +4) - \(\text{S}\) (sulfur in 0, yellow solid) - \(\text{H}_2\text{S}\) (sulfur in -2, foul-smelling gas)
In these redox processes, iodide is oxidised to iodine, \(\text{I}_2\) (purple vapour / dark solid). (Note: \(\text{HI}\) and \(\text{NaHSO}_4\) are products of the acid-base reaction, not redox products).
Therefore, the set of redox products is \(\text{I}_2\), \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\).
Marking scheme
Award 1 mark for option D. [1] M1: Correctly identifies all reduction products of sulfuric acid by iodide along with elemental iodine.
Question 5 · multiple_choice
1 marks
The first six successive ionisation energies of an element \(\text{Q}\), in \(\text{kJ}\,\text{mol}^{-1}\), are: \[578, \quad 1817, \quad 2745, \quad 11578, \quad 14842, \quad 18379\]
What is the formula of the chloride formed when element \(\text{Q}\) reacts completely with excess chlorine gas?
A.\(\text{QCl}_2\)
B.\(\text{QCl}_3\)
C.\(\text{QCl}_4\)
D.\(\text{Q}_2\text{Cl}_3\)
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Worked solution
Looking at the successive ionisation energies: - \(IE_1 = 578\) - \(IE_2 = 1817\) - \(IE_3 = 2745\) - \(IE_4 = 11578\) (large jump between the 3rd and 4th ionisation energies, an increase of \(8833\text{ kJ}\,\text{mol}^{-1}\))
The large jump after the 3rd ionisation energy indicates that the 4th electron is removed from an inner, fully-filled electron shell. Thus, element \(\text{Q}\) has 3 valence electrons and belongs to Group 13 of the Periodic Table.
Elements in Group 13 (such as aluminium) form stable chlorides with the formula \(\text{QCl}_3\).
Marking scheme
Award 1 mark for option B. [1] M1: Identifies large jump between 3rd and 4th IE, placing element Q in Group 13 to give QCl3.
Question 6 · multiple_choice
1 marks
An alkene \(\text{W}\) has the molecular formula \(\text{C}_6\text{H}_{12}\). When \(\text{W}\) is heated under reflux with acidified potassium manganate(VII), the only organic product obtained is propanone, \(\text{CH}_3\text{COCH}_3\).
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Worked solution
Hot concentrated acidified \(\text{KMnO}_4\) cleaves carbon-carbon double bonds: - \(=\text{CH}_2\) is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\) - \(=\text{CHR}\) is oxidised to a carboxylic acid, \(\text{RCOOH}\) - \(=\text{CR}_1\text{R}_2\) is oxidised to a ketone, \(\text{R}_1\text{COR}_2\)
Since propanone, \((\text{CH}_3)_2\text{C}=\text{O}\), is the ONLY organic product formed, both sides of the double bond must be identical and have two methyl groups attached to the double-bonded carbon. Thus, \(\text{W}\) must be 2,3-dimethylbut-2-ene: \[(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\]
Cleavage yields 2 molecules of \(\text{CH}_3\text{COCH}_3\).
Marking scheme
Award 1 mark for option C. [1] M1: Identifies 2,3-dimethylbut-2-ene as the symmetrical alkene that cleaves to give only propanone.
Question 7 · multiple_choice
1 marks
An organic compound \(\text{Z}\), \(\text{C}_4\text{H}_{10}\text{O}\), gives a yellow precipitate when warmed with alkaline aqueous iodine.
B.Compound \(\text{Z}\) can be oxidised to a carboxylic acid under reflux with acidified potassium dichromate(VI).
C.Dehydration of \(\text{Z}\) gives only one alkene product.
D.Compound \(\text{Z}\) is 2-methylpropan-2-ol.
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Worked solution
A positive tri-iodomethane (iodoform) test with alkaline aqueous iodine requires the presence of a \(\text{CH}_3\text{CH(OH)}-\) group (in alcohols) or a \(\text{CH}_3\text{C(=O)}-\) group (in carbonyls).
Since \(\text{Z}\) has formula \(\text{C}_4\text{H}_{10}\text{O}\), it is a saturated alcohol (alkanol). Containing \(\text{CH}_3\text{CH(OH)}-\), the 4-carbon alcohol must be butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
Let us evaluate the statements: A: Butan-2-ol has a chiral centre at carbon-2 (attached to \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\)), so it exhibits optical isomerism. (Correct) B: Butan-2-ol is a secondary alcohol, which oxidises to butanone (a ketone), not a carboxylic acid. (Incorrect) C: Dehydration of butan-2-ol over hot \(\text{Al}_2\text{O}_3\) yields but-1-ene, \(\text{cis}\)-but-2-ene, and \(\text{trans}\)-but-2-ene (three alkene products). (Incorrect) D: 2-methylpropan-2-ol is tertiary and does not give a positive tri-iodomethane test. (Incorrect)
Marking scheme
Award 1 mark for option A. [1] M1: Identifies compound Z as butan-2-ol and deduces that it possesses a chiral centre and exhibits optical isomerism.
Question 8 · multiple_choice
1 marks
The mass spectrum of an unknown organic compound shows a molecular ion peak, \(\text{M}^+\), at \(m/e = 72\) with a relative abundance of \(28.4\%\), and an \([\text{M}+1]^+\) peak at \(m/e = 73\) with a relative abundance of \(1.25\%\).
How many carbon atoms are present in one molecule of this compound?
A.3
B.4
C.5
D.6
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Worked solution
The number of carbon atoms, \(n\), in an organic compound can be calculated using the relative abundances of the \(\text{M}^+\) and \([\text{M}+1]^+\) peaks: \[n = \frac{100 \times \text{abundance of }[\text{M}+1]^+}{1.1 \times \text{abundance of }\text{M}^+}\]
Substituting the given values: \[n = \frac{100 \times 1.25}{1.1 \times 28.4} = \frac{125}{31.24} = 4.001\]
Therefore, there are 4 carbon atoms in the molecule.
Marking scheme
Award 1 mark for option B. [1] M1: Correct application of the formula \(n = \frac{100 \times 1.25}{1.1 \times 28.4} = 4\).
Question 9 · Multiple Choice
1 marks
A \(20.0\text{ cm}^3\) sample of a gaseous hydrocarbon was mixed with \(150\text{ cm}^3\) of oxygen (an excess) and completely combusted.
After cooling to room temperature, the volume of the remaining gaseous mixture was \(110\text{ cm}^3\).
When this mixture was shaken with excess aqueous sodium hydroxide, the volume decreased to \(30.0\text{ cm}^3\).
All gas volumes were measured at the same temperature and pressure.
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_6\)
B.\(\text{C}_3\text{H}_8\)
C.\(\text{C}_4\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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Worked solution
Let the hydrocarbon formula be \(\text{C}_x\text{H}_y\).
Which formula represents the chloride formed when element \(X\) reacts with excess chlorine?
A.\(X\text{Cl}\)
B.\(X\text{Cl}_2\)
C.\(X\text{Cl}_3\)
D.\(X\text{Cl}_4\)
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Worked solution
1. Look for the largest ratio / jump between successive ionisation energies: - Between 1st and 2nd: \(1817 / 578 \approx 3.1\) - Between 2nd and 3rd: \(2745 / 1817 \approx 1.5\) - Between 3rd and 4th: \(11578 / 2745 \approx 4.2\) (large jump) 2. The large jump between the 3rd and 4th ionisation energies indicates that the 4th electron is being removed from an inner, fully occupied principal quantum shell that experiences significantly less shielding and greater electrostatic attraction to the nucleus. 3. This confirms that element \(X\) has 3 valence electrons and belongs to Group 13. 4. Group 13 elements form chlorides with the empirical formula \(X\text{Cl}_3\).
Marking scheme
[1 mark] C is correct. - Identify 3 outer shell electrons from the large increase between 3rd and 4th IEs, deducing Group 13 chloride XCl3.
Question 12 · Multiple Choice
1 marks
Which species has a trigonal pyramidal molecular geometry and a bond angle of approximately \(107^\circ\)?
A.\(\text{CO}_3^{2-}\)
B.\(\text{H}_3\text{O}^+\)
C.\(\text{NH}_4^+\)
D.\(\text{AlCl}_3\)
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Worked solution
- \(\text{CO}_3^{2-}\) has 3 regions of electron density (trigonal planar, bond angle \(120^\circ\)). - \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair on the central oxygen atom (total 4 electron pairs, tetrahedral electron arrangement). Due to lone pair–bond pair repulsion being greater than bond pair–bond pair repulsion, the bond angle is compressed from \(109.5^\circ\) to approximately \(107^\circ\), giving a trigonal pyramidal shape. - \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs (tetrahedral, bond angle \(109.5^\circ\)). - \(\text{AlCl}_3\) has 3 bonding pairs and 0 lone pairs around the aluminium atom (trigonal planar, bond angle \(120^\circ\)).
Marking scheme
[1 mark] B is correct. - H3O+ has 3 bonding pairs and 1 lone pair, leading to a trigonal pyramidal shape with ~107° bond angles.
Question 13 · Multiple Choice
1 marks
Separate samples of four Period 3 chlorides are added to equal volumes of cold water at \(298\text{ K}\).
Which row correctly shows the approximate pH of the resulting solutions or mixtures?
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Worked solution
- \(\text{NaCl}\) undergoes simple hydration to form \(\text{Na}^+\text{(aq)}\) and \(\text{Cl}^-\text{(aq)}\), producing a neutral solution with \(\text{pH} \approx 7\). - \(\text{MgCl}_2\) dissolves with slight hydrolysis of the \([\text{Mg}(\text{H}_2\text{O})_6]^{2+}\) complex, yielding a very weakly acidic solution with \(\text{pH} \approx 6.5\). - \(\text{AlCl}_3\) undergoes significant hydrolysis because \(\text{Al}^{3+}\) has a high charge density, polarising coordinated water molecules to release \(\text{H}^+\text{(aq)}\), giving \(\text{pH} \approx 3\). - \(\text{SiCl}_4\) undergoes complete hydrolysis to form \(\text{SiO}_2\cdot x\text{H}_2\text{O(s)}\) and \(\text{HCl(aq)}\), producing a strongly acidic solution with \(\text{pH} \approx 1\).
When separate solid samples of sodium halides, \(\text{Na}X\), react with concentrated sulfuric acid: - \(\text{NaCl}\) gives steamy white fumes only; - \(\text{NaBr}\) gives steamy white fumes, a brown vapour, and a colourless gas that turns acidified potassium dichromate(VI) from orange to green; - \(\text{NaI}\) gives a purple vapour, a yellow solid, and a gas that turns moist lead(II) ethanoate paper black.
Which statement is correct?
A.The oxidation state of sulfur changes from \(+6\) to \(+4\) when concentrated sulfuric acid oxidises bromide ions.
B.Concentrated sulfuric acid is reduced to \(\text{H}_2\text{S}\) by bromide ions.
C.Chloride ions reduce sulfuric acid to sulfur dioxide.
D.Iodide ions reduce sulfuric acid to sulfur dioxide only.
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Worked solution
- \(\text{Br}^-\) is a moderately strong reducing agent that reduces concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\), where sulfur has an oxidation number of \(+6\)) to sulfur dioxide (\(\text{SO}_2\), where sulfur has an oxidation number of \(+4\)). Thus, statement A is correct. - \(\text{Br}^-\) is not strong enough to reduce \(\text{H}_2\text{SO}_4\) to hydrogen sulfide (\(\text{H}_2\text{S}\)); this only occurs with \(\text{I}^-\). - \(\text{Cl}^-\) cannot reduce concentrated sulfuric acid. - \(\text{I}^-\) reduces \(\text{H}_2\text{SO}_4\) to \(\text{SO}_2\) (oxidation state \(+4\)), \(\text{S}\) (yellow solid, oxidation state \(0\)), and \(\text{H}_2\text{S}\) (oxidation state \(-2\)).
Marking scheme
[1 mark] A is correct. - In the reaction of Br- with conc. H2SO4, H2SO4 (+6) is reduced to SO2 (+4).
