An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge International AS Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
卷一 (選擇題)
Answer all 40 multiple-choice questions. Each question has four options (A, B, C, D) and carries one mark.
40 題目 · 40 分
題目 1 · 選擇題
1 分
A gaseous hydrocarbon, \(\text{W}\), with a volume of \(20\text{ cm}^3\) is completely combusted in \(150\text{ cm}^3\) of excess oxygen. After cooling to room temperature, the total remaining gas volume is \(110\text{ cm}^3\). When this gaseous mixture is passed through an excess of aqueous sodium hydroxide, the volume decreases to \(30\text{ cm}^3\).
All gas volumes are measured under the same conditions of temperature and pressure.
What is the molecular formula of hydrocarbon \(\text{W}\)?
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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解題
Let the hydrocarbon be \(\text{C}_x\text{H}_y\).
The combustion equation is: \[ \text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l}) \]
1. Initial volume of \(\text{C}_x\text{H}_y = 20\text{ cm}^3\). 2. Initial volume of \(\text{O}_2 = 150\text{ cm}^3\). 3. After cooling to room temperature, water is liquid, so the remaining gas consists of unreacted \(\text{O}_2\) and produced \(\text{CO}_2\). Total volume \(= 110\text{ cm}^3\). 4. Aqueous \(\text{NaOH}\) absorbs \(\text{CO}_2\). The remaining gas is unreacted \(\text{O}_2 = 30\text{ cm}^3\). 5. Volume of \(\text{CO}_2\) produced \(= 110 - 30 = 80\text{ cm}^3\). 6. Therefore, \(20x = 80 \implies x = 4\). 7. Volume of \(\text{O}_2\) reacted \(= 150 - 30 = 120\text{ cm}^3\). 8. Using the stoichiometry: \(20\left(x + \frac{y}{4}\right) = 120 \implies x + \frac{y}{4} = 6\). 9. Since \(x = 4\), \(4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\).
Thus, the molecular formula is \(\text{C}_4\text{H}_8\).
Which species contains bond angles that are all equal to \(120^\circ\)?
A.\(\text{NH}_3\)
B.\(\text{CO}_3^{2-}\)
C.\(\text{ClF}_3\)
D.\(\text{H}_3\text{O}^+\)
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解題
A: \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair on nitrogen (trigonal pyramidal shape, bond angle \(\approx 107^\circ\)). B: \(\text{CO}_3^{2-}\) has 3 electron regions (resonance hybrid of two single bonds and one double bond) with 0 lone pairs on carbon. Its shape is trigonal planar with all bond angles exactly equal to \(120^\circ\). C: \(\text{ClF}_3\) has 3 bonding pairs and 2 lone pairs on chlorine (T-shaped, bond angles \(< 90^\circ\) and \(< 180^\circ\)). D: \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair on oxygen (trigonal pyramidal shape, bond angle \(\approx 107^\circ\)).
Therefore, \(\text{CO}_3^{2-}\) is the correct species.
評分準則
Award [1] for B (\(\text{CO}_3^{2-}\) has trigonal planar geometry with no lone pairs on the central atom).
題目 3 · 選擇題
1 分
Solid sodium iodide is treated with concentrated sulfuric acid.
Which row correctly identifies all the reduction products formed by the reduction of sulfur in this reaction?
A.\(\text{SO}_2\) only
B.\(\text{SO}_2\) and \(\text{S}\) only
C.\(\text{H}_2\text{S}\) only
D.\(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\)
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解題
Iodide ions (\(\text{I}^-\)) are strong reducing agents. When solid \(\text{NaI}\) reacts with concentrated \(\text{H}_2\text{SO}_4\), sulfur (oxidation state \(+6\) in \(\text{H}_2\text{SO}_4\)) is reduced to: 1. Sulfur dioxide, \(\text{SO}_2\) (oxidation state \(+4\)) 2. Elemental sulfur, \(\text{S}\) (oxidation state \(0\)) 3. Hydrogen sulfide, \(\text{H}_2\text{S}\) (oxidation state \(-2\))
Therefore, the reduction products containing sulfur are \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\).
評分準則
Award [1] for D (\(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\)).
A/B/C are incorrect because iodide reduces concentrated sulfuric acid through multiple steps forming \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\).
題目 4 · 選擇題
1 分
An alkene, \(\text{Q}\), with the molecular formula \(\text{C}_6\text{H}_{12}\) is heated under reflux with acidified potassium manganate(VII). Carbon dioxide and a single carboxylic acid, \(\text{R}\), are formed in a \(1:1\) molar ratio.
What is the identity of alkene \(\text{Q}\)?
A.hex-3-ene
B.hex-2-ene
C.hex-1-ene
D.2-methylpent-1-ene
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解題
Hot concentrated acidified \(\text{KMnO}_4\) causes oxidative cleavage of alkene double bonds: - A terminal \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - 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 \(\text{CO}_2\) and a single carboxylic acid with 5 carbons (to keep the total at 6 carbons) are produced in a \(1:1\) ratio, the alkene must have the structure \(\text{CH}_2=\text{CH}-\text{R}\) where \(\text{R}\) has 4 carbons (i.e., a hex-1-ene derivative without branching at the \(\text{C}=2\) carbon).
- Hex-1-ene (\(\text{CH}_2=\text{CH(CH}_2)_3\text{CH}_3\)) oxidises to give \(\text{CO}_2\) and pentanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH}\)). This fits all criteria. - Hex-2-ene gives ethanoic acid and butanoic acid (two acids, no \(\text{CO}_2\)). - Hex-3-ene gives propanoic acid only (no \(\text{CO}_2\)). - 2-Methylpent-1-ene has \(=\text{C(CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\), which gives \(\text{CO}_2\) and pentan-2-one (a ketone, not a carboxylic acid).
評分準則
Award [1] for C (hex-1-ene).
A, B: do not produce \(\text{CO}_2\). D: produces a ketone and \(\text{CO}_2\).
題目 5 · 選擇題
1 分
Successive ionisation energies for element \(\text{E}\) are shown in the table.
In which group of the Periodic Table is element \(\text{E}\) located?
A.Group 2
B.Group 14
C.Group 15
D.Group 16
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解題
Inspect the jumps in successive ionisation energies: - \(1\text{st}\) to \(6\text{th}\): gradual increases from \(1000\) to \(8496\text{ kJ mol}^{-1}\). - \(6\text{th}\) to \(7\text{th}\): a very large jump from \(8496\text{ kJ mol}^{-1}\) to \(27107\text{ kJ mol}^{-1}\) (an increase of \(\approx 18600\text{ kJ mol}^{-1}\)).
This huge increase between the 6th and 7th ionisation energies indicates that the 7th electron is removed from an inner quantum shell. Therefore, element \(\text{E}\) has 6 outer-shell (valence) electrons and belongs to Group 16.
評分準則
Award [1] for D (Group 16).
Key reasoning: Large jump between 6th and 7th ionisation energy values confirms 6 valence electrons.
題目 6 · 選擇題
1 分
Nitrogen oxides, \(\text{NO}_x\), act as homogeneous catalysts in the atmosphere.
Which reaction demonstrates the role of nitrogen dioxide, \(\text{NO}_2\), as a catalyst in the oxidation of atmospheric sulfur dioxide?
In the atmosphere, \(\text{NO}_2\) oxidises \(\text{SO}_2\) to \(\text{SO}_3\) and is reduced to \(\text{NO}\): \[ \text{SO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{SO}_3(\text{g}) + \text{NO}(\text{g}) \]
The catalyst \(\text{NO}_2\) is then regenerated by the reaction of \(\text{NO}\) with atmospheric oxygen: \[ \text{NO}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}_2(\text{g}) \]
Thus, the step showing \(\text{NO}_2\) acting as the oxidant for \(\text{SO}_2\) is \(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\).
評分準則
Award [1] for A (\(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\)).
題目 7 · 選擇題
1 分
At a temperature of \(500\text{ K}\), \(0.400\text{ mol}\) of \(\text{PCl}_5(\text{g})\) is placed into a sealed container of volume \(2.00\text{ dm}^3\). The system is allowed to reach equilibrium:
Common errors: - Omitting the volume (using moles directly gives \(\frac{0.160^2}{0.240} = 0.107\), option C). - Incorrect equilibrium mole of \(\text{PCl}_5\).