Question 15 · Multiple Choice
1 marks
An unbranched alkene with molecular formula \(\text{C}_6\text{H}_{12}\) is heated under reflux with acidified potassium manganate(VII).
The reaction produces only one organic compound, \(Y\).
Compound \(Y\) reacts with aqueous sodium carbonate to produce carbon dioxide gas.
What is the identity of the original alkene?
A.hex-1-ene
B.hex-2-ene
C.hex-3-ene
D.3-methylpent-2-ene
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Worked solution
Let us consider the oxidative cleavage of unbranched \(\text{C}_6\text{H}_{12}\) alkenes by hot concentrated acidified \(\text{KMnO}_4\): - \(\text{hex-1-ene}\) (\(\text{CH}_2=\text{CH(CH}_2)_3\text{CH}_3\)) yields \(\text{CO}_2\) and pentanoic acid (two different products). - \(\text{hex-2-ene}\) (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_3\)) yields ethanoic acid and butanoic acid (two different organic compounds). - \(\text{hex-3-ene}\) (\(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_2\text{CH}_3\)) is symmetrical about the double bond and cleaves to form two moles of propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)), which is a single organic compound. - Propanoic acid contains a carboxylic acid group, so it reacts with \(\text{Na}_2\text{CO}_3\text{(aq)}\) with effervescence to release \(\text{CO}_2\).
Therefore, the original alkene is hex-3-ene.
Marking scheme
[1 mark] C is correct. - Hex-3-ene is symmetrically cleaved by hot acidified KMnO4 into two moles of propanoic acid (only one organic compound), which effervesces with carbonate.
Question 16 · Multiple Choice
1 marks
Compound \(Z\) is an alcohol with molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
- \(Z\) produces a yellow precipitate when warmed with alkaline aqueous iodine. - When \(Z\) is heated with an excess of concentrated sulfuric acid, a mixture of three isomeric alkenes of formula \(\text{C}_4\text{H}_8\) is formed, including a pair of stereoisomers.
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Worked solution
1. A positive tri-iodomethane (iodoform) test indicates the presence of a \(\text{CH}_3\text{CH(OH)}-\) group. 2. Among the four structural isomers of \(\text{C}_4\text{H}_{10}\text{O}\): - \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (butan-1-ol): negative iodoform test. - \((\text{CH}_3)_2\text{CHCH}_2\text{OH}\) (2-methylpropan-1-ol): negative iodoform test. - \((\text{CH}_3)_3\text{COH}\) (2-methylpropan-2-ol): negative iodoform test, dehydrates to 2-methylpropene only. - \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) (butan-2-ol): positive iodoform test (forms yellow \(\text{CHI}_3\)). 3. Acid-catalysed dehydration of butan-2-ol can eliminate a proton from C1 to give but-1-ene, or from C3 to give but-2-ene. But-2-ene exhibits geometric (cis-trans) isomerism, yielding cis-but-2-ene and trans-but-2-ene. Thus, three isomeric alkenes are obtained.
Hence, \(Z\) is \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
Marking scheme
[1 mark] B is correct. - CH3CH(OH)CH2CH3 contains the CH3CH(OH)- group (positive iodoform test) and dehydrates to but-1-ene, (Z)-but-2-ene, and (E)-but-2-ene.
Question 17 · Multiple Choice
1 marks
A gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), with a volume of \(15.0\text{ cm}^3\) was completely combusted in \(100.0\text{ cm}^3\) of oxygen gas (an excess).
After cooling to room temperature and pressure, the total volume of gas remaining was \(77.5\text{ cm}^3\). Passing this remaining gas through aqueous potassium hydroxide reduced the volume to \(17.5\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_8\)
B.\(\text{C}_4\text{H}_6\)
C.\(\text{C}_4\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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Worked solution
1. Volume of \(\text{CO}_2\) produced: Passing the mixture through aqueous \(\text{KOH}\) absorbs \(\text{CO}_2\). \(\text{Volume of } \text{CO}_2 = 77.5 - 17.5 = 60.0\text{ cm}^3\). Since \(15.0\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) gives \(60.0\text{ cm}^3\) of \(\text{CO}_2\), the number of carbon atoms is: \(x = \frac{60.0}{15.0} = 4\).
What is the standard enthalpy change of formation, \(\Delta H^\circ_{\text{f}}\), of propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}(\text{l})\)?
A.\(-513\text{ kJ mol}^{-1}\)
B.\(-847\text{ kJ mol}^{-1}\)
C.\(+513\text{ kJ mol}^{-1}\)
D.\(+2040\text{ kJ mol}^{-1}\)
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Worked solution
The equation for the standard enthalpy change of formation of propanoic acid is: \(3\text{C}(\text{graphite}) + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(\text{l})\)
Using Hess's Law with standard enthalpy changes of combustion: \(\Delta H^\circ_{\text{f}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products})\) \(\Delta H^\circ_{\text{f}} = [3(-394) + 3(-286)] - [-1527]\) \(\Delta H^\circ_{\text{f}} = [-1182 - 858] + 1527\) \(\Delta H^\circ_{\text{f}} = -2040 + 1527 = -513\text{ kJ mol}^{-1}\).
Marking scheme
A is correct [1 mark]. B, C, and D arise from arithmetic errors or incorrect signs in applying Hess's Law.
Question 19 · Multiple Choice
1 marks
The first eight successive ionisation energies of an element \(X\) are given in the table.
What is the formula of the chloride formed when element \(X\) reacts completely with chlorine?
A.\(X\text{Cl}\)
B.\(X\text{Cl}_2\)
C.\(X\text{Cl}_3\)
D.\(X_2\text{Cl}_3\)
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Worked solution
A large jump in ionisation energy occurs between the 2nd and 3rd ionisation energies (from \(1450\) to \(7730\text{ kJ mol}^{-1}\)). This indicates that element \(X\) has 2 valence electrons in its outer shell and is in Group 2 of the Periodic Table. Therefore, \(X\) forms a \(2+\) ion, \(X^{2+}\), and combines with chloride ions, \(\text{Cl}^-\), to form \(X\text{Cl}_2\).
Marking scheme
B is correct [1 mark]. A corresponds to Group 1, C to Group 13, D to Group 13/transition.
Question 20 · Multiple Choice
1 marks
Which species has the smallest bond angle?
A.\(\text{NH}_4^+\)
B.\(\text{NH}_3\)
C.\(\text{NH}_2^-\)
D.\(\text{NO}_2^+\)
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Worked solution
Evaluate the electron geometry and bond angles of each species: - \(\text{NO}_2^+\): 2 bonding pairs, 0 lone pairs \(\rightarrow\) linear, bond angle = \(180^\circ\). - \(\text{NH}_4^+\): 4 bonding pairs, 0 lone pairs \(\rightarrow\) tetrahedral, bond angle = \(109.5^\circ\). - \(\text{NH}_3\): 3 bonding pairs, 1 lone pair \(\rightarrow\) trigonal pyramidal, bond angle \(\approx 107^\circ\). - \(\text{NH}_2^-\): 2 bonding pairs, 2 lone pairs \(\rightarrow\) non-linear (bent), bond angle \(\approx 104.5^\circ\).
Lone pair-lone pair repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion, so \(\text{NH}_2^-\), with two lone pairs, has the smallest bond angle.
Marking scheme
C is correct [1 mark]. Bond angles: \(\text{NH}_2^- \approx 104.5^\circ\), \(\text{NH}_3 \approx 107^\circ\), \(\text{NH}_4^+ = 109.5^\circ\), \(\text{NO}_2^+ = 180^\circ\).
Question 21 · Multiple Choice
1 marks
Which row correctly describes the reactions of silicon(IV) chloride, \(\text{SiCl}_4\), and phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), when each is added to water at room temperature?
A.\(\text{SiCl}_4\): dissolves to form a neutral solution; \(\text{P}_4\text{O}_{10}\): dissolves to form an alkaline solution
B.\(\text{SiCl}_4\): hydrolyses to form an acidic solution; \(\text{P}_4\text{O}_{10}\): dissolves to form a neutral solution
C.\(\text{SiCl}_4\): hydrolyses to form an acidic solution; \(\text{P}_4\text{O}_{10}\): reacts to form an acidic solution
D.\(\text{SiCl}_4\): does not react with water; \(\text{P}_4\text{O}_{10}\): reacts to form an acidic solution
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Worked solution
- \(\text{SiCl}_4\) is a simple covalent liquid that undergoes rapid hydrolysis with water: \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\) This produces a strongly acidic solution (pH 1–2) and steamy fumes of \(\text{HCl}\).
- \(\text{P}_4\text{O}_{10}\) is an acidic oxide that reacts vigorously with water to form phosphoric(V) acid: \(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\) This also produces a strongly acidic solution (pH 1–2).
Marking scheme
C is correct [1 mark]. Both Period 3 compounds react/hydrolyse to form acidic solutions.
Question 22 · Multiple Choice
1 marks
Solid sodium iodide reacts with concentrated sulfuric acid to produce a mixture of products because iodide ions act as a strong reducing agent.
Which sulfur-containing product is formed when iodide ions reduce sulfur from oxidation state \(+6\) to oxidation state \(-2\)?
A.\(\text{SO}_2\)
B.\(\text{S}\)
C.\(\text{H}_2\text{S}\)
D.\(\text{NaHSO}_4\)
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Worked solution
In concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)), the oxidation state of sulfur is \(+6\). When iodide ions reduce sulfur: - In \(\text{SO}_2\), the oxidation state of \(\text{S}\) is \(+4\). - In \(\text{S}\), the oxidation state of \(\text{S}\) is \(0\). - In \(\text{H}_2\text{S}\), the oxidation state of \(\text{S}\) is \(-2\). - In \(\text{NaHSO}_4\), the oxidation state of \(\text{S}\) is \(+6\) (acid-base reaction, not redox).
Therefore, the reduction of \(\text{S}\) from \(+6\) to \(-2\) produces hydrogen sulfide, \(\text{H}_2\text{S}\).
Marking scheme
C is correct [1 mark]. Oxidation numbers: \(\text{H}_2\text{S}\) (-2), \(\text{S}\) (0), \(\text{SO}_2\) (+4), \(\text{NaHSO}_4\) (+6).
Question 23 · Multiple Choice
1 marks
An alkene, \(Y\), is heated under reflux with hot, concentrated, acidified potassium manganate(VII).
The only organic compound detected in the reaction product mixture is propanone, \(\text{CH}_3\text{COCH}_3\).
What is the identity of alkene \(Y\)?
A.2-methylbut-2-ene
B.2,3-dimethylbut-2-ene
C.2-methylpropene
D.but-2-ene
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Worked solution
Under hot, concentrated, acidified \(\text{KMnO}_4\), oxidative cleavage of the \(\text{C}=\text{C}\) double bond occurs: - \(=\text{CH}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\) - \(=\text{CHR} \rightarrow \text{RCOOH}\) - \(=\text{CR}_1\text{R}_2 \rightarrow \text{R}_1\text{COR}_2\) (ketone)
Let us analyse each option: - 2-methylbut-2-ene: \((\text{CH}_3)_2\text{C}=\text{CHCH}_3 \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}\) (two organic products). - 2,3-dimethylbut-2-ene: \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 \rightarrow 2\,\text{CH}_3\text{COCH}_3\) (propanone only). - 2-methylpropene: \((\text{CH}_3)_2\text{C}=\text{CH}_2 \rightarrow \text{CH}_3\text{COCH}_3 + \text{CO}_2 + \text{H}_2\text{O}\) (carbon dioxide is inorganic, but \(Y\) must give only propanone from both ends). - but-2-ene: \(\text{CH}_3\text{CH}=\text{CHCH}_3 \rightarrow 2\,\text{CH}_3\text{COOH}\) (ethanoic acid only).
Hence, 2,3-dimethylbut-2-ene gives propanone as the only organic compound.
Marking scheme
B is correct [1 mark]. Symmetrical disubstituted alkene \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\) cleaves exclusively into propanone molecules.