題目 8 · 選擇題
1 分
An organic compound, \(\text{X}\), gives an orange precipitate with 2,4-dinitrophenylhydrazine, but does not form a red precipitate when warmed with Fehling's solution. \(\text{X}\) also forms a yellow precipitate when reacted with alkaline aqueous iodine.
Which structure could represent compound \(\text{X}\)?
1. Orange precipitate with 2,4-DNPH indicates the presence of a carbonyl group (aldehyde or ketone). 2. No reaction (no red precipitate) with Fehling's solution indicates that \(\text{X}\) is a ketone, not an aldehyde. 3. Yellow precipitate (tri-iodomethane, \(\text{CHI}_3\)) with alkaline aqueous iodine indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C}=\text{O}\).
Let's analyse the options: A: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (butan-2-one) is a ketone containing the \(\text{CH}_3\text{C}=\text{O}\) group. It meets all three criteria. B: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\) (butanal) is an aldehyde, so it would react with Fehling's solution. C: \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\) (pentan-3-one) is a ketone but does not contain a \(\text{CH}_3\text{C}=\text{O}\) group, so it gives a negative tri-iodomethane test. D: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) (butan-2-ol) is an alcohol, so it would not react with 2,4-DNPH.
評分準則
Award [1] for A (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)).
- B is rejected because it reduces Fehling's. - C is rejected because it lacks \(\text{CH}_3\text{C}=\text{O}\). - D is rejected because it is not a carbonyl.
題目 9 · 選擇題
1 分
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(150\text{ cm}^3\) of oxygen gas (an excess).
After cooling the mixture to room temperature, the total volume of remaining gas is \(110\text{ cm}^3\). When this remaining gas is shaken with an excess of aqueous sodium hydroxide, the volume decreases to \(30\text{ cm}^3\).
All gas volumes are 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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解題
1. The combustion reaction is: \[\text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})\]
2. At room temperature, water is a liquid, so the \(110\text{ cm}^3\) of gas remaining consists of unreacted \(\text{O}_2\) and produced \(\text{CO}_2\).
3. Aqueous \(\text{NaOH}\) absorbs \(\text{CO}_2\). The remaining gas is unreacted \(\text{O}_2 = 30\text{ cm}^3\). Therefore, the volume of \(\text{CO}_2\) produced is: \[110\text{ cm}^3 - 30\text{ cm}^3 = 80\text{ cm}^3\]
4. The volume of \(\text{O}_2\) reacted is: \[150\text{ cm}^3 - 30\text{ cm}^3 = 120\text{ cm}^3\]
5. Using mole/volume ratios from \(20\text{ cm}^3\) of \(\text{C}_x\text{H}_y\): \[x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{80}{20} = 4\] \[x + \frac{y}{4} = \frac{V(\text{O}_2\text{ reacted})}{V(\text{C}_x\text{H}_y)} = \frac{120}{20} = 6\] \[4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\]
Thus, the formula is \(\text{C}_4\text{H}_8\).
評分準則
M1: Correct deduction of \(x = 4\) and \(y = 8\) leading to option C.
題目 10 · 選擇題
1 分
Which species is non-linear and has exactly two lone pairs of electrons on its central atom?
A.\(\text{NH}_2^-\)
B.\(\text{CO}_2\)
C.\(\text{SF}_4\)
D.\(\text{ICl}_2^-\)
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解題
In the amide ion, \(\text{NH}_2^-\): - Nitrogen has 5 valence electrons + 1 electron from the negative charge = 6 valence electrons. - Forms 2 single covalent bonds with hydrogen atoms (2 bonding pairs). - The remaining 4 electrons form 2 lone pairs. - With 2 bonding pairs and 2 lone pairs, the electron pair arrangement is tetrahedral, giving a non-linear (bent) molecular shape.
Other options: - \(\text{CO}_2\): 2 double bonds, 0 lone pairs on carbon (linear). - \(\text{SF}_4\): 4 bonding pairs, 1 lone pair on sulfur (see-saw). - \(\text{ICl}_2^-\): 2 bonding pairs, 3 lone pairs on iodine (linear).
評分準則
M1: Correct identification of \(\text{NH}_2^-\), which has 2 bond pairs and 2 lone pairs on nitrogen.
題目 11 · 選擇題
1 分
Solid samples of potassium halides are reacted separately with concentrated sulfuric acid.
Which statement describes a correct observation or deduction?
A.Potassium chloride reduces concentrated sulfuric acid to sulfur dioxide.
B.Potassium bromide produces a mixture of acidic steamy fumes and brown vapour.
C.Potassium iodide produces only hydrogen iodide and potassium hydrogensulfate.
D.The reducing ability of the halide ions decreases from chloride to iodide.
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解題
- \(\text{KCl}\) reacts with concentrated \(\text{H}_2\text{SO}_4\) in an acid-base reaction forming \(\text{HCl}\) (steamy fumes) and \(\text{KHSO}_4\); \(\text{Cl}^-\) cannot reduce \(\text{H}_2\text{SO}_4\). - \(\text{KBr}\) forms \(\text{HBr}\) (steamy acidic fumes), and \(\text{Br}^-\) is a strong enough reducing agent to reduce \(\text{H}_2\text{SO}_4\) to \(\text{SO}_2\), producing orange-brown \(\text{Br}_2\) vapour. - \(\text{KI}\) is strongly reducing and produces \(\text{I}_2\), \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\), not just \(\text{HI}\). - The reducing ability of halide ions increases down Group 17 (from \(\text{Cl}^-\) to \(\text{I}^-\)).
評分準則
M1: Recognition that \(\text{KBr}\) produces both steamy fumes (\(\text{HBr}\)) and brown vapour (\(\text{Br}_2\)).
題目 12 · 選擇題
1 分
An alkene \(\text{X}\) is heated under reflux with acidified potassium manganate(VII).
The only organic products obtained are propanone, \(\text{CH}_3\text{COCH}_3\), and propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
Oxidative cleavage of alkenes with hot concentrated acidified \(\text{KMnO}_4\): - A \(=\text{C(CH}_3)_2\) group oxidises to propanone, \(\text{CH}_3\text{COCH}_3\). - A \(=\text{CH}-\text{CH}_2\text{CH}_3\) group oxidises to propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
Combining these two fragments across the double bond gives: \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\) (2-methylpent-2-ene).
評分準則
M1: Correct identification of \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\).
題目 13 · 選擇題
1 分
The first eight successive ionisation energies of an element \(\text{Z}\), in \(\text{kJ mol}^{-1}\), are shown.
Which formula represents a stable binary compound of element \(\text{Z}\) with nitrogen?
A.\(\text{ZN}\)
B.\(\text{ZN}_2\)
C.\(\text{Z}_3\text{N}_2\)
D.\(\text{Z}_2\text{N}_3\)
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解題
1. Look at the jump between successive ionisation energies: - \(\text{IE}_1 = 738\text{ kJ mol}^{-1}\) - \(\text{IE}_2 = 1451\text{ kJ mol}^{-1}\) - \(\text{IE}_3 = 7733\text{ kJ mol}^{-1}\)
2. The large jump occurs between \(\text{IE}_2\) and \(\text{IE}_3\), indicating that element \(\text{Z}\) has 2 valence electrons and belongs to Group 2 of the Periodic Table (forming \(\text{Z}^{2+}\) ions).
3. Nitrogen forms nitride ions, \(\text{N}^{3-}\).
4. Combining \(\text{Z}^{2+}\) and \(\text{N}^{3-}\) gives the empirical formula \(\text{Z}_3\text{N}_2\).
評分準則
M1: Deduction of Group 2 from the jump between \(\text{IE}_2\) and \(\text{IE}_3\) leading to \(\text{Z}_3\text{N}_2\).
題目 14 · 選擇題
1 分
In the atmosphere, sulfur dioxide is oxidised to sulfur trioxide in a reaction catalysed by nitrogen dioxide.
Which pair of equations represents the mechanism for this homogeneous catalytic process?
1. Positive 2,4-DNPH test indicates a carbonyl compound (aldehyde or ketone). 2. Negative Tollens' test confirms that \(\text{Q}\) is a ketone (not an aldehyde). 3. Positive tri-iodomethane test (pale yellow precipitate with alkaline \(\text{I}_2\)) indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C=O}\). 4. Butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) is a methyl ketone and satisfies all three criteria.
Analysis of other options: - Propanal: reacts with Tollens' reagent. - Propan-2-ol: does not react with 2,4-DNPH. - Pentan-3-one: does not contain a \(\text{CH}_3\text{C=O}\) group, so gives a negative tri-iodomethane test.
評分準則
M1: Correct identification of butan-2-one matching all qualitative test results.
題目 16 · 選擇題
1 分
Ammonia is manufactured by the Haber process according to the following reversible reaction: \[\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H^\circ = -92\text{ kJ mol}^{-1}\]
Which row correctly describes the effect of increasing the total pressure of the system at constant temperature on the equilibrium yield of \(\text{NH}_3\) and on the value of \(K_p\)?
1. Equilibrium yield: According to Le Chatelier's principle, an increase in total pressure favours the side of the reaction with fewer moles of gas. There are 4 moles of gas on the left and 2 moles of gas on the right, so the position of equilibrium shifts to the right, increasing the yield of \(\text{NH}_3\).
2. Equilibrium constant: \(K_p\) depends only on temperature. Since temperature is kept constant, the value of \(K_p\) remains unchanged.
評分準則
M1: Selection of row B.
題目 17 · 選擇題
1 分
A \(0.150\text{ g}\) sample of an unbranched hydrocarbon gas occupies \(120\text{ cm}^3\) at room temperature and pressure (r.t.p.).
Under identical conditions of temperature and pressure, the molar volume of a gas is \(24.0\text{ dm}^3\text{ mol}^{-1}\).
What is the molecular formula of the hydrocarbon?
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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解題
First, calculate the amount of the gas in moles: \[n = \frac{V}{V_m} = \frac{120 \times 10^{-3}\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 5.00 \times 10^{-3}\text{ mol}\]
Next, determine the relative molecular mass (\(M_r\)): \[M_r = \frac{\text{mass}}{n} = \frac{0.150\text{ g}}{5.00 \times 10^{-3}\text{ mol}} = 30.0\text{ g mol}^{-1}\]
What is the formula of the chloride formed by element \(X\)?
A.\(X\text{Cl}\)
B.\(X\text{Cl}_2\)
C.\(X\text{Cl}_3\)
D.\(X\text{Cl}_4\)
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解題
Look at the differences between successive ionisation energies: - 1st to 2nd: \(1817 - 578 = 1239\text{ kJ mol}^{-1}\) - 2nd to 3rd: \(2745 - 1817 = 928\text{ kJ mol}^{-1}\) - 3rd to 4th: \(11578 - 2745 = 8833\text{ kJ mol}^{-1}\) (very large jump)
The large jump occurs between the 3rd and 4th ionisation energies, which indicates that the 4th electron is being removed from an inner quantum shell. Thus, element \(X\) has 3 valence electrons and belongs to Group 13 (Period 3 element is aluminium, \(\text{Al}\)).
The chloride formed by a Group 13 element has the empirical/molecular formula \(X\text{Cl}_3\) (e.g. \(\text{AlCl}_3\)).
評分準則
C is correct [1].
Identification of Group 13 from the large jump between the 3rd and 4th ionisation energies, giving formula \(X\text{Cl}_3\).
題目 19 · 選擇題
1 分
Which species has the smallest bond angle?
A.\(\text{NH}_4^+\)
B.\(\text{H}_2\text{S}\)
C.\(\text{BF}_3\)
D.\(\text{NH}_3\)
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解題
Apply VSEPR theory: - \(\text{NH}_4^+\): Central \(\text{N}\) has 4 bonding pairs, 0 lone pairs. Shape is tetrahedral with bond angle \(109.5^\circ\). - \(\text{H}_2\text{S}\): Central \(\text{S}\) has 2 bonding pairs, 2 lone pairs. Non-linear/bent with bond angle \(\approx 92^\circ\) to \(104.5^\circ\) (due to lone pair-lone pair repulsion). - \(\text{BF}_3\): Central \(\text{B}\) has 3 bonding pairs, 0 lone pairs. Trigonal planar with bond angle \(120^\circ\). - \(\text{NH}_3\): Central \(\text{N}\) has 3 bonding pairs, 1 lone pair. Trigonal pyramidal with bond angle \(107^\circ\).
Among the choices, \(\text{H}_2\text{S}\) has 2 lone pairs on the central atom and therefore the smallest bond angle (less than \(104.5^\circ\)).
評分準則
B is correct [1].
\(\text{H}_2\text{S}\) has two bonding pairs and two lone pairs around sulfur, resulting in the greatest repulsion and the smallest bond angle.
題目 20 · 選擇題
1 分
An aqueous solution of sodium halide \(Y\) gives a pale cream precipitate when treated with aqueous silver nitrate acidified with dilute nitric acid.
This precipitate is partially soluble in dilute aqueous ammonia and dissolves completely in concentrated aqueous ammonia.
What is the identity of halogen halide ion in \(Y\)?
A.Chloride, \(\text{Cl}^-\)
B.Bromide, \(\text{Br}^-\)
C.Iodide, \(\text{I}^-\)
D.Fluoride, \(\text{F}^-\)
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解題
The qualitative analysis test for halides: - \(\text{Cl}^-\) forms a white precipitate of \(\text{AgCl}\), soluble in dilute \(\text{NH}_3\text{(aq)}\). - \(\text{Br}^-\) forms a cream / pale cream precipitate of \(\text{AgBr}\), insoluble or partially soluble in dilute \(\text{NH}_3\text{(aq)}\), but soluble in concentrated \(\text{NH}_3\text{(aq)}\). - \(\text{I}^-\) forms a yellow precipitate of \(\text{AgI}\), insoluble in both dilute and concentrated \(\text{NH}_3\text{(aq)}\). - \(\text{F}^-\) forms no precipitate because \(\text{AgF}\) is soluble.
Therefore, halide \(Y\) is bromide, \(\text{Br}^-\).
評分準則
B is correct [1].
Cream precipitate of \(\text{AgBr}\) dissolving in concentrated \(\text{NH}_3\text{(aq)}\) confirms bromide.
題目 21 · 選擇題
1 分
An alkene \(Z\) 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, \(\text{CH}_3\text{CH}_2\text{COOH}\).
Under hot, concentrated acidified \(\text{KMnO}_4\) oxidation, the carbon-carbon double bond is cleaved: - A \(=\text{C(R)(R')}\) group forms a ketone (here, \(=\text{C(CH}_3\text{)}_2\) gives propanone). - A \(=\text{CH(R)}\) group forms a carboxylic acid (here, \(=\text{CH(CH}_2\text{CH}_3\text{)}\) gives propanoic acid).
Reassembling these two fragments by replacing the carbonyl oxygens with a double bond: \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\), which is 2-methylpent-2-ene.
評分準則
C is correct [1].
Oxidative cleavage of \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\) produces propanone from the \((\text{CH}_3)_2\text{C}=\) end and propanoic acid from the \(=\text{CHCH}_2\text{CH}_3\) end.
題目 22 · 選擇題
1 分
Which compound forms a yellow precipitate when warmed with alkaline aqueous iodine AND can be oxidised by acidified potassium dichromate(VI) to form a carboxylic acid?
A.\(\text{CH}_3\text{CHO}\)
B.\(\text{CH}_3\text{CH(OH)CH}_3\)
C.\(\text{CH}_3\text{COCH}_3\)
D.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)
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解題
1. To give a yellow precipitate (tri-iodomethane, \(\text{CHI}_3\)) with alkaline aqueous iodine, the molecule must contain either the \(\text{CH}_3\text{CO}-\) group or the \(\text{CH}_3\text{CH(OH)}-\) group. 2. To be oxidised to a carboxylic acid with acidified potassium dichromate(VI), the molecule must be a primary alcohol (\(-\text{CH}_2\text{OH}\)) or an aldehyde (\(-\text{CHO}\)).