Question 24 · Multiple Choice
1 marks
An alcohol \(Z\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
When \(Z\) is warmed with alkaline aqueous iodine, a pale yellow precipitate is observed. When \(Z\) is heated in the presence of an aluminium oxide catalyst, it dehydrates to form a mixture containing three isomeric alkenes of formula \(\text{C}_4\text{H}_8\) (including stereoisomers).
What is the identity of compound \(Z\)?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
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Worked solution
1. The positive tri-iodomethane (iodoform) reaction (formation of pale yellow precipitate \(\text{CHI}_3\)) requires the structural unit \(\text{CH}_3\text{CH(OH)}-\) in alcohols. Among the \(\text{C}_4\text{H}_{10}\text{O}\) alcohols, only butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\), contains this group.
2. Dehydration of butan-2-ol over \(\text{Al}_2\text{O}_3\) produces: - but-1-ene (from elimination towards the \(\text{C}_1\) methyl group) - cis-but-2-ene (\((Z)\)-but-2-ene) - trans-but-2-ene (\((E)\)-but-2-ene) This gives a total of three isomeric alkenes.
Therefore, compound \(Z\) is butan-2-ol.
Marking scheme
B is correct [1 mark]. Butan-2-ol gives a positive iodoform test and dehydrates to give but-1-ene, cis-but-2-ene, and trans-but-2-ene.
Question 25 · multiple_choice
1 marks
A gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), has a volume of \(20\text{ cm}^3\). It requires \(130\text{ cm}^3\) of oxygen gas for complete combustion. Complete combustion produces \(80\text{ cm}^3\) of carbon dioxide.
All gas volumes are measured at the same temperature and pressure.
What is the formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_8\)
B.\(\text{C}_4\text{H}_8\)
C.\(\text{C}_4\text{H}_{10}\)
D.\(\text{C}_5\text{H}_{12}\)
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Worked solution
The balanced equation for the complete combustion of a hydrocarbon is:
Award 1 mark for A [\(-486\text{ kJ mol}^{-1}\)]. [Total: 1 mark]
Question 27 · multiple_choice
1 marks
An anhydrous chloride of a Period 3 element is a white solid that reacts vigorously with water to produce an acidic solution and white steamy fumes.
When aqueous sodium hydroxide is added dropwise to the resulting solution, a white precipitate forms that dissolves when excess aqueous sodium hydroxide is added.
Which element forms this chloride?
A.magnesium
B.aluminium
C.silicon
D.phosphorus
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Worked solution
1. Period 3 chlorides: - \(\text{NaCl}\): dissolves neutrally, no fumes. - \(\text{MgCl}_2\): dissolves to give a weakly acidic solution; with \(\text{NaOH(aq)}\) it gives a white precipitate of \(\text{Mg(OH)}_2\) that does not dissolve in excess \(\text{NaOH}\). - \(\text{AlCl}_3\): a white solid that hydrolyses vigorously in water to release steamy fumes of \(\text{HCl}\) and forms \([\text{Al(H}_2\text{O)}_6]^{3+}\). Adding \(\text{NaOH(aq)}\) forms a white precipitate of amphoteric \(\text{Al(OH)}_3\), which dissolves in excess \(\text{NaOH(aq)}\) to give \([\text{Al(OH)}_4]^-\). - \(\text{SiCl}_4\): a colourless liquid at rtp. - \(\text{PCl}_5\): reacts vigorously with water to form phosphoric acid and \(\text{HCl}\), but forms no metal hydroxide precipitate with \(\text{NaOH(aq)}\).
Thus, the element is aluminium.
Marking scheme
Award 1 mark for B [aluminium]. [Total: 1 mark]
Question 28 · multiple_choice
1 marks
An alkene \(\text{X}\) with the molecular formula \(\text{C}_6\text{H}_{12}\) is heated under reflux with hot concentrated acidified potassium manganate(VII).
Propanone is the only organic product obtained.
What is the identity of alkene \(\text{X}\)?
A.2,3-dimethylbut-1-ene
B.2,3-dimethylbut-2-ene
C.2-methylpent-2-ene
D.3-methylpent-2-ene
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Worked solution
Hot concentrated acidified \(\text{KMnO}_4\) cleaves the \(\text{C}=\text{C}\) double bond: - A \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\). - A \(=\text{CHR}\) group is oxidised to a carboxylic acid, \(\text{RCOOH}\). - A \(=\text{CR}_1\text{R}_2\) group is oxidised to a ketone, \(\text{R}_1\text{COR}_2\).
Since propanone, \((\text{CH}_3)_2\text{C}=\text{O}\), is the only organic product, the alkene must be symmetrical and both carbons of the double bond must be bonded to two methyl groups:
Which formula represents a stable compound of element \(\text{E}\)?
A.\(\text{E}_2\text{O}_3\)
B.\(\text{EO}_2\)
C.\(\text{ECl}_2\)
D.\(\text{E}_3\text{N}_2\)
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Worked solution
Examine the successive ionisation energies: - 1st to 2nd: increase of \(1239\text{ kJ mol}^{-1}\) - 2nd to 3rd: increase of \(928\text{ kJ mol}^{-1}\) - 3rd to 4th: massive increase of \(8833\text{ kJ mol}^{-1}\)
The huge jump occurs between the 3rd and 4th ionisation energies, indicating that the 4th electron is removed from an inner, fully-filled principal quantum shell. Therefore, element \(\text{E}\) has 3 valence electrons and belongs to Group 13.
Group 13 elements form \(3+\) ions (e.g. \(\text{E}^{3+}\)), which combine with oxide ions (\(\text{O}^{2-}\)) to give the compound \(\text{E}_2\text{O}_3\).
Marking scheme
Award 1 mark for A [\(\text{E}_2\text{O}_3\)]. [Total: 1 mark]
Question 30 · multiple_choice
1 marks
When solid potassium halide \(\text{KX}\) is treated with concentrated sulfuric acid, steamy fumes, a purple vapour, and a yellow solid are observed.
Which row correctly identifies the halide ion \(\text{X}^-\) and the yellow solid?
A.\(\text{X}^-\) is \(\text{Br}^-\); yellow solid is \(\text{S}\)
B.\(\text{X}^-\) is \(\text{Br}^-\); yellow solid is \(\text{SO}_2\)
C.\(\text{X}^-\) is \(\text{I}^-\); yellow solid is \(\text{S}\)
D.\(\text{X}^-\) is \(\text{I}^-\); yellow solid is \(\text{SO}_2\)
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Worked solution
When concentrated sulfuric acid reacts with potassium iodide (\(\text{KI}\)): 1. \(\text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HI}\) (steamy fumes of \(\text{HI}\)). 2. Iodide is a strong reducing agent and reduces \(\text{H}_2\text{SO}_4\) to: - \(\text{I}_2\) (purple vapour / dark solid), - \(\text{S}\) (yellow solid), - \(\text{H}_2\text{S}\) (toxic gas with bad egg smell), - \(\text{SO}_2\) (colourless choking gas).
Bromide ions only reduce sulfuric acid to sulfur dioxide (\(\text{SO}_2\)) and orange-brown bromine (\(\text{Br}_2\)), not to elemental sulfur.
Therefore, \(\text{X}^-\) is \(\text{I}^-\) and the yellow solid is sulfur (\(\text{S}\)).
Marking scheme
Award 1 mark for C [\(\text{X}^-\) is \(\text{I}^-\); yellow solid is \(\text{S}\)]. [Total: 1 mark]
Question 31 · multiple_choice
1 marks
Which species has a trigonal planar shape with bond angles of \(120^\circ\)?
A.\(\text{BF}_3\)
B.\(\text{NH}_3\)
C.\(\text{H}_3\text{O}^+\)
D.\(\text{PF}_3\)
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Worked solution
- \(\text{BF}_3\): Boron has 3 valence electrons, forming 3 bonding pairs and 0 lone pairs. According to VSEPR theory, 3 electron pairs adopt a trigonal planar geometry with \(120^\circ\) bond angles. - \(\text{NH}_3\): 3 bonding pairs and 1 lone pair \(\rightarrow\) trigonal pyramidal (\(107^\circ\)). - \(\text{H}_3\text{O}^+\): 3 bonding pairs and 1 lone pair \(\rightarrow\) trigonal pyramidal (\(107^\circ\)). - \(\text{PF}_3\): 3 bonding pairs and 1 lone pair \(\rightarrow\) trigonal pyramidal (approx. \(97^\circ\)).
Marking scheme
Award 1 mark for A [\(\text{BF}_3\)]. [Total: 1 mark]
Question 32 · multiple_choice
1 marks
An alcohol \(\text{Y}\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). When \(\text{Y}\) is warmed with alkaline aqueous iodine, a yellow precipitate is observed.
When \(\text{Y}\) is heated under reflux with acidified potassium dichromate(VI), compound \(\text{Z}\) is formed. Compound \(\text{Z}\) gives no reaction with Tollens' reagent.
What is the identity of compound \(\text{Z}\)?
A.butanal
B.butanoic acid
C.butanone
D.2-methylpropan-2-ol
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Worked solution
1. Alcohol \(\text{Y}\) has formula \(\text{C}_4\text{H}_{10}\text{O}\). 2. It gives a yellow precipitate with alkaline aqueous iodine (tri-iodomethane / iodoform test), so it must contain the \(\text{CH}_3\text{CH(OH)}-\) group. 3. The only \(\text{C}_4\text{H}_{10}\text{O}\) alcohol containing this group is butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\). 4. Oxidation of secondary alcohol butan-2-ol with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) forms the ketone butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\). 5. Ketones do not react with Tollens' reagent (they cannot be oxidised further under mild conditions).
Therefore, compound \(\text{Z}\) is butanone.
Marking scheme
Award 1 mark for C [butanone]. [Total: 1 mark]
Question 33 · multiple_choice
1 marks
A \(10.0\text{ cm}^3\) sample of a gaseous hydrocarbon \(\text{C}_x\text{H}_y\) is completely burned in \(70.0\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature, the residual gas volume is \(50.0\text{ cm}^3\). On passing this residual gas through excess aqueous sodium hydroxide, the volume decreases by \(30.0\text{ cm}^3\). All gas volumes are measured at room temperature and pressure.
What is the formula of the hydrocarbon?
A.\(\text{C}_2\text{H}_6\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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Worked solution
1. The volume of \(\text{CO}_2\) produced is absorbed by aqueous sodium hydroxide, so \(V(\text{CO}_2) = 30.0\text{ cm}^3\).
2. The combustion reaction is: \[\text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(l)}\]
3. From the mole ratio of hydrocarbon to \(\text{CO}_2\): \[x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{30.0}{10.0} = 3\]
What is the standard enthalpy change of formation of ethanol, \(\Delta H_\text{f}^\ominus [\text{CH}_3\text{CH}_2\text{OH(l)}]\)?
A.\(-687\text{ kJ mol}^{-1}\)
B.\(-279\text{ kJ mol}^{-1}\)
C.\(+279\text{ kJ mol}^{-1}\)
D.\(+687\text{ kJ mol}^{-1}\)
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Worked solution
The formation equation for ethanol is: \[2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(l)}\]
Using Hess's Law with standard enthalpy changes of combustion: \[\Delta H_\text{f}^\ominus = 2\Delta H_\text{c}^\ominus(\text{C}) + 3\Delta H_\text{c}^\ominus(\text{H}_2) - \Delta H_\text{c}^\ominus(\text{ethanol})\] \[\Delta H_\text{f}^\ominus = 2(-394) + 3(-286) - (-1367)\] \[\Delta H_\text{f}^\ominus = -788 - 858 + 1367 = -279\text{ kJ mol}^{-1}\]
Marking scheme
[1] B is correct. (M1 for correct application of Hess's cycle giving -279 kJ mol^-1).
Question 35 · multiple_choice
1 marks
The first six successive ionisation energies of an element \(X\) in Period 3 of the Periodic Table are shown below.
What is the formula of the chloride of element \(X\) and the approximate pH of the solution formed when this chloride is added to excess water?