Let us evaluate the options: - \(\text{CH}_3\text{CHO}\) (ethanal): Contains \(\text{CH}_3\text{CO}-\) (positive tri-iodomethane test) and is an aldehyde, so it oxidises to ethanoic acid (a carboxylic acid). This fits both conditions. - \(\text{CH}_3\text{CH(OH)CH}_3\) (propan-2-ol): Positive tri-iodomethane test, but oxidises to a ketone (propanone), not a carboxylic acid. - \(\text{CH}_3\text{CH}_2\text{OH}\) (ethanol): Contains \(\text{CH}_3\text{CH(OH)}-\) and is a primary alcohol, but on oxidation with dichromate it yields ethanoic acid. However, \(\text{CH}_3\text{CHO}\) is listed. - \(\text{CH}_3\text{COCH}_3\) (propanone): Positive tri-iodomethane test, but resistant to oxidation by acidified potassium dichromate(VI).
Both \(\text{CH}_3\text{CHO}\) and \(\text{CH}_3\text{CH}_2\text{OH}\) can oxidise to carboxylic acids; checking the options given: ethanal is option A.
評分準則
A is correct [1].
Ethanal contains the \(\text{CH}_3\text{C=O}\) group giving a positive iodoform test, and as an aldehyde it oxidises to ethanoic acid.
題目 23 · 選擇題
1 分
In the atmosphere, nitrogen monoxide, \(\text{NO}\), acts as a catalyst in the oxidation of sulfur dioxide, \(\text{SO}_2\), to sulfur trioxide, \(\text{SO}_3\).
Which reaction represents a propagation step in this catalytic mechanism?
\(\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3\) represents the reduction of nitrogen dioxide while oxidising sulfur dioxide.
題目 24 · 選擇題
1 分
Consider the following reversible reaction at dynamic equilibrium: \[2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)} \qquad \Delta H = -196\text{ kJ mol}^{-1}\]
Which change will increase the equilibrium yield of \(\text{SO}_3\text{(g)}\)?
A.Decreasing the temperature at constant pressure
B.Increasing the temperature at constant volume
C.Decreasing the total pressure at constant temperature
D.Adding a vanadium(V) oxide catalyst
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解題
To increase the equilibrium yield of \(\text{SO}_3\text{(g)}\), the equilibrium position must shift to the right (forward direction): - The forward reaction is exothermic (\(\Delta H < 0\)). Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right), increasing the yield of \(\text{SO}_3\). - Increasing pressure favours the side with fewer gas moles (3 moles of gas on left \(\rightarrow\) 2 moles of gas on right), which also shifts the equilibrium to the right. - Adding a catalyst does not change the position of equilibrium or the yield.
Option A states 'decreasing the temperature at constant pressure', which shifts the equilibrium to the right and increases yield.
評分準則
A is correct [1].
Decreasing the temperature shifts the equilibrium in the exothermic (forward) direction, increasing the yield of \(\text{SO}_3\).
題目 25 · 選擇題
1 分
A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(70\text{ cm}^3\) of oxygen (an excess).
After cooling to room temperature and pressure, the remaining gas volume is \(50\text{ cm}^3\).
When this remaining gas is shaken with excess aqueous potassium hydroxide, the total volume decreases by \(30\text{ cm}^3\).
What is the molecular 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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解題
1. The decrease in volume upon treatment with \(\text{KOH(aq)}\) corresponds to the volume of \(\text{CO}_2\text{(g)}\) absorbed: \[ V(\text{CO}_2) = 30\text{ cm}^3 \]
2. Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(30\text{ cm}^3\) of \(\text{CO}_2\): \[ x = \frac{30}{10} = 3 \]
3. The remaining gas after \(\text{KOH}\) absorption is unreacted \(\text{O}_2\): \[ V(\text{unreacted } \text{O}_2) = 50 - 30 = 20\text{ cm}^3 \]
4. Therefore, the volume of \(\text{O}_2\) consumed in the combustion is: \[ V(\text{reacted } \text{O}_2) = 70 - 20 = 50\text{ cm}^3 \]
5. For the general combustion reaction: \[ \text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O(l)} \] \[ x + \frac{y}{4} = \frac{50}{10} = 5 \] \[ 3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8 \]
Thus, the hydrocarbon is \(\text{C}_3\text{H}_8\).
評分準則
C is correct [1].
Method: - Identification of \(\text{CO}_2\) volume as \(30\text{ cm}^3\) to deduce \(x = 3\). - Calculation of reacted \(\text{O}_2 = 50\text{ cm}^3\) to deduce \(y = 8\).
題目 26 · 選擇題
1 分
Which species has bond angles that are all exactly \(120^\circ\)?
A.\(\text{ClF}_3\)
B.\(\text{NO}_3^-\
C.\(\text{PCl}_3\)
D.\(\text{SO}_3^{2-}\)
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解題
- \(\text{ClF}_3\) has 3 bonding pairs and 2 lone pairs around chlorine (T-shaped, bond angles \(<90^\circ\)). - \(\text{NO}_3^-\) has 3 regions of electron density around the central nitrogen atom with zero lone pairs, giving a trigonal planar shape with bond angles of exactly \(120^\circ\). - \(\text{PCl}_3\) has 3 bonding pairs and 1 lone pair around phosphorus (trigonal pyramidal, bond angle \(\approx 107^\circ\)). - \(\text{SO}_3^{2-}\) has 3 bonding pairs and 1 lone pair around sulfur (trigonal pyramidal, bond angle \(\approx 107^\circ\)).
評分準則
B is correct [1]. \(\text{NO}_3^-\) has 3 electron regions and no lone pairs on the central N atom, hence trigonal planar with \(120^\circ\) bond angles.
題目 27 · 選擇題
1 分
The first six successive ionisation energies of a Period 3 element, \(\text{X}\), are given in the table.
Which statement about element \(\text{X}\) is correct?
A.The oxide of \(\text{X}\) dissolves in water to form an alkaline solution.
B.Element \(\text{X}\) forms a chloride with the formula \(\text{XCl}_3\) that reacts with water to form an acidic solution.
C.Element \(\text{X}\) is a good electrical conductor in the solid state.
D.The first ionisation energy of \(\text{X}\) is lower than that of the element immediately to its left in Period 3.
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解題
1. There is a very large jump in ionisation energy between the 5th and 6th ionisation energies (from \(6274\text{ kJ mol}^{-1}\) to \(21268\text{ kJ mol}^{-1}\)). 2. This indicates that element \(\text{X}\) has 5 valence electrons and belongs to Group 15 of the Periodic Table. In Period 3, element \(\text{X}\) is phosphorus (\(\text{P}\)). 3. Phosphorus forms chlorides such as \(\text{PCl}_3\) and \(\text{PCl}_5\). \(\text{PCl}_3\) hydrolyses in water according to \(\text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3\text{HCl}\), producing a strongly acidic solution. 4. Statements A, C, and D are incorrect: the oxide \(\text{P}_4\text{O}_{10}\) is acidic, \(\text{P}\) is a non-metal, and the first ionisation energy of \(\text{P}\) is higher than that of silicon due to higher nuclear charge (and higher than sulfur due to paired electron repulsion in sulfur).
評分準則
B is correct [1]. - Deduces Group 15 from large jump between 5th and 6th IEs. - Identifies \(\text{X}\) as phosphorus and \(\text{PCl}_3\) yielding an acidic solution.
題目 28 · 選擇題
1 分
Solid potassium bromide, \(\text{KBr}\), is heated with concentrated sulfuric acid.
Which substance is not produced in this reaction?
A.\(\text{Br}_2\)
B.\(\text{HBr}\)
C.\(\text{H}_2\text{S}\)
D.\(\text{SO}_2\)
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解題
1. When solid \(\text{KBr}\) reacts with concentrated \(\text{H}_2\text{SO}_4\), two reactions occur: - Acid-base reaction: \(\text{KBr} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HBr}\) - Redox reaction: Bromide ions reduce sulfuric acid to sulfur dioxide: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) 2. Bromide ions (\(\text{Br}^-\)) are not strong enough reducing agents to reduce sulfur from \(+6\) to \(-2\) (as \(\text{H}_2\text{S}\)); only iodide ions (\(\text{I}^-\)) are sufficiently strong reducing agents to produce \(\text{H}_2\text{S}\). 3. Therefore, \(\text{H}_2\text{S}\) is not formed.
評分準則
C is correct [1]. \(\text{Br}^-\) only reduces \(\text{H}_2\text{SO}_4\) as far as \(\text{SO}_2\); it cannot form \(\text{H}_2\text{S}\).
題目 29 · 選擇題
1 分
A reaction between two gases is carried out at temperature \(T_1\) and then repeated at a higher temperature \(T_2\).
Which statement correctly explains the increase in reaction rate at \(T_2\) using the Maxwell–Boltzmann distribution?
A.The activation energy of the reaction is reduced at \(T_2\).
B.The most probable kinetic energy of the molecules decreases and the peak shifts to the left.
C.A greater proportion of molecules have kinetic energy equal to or greater than the activation energy.
D.The total area under the Maxwell–Boltzmann distribution curve increases.
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解題
1. Increasing the temperature does not change the activation energy (\(E_a\)) of the reaction; \(E_a\) is a constant for an uncatalysed pathway. 2. At a higher temperature \(T_2\), the Maxwell–Boltzmann distribution curve flattens and its peak shifts to the right (higher energy). 3. The total area under the curve represents the total number of particles, which remains constant. 4. The fraction (or proportion) of molecules with kinetic energy greater than or equal to \(E_a\) (the area under the curve to the right of \(E_a\)) increases significantly, leading to a higher frequency of successful collisions.
評分準則
C is correct [1]. A greater proportion of molecules have \(E \ge E_a\) at higher temperature.
題目 30 · 選擇題
1 分
An organic compound \(\text{W}\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
- Compound \(\text{W}\) is oxidised by acidified potassium dichromate(VI) to form compound \(\text{X}\). - Compound \(\text{X}\) forms an orange precipitate with 2,4-dinitrophenylhydrazine. - Compound \(\text{X}\) gives no visible change when warmed with Tollens' reagent. - Compound \(\text{W}\) forms a yellow precipitate when warmed with alkaline aqueous iodine.
1. Compound \(\text{W}\) is an alcohol of formula \(\text{C}_4\text{H}_{10}\text{O}\). 2. Oxidation of \(\text{W}\) gives \(\text{X}\), which reacts with 2,4-DNPH, indicating \(\text{X}\) is a carbonyl compound (aldehyde or ketone). 3. Compound \(\text{X}\) does not react with Tollens' reagent, so \(\text{X}\) is a ketone. This proves \(\text{W}\) must be a secondary alcohol. 4. Compound \(\text{W}\) forms a yellow precipitate (\(\text{CHI}_3\)) with alkaline aqueous iodine, confirming it contains the \(\text{CH}_3\text{CH(OH)}-\) group. 5. The only 4-carbon secondary alcohol is butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
評分準則
B is correct [1]. - \(\text{W}\) is a secondary alcohol oxidised to a ketone (no reaction with Tollens'). - Positive tri-iodomethane test confirms \(\text{CH}_3\text{CH(OH)}-\) structure.
題目 31 · 選擇題
1 分
Two separate samples of Period 3 chlorides, \(\text{Y}\) and \(\text{Z}\), are added to water at room temperature.
- Chloride \(\text{Y}\) dissolves readily to give a solution with a \(\text{pH}\) of approximately 7. - Chloride \(\text{Z}\) reacts vigorously with water, producing dense steamy fumes and a solution with a \(\text{pH}\) of approximately 2.
Which row correctly identifies \(\text{Y}\) and \(\text{Z}\)?
1. Period 3 chlorides and their reactions with water: - \(\text{NaCl}\): Giant ionic lattice, dissolves without hydrolysis to form a neutral solution (\(\text{pH} \approx 7\)). Thus, \(\text{Y}\) is \(\text{NaCl}\). - \(\text{MgCl}_2\): Dissolves with slight hydrolysis to give a slightly acidic solution (\(\text{pH} \approx 6.5\)). - \(\text{AlCl}_3\): Hydrolyses to give an acidic solution (\(\text{pH} \approx 3\)). - \(\text{SiCl}_4\): Simple covalent liquid, undergoes vigorous hydrolysis with water: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\), evolving dense steamy fumes of \(\text{HCl}\text{(g)}\) and forming a strongly acidic solution (\(\text{pH} \approx 1-2\)). Thus, \(\text{Z}\) is \(\text{SiCl}_4\). 2. Therefore, row B correctly identifies \(\text{Y}\) as \(\text{NaCl}\) and \(\text{Z}\) as \(\text{SiCl}_4\).
評分準則
B is correct [1]. \(\text{NaCl}\) gives a neutral solution (\(\text{pH} \approx 7\)) and \(\text{SiCl}_4\) hydrolyses vigorously to release \(\text{HCl}\) fumes giving \(\text{pH} \approx 2\).
題目 32 · 選擇題
1 分
An alkene \(\text{J}\) is heated under reflux with hot concentrated acidified potassium manganate(VII).
The only organic product formed is propanone, \(\text{CH}_3\text{COCH}_3\).
1. Oxidative cleavage of alkenes with hot concentrated acidified \(\text{KMnO}_4\): - \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - \(=\text{CHR}\) group is oxidised to a carboxylic acid, \(\text{RCOOH}\). - \(=\text{CR}_1\text{R}_2\) group is oxidised to a ketone, \(\text{R}_1\text{COR}_2\). 2. Since the only organic product obtained is propanone (\(\text{CH}_3\text{COCH}_3\)), both carbons of the \(\text{C}=\text{C}\) double bond must be bonded to two methyl groups. 3. The alkene is 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\). 4. Reaction: \[ (\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 + 2[\text{O}] \rightarrow 2\text{CH}_3\text{COCH}_3 \]
評分準則
D is correct [1]. Cleavage of \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\) yields two molecules of propanone as the sole organic product.
題目 33 · multiple_choice
1 分
A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(70\text{ cm}^3\) of oxygen gas (an excess). After cooling to room conditions, the remaining gaseous volume is \(50\text{ cm}^3\). When this residual gas is passed through aqueous sodium hydroxide, the volume decreases by \(30\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_4\)
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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解題
The general equation for the complete combustion of a hydrocarbon is: $$\text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})$$
1. At room temperature and pressure, water is a liquid with negligible volume. 2. The reduction in volume upon passing through aqueous \(\text{NaOH}\) represents the volume of acidic \(\text{CO}_2\) gas absorbed: $$\text{Volume of } \text{CO}_2 = 30\text{ cm}^3$$ Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) reacted: $$x = \frac{30\text{ cm}^3}{10\text{ cm}^3} = 3$$
3. The residual gas before absorption (\(50\text{ cm}^3\)) consists of \(\text{CO}_2\) and unreacted \(\text{O}_2\): $$\text{Volume of unreacted } \text{O}_2 = 50 - 30 = 20\text{ cm}^3$$ $$\text{Volume of } \text{O}_2 \text{ consumed} = 70 - 20 = 50\text{ cm}^3$$
4. Using the stoichiometry for \(\text{O}_2\): $$\left(x + \frac{y}{4}\right) \times 10 = 50 \implies 3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8$$
Thus, the hydrocarbon is \(\text{C}_3\text{H}_8\).
評分準則
[1] for correct deduction of the molecular formula (C).
題目 34 · multiple_choice
1 分
Which species has exactly two lone pairs of electrons on its central atom and a non-linear (bent) shape?
A.\(\text{CO}_2\)
B.\(\text{SO}_2\)
C.\(\text{NH}_2^-\)
D.\(\text{ClF}_3\)
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解題
1. \(\text{CO}_2\): The central C atom has 4 valence electrons, forms 2 double bonds to oxygen, and has 0 lone pairs. Shape: linear. 2. \(\text{SO}_2\): The central S atom has 6 valence electrons, forms 2 bonds to oxygen atoms, and has 1 lone pair. Shape: non-linear (bent). 3. \(\text{NH}_2^-\): The central N atom has 5 valence electrons + 1 electron from the negative charge = 6 electrons in total. It forms 2 single covalent bonds to H atoms (using 2 electrons), leaving 4 non-bonding electrons, which make 2 lone pairs. With 2 bonding pairs and 2 lone pairs, the electron geometry is tetrahedral and the molecular shape is non-linear (bent). 4. \(\text{ClF}_3\): The central Cl atom has 7 valence electrons, forms 3 single bonds to F, and has 2 lone pairs. Shape: T-shaped.
評分準則
[1] for identifying NH2- as the correct species (C).