A.formula \(\text{XCl}_3\) and pH 2
B.formula \(\text{XCl}_3\) and pH 7
C.formula \(\text{XCl}_4\) and pH 2
D.formula \(\text{XCl}_4\) and pH 7
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Worked solution
1. The successive ionisation energies show a large jump between the 4th and 5th values (from \(4356\) to \(16091\text{ kJ mol}^{-1}\)). This indicates that element \(X\) has 4 valence electrons in its outer shell and belongs to Group 14 (silicon).
2. The chloride formed is \(\text{XCl}_4\) (silicon tetrachloride, \(\text{SiCl}_4\)).
3. When \(\text{SiCl}_4\) reacts with water, it hydrolyses vigorously to produce \(\text{SiO}_2\) and \(\text{HCl}\): \[\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}\] This produces an acidic solution with an approximate pH of 1 to 2.
Marking scheme
[1] C is correct. (M1 for identifying Group 14 from the IE jump and the resulting hydrolysed acidic pH of ~2).
Question 36 · multiple_choice
1 marks
Which pair contains two species that both have a non-linear (bent) molecular shape?
A.\(\text{CO}_2\) and \(\text{BeCl}_2\)
B.\(\text{H}_2\text{O}\) and \(\text{HCN}\)
C.\(\text{NO}_2^+\) and \(\text{BeCl}_2\)
D.\(\text{SO}_2\) and \(\text{NH}_2^-\)
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Worked solution
1. In \(\text{SO}_2\), the central sulfur atom has 6 valence electrons, forming two bonding regions to oxygen atoms and possessing 1 lone pair (3 electron domains \(\rightarrow\) bent shape with a bond angle of approximately \(119^\circ\)).
2. In \(\text{NH}_2^-\), the central nitrogen atom has \(5 + 1 = 6\) valence electrons, forming two single bonds to hydrogen atoms and possessing 2 lone pairs (4 electron domains \(\rightarrow\) bent shape with a bond angle of approximately \(104.5^\circ\)).
Other options contain linear species: \(\text{CO}_2\), \(\text{BeCl}_2\), \(\text{HCN}\), and \(\text{NO}_2^+\) are all linear.
Marking scheme
[1] D is correct. (M1 for identifying that both SO2 and NH2^- possess lone pairs leading to non-linear geometries).
Question 37 · multiple_choice
1 marks
Solid samples of sodium halides are added separately to concentrated sulfuric acid at room temperature.
Which statement is correct?
A.Steamy fumes are produced with sodium chloride, and no sulfur-containing gases are formed.
B.Sodium chloride reduces sulfuric acid to sulfur dioxide gas.
C.Sodium bromide reduces sulfuric acid exclusively to hydrogen sulfide gas.
D.Sodium iodide reacts solely as a Brønsted–Lowry base and produces no elemental iodine.
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Worked solution
1. Chloride ions (\(\text{Cl}^-\)) are not strong enough reducing agents to reduce sulfuric acid. \(\text{NaCl}\) undergoes an acid–base reaction to produce steamy fumes of hydrogen chloride (\(\text{HCl}\)): \[\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\] No reduction products of sulfur (such as \(\text{SO}_2\), \(\text{S}\), or \(\text{H}_2\text{S}\)) are formed.
2. Bromide ions reduce sulfuric acid to sulfur dioxide (\(\text{SO}_2\)), not hydrogen sulfide. 3. Iodide ions are powerful reducing agents and reduce sulfuric acid to \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\), while forming solid purple/black iodine.
Marking scheme
[1] A is correct. (M1 for identifying that chloride does not reduce concentrated sulfuric acid, producing only HCl fumes).
Question 38 · multiple_choice
1 marks
An alkene \(Y\) is heated under reflux with hot concentrated acidified potassium manganate(VII). The only organic products obtained are propanone, \(\text{CH}_3\text{COCH}_3\), and propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
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Worked solution
Under hot concentrated acidified \(\text{KMnO}_4\): - A \(=\text{C(R)}_1\text{R}_2\) group is oxidised to a ketone, \(\text{R}_1\text{COR}_2\). Therefore, the propanone fragment originates from \(=C(\text{CH}_3)_2\). - A \(=\text{CHR}\) group is oxidised to a carboxylic acid, \(\text{RCOOH}\). Therefore, the propanoic acid fragment originates from \(=\text{CHCH}_2\text{CH}_3\).
Combining these fragments across the double bond gives \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\) (2-methylpent-2-ene).
Marking scheme
[1] C is correct. (M1 for deducing the alkene structure from oxidative cleavage products).
Question 39 · multiple_choice
1 marks
The temperature of a gaseous reaction mixture is increased from \(T_1\) to \(T_2\) in the absence of a catalyst.
Which row correctly describes the changes to the Maxwell–Boltzmann distribution curve and the activation energy, \(E_\text{a}\)?
A.total area under curve is unchanged; peak shifts to the right with lower height; \(E_\text{a}\) is unchanged
B.total area under curve increases; peak shifts to the right with higher height; \(E_\text{a}\) decreases
C.total area under curve is unchanged; peak shifts to the left with higher height; \(E_\text{a}\) decreases
D.total area under curve increases; peak shifts to the right with lower height; \(E_\text{a}\) is unchanged
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Worked solution
1. The total area under the Maxwell–Boltzmann distribution curve represents the total number of particles in the system, which remains constant when temperature increases.
2. At a higher temperature (\(T_2 > T_1\)), the curve broadens, the peak shifts to the right (higher energy), and the maximum height of the peak decreases.
3. Activation energy, \(E_\text{a}\), is an intrinsic property of the reaction pathway and is not affected by changes in temperature.
Marking scheme
[1] A is correct. (M1 for recognising that area is constant, peak shifts right/down, and Ea is unchanged).
Question 40 · multiple_choice
1 marks
Compound \(Z\) is an organic liquid with the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
- When \(Z\) is heated with acidified potassium dichromate(VI), the solution turns from orange to green. - When \(Z\) is warmed with alkaline aqueous iodine, a yellow precipitate is formed.
What is the identity of compound \(Z\)?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
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Worked solution
1. Reaction with acidified potassium dichromate(VI) (orange to green) indicates that \(Z\) is an oxidisable alcohol (either a primary or secondary alcohol; tertiary alcohols do not react).
2. Formation of a yellow precipitate with alkaline aqueous iodine (the tri-iodomethane test) requires a \(\text{CH}_3\text{CH(OH)}-\) group.
3. Evaluating the isomers of \(\text{C}_4\text{H}_{10}\text{O}\): - butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)): primary alcohol, oxidised, but negative tri-iodomethane test. - butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)): secondary alcohol, oxidised, contains \(\text{CH}_3\text{CH(OH)}-\), positive tri-iodomethane test. - 2-methylpropan-1-ol: primary alcohol, negative tri-iodomethane test. - 2-methylpropan-2-ol: tertiary alcohol, not oxidised.
Therefore, \(Z\) is butan-2-ol.
Marking scheme
[1] B is correct. (M1 for deducing that only butan-2-ol is oxidisable and contains the CH3CH(OH)- group).
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Answer all structured questions in the spaces provided on the question paper. Show all working in calculations and give answers to appropriate significant figures.
25 Question · 53 marks
Question 1 · short_answer
1 marks
Calculate the total number of chloride ions present in \(0.0250\text{ mol}\) of aluminium chloride, \(\text{AlCl}_3\). Give your answer to 3 significant figures. [\(L = 6.022 \times 10^{23}\text{ mol}^{-1}\)]
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Worked solution
1 mole of \(\text{AlCl}_3\) contains 3 moles of \(\text{Cl}^-\). Amount of \(\text{Cl}^-\text{ ions} = 3 \times 0.0250 = 0.0750\text{ mol}\). Total number of \(\text{Cl}^-\text{ ions} = 0.0750 \times 6.022 \times 10^{23} = 4.5165 \times 10^{22} = 4.52 \times 10^{22}\) (to 3 s.f.).
Marking scheme
[1] \(4.52 \times 10^{22}\) (Allow \(4.51 \times 10^{22}\) if using \(6.02 \times 10^{23}\)). Award 1 mark for correct value to 3 significant figures.
Question 2 · short-answer
1 marks
Calculate the volume, in \(\text{cm}^3\), of \(0.0250\text{ mol}\,\text{dm}^{-3}\text{ Ba(OH)}_2\text{(aq)}\) required to neutralise completely \(20.0\text{ cm}^3\) of \(0.0400\text{ mol}\,\text{dm}^{-3}\text{ HNO}_3\text{(aq)}\).
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Use the following average bond energy values: - \(E(\text{C}=\text{C}) = 612\text{ kJ}\,\text{mol}^{-1}\) - \(E(\text{C}-\text{C}) = 347\text{ kJ}\,\text{mol}^{-1}\) - \(E(\text{C}-\text{H}) = 413\text{ kJ}\,\text{mol}^{-1}\) - \(E(\text{H}-\text{H}) = 436\text{ kJ}\,\text{mol}^{-1}\)
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[1] Award 1 mark for \(-125\) (with negative sign explicit or implied by \(-125\text{ kJ}\,\text{mol}^{-1}\)). [R] \(+125\) / \(125\) without sign.
Question 4 · short-answer
1 marks
Construct a balanced chemical equation for the reaction of solid phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), with water.
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Worked solution
Phosphorus(V) oxide reacts vigorously with water to produce phosphoric(V) acid: \[\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\]
Marking scheme
[1] Award 1 mark for the correct balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\). [I] State symbols. [A] Correct multiples.
Question 5 · short-answer
1 marks
An alkene, \(\mathbf{W}\), with molecular formula \(\text{C}_5\text{H}_{10}\) is oxidised by heating under reflux with hot, concentrated acidified \(\text{KMnO}_4\). The only organic products formed are propanone, \(\text{CH}_3\text{COCH}_3\), and ethanoic acid, \(\text{CH}_3\text{COOH}\).
State the systematic IUPAC name of alkene \(\mathbf{W}\).
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Worked solution
Oxidative cleavage of a carbon-carbon double bond with hot acidified \(\text{KMnO}_4\): - A \(=\text{C(CH}_3)_2\) group forms propanone, \(\text{CH}_3\text{COCH}_3\). - A \(=\text{CHCH}_3\) group forms ethanoic acid, \(\text{CH}_3\text{COOH}\).
Combining these fragments gives \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\), which is 2-methylbut-2-ene.
Marking scheme
[1] Award 1 mark for 2-methylbut-2-ene (or 2-methyl-2-butene). [R] methylbut-2-ene (missing locant).
Question 6 · Chemical Equations & Dot-Cross Diagrams
2 marks
Hydroxylamine has the molecular formula \(\text{NH}_2\text{OH}\).
(i) Draw a dot-and-cross diagram to show the bonding in a molecule of hydroxylamine. Show outer-shell electrons only. [1] (ii) State the total number of lone pairs of electrons in one molecule of \(\text{NH}_2\text{OH}\). [1]
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Worked solution
In \(\text{NH}_2\text{OH}\): - The central nitrogen atom shares single covalent bonds with two hydrogen atoms and one oxygen atom (3 bonding pairs = 6 electrons) and retains one lone pair (2 electrons), completing its octet (8 electrons). - The oxygen atom forms a single covalent bond with the nitrogen atom and a single covalent bond with the hydrogen atom (2 bonding pairs = 4 electrons) and retains two lone pairs (4 electrons), completing its octet (8 electrons). - Each hydrogen atom shares one pair of electrons (2 electrons). Total lone pairs = 1 (on N) + 2 (on O) = 3.
Marking scheme
M1: Correct dot-and-cross representation of all bonding pairs (two \(\text{N}-\text{H}\), one \(\text{N}-\text{O}\), one \(\text{O}-\text{H}\)) with no extra bonds. M2: Correct depiction of non-bonding outer electrons (one lone pair on \(\text{N}\), two lone pairs on \(\text{O}\)) AND total stated as 3.
Question 7 · Chemical Equations & Dot-Cross Diagrams
2 marks
Phosphorus(V) oxide reacts vigorously with water in an exothermic reaction.