題目 35 · multiple_choice
1 分
The first six successive ionisation energies of element \(X\), in \(\text{kJ mol}^{-1}\), are shown:
578, 1817, 2745, 11578, 14842, 18379
What are the formulas of the chloride and sulfate formed by element \(X\)?
1. Look for the largest ratio / jump between successive ionisation energies: - \(\text{IE}_1 = 578\) - \(\text{IE}_2 = 1817\) - \(\text{IE}_3 = 2745\) - \(\text{IE}_4 = 11578\)
The large jump occurs between the 3rd and 4th ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)), showing that the 4th electron is removed from an inner quantum shell. Therefore, element \(X\) has 3 valence electrons and forms stable \(X^{3+}\) ions (Group 13 element).
2. The chloride ion is \(\text{Cl}^-\), so the chloride formed is \(X\text{Cl}_3\). 3. The sulfate ion is \(\text{SO}_4^{2-}\), so the sulfate formed is \(X_2(\text{SO}_4)_3\).
評分準則
[1] for correct deduction of group number and formula of both compounds (B).
題目 36 · multiple_choice
1 分
A solid sodium halide, \(\text{Na}Y\), reacts with concentrated sulfuric acid to produce steamy fumes, a red-brown liquid/vapour, and a choking gas that turns acidified potassium dichromate(VI) paper from orange to green.
What is the identity of \(Y^-\) and what role is played by concentrated sulfuric acid in producing the red-brown substance?
A.\(Y^-\): \(\text{Cl}^-\); Role of \(\text{H}_2\text{SO}_4\): acid
B.\(Y^-\): \(\text{Br}^-\); Role of \(\text{H}_2\text{SO}_4\): oxidising agent
C.\(Y^-\): \(\text{Br}^-\); Role of \(\text{H}_2\text{SO}_4\): reducing agent
D.\(Y^-\): \(\text{I}^-\); Role of \(\text{H}_2\text{SO}_4\): oxidising agent
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解題
1. Bromide ions (\(\text{Br}^-\)) react with concentrated sulfuric acid in two ways: - Acid-base reaction: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (steamy fumes of \(\text{HBr}\)) - Redox reaction: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)
2. \(\text{Br}_2\) is a red-brown liquid/vapour, and \(\text{SO}_2\) is a choking gas which acts as a reducing agent and turns acidified potassium dichromate(VI) from orange (\(\text{Cr}^{6+}\)) to green (\(\text{Cr}^{3+}\)).
3. In the redox reaction, \(\text{H}_2\text{SO}_4\) oxidises \(\text{Br}^-\) to \(\text{Br}_2\) while itself being reduced to \(\text{SO}_2\). Thus, concentrated \(\text{H}_2\text{SO}_4\) acts as an oxidising agent.
評分準則
[1] for identifying bromide and the role as oxidising agent (B).
題目 37 · multiple_choice
1 分
An alkene, \(Q\), is heated under reflux with hot concentrated acidified potassium manganate(VII). Only one organic product, butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)), is formed.
What is the identity of alkene \(Q\)?
A.2,3-dimethylbut-2-ene
B.3,4-dimethylhex-3-ene
C.2-methylhex-2-ene
D.3-methylpent-2-ene
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解題
Under hot concentrated acidified \(\text{KMnO}_4\), the \(\text{C}=\text{C}\) double bond is cleaved oxidatively: - \(=\text{CH}_2\) fragments oxidise to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - \(=\text{CH}R\) fragments oxidise to carboxylic acids \(R\text{COOH}\). - \(=\text{C}R_1R_2\) fragments oxidise to ketones \(R_1\text{CO}R_2\).
Since the only organic product is the ketone butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)), alkene \(Q\) must be symmetrical and composed of two \(=\text{C}(\text{CH}_3)(\text{CH}_2\text{CH}_3)\) units linked by the double bond: $$\text{CH}_3\text{CH}_2\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3$$ This is 3,4-dimethylhex-3-ene.
評分準則
[1] for identifying 3,4-dimethylhex-3-ene (B).
題目 38 · multiple_choice
1 分
In atmospheric chemistry, nitrogen oxides catalyse the oxidation of sulfur dioxide to sulfur trioxide as shown:
Which statement concerning this catalytic cycle is correct?
A.\(\text{NO}_2\) acts as a heterogeneous catalyst.
B.The oxidation number of nitrogen in the catalyst decreases from \(+4\) to \(+2\) in Step 1.
C.\(\text{SO}_2\) acts as an oxidising agent in Step 1.
D.\(\text{NO}\) is an intermediate that contains a nitrogen atom with an oxidation number of \(+4\).
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解題
1. In Step 1, \(\text{NO}_2\) is converted to \(\text{NO}\). The oxidation state of nitrogen in \(\text{NO}_2\) is \(+4\) and in \(\text{NO}\) it is \(+2\). Thus, the oxidation state of nitrogen decreases from \(+4\) to \(+2\) (it is reduced). 2. Both the catalyst (\(\text{NO}_2\)) and reactants (\(\text{SO}_2, \text{O}_2\)) are in the gas phase, so it is a homogeneous catalyst, not heterogeneous. 3. In Step 1, \(\text{SO}_2\) is oxidised to \(\text{SO}_3\) (oxidation state of S increases from \(+4\) to \(+6\)), so \(\text{SO}_2\) acts as a reducing agent. 4. \(\text{NO}\) contains nitrogen with an oxidation state of \(+2\), not \(+4\).
評分準則
[1] for selecting B.
題目 39 · multiple_choice
1 分
An organic compound \(W\) gives a pale yellow precipitate when gently warmed with alkaline aqueous iodine.
When \(W\) is heated under reflux with acidified potassium dichromate(VI), the solution turns from orange to green to give organic compound \(Z\). Compound \(Z\) gives no reaction when heated with Fehling's solution.
What could be the structure of compound \(W\)?
A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)
B.\(\text{CH}_3\text{CH(OH)}\text{CH}_3\)
C.\(\text{CH}_3\text{CH}_2\text{CHO}\)
D.(\(\text{CH}_3)_3\text{COH}\)
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解題
1. Positive tri-iodomethane test (pale yellow precipitate of \(\text{CHI}_3\) with alkaline \(\text{I}_2\)): \(W\) must contain either a \(\text{CH}_3\text{C}=\text{O}\) group or a \(\text{CH}_3\text{CH(OH)}-\) group. 2. \(W\) is oxidised by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) (orange to green), confirming it is an alcohol (a secondary methyl alcohol, \(\text{CH}_3\text{CH(OH)}R\)). 3. The oxidation product \(Z\) is a ketone, \(\text{CH}_3\text{CO}R\), which does not reduce Fehling's solution. 4. Evaluating the options: - A: propan-1-ol (primary alcohol, negative tri-iodomethane test). - B: propan-2-ol (contains \(\text{CH}_3\text{CH(OH)}-\), positive tri-iodomethane test, oxidised to propanone which gives no reaction with Fehling's). - C: propanal (an aldehyde, reacts with Fehling's). - D: 2-methylpropan-2-ol (tertiary alcohol, cannot be oxidised by acidified dichromate).
評分準則
[1] for identifying propan-2-ol (B).
題目 40 · multiple_choice
1 分
The reversible reaction between gases \(\text{A}\) and \(\text{B}\) to form gas \(\text{C}\) reaches dynamic equilibrium:
Which single change increases both the equilibrium yield of \(\text{C}(\text{g})\) and the value of the equilibrium constant, \(K_p\)?
A.increasing the total pressure at constant temperature
B.decreasing the total pressure at constant temperature
C.decreasing the temperature at constant pressure
D.adding a suitable catalyst at constant temperature and pressure
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解題
1. Effect of temperature: The forward reaction is exothermic (\(\Delta H^\circ < 0\)). According to Le Chatelier's principle, decreasing temperature favours the exothermic direction, shifting the equilibrium to the right and increasing the yield of \(\text{C}(\text{g})\). Furthermore, temperature is the only factor that alters the value of an equilibrium constant (\(K_p\)); for an exothermic forward reaction, lowering temperature increases \(K_p\). 2. Effect of pressure: Increasing total pressure shifts the position of equilibrium to the side with fewer moles of gas (forward direction, increasing the yield of \(\text{C}\)), but it does not change the value of \(K_p\). 3. Adding a catalyst increases the rate of both forward and reverse reactions equally, having no effect on equilibrium yield or \(K_p\).
Answer all questions in the spaces provided. Show all working in calculations and include appropriate units and state symbols where required.
6 題目 · 60 分
題目 1 · structured
17 分
Silicon and phosphorus form several covalent hydrides and chlorides with distinctive structural, physical, and chemical properties.
(a) Silicon tetrachloride, $\text{SiCl}_4$, is a volatile liquid at room temperature.
(i) Draw a 'dot-and-cross' diagram of a molecule of $\text{SiCl}_4$. Show outer shell electrons only. [1]
(ii) State the shape of the $\text{SiCl}_4$ molecule and deduce the bond angle. Explain this shape in terms of the Valence Shell Electron Pair Repulsion (VSEPR) theory. [3]
(iii) Silicon dioxide, $\text{SiO}_2$, is a solid with a melting point of over $1600\text{ }^\circ\text{C}$, whereas silicon tetrachloride, $\text{SiCl}_4$, boils at $57\text{ }^\circ\text{C}$. Explain this difference in boiling / melting points in terms of structure and bonding. [3]
(b) A gaseous sample of silane, $\text{SiH}_4$, has a mass of $0.481\text{ g}$ and occupies a volume of $375\text{ cm}^3$ at a temperature of $35.0\text{ }^\circ\text{C}$ and a pressure of $102\text{ kPa}$.
(i) Use the ideal gas equation, $pV = nRT$, to calculate the relative molecular mass, $M_r$, of this sample of silane. Give your answer to 3 significant figures. ($R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$) [3]
(ii) State two assumptions of the kinetic theory of ideal gases that are not fully obeyed by real gases such as silane. [2]
(c) Gaseous silane reacts vigorously with chlorine gas according to the equation: $$\text{SiH}_4(\text{g}) + 4\text{Cl}_2(\text{g}) \rightarrow \text{SiCl}_4(\text{g}) + 4\text{HCl}(\text{g})$$
Use the bond energy values in the table to calculate the standard enthalpy change of reaction, $\Delta H_r$, in $\text{kJ mol}^{-1}$, for this reaction.
(d) Phosphine, $\text{PH}_3$, is the hydride of phosphorus.
(i) State the shape and bond angle of a molecule of phosphine, $\text{PH}_3$. [1]
(ii) The boiling point of ammonia, $\text{NH}_3$, is $240\text{ K}$, whereas the boiling point of phosphine, $\text{PH}_3$, is $185\text{ K}$. Explain this difference in boiling points with reference to the intermolecular forces present in each compound. [2]
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解題
(a)(i) A central $\text{Si}$ atom sharing 4 pairs of electrons (one with each $\text{Cl}$ atom). Each $\text{Cl}$ has 6 non-bonding valence electrons (3 lone pairs). Total 8 electrons around $\text{Si}$ and 8 around each $\text{Cl}$.
(a)(ii) Shape: tetrahedral. Bond angle: $109.5^\circ$. Explanation: There are 4 bonding pairs (regions of electron density) and 0 lone pairs around the central silicon atom. Electron pairs repel to achieve minimum repulsion / maximum separation.
(a)(iii) $\text{SiO}_2$ has a giant covalent (macromolecular) lattice with many strong covalent bonds throughout the structure, requiring a very large amount of thermal energy to break. $\text{SiCl}_4$ has a simple molecular structure with weak instantaneous dipole-induced dipole (London dispersion) forces between separate molecules, requiring much less energy to overcome.
(b)(ii) Real gas particles occupy a non-negligible volume (the volume of particles is not zero relative to container volume), and there are attractive/intermolecular forces between the gas molecules (collisions are not purely elastic / forces exist between particles).
(d)(i) Shape: trigonal pyramidal (or pyramidal). Bond angle: $93.5^\circ$ (accept any value in range $92^\circ - 107^\circ$).
(d)(ii) $\text{NH}_3$ has hydrogen bonding between molecules (as nitrogen is highly electronegative and has a lone pair bonded to H). $\text{PH}_3$ cannot form hydrogen bonds and only has permanent dipole-dipole forces and London dispersion forces between molecules. Hydrogen bonds are stronger than permanent dipole-dipole / London dispersion forces, requiring more energy to overcome.
評分準則
(a)(i) [1 mark] - Correct dot-and-cross diagram showing 4 shared pairs between central $\text{Si}$ and four $\text{Cl}$ atoms, and 6 non-bonding electrons (3 lone pairs) on each $\text{Cl}$ atom. [1]
(a)(ii) [3 marks] - Tetrahedral AND $109.5^\circ$ [1] - States that there are 4 bonding pairs / 4 electron pairs / regions of electron density around Si and 0 lone pairs [1] - Electron pairs repel equally to achieve maximum separation / minimum repulsion [1]
(a)(iii) [3 marks] - $\text{SiO}_2$ is giant covalent / macromolecular AND $\text{SiCl}_4$ is simple molecular [1] - Melting $\text{SiO}_2$ requires breaking strong covalent bonds [1] - Boiling $\text{SiCl}_4$ only requires overcoming weak intermolecular / London dispersion forces [1]
(b)(i) [3 marks] - Correct conversions of $p$ to $\text{Pa}$, $V$ to $\text{m}^3$, and $T$ to $\text{K}$ ($102000\text{ Pa}$, $3.75 \times 10^{-4}\text{ m}^3$, $308\text{ K}$) [1] - Correct rearrangement / substitution into $pV = \frac{m}{M_r}RT$ [1] - Final answer: $32.2$ (must be 3 sig figs) [1]
(b)(ii) [2 marks] - Gas molecules/particles have zero / negligible volume (compared to volume of gas) [1] - There are no intermolecular forces / no forces of attraction or repulsion between molecules [1]
(c) [2 marks] - Correct energy for bonds broken ($+2240\text{ kJ}$) AND bonds formed ($3312\text{ kJ}$) [1] - Correct value and negative sign: $-1072\text{ kJ mol}^{-1}$ [1]
(d)(i) [1 mark] - Trigonal pyramidal (or pyramidal) AND bond angle in range $92^\circ - 107^\circ$ [1]
(d)(ii) [2 marks] - $\text{NH}_3$ has hydrogen bonding between molecules, whereas $\text{PH}_3$ has permanent dipole-dipole / London dispersion forces [1] - Hydrogen bonds are stronger than dipole-dipole / London dispersion forces (so more thermal energy is needed to overcome them) [1]
題目 2 · structured
9 分
Methanol can be manufactured industrially by the reversible reaction between carbon monoxide and hydrogen.
(a)(i) Write the expression for the equilibrium constant, $K_c$, for this reaction and state its units.
$K_c =$
units = [2]
(ii) A mixture containing $0.400\text{ mol}$ of $\text{CO(g)}$ and $0.800\text{ mol}$ of $\text{H}_2\text{(g)}$ is sealed in a container of volume $2.50\text{ dm}^3$ and allowed to reach equilibrium at temperature $T_1$.
At equilibrium, the mixture contains $0.150\text{ mol}$ of $\text{CH}_3\text{OH(g)}$.
Calculate the value of $K_c$ at temperature $T_1$.
$K_c =$ [3]
(b) State and explain the effect on the equilibrium yield of methanol when the overall pressure of the system is increased at constant temperature. [2]
(c) A sample of pure methanol vapour, $\text{CH}_3\text{OH(g)}$, of mass $0.768\text{ g}$ exerts a pressure of $1.05 \times 10^5\text{ Pa}$ in a rigid vessel of volume $6.20 \times 10^{-4}\text{ m}^3$.
Use the ideal gas equation, $pV = nRT$, to calculate the temperature of the methanol vapour in kelvin, $\text{K}$.
Give your answer to 3 significant figures. ($R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$; $M_r(\text{CH}_3\text{OH}) = 32.0$)
(b) - Yield of methanol increases / greater. - The forward reaction results in a decrease in the number of moles of gas (3 moles of gas on the left $\rightarrow$ 1 mole of gas on the right), so an increase in pressure shifts the equilibrium to the right to oppose the increase in pressure.