(i) Construct the balanced chemical equation for the reaction of solid tetraphosphorus decaoxide, \(\text{P}_4\text{O}_{10}\), with water. [1] (ii) Suggest the approximate pH of the resulting solution. [1]
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Worked solution
(i) \(\text{P}_4\text{O}_{10}\) reacts with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\): \[\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\] (ii) Phosphoric(V) acid is a moderately strong acid in aqueous solution, yielding a solution with a pH around 0 to 2.
Marking scheme
M1: Correctly balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) (Allow multiples; state symbols not required). M2: Any value or range in the interval \(0 \le \text{pH} \le 2\).
Question 8 · Chemical Equations & Dot-Cross Diagrams
2 marks
In the vapour phase at moderate temperatures, aluminium chloride exists as a dimer, \(\text{Al}_2\text{Cl}_6\).
(i) Draw a dot-and-cross diagram to represent the bonding in \(\text{Al}_2\text{Cl}_6\). Show outer-shell electrons only. [1] (ii) Identify the type of bond that holds the two monomer units together in the dimer. [1]
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Worked solution
In \(\text{Al}_2\text{Cl}_6\): - Each Al atom forms three standard single covalent bonds to chlorine atoms. - To achieve a stable octet, each Al atom accepts a lone pair of electrons from a bridging chlorine atom attached to the other aluminium atom, forming two coordinate (dative covalent) bonds. - All six chlorine atoms possess complete outer octets (3 lone pairs on terminal chlorines, 2 lone pairs on bridging chlorines).
Marking scheme
M1: Fully correct dot-and-cross diagram with two bridging \(\text{Cl}\) atoms, all four terminal \(\text{Cl}\) atoms showing complete outer shells (6 non-bonding electrons each), and bridging \(\text{Cl}\) atoms showing 4 non-bonding electrons and 2 shared pairs with \(\text{Al}\). M2: Dative covalent bond / coordinate bond.
Question 9 · Chemical Equations & Dot-Cross Diagrams
2 marks
When chlorine gas is bubbled into hot concentrated aqueous sodium hydroxide, a disproportionation reaction occurs.
(i) Construct the balanced chemical equation for this reaction. [1] (ii) State the initial and two final oxidation states of chlorine in this reaction. [1]
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Worked solution
(i) Chlorine reacts with hot aqueous \(\text{NaOH}\) according to: \[3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\] (ii) In \(\text{Cl}_2\), the oxidation state is 0. In \(\text{NaCl}\) (\(\text{Cl}^-\)), the oxidation state is -1. In \(\text{NaClO}_3\) (\(\text{ClO}_3^-\)), the oxidation state is +5.
Marking scheme
M1: Balanced equation: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\) OR ionic form \(3\text{Cl}_2 + 6\text{OH}^- \rightarrow 5\text{Cl}^- + \text{ClO}_3^- + 3\text{H}_2\text{O}\). M2: Initial = 0 AND Final = -1 and +5 (both final values required).
Question 10 · Chemical Equations & Dot-Cross Diagrams
2 marks
Hydrogen cyanide, \(\text{HCN}\), is a volatile liquid used in organic synthesis.
(i) Draw a dot-and-cross diagram of a molecule of \(\text{HCN}\). Show outer-shell electrons only. [1] (ii) Predict the \(\text{H}-\text{C}-\text{N}\) bond angle in \(\text{HCN}\). [1]
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(i) Carbon has 4 valence electrons, nitrogen has 5, and hydrogen has 1. - H shares 1 electron pair with C. - C shares 3 electron pairs with N to form a triple bond \(\text{C}\equiv\text{N}\). - N retains 1 lone pair (2 electrons). (ii) The central carbon atom has two \(\sigma\)-bonding regions and no lone pairs, giving a linear geometry with a bond angle of \(180^\circ\).
Marking scheme
M1: Correct dot-and-cross diagram showing one shared pair between \(\text{H}\) and \(\text{C}\), three shared pairs between \(\text{C}\) and \(\text{N}\), and one lone pair on \(\text{N}\). M2: \(180^\circ\) (or \(180\) degrees).
Question 11 · Chemical Equations & Dot-Cross Diagrams
2 marks
Group 2 nitrates undergo thermal decomposition when heated strongly.
(i) Construct the balanced chemical equation for the thermal decomposition of anhydrous barium nitrate, \(\text{Ba(NO}_3)_2\). [1] (ii) Identify the brown gas evolved during this reaction. [1]
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Worked solution
(i) Group 2 nitrates decompose upon heating to form the metal oxide, nitrogen dioxide gas, and oxygen gas: \[2\text{Ba(NO}_3)_2 \rightarrow 2\text{BaO} + 4\text{NO}_2 + \text{O}_2\] (ii) Nitrogen dioxide (\(\text{NO}_2\)) is a characteristic brown gas.
Question 12 · Chemical Equations & Dot-Cross Diagrams
2 marks
Magnesium reacts with nitrogen gas at elevated temperatures to form magnesium nitride, \(\text{Mg}_3\text{N}_2\).
(i) Draw a dot-and-cross diagram to show the arrangement of outer-shell electrons and charges in magnesium nitride. [1] (ii) State the electronic configuration of the nitride ion, \(\text{N}^{3-}\). [1]
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Worked solution
(i) Magnesium nitride consists of \(\text{Mg}^{2+}\) and \(\text{N}^{3-}\) ions in a 3:2 ratio. - Each \(\text{Mg}^{2+}\) ion has lost two electrons: \([\text{Mg}]^{2+}\) (or showing outer octet). - Each \(\text{N}^{3-}\) ion has gained three electrons to complete its valence shell: \([\text{N}]^{3-}\) with 8 outer electrons. - Correct stoichiometry \(3\times \text{Mg}^{2+}\) and \(2\times \text{N}^{3-}\). (ii) Nitrogen has atomic number 7. With three added electrons, \(\text{N}^{3-}\) has 10 electrons: \(1s^2 2s^2 2p^6\).
Marking scheme
M1: Correct dot-and-cross diagram showing three \(\text{Mg}^{2+}\) ions and two \(\text{N}^{3-}\) ions with correct charges and correct outer electrons on \(\text{N}^{3-}\). M2: \(1s^2 2s^2 2p^6\) (Allow \([\text{Ne}]\)).
Question 13 · Chemical Equations & Dot-Cross Diagrams
2 marks
Silicon tetrachloride, \(\text{SiCl}_4\), reacts vigorously with water in a hydrolysis reaction.
(i) Construct the balanced chemical equation for the reaction between liquid silicon tetrachloride and water. [1] (ii) Describe one observation that can be made during this reaction. [1]
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(i) \(\text{SiCl}_4\) undergoes complete hydrolysis upon contact with water: \[\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g}/\text{aq})\] (or \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\) / \(\text{SiO}_2\cdot x\text{H}_2\text{O}\)). (ii) Observations include steamy/white acidic fumes of \(\text{HCl}\) gas or formation of an off-white/white solid precipitate of hydrated silicon dioxide.
Marking scheme
M1: Balanced equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) OR \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\). M2: Steamy / misty / white fumes OR white precipitate / white solid.
When 2-methylbut-2-ene, \(\text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_3\), reacts with gaseous hydrogen bromide, \(\text{HBr}\), 2-bromo-2-methylbutane is obtained as the major organic product and 2-bromo-3-methylbutane is obtained as the minor product.
Explain why 2-bromo-2-methylbutane is the major product in this reaction.
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Worked solution
During the electrophilic addition of \(\text{HBr}\) to 2-methylbut-2-ene, protonation of the double bond can generate two possible carbocation intermediates: 1. A tertiary carbocation: \(\text{CH}_3\text{C}^+(\text{CH}_3)\text{CH}_2\text{CH}_3\) 2. A secondary carbocation: \(\text{CH}_3\text{CH}(\text{CH}_3)\text{C}^+\text{HCH}_3\)
The tertiary carbocation has three electron-donating alkyl groups attached to the positively charged carbon atom, exerting a greater positive inductive effect (electron-releasing effect) that delocalises the positive charge and stabilises the intermediate. In contrast, the secondary carbocation is attached to only two alkyl groups and is therefore less stable. The reaction proceeds predominantly via the more stable tertiary intermediate, making 2-bromo-2-methylbutane the major product.
Marking scheme
M1: Identifies the two intermediate carbocations: formation of a tertiary carbocation for the major product AND a secondary carbocation for the minor product [1] [Accept drawn structural formulas or systematic names for both carbocations]
M2: States that the tertiary carbocation is more stable than the secondary carbocation [1]
M3: Explains that the tertiary carbocation has three electron-releasing / electron-donating alkyl groups compared to two in the secondary carbocation / greater positive inductive effect stabilising the positive charge [1] [Reject 'electronegativity' instead of inductive effect]
Ethane reacts with chlorine in the presence of ultraviolet light via a free-radical substitution mechanism.
(i) Construct equations for the two propagation steps leading to the formation of chloroethane. (ii) Suggest how a small amount of butane is formed during this reaction.
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Formation of butane: During termination, two ethyl free radicals collide and combine: \(\text{CH}_3\text{CH}_2^\bullet + \text{CH}_3\text{CH}_2^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)
Marking scheme
M1: Propagation step 1: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\) [1] [Radical dot must be clearly shown on Cl and on the C of ethyl]
2-bromo-2-methylpropane reacts with aqueous sodium hydroxide predominantly via an \(\text{S}_\text{N}1\) mechanism, whereas 1-bromobutane reacts predominantly via an \(\text{S}_\text{N}2\) mechanism.
Describe the initial step of the \(\text{S}_\text{N}1\) mechanism for 2-bromo-2-methylpropane and explain why 2-bromo-2-methylpropane reacts via \(\text{S}_\text{N}1\) rather than \(\text{S}_\text{N}2\).
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Worked solution
In the \(\text{S}_\text{N}1\) mechanism: Step 1 is the slow, rate-determining heterolytic cleavage of the \(\text{C}-\text{Br}\) bond in 2-bromo-2-methylpropane, producing the tertiary carbocation \((\text{CH}_3)_3\text{C}^+\) and \(\text{Br}^-\).
2-bromo-2-methylpropane favours \(\text{S}_\text{N}1\) over \(\text{S}_\text{N}2\) because: 1. The tertiary carbocation intermediate is strongly stabilised by the positive inductive (+I) effect of three methyl groups releasing electron density towards the positive carbon atom. 2. In \(\text{S}_\text{N}2\), backside attack by the \(\text{OH}^-\) nucleophile is prevented by steric hindrance from the three bulky methyl groups surrounding the central carbon atom.
Marking scheme
M1: Heterolytic fission / breaking of the \(\text{C}-\text{Br}\) bond to produce a tertiary carbocation \((\text{CH}_3)_3\text{C}^+\) (and \(\text{Br}^-\)) [1]
M2: Tertiary carbocation is stabilised by the positive inductive effect / electron-donating effect of three methyl / alkyl groups [1]
M3: Steric hindrance from three methyl groups prevents backside attack by the nucleophile / \(\text{OH}^-\) (ruling out \(\text{S}_\text{N}2\)) [1]
Propanone reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a small amount of sodium cyanide, \(\text{NaCN}\), to form 2-hydroxy-2-methylpropanenitrile.
Describe the role of the cyanide ion, \(\text{CN}^-\), and the two mechanistic steps involved in this reaction.
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The reaction is nucleophilic addition: 1. \(\text{CN}^-\) acts as a nucleophile. The lone pair on the carbon of \(:\text{C}\equiv\text{N}^-\) attacks the partially positive carbon atom (\(\text{C}^{\delta+}\)) of the carbonyl group (\(\text{C}=\text{O}\)). The \(\text{C}=\text{O}\) \(\pi\)-bond breaks heterolytically, transferring electron density onto the oxygen atom to form an intermediate alkoxide ion, \((\text{CH}_3)_2\text{C}(\text{O}^-)\text{CN}\). 2. The alkoxide oxygen atom \(\text{O}^-\) attacks an \(\text{H}\) atom on an \(\text{HCN}\) molecule (or accepts a proton \(\text{H}^+\)), forming the product 2-hydroxy-2-methylpropanenitrile, \((\text{CH}_3)_2\text{C}(\text{OH})\text{CN}\). 3. A \(\text{CN}^-\) ion is reformed in this protonation step, showing that \(\text{NaCN}\)/\(\text{CN}^-\) acts as a catalyst.
Marking scheme
M1: \(\text{CN}^-\) acts as a nucleophile and attacks the \(\text{C}^{\delta+}\) / carbonyl carbon atom, causing heterolytic cleavage of the \(\text{C}=\text{O}\) bond [1]
M2: Formation of the intermediate anion \((\text{CH}_3)_2\text{C}(\text{O}^-)\text{CN}\) / alkoxide intermediate [1]
M3: Intermediate is protonated by \(\text{HCN}\) (or \(\text{H}^+\)) to form the hydroxynitrile AND \(\text{CN}^-\) is regenerated / acts as a catalyst [1]
Separate samples of 1-chlorobutane and 1-iodobutane are heated with aqueous silver nitrate in ethanol under identical conditions.
State which halogenoalkane produces a precipitate more rapidly. Explain your answer in terms of bond strength and reaction kinetics.
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Worked solution
When halogenoalkanes are heated with aqueous silver nitrate in ethanol, hydrolysis occurs to release halide ions, which then form a precipitate with \(\text{Ag}^+\). 1-iodobutane reacts faster than 1-chlorobutane, forming a yellow precipitate of \(\text{AgI}\) much faster than the white precipitate of \(\text{AgCl}\).
Explanation: - The bond enthalpy of \(\text{C}-\text{I}\) (approx. \(240\,\text{kJ}\,\text{mol}^{-1}\)) is lower than that of \(\text{C}-\text{Cl}\) (approx. \(340\,\text{kJ}\,\text{mol}^{-1}\)) due to poorer orbital overlap with the larger iodine atom. - The rate of nucleophilic substitution depends on the strength of the \(\text{C}-\text{X}\) bond being cleaved in the rate-determining step. - A lower bond enthalpy means lower activation energy (\(E_a\)), resulting in a faster rate of reaction for 1-iodobutane.
Marking scheme
M1: States that 1-iodobutane reacts faster / produces a precipitate faster (or \(\text{AgI}\) forms faster than \(\text{AgCl}\)) [1]
M2: States that the \(\text{C}-\text{I}\) bond is weaker / has a lower bond enthalpy / bond energy than the \(\text{C}-\text{Cl}\) bond [1]
M3: States that breaking the \(\text{C}-\text{X}\) bond is the rate-determining step / lower activation energy for \(\text{C}-\text{I}\) bond cleavage [1]
When butan-2-ol is heated with concentrated sulfuric acid, an elimination reaction occurs yielding a mixture of three isomeric alkenes of molecular formula \(\text{C}_4\text{H}_8\).
(i) Name the three isomeric alkenes formed. (ii) Explain why two of these isomers exhibit stereoisomerism.
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(i) Elimination of water from butan-2-ol can remove a hydrogen atom from C1 or C3: - Removal from C1 gives but-1-ene. - Removal from C3 gives but-2-ene, which exists as two stereoisomers: \((E)\)-but-2-ene (trans-but-2-ene) and \((Z)\)-but-2-ene (cis-but-2-ene).
(ii) The requirements for cis-trans / stereoisomerism in but-2-ene are: 1. Restricted rotation about the \(\text{C}=\text{C}\) double bond due to the overlap of p orbitals forming the \(\pi\) bond. 2. Each carbon atom involved in the \(\text{C}=\text{C}\) double bond is attached to two different groups: \(-\text{H}\) and \(-\text{CH}_3\).
Marking scheme
M1: Names all three alkenes: but-1-ene AND \((E)\)-but-2-ene / trans-but-2-ene AND \((Z)\)-but-2-ene / cis-but-2-ene [1] [Accept but-1-ene, cis-but-2-ene, trans-but-2-ene]
M2: Restricted rotation about the \(\text{C}=\text{C}\) bond / presence of \(\pi\) bond prevents rotation [1]
M3: Each carbon atom of the \(\text{C}=\text{C}\) bond is attached to two different groups (\(-\text{H}\) and \(-\text{CH}_3\)) [1]
Propene reacts with aqueous bromine, \(\text{Br}_2(\text{aq})\), via an electrophilic addition mechanism.
Describe the stages of this mechanism for the reaction between propene and \(\text{Br}_2\), explaining how the dipole in the bromine molecule is induced and how the final product, 1,2-dibromopropane, is formed.
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Worked solution
1. Induction of dipole: As the non-polar \(\text{Br}_2\) molecule approaches the electron-rich \(\text{C}=\text{C}\) double bond of propene, the high electron density in the \(\pi\)-bond repels electron density in the \(\text{Br}-\text{Br}\) bond, inducing a temporary dipole (\(\text{Br}^{\delta+}-\text{Br}^{\delta-}\)). 2. First step: The electron pair from the \(\text{C}=\text{C}\) \(\pi\)-bond attacks the \(\text{Br}^{\delta+}\) atom. Heterolytic fission of the \(\text{Br}-\text{Br}\) bond occurs, releasing a bromide ion, \(:\text{Br}^-\), and forming the more stable secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+\text{CH}_2\text{Br}\). 3. Second step: The \(:\text{Br}^-\) ion uses a lone pair of electrons to attack the positively charged carbon atom of the carbocation, forming a covalent \(\text{C}-\text{Br}\) bond and yielding 1,2-dibromopropane, \(\text{CH}_3\text{CHBrCH}_2\text{Br}\).
Marking scheme
M1: Explains that the high electron density of the \(\text{C}=\text{C}\) / \(\pi\)-bond induces a dipole in the \(\text{Br}_2\) molecule (\(\text{Br}^{\delta+}-\text{Br}^{\delta-}\)) [1]
M2: Attack of \(\pi\)-electrons on \(\text{Br}^{\delta+}\) with heterolytic cleavage of \(\text{Br}-\text{Br}\) to form a carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{Br}\)) AND a bromide ion (\(:\text{Br}^-\)) [1]
M3: Nucleophilic attack of \(:\text{Br}^-\) / donation of lone pair from \(:\text{Br}^-\) to the carbocation carbon to form 1,2-dibromopropane [1]
1-bromopropane reacts with warm aqueous sodium hydroxide to form propan-1-ol via an \(\text{S}_\text{N}2\) mechanism.
Describe the key features of this \(\text{S}_\text{N}2\) mechanism, including the direction of nucleophilic attack, the structure of the transition state, and the stereochemical outcome at the reacting carbon atom.
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In the \(\text{S}_\text{N}2\) mechanism for 1-bromopropane: 1. The \(\text{OH}^-\) nucleophile attacks the partially positive carbon atom (\(\text{C}^{\delta+}\)) from the backside (\(180^\circ\) away from the leaving \(\text{Br}\) atom) to minimise electrostatic repulsion with the electronegative bromine atom. 2. The reaction proceeds through a single transition state in which the central carbon forms partial bonds simultaneously to both the incoming \(\text{OH}\) group and the departing \(\text{Br}\) atom (represented as \([\text{HO}\cdots\text{CH}_2(\text{CH}_2\text{CH}_3)\cdots\text{Br}]^-\)). 3. The departure of the bromide ion (\(\text{Br}^-\)) results in inversion of stereochemical configuration at the carbon centre (umbrella inversion).
Marking scheme
M1: \(\text{OH}^-\) attacks the \(\text{C}^{\delta+}\) carbon from the opposite side / backside relative to the \(\text{Br}\) atom [1]
M2: Formation of a transition state with partial bonds to both \(\text{OH}\) and \(\text{Br}\) (pentacoordinate carbon) AND with a negative charge [1]
M3: Departure of \(\text{Br}^-\) resulting in inversion of configuration at the carbon atom [1]
Question 22 · structured
2 marks
An unknown carbonyl compound J with the molecular formula \(\text{C}_n\text{H}_{2n}\text{O}\) gives a mass spectrum with a molecular ion peak, \(\text{M}^+\), at \(m/e = 58\) (relative abundance \(28.4\)) and an \([\text{M}+1]^+\) peak at \(m/e = 59\) (relative abundance \(0.94\)).
(a) Show by calculation that compound J contains 3 carbon atoms.
(b) Compound J does not react when warmed with Tollens' reagent. Draw the displayed formula of compound J.
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(a) Using the formula for the number of carbon atoms: \[n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+} = \frac{100 \times 0.94}{1.1 \times 28.4} = \frac{94}{31.24} = 3.01 \approx 3\]
(b) Because J does not react with Tollens' reagent, it is a ketone rather than an aldehyde. With 3 carbon atoms and formula \(\text{C}_3\text{H}_6\text{O}\), J is propanone. The displayed formula must show every single atom and bond, with a central carbon double-bonded to an oxygen atom and single-bonded to two methyl carbons, each bonded to three hydrogen atoms.
Marking scheme
[M1] Correct calculation of number of carbon atoms: \(n = \frac{100 \times 0.94}{1.1 \times 28.4} = 3.01\) (evaluates to 3) [A] full expression shown leading clearly to 3
[M2] Displayed formula of propanone with all bonds explicitly drawn (including all \(\text{C}-\text{H}\) and \(\text{C}=\text{O}\) bonds). [R] condensed structural formula \(\text{CH}_3\text{COCH}_3\) or missing bonds
Question 23 · structured
2 marks
Compound K has the molecular formula \(\text{C}_3\text{H}_6\text{O}_2\). The infrared spectrum of K shows strong absorptions at \(1735\text{ cm}^{-1}\) and \(1240\text{ cm}^{-1}\), but shows no absorption in the region \(2500\text{--}3600\text{ cm}^{-1}\).
(a) Identify the functional group present in K.
(b) Suggest the structural formula of an isomer of K that exhibits a very broad absorption band in the region \(2500\text{--}3000\text{ cm}^{-1}\).
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Worked solution
(a) The strong absorption at \(1735\text{ cm}^{-1}\) corresponds to a carbonyl group (\(\text{C}=\text{O}\)), and the absorption at \(1240\text{ cm}^{-1}\) corresponds to a \(\text{C}-\text{O}\) single bond. The absence of any broad absorption above \(2500\text{ cm}^{-1}\) rules out carboxylic acids and alcohols, indicating that K is an ester.
(b) A broad absorption at \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the \(\text{O}-\text{H}\) group of a carboxylic acid. An isomer of \(\text{C}_3\text{H}_6\text{O}_2\) containing a carboxylic acid group is propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
[M2] \(\text{CH}_3\text{CH}_2\text{COOH}\) or \(\text{CH}_3\text{CH}_2\text{CO}_2\text{H}\) [A] displayed or skeletal formula of propanoic acid
Question 24 · structured
2 marks
Compound L is a mono-halogenated alkane. In the mass spectrum of L, two molecular ion peaks are observed at \(m/e = 122\) and \(m/e = 124\) with an intensity ratio of approximately \(1:1\). A major fragment ion peak is recorded at \(m/e = 43\).
(a) Deduce the identity of the halogen atom present in L. Explain your answer using the peak data provided.
(b) Give the formula of the fragment ion responsible for the peak at \(m/e = 43\), including its charge.
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Worked solution
(a) The presence of \([\text{M}]^+\) and \([\text{M}+2]^+\) peaks at \(m/e = 122\) and \(124\) in an approximate \(1:1\) ratio indicates the presence of a bromine atom, due to the natural isotopes \(^{79}\text{Br}\) and \(^{81}\text{Br}\) existing in approximately equal abundance.
(b) Subtracting the mass of bromine (\(79\)) from \(122\) gives \(122 - 79 = 43\). A mass of \(43\) corresponds to a propyl group, \(\text{C}_3\text{H}_7^+\) (\(3 \times 12 + 7 \times 1 = 43\)).
Marking scheme
[M1] Bromine / \(\text{Br}\) AND \([\text{M}]^+\) and \([\text{M}+2]^+\) peaks are of equal height / \(1:1\) ratio due to \(^{79}\text{Br}\) and \(^{81}\text{Br}\) having equal abundance. [R] Chlorine / \(\text{Cl}\)
[M2] \(\text{C}_3\text{H}_7^+\) or \(\text{CH}_3\text{CH}_2\text{CH}_2^+\) or \((\text{CH}_3)_2\text{CH}^+\) or \([\text{C}_3\text{H}_7]^+\) [R] formula without the positive charge
Question 25 · structured
2 marks
An organic compound W containing carbon, hydrogen, and oxygen gives a molecular ion peak at \(m/e = 74\) in its mass spectrum. The infrared spectrum of W shows a broad absorption at \(3200\text{--}3650\text{ cm}^{-1}\) but no absorption between \(1650\text{ cm}^{-1}\) and \(1750\text{ cm}^{-1}\). When W is heated under reflux with acidified potassium dichromate(VI), no reaction occurs and the solution remains orange.
(a) Identify the bond responsible for the absorption at \(3200\text{--}3650\text{ cm}^{-1}\) and state the classification of alcohol (primary, secondary, or tertiary) that compound W represents.
(b) Deduce the structural formula of W.
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Worked solution
(a) The broad absorption at \(3200\text{--}3650\text{ cm}^{-1}\) corresponds to the \(\text{O}-\text{H}\) bond of an alcohol. Since W is not oxidised by acidified potassium dichromate(VI) (solution remains orange), it must be a tertiary alcohol.
(b) An alcohol with \(M_r = 74\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) (\(4 \times 12 + 10 \times 1 + 16 = 74\)). The only tertiary alcohol with 4 carbons is 2-methylpropan-2-ol, which has the structural formula \((\text{CH}_3)_3\text{COH}\) or \(\text{CH}_3\text{C}(\text{CH}_3)(\text{OH})\text{CH}_3\).
Marking scheme
[M1] \(\text{O}-\text{H}\) (alcohol / hydroxy) bond AND tertiary (alcohol). [A] both required for 1 mark
[M2] \((\text{CH}_3)_3\text{COH}\) or \(\text{CH}_3\text{C}(\text{CH}_3)(\text{OH})\text{CH}_3\) or \(\text{C}(\text{CH}_3)_3\text{OH}\) [A] correct displayed or skeletal formula of 2-methylpropan-2-ol
Paper 3 (Advanced Practical Skills)
Complete all practical tasks in the laboratory. Record all raw data and observations systematically in tables provided. Carry out required quantitative calculations and qualitative deductions.
3 Question · 40 marks
Question 1 · practical
16 marks
In this experiment, you will determine the enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), for the reaction between aqueous sodium hydroxide and aqueous hydrochloric acid.
FA 1 is \(2.00\text{ mol dm}^{-3}\text{ NaOH(aq)}\). FA 2 is \(2.20\text{ mol dm}^{-3}\text{ HCl(aq)}\).
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### Method
1. Support an expanded polystyrene cup inside a \(250\text{ cm}^3\) glass beaker. 2. Using a measuring cylinder, transfer \(30.0\text{ cm}^3\) of FA 1 into the polystyrene cup. 3. Place the thermometer in the solution and start the stopwatch. Record the temperature of the solution in the cup every \(30\text{ s}\) for \(2.5\text{ min}\). 4. At \(t = 3.0\text{ min}\), use a second measuring cylinder to rapidly add \(30.0\text{ cm}^3\) of FA 2 to the cup. Do not record the temperature at \(t = 3.0\text{ min}\). 5. Stir the mixture thoroughly using the thermometer and continue to record the temperature every \(30\text{ s}\) from \(t = 3.5\text{ min}\) to \(t = 8.0\text{ min}\).
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(a) In the space below, prepare a single table to record all your temperature and time readings. Record all raw temperature measurements to the nearest \(0.5\,^\circ\text{C}\).
[4 marks]
(b) On the grid provided in the examination, plot a graph of temperature (\(y\)-axis) against time (\(x\)-axis).
1. Choose suitable scales so that your plotted points occupy at least half the grid in both directions. 2. Draw two straight lines of best fit: - one line through the points from \(t = 0\text{ min}\) to \(t = 2.5\text{ min}\); - one line through the cooling points from \(t = 4.5\text{ min}\) to \(t = 8.0\text{ min}\). 3. Extrapolate both lines of best fit to \(t = 3.0\text{ min}\). 4. Determine the theoretical temperature change, \(\Delta T\), at \(t = 3.0\text{ min}\).
$$\Delta T = \dots\dots\dots\dots\dots\,^\circ\text{C}$$
[4 marks]
(c) Calculations
Assume that the specific heat capacity of the final reaction mixture is \(c = 4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1}\) and that the density of the solution is \(1.00\text{ g cm}^{-3}\).
(i) Calculate the heat energy released, \(q\), in \(\text{J}\), during this neutralisation reaction. Show your working.
$$q = \dots\dots\dots\dots\dots\text{ J}$$
[2 marks]
(ii) Calculate the amount, in \(\text{mol}\), of FA 1 (\(\text{NaOH}\)) and of FA 2 (\(\text{HCl}\)) mixed together. Hence, state the amount, in \(\text{mol}\), of water formed in this reaction.
$$\text{amount of FA 1} = \dots\dots\dots\dots\dots\text{ mol}$$ $$\text{amount of FA 2} = \dots\dots\dots\dots\dots\text{ mol}$$ $$\text{amount of H}_2\text{O}\text{ formed} = \dots\dots\dots\dots\dots\text{ mol}$$
[2 marks]
(iii) Calculate the molar enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), in \(\text{kJ mol}^{-1}\), for the reaction between \(\text{NaOH(aq)}\) and \(\text{HCl(aq)}\).
(i) The thermometer used has graduations of \(1\,^\circ\text{C}\), and the maximum uncertainty for each reading is \(\pm 0.5\,^\circ\text{C}\). Calculate the maximum percentage uncertainty in your value of \(\Delta T\) (assume a temperature rise of \(13.5\,^\circ\text{C}\)).
(ii) A student repeats the experiment using \(30.0\text{ cm}^3\) of \(2.00\text{ mol dm}^{-3}\text{ CH}_3\text{COOH(aq)}\) (ethanoic acid) instead of \(30.0\text{ cm}^3\) of FA 2.
Predict and explain whether the magnitude of \(\Delta H_{\text{neut}}\) obtained for ethanoic acid will be greater than, less than, or the same as that for \(\text{HCl(aq)}\).
[1 mark]
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Worked solution
(a) Table format and raw readings: - Clearly structured single table with unambiguous headings and units: Time / min (or t / min) and Temperature / °C (or T / °C). - Readings recorded at exact half-minute intervals from \(0.0\) to \(2.5\text{ min}\) and from \(3.5\) to \(8.0\text{ min}\) (with \(3.0\text{ min}\) left blank or marked with a dash). - All raw temperature values recorded consistently to \(0.5\,^\circ\text{C}\).
(b) Graph plotting: - Both axes linear, scales chosen such that data points occupy \(\ge 50\%\) of grid along both \(x\) and \(y\) directions, clearly labeled with quantities and units. - All points plotted accurately to within half a small square. - Two smooth, straight lines of best fit drawn: horizontal/near-constant pre-addition line (\(0.0\text{--}2.5\text{ min}\)) and cooling line (\(4.5\text{--}8.0\text{ min}\)), both extrapolated to \(t = 3.0\text{ min}\). - Correct deduction of \(\Delta T = T_{\text{extrapolated, max}} - T_{\text{extrapolated, initial}}\). *(Typical exemplar value: \(T_1 = 20.5\,^\circ\text{C}\), \(T_2 = 34.0\,^\circ\text{C}\) \(\Rightarrow \Delta T = 13.5\,^\circ\text{C}\)).*
(c) Calculations (i) Total mass of reaction mixture: $$m = (30.0 + 30.0)\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 60.0\text{ g}$$ $$q = m \times c \times \Delta T = 60.0\text{ g} \times 4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1} \times 13.5\,^\circ\text{C} = 3385.8\text{ J} = 3.39\text{ kJ}$$
(ii) Amount of reactants: $$n(\text{FA 1, NaOH}) = 2.00\text{ mol dm}^{-3} \times \frac{30.0}{1000}\text{ dm}^3 = 0.0600\text{ mol}$$ $$n(\text{FA 2, HCl}) = 2.20\text{ mol dm}^{-3} \times \frac{30.0}{1000}\text{ dm}^3 = 0.0660\text{ mol}$$ Since \(\text{NaOH}\) is the limiting reactant: $$n(\text{H}_2\text{O}\text{ formed}) = 0.0600\text{ mol}$$
(iii) Enthalpy change of neutralisation: $$\Delta H_{\text{neut}} = -\frac{q}{1000 \times n(\text{H}_2\text{O})} = -\frac{3385.8}{1000 \times 0.0600} = -56.43\text{ kJ mol}^{-1}$$ Quoted to 3 significant figures with sign: **\(-56.4\text{ kJ mol}^{-1}\).
(d) Evaluation and Extension** (i) Total absolute uncertainty for two temperature readings \(= 2 \times 0.5 = 1.0\,^\circ\text{C}\). $$\text{Percentage uncertainty} = \frac{1.0}{13.5} \times 100 = 7.41\%$$ (allow \(3.70\%\) if based on single reading uncertainty of \(0.5\,^\circ\text{C}\)).
(ii) Less exothermic / smaller magnitude / less negative, because ethanoic acid is a weak acid (partially dissociated) and energy is taken in / absorbed to break \(\text{O-H}\) bonds / fully ionise the acid molecules.
Marking scheme
(a) [Total: 4 marks] - M1: Structured table containing unambiguous column/row headings with units: Time / min and Temperature / °C [1] - M2: All raw temperatures recorded consistently to the nearest \(0.5\,^\circ\text{C}\) [1] - M3: Complete set of readings taken at regular \(30\text{ s}\) intervals with no reading recorded at \(3.0\text{ min}\) [1] - M4: Initial temperature readings from \(0.0\) to \(2.5\text{ min}\) are constant or show steady baseline drift (within \(\pm 0.5\,^\circ\text{C}\)) [1]
(b) [Total: 4 marks] - M1: Axes correctly labelled with quantities/units and scales chosen such that plotted points occupy at least half the graph grid in both directions [1] - M2: All points plotted accurately to within half a small square [1] - M3: Two straight lines of best fit drawn correctly: initial baseline and cooling curve line from \(4.5\) to \(8.0\text{ min}\), both extrapolated to \(t = 3.0\text{ min}\) [1] - M4: Intercepts at \(t = 3.0\text{ min}\) correctly read from graph and \(\Delta T\) correctly calculated as \(T_{\text{final}} - T_{\text{initial}}\) [1]
(c) [Total: 6 marks] - (i)M1: Correct expression for \(q = 60.0 \times 4.18 \times \Delta T\) [1] - (i)M2: Correct evaluation of \(q\) in \(\text{J}\) (or \(\text{kJ}\) if unit is clearly stated) to 3 or 4 significant figures [1] - (ii)M1: Both moles calculated correctly: \(n(\text{NaOH}) = 0.0600\text{ mol}\) and \(n(\text{HCl}) = 0.0660\text{ mol}\) [1] - (ii)M2: Identifies \(\text{NaOH}\) as limiting reagent, hence \(n(\text{H}_2\text{O}) = 0.0600\text{ mol}\) [1] - (iii)M1: Calculation of \(\Delta H_{\text{neut}} = \frac{q}{n(\text{H}_2\text{O})}\) with conversion of \(\text{J}\) to \(\text{kJ}\) (ecf from (c)(i) and (c)(ii)) [1] - (iii)M2: Final answer given with negative sign (\(-\)) and quoted to 3 significant figures (e.g. \(-56.4\text{ kJ mol}^{-1}\)) [1]
(d) [Total: 2 marks] - (i)M1: Correct calculation of percentage uncertainty: \(\frac{1.0}{\Delta T} \times 100 = 7.41\%\) (or \(\frac{0.5}{\Delta T} \times 100 = 3.70\%\)) [1] - (ii)M1: States less exothermic / smaller magnitude / less negativeAND explains that ethanoic acid is a weak acid / partially dissociated, requiring energy/heat to complete ionisation / dissociate [1]
Question 2 · practical
11 marks
Read the entire question and plan your method before starting.
FB 1 is hydrated magnesium sulfate, \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\).
You will determine the value of \(x\) by heating FB 1 in a crucible to remove all water of crystallisation.
1. Weigh an empty, clean and dry crucible with its lid. 2. Add between 2.50\text{ g} and 3.00\text{ g} of FB 1 into the crucible. 3. Reweigh the crucible, lid and FB 1. 4. Place the crucible containing FB 1 with its lid on a pipe-clay triangle supported by a tripod. 5. Leave the lid slightly displaced to allow steam to escape. 6. Heat the crucible gently for 2 minutes, then heat strongly for 5 minutes. 7. Cover the crucible completely with the lid, place it on a heatproof mat and allow it to cool for at least 5 minutes. 8. Reweigh the crucible, lid and anhydrous residue. 9. Reheat strongly for a further 2 minutes, cool with the lid on, and reweigh. 10. Repeat heating, cooling and reweighing until constant mass is attained.
(a) In an appropriate format in the space below, record all mass measurements, including the final heating to constant mass, with appropriate headings and units.
(b) Calculations Show your working and appropriate units in all parts of your calculations. Give your answers to 3 significant figures. [\(A_{\text{r}}\): \(\text{H}, 1.0\); \(\text{O}, 16.0\); \(\text{Mg}, 24.3\); \(\text{S}, 32.1\)]
(i) Calculate the mass of FB 1 used, the mass of anhydrous \(\text{MgSO}_4\) residue formed, and the mass of \(\text{H}_2\text{O}\) lost.
(ii) Calculate the percentage by mass of water of crystallisation in FB 1.
(iii) Calculate the amount, in \(\text{mol}\), of anhydrous \(\text{MgSO}_4\) residue obtained.
(iv) Calculate the amount, in \(\text{mol}\), of \(\text{H}_2\text{O}\) driven off during heating.
(v) Calculate the experimental value of \(x\) in \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\) to the nearest integer.
(c) Explain why it is necessary to reheat the crucible and residue until two successive masses are constant to within 0.02\text{ g}.
(d) The balance used has an uncertainty of \(\pm 0.01\text{ g}\) for each individual reading. Calculate the percentage uncertainty in the mass of FB 1 measured.
(e) A student performing this experiment heated the crucible with an intense roaring flame for 15 minutes without a lid. The value of \(x\) calculated by the student was significantly greater than 7.
Suggest one procedural reason for this higher value of \(x\) and explain how this error led to the overestimated value.
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Worked solution
(a) Table of Results (Specimen Data)
| Measurement | Mass / g | | :--- | :---: | | Mass of empty crucible + lid | 21.45 | | Mass of crucible + lid + FB 1 | 24.20 | | Mass of crucible + lid + residue after 1st heating | 22.82 | | Mass of crucible + lid + residue after 2nd heating | 22.79 | | Mass of crucible + lid + residue after 3rd heating | 22.79 |
(All balance readings recorded consistently to 2 decimal places with clear headings and units.)
(c) To ensure that all the water of crystallisation has been completely driven off / to ensure thermal decomposition/dehydration is complete.
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(d) Since the mass of FB 1 requires two balance readings (empty crucible and crucible + salt): \[\text{Total absolute uncertainty} = 2 \times 0.01\text{ g} = 0.02\text{ g}\] \[\text{Percentage uncertainty} = \left( \frac{0.02}{2.75} \right) \times 100 = 0.727\%\ (3\text{ s.f.})\]
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(e) - Reason: Some solid jumped / spat out of the uncovered crucible during strong heating (OR: \(\text{MgSO}_4\) partially decomposed into \(\text{MgO}\) and \(\text{SO}_3\) at extremely high temperatures). - Explanation: This loss of solid increases the apparent difference in mass on heating, making the calculated mass of water lost larger than it actually was, thus yielding a higher mole ratio of water to anhydrous salt.
Marking scheme
Part (a) [2 marks] - M1: Table constructed with clear headings (mass of crucible + lid, mass with sample, successive masses after heating) and correct units recorded in standard notation (e.g. / g or (g)). [1] - M2: All balance readings recorded consistently to at least 2 decimal places and showing at least two reheatings demonstrating constant mass (within 0.02 g). [1]
Part (b) [5 marks] - (i) M1: Correct calculation of mass of FB 1, mass of anhydrous residue, and mass of water lost from candidate's readings. [1] - (ii) M2: Correct calculation of percentage by mass of water to 3 significant figures: \((\text{mass of water} / \text{mass of FB 1}) \times 100\). [1] - (iii) M3: Correct calculation of \(n(\text{MgSO}_4) = \text{mass of residue} / 120.4\), quoted to 3 s.f. (allow ecf). [1] - (iv) M4: Correct calculation of \(n(\text{H}_2\text{O}) = \text{mass of water} / 18.0\), quoted to 3 s.f. (allow ecf). [1] - (v) M5: Correct calculation of \(x = n(\text{H}_2\text{O}) / n(\text{MgSO}_4)\) and rounded to the nearest integer. [1]
Part (c) [1 mark] - M1: State that heating to constant mass ensures complete removal/evaporation of all water of crystallisation (or that the reaction/dehydration has gone to completion). [1] (Reject: 'to make it accurate' without qualification)
Part (d) [1 mark] - M1: Correct calculation of percentage uncertainty taking into account two weighings: \(((2 \times 0.01) / \text{mass of FB 1}) \times 100\). [1]
Part (e) [2 marks] - M1: Valid procedural reason: solid spat out / lost from crucible due to lack of lid / vigorous heating OR thermal decomposition of \(\text{MgSO}_4\) into \(\text{MgO}\) / gaseous sulfur oxides due to overheating. [1] - M2: Explaining that mass of residue is smaller than expected, which makes the measured mass loss (attributed to water) too large, increasing the calculated moles of water and hence increasing \(x\). [1]
Read through the whole question before starting and prepare a table for your observations where indicated.
(a) $\text{FA 6}$ is an aqueous solution containing two cations and one anion, all of which are listed in the Qualitative Analysis Notes.
Carry out the following tests on $\text{FA 6}$ and record your observations in a single table.
• Test 1: To a $1\text{ cm}$ depth of $\text{FA 6}$ in a test-tube, add aqueous sodium hydroxide dropwise until in excess. Then, warm the mixture gently and test any gas evolved with damp red litmus paper. • Test 2: To a $1\text{ cm}$ depth of $\text{FA 6}$ in a test-tube, add aqueous ammonia dropwise until in excess. • Test 3: To a $1\text{ cm}$ depth of $\text{FA 6}$ in a test-tube, add a few drops of aqueous barium nitrate followed by dilute nitric acid. • Test 4: To a $1\text{ cm}$ depth of $\text{FA 6}$ in a test-tube, add an equal volume of dilute sulfuric acid, then add a few drops of aqueous potassium manganate(VII).
(i) Record all your observations for Tests 1–4 in a suitable table.
(ii) Identify the ions present in $\text{FA 6}$. Cations: ....................................... and ....................................... Anion: .......................................
(iii) Construct an ionic equation, including state symbols, for the reaction occurring when aqueous sodium hydroxide is first added to $\text{FA 6}$.
(b) $\text{FA 7}$ is an aqueous solution containing one cation and one anion.
(i) Carry out the following tests on $\text{FA 7}$ and record your observations.
• Test 1: To a $1\text{ cm}$ depth of $\text{FA 7}$ in a test-tube, add aqueous sodium hydroxide dropwise until in excess. Observations: .........................................................................................................................
• Test 2: To a $1\text{ cm}$ depth of $\text{FA 7}$ in a test-tube, add a few drops of dilute nitric acid followed by aqueous silver nitrate. Then add aqueous ammonia (both dilute and concentrated) to the mixture. Observations: .........................................................................................................................
(ii) Identify the cation and anion present in $\text{FA 7}$. Cation: ....................................... Anion: .......................................
(iii) Suggest the type of reaction that occurs when excess aqueous sodium hydroxide is added to the precipitate formed in Test 1 on $\text{FA 7}$.
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Worked solution
(a)(i) Observations: • Test 1 (with $\text{NaOH}(\text{aq})$): Green precipitate formed, insoluble in excess $\text{NaOH}(\text{aq})$. On warming, a gas is evolved that turns damp red litmus paper blue. • Test 2 (with $\text{NH}_3(\text{aq})$): Green precipitate formed, insoluble in excess $\text{NH}_3(\text{aq})$. • Test 3 (with $\text{Ba}(\text{NO}_3)_2(\text{aq}) / \text{HNO}_3(\text{aq})$): White precipitate formed, insoluble in dilute $\text{HNO}_3(\text{aq})$. • Test 4 (with acidified $\text{KMnO}_4(\text{aq})$): Purple solution decolourises / turns colourless (or pale yellow/brown due to $\text{Fe}^{3+}$ formation).
(a)(ii) • $\text{Fe}^{2+}$ is identified by the green precipitate insoluble in excess $\text{NaOH}$ and excess $\text{NH}_3$. • $\text{NH}_4^+$ is identified by ammonia gas evolved upon warming with $\text{NaOH}$, which turns red litmus blue. • $\text{SO}_4^{2-}$ is identified by the white precipitate with $\text{Ba}^{2+}$ that does not dissolve in dilute acid.
(b)(i) • Test 1: Grey-green precipitate soluble in excess $\text{NaOH}(\text{aq})$ to give a dark green solution. • Test 2: Cream / off-white precipitate formed with $\text{AgNO}_3(\text{aq})$, which remains insoluble in dilute $\text{NH}_3(\text{aq})$ but dissolves in concentrated $\text{NH}_3(\text{aq})$.
(b)(ii) • Cation: $\text{Cr}^{3+}$ (from grey-green precipitate soluble in excess $\text{NaOH}$). • Anion: $\text{Br}^-$ (from cream precipitate with $\text{Ag}^+$ soluble in concentrated $\text{NH}_3$).
(a)(i) [5 marks total for table and observations] M1: Clearly tabulated observations for all 4 tests. M2: Test 1: Green precipitate insoluble in excess $\text{NaOH}(\text{aq})$ AND gas turns damp red litmus paper blue on warming. M3: Test 2: Green precipitate insoluble in excess $\text{NH}_3(\text{aq})$. M4: Test 3: White precipitate insoluble in dilute $\text{HNO}_3(\text{aq})$. M5: Test 4: Purple / pink solution turns colourless / decolourises (or turns pale yellow/orange). [Reject 'clear' for colourless]
(a)(ii) [3 marks] M6: Cation 1: $\text{Fe}^{2+}$ / $\text{iron(II)}$ (dependent on green precipitate observations). M7: Cation 2: $\text{NH}_4^+$ / $\text{ammonium}$ (dependent on alkaline gas / litmus turning blue). M8: Anion: $\text{SO}_4^{2-}$ / $\text{sulfate}$ (dependent on insoluble white precipitate with barium nitrate).
(a)(iii) [1 mark] M9: $\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})$ [Both correct species, balancing, and state symbols required]
(b)(i) [2 marks] M10: Test 1: Grey-green precipitate soluble in excess $\text{NaOH}(\text{aq})$ to form a dark green solution. M11: Test 2: Cream precipitate insoluble in dilute $\text{NH}_3(\text{aq})$ AND dissolves / soluble in concentrated $\text{NH}_3(\text{aq})$. [Accept 'off-white' for cream; reject 'yellow' or 'white']
(b)(ii) [1 mark] M12: Cation: $\text{Cr}^{3+}$ AND Anion: $\text{Br}^-$. [Both required for 1 mark; allow ecf from candidate's recorded observations in (b)(i)]
(b)(iii) [1 mark] M13: Amphoteric (behaviour / reaction) OR formation of a complex ion / complexation.
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