(a)(ii) - M1: Deduces correct equilibrium amounts of $\text{CO} = 0.250\text{ mol}$ and $\text{H}_2 = 0.500\text{ mol}$ [1] - M2: Divides all equilibrium amounts by $2.50$ to obtain concentrations ($[\text{CO}] = 0.100$, $[\text{H}_2] = 0.200$, $[\text{CH}_3\text{OH}] = 0.0600$) and correctly substitutes into $K_c$ expression [1] - M3: Correct final answer: $15.0$ (or $15$) [1] (Allow ECF from arithmetic/mole errors if logic is consistent)
(b) - M1: (Yield of methanol) increases [1] - M2: Equilibrium shifts to the right / product side because there are fewer moles/molecules of gas on the right-hand side (or 3 moles on left vs 1 mole on right) [1]
(c) - M1: Correct calculation of $n = \frac{0.768}{32.0} = 0.0240\text{ mol}$ (or $0.02397\text{ mol}$) OR correctly arranges $T = \frac{pV}{nR}$ [1] - M2: Correct calculation of $T = 326\text{ K}$ (accept $326.4$ or $327$) to 3 significant figures [1]
題目 3 · structured
10 分
Solid sodium halides undergo characteristic reactions when treated with concentrated sulfuric acid, \(\text{H}_2\text{SO}_4\).
(a) (i) State what is observed when concentrated sulfuric acid is added to solid sodium chloride, and construct a balanced chemical equation for the reaction that takes place. [2]
(ii) When concentrated sulfuric acid is added to solid sodium iodide, redox reactions occur producing sulfur dioxide, \(\text{SO}_2\), and hydrogen sulfide, \(\text{H}_2\text{S}\). - State the oxidation state of sulfur in \(\text{H}_2\text{SO}_4\) and in \(\text{H}_2\text{S}\). - Construct an ionic half-equation for the reduction of \(\text{H}_2\text{SO}_4\) to form \(\text{H}_2\text{S}\). - Explain, in terms of atomic structure, why sodium iodide is able to reduce concentrated sulfuric acid to \(\text{H}_2\text{S}\), whereas sodium bromide only reduces it to \(\text{SO}_2\). [4]
(b) Chlorine undergoes a disproportionation reaction when bubbled into cold dilute aqueous sodium hydroxide. (i) Write a balanced equation for the reaction of chlorine with cold, dilute aqueous sodium hydroxide. [1] (ii) Explain, in terms of changes in oxidation numbers, why this reaction is classified as a disproportionation reaction. [1]
(c) A solution contains both chloride ions, \(\text{Cl}^-\), and iodide ions, \(\text{I}^-\). Describe a testing procedure using aqueous silver nitrate and aqueous ammonia (dilute and concentrated) that allows the presence of both halide ions to be confirmed. [2]
(a)(ii) - Oxidation states: - In \(\text{H}_2\text{SO}_4\): \(+6\) - In \(\text{H}_2\text{S}\): \(-2\) - Ionic half-equation: \[\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\] *(or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\))* - Explanation: - Iodide ions, \(\text{I}^-\), have a larger ionic radius / more electron shells / more shielding than bromide ions, \(\text{Br}^-\). - There is a weaker electrostatic attraction between the nucleus and the outermost electrons in \(\text{I}^-\), making it easier for \(\text{I}^-\) to lose an electron (it is a stronger reducing agent).
(b)(ii) - The oxidation state of chlorine in \(\text{Cl}_2\) is \(0\). - Chlorine is simultaneously oxidised to \(+1\) in \(\text{NaClO}\) (or \(\text{ClO}^-\)) and reduced to \(-1\) in \(\text{NaCl}\) (or \(\text{Cl}^-\)).
(c) - Add aqueous silver nitrate, \(\text{AgNO}_3(\text{aq})\), (acidified with dilute \(\text{HNO}_3\)) to produce a mixture of precipitates (white \(\text{AgCl}\) and yellow \(\text{AgI}\)). - Add dilute \(\text{NH}_3(\text{aq})\): \(\text{AgCl}\) dissolves (confirming chloride), leaving the yellow precipitate of \(\text{AgI}\) which is insoluble in both dilute and concentrated \(\text{NH}_3(\text{aq})\) (confirming iodide).
(a)(ii) [4 marks] - M1: Both oxidation numbers correct: \(+6\) in \(\text{H}_2\text{SO}_4\) AND \(-2\) in \(\text{H}_2\text{S}\). [1] - M2: Correct balanced half-equation: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) OR \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\). [1] - M3: \(\text{I}^-\) is larger / has more shells / greater shielding than \(\text{Br}^-\). [1] - M4: Weaker nuclear attraction for valence electrons / outer electron is lost more easily in \(\text{I}^-\) (so \(\text{I}^-\) is a stronger reducing agent). [1]
(b)(ii) [1 mark] - M1: Chlorine is simultaneously oxidised (from \(0\) to \(+1\)) and reduced (from \(0\) to \(-1\)) in the same reaction. [1]
(c) [2 marks] - M1: Add \(\text{AgNO}_3(\text{aq})\) to obtain a precipitate / yellow and white precipitate. [1] - M2: Add dilute \(\text{NH}_3(\text{aq})\) where \(\text{AgCl}\) dissolves / precipitate partially dissolves, AND yellow \(\text{AgI}\) remains insoluble (in both dilute and concentrated \(\text{NH}_3\)). [1]
題目 4 · structured
9 分
Propene, \(\text{CH}_3\text{CH}=\text{CH}_2\), can be used as the starting material in the organic synthesis shown below.
(a) When propene reacts with \(\text{HBr}\) in Step 1, compound B is formed as the major organic product. - State the structural formula of B. - Explain why B is formed as the major product rather than 1-bromopropane. [3]
(b) Propan-2-ol is converted into propanone, \(\text{CH}_3\text{COCH}_3\), in Step 3. (i) State the reagents and conditions required for Step 3. [1] (ii) Describe the colour change observed during the reaction in Step 3. [1]
(c) In Step 4, propanone reacts with hydrogen cyanide in the presence of a catalytic amount of sodium cyanide to form compound E. (i) State the role of \(\text{NaCN}\) in this reaction. [1] (ii) State the name of the mechanism of this reaction. [1]
(d) Describe a chemical test to distinguish between propanone, \(\text{CH}_3\text{COCH}_3\), and propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\). State the reagent used and the expected observation with each compound. [2]
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解題
(a) - Formula of B: \(\text{CH}_3\text{CHBrCH}_3\) (2-bromopropane). - Explanation: The reaction proceeds via a carbocation intermediate. The secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_3\)) is more stable than the primary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)) because of the greater positive inductive (electron-donating) effect of two alkyl / methyl groups compared to one.
(b) (i) Acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\), with heat (or heat under reflux). (ii) The solution changes from orange to green (due to the reduction of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\)).
(c) (i) It acts as a catalyst / provides the nucleophile (\(\text{CN}^-\)). (ii) Nucleophilic addition.
(d) - Reagent: Tollens' reagent (ammoniacal silver nitrate) [or Fehling's solution]. - Observation with propanal: Silver mirror / grey-black precipitate formed [or brick-red precipitate with Fehling's]. - Observation with propanone: No reaction / remains colourless [or remains blue with Fehling's].
評分準則
(a) - M1: Structural formula of B: \(\text{CH}_3\text{CHBrCH}_3\) [1] - M2: (Forms via a) secondary carbocation which is more stable than the primary carbocation [1] - M3: Due to the electron-donating / positive inductive effect of two alkyl / methyl groups (compared to one) [1]
(b) - (i) Acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4\) AND heat / reflux [1] *(Accept \(\text{H}^+ / \text{Cr}_2\text{O}_7^{2-}\). Reject \(\text{KMnO}_4\) if specified for mild oxidation to ketone without acid, or if conditions missing.)* - (ii) Orange to green [1]
(c) - (i) Catalyst / provides \(\text{CN}^-\) (ions) / generates nucleophile [1] - (ii) Nucleophilic addition [1] (Both words required. Do not accept electrophilic addition.)
(d) - M1: Tollens' reagent (or ammoniacal silver nitrate) AND silver mirror / grey-black precipitate with propanal [1] (OR Fehling's solution / Benedict's solution AND red / brick-red precipitate with propanal) (OR Acidified potassium dichromate(VI) AND orange to green with propanal) - M2: No change / no reaction / remains colourless (or remains blue / orange) with propanone [1]
題目 5 · null
0 分
題目 6 · structured
15 分
Compound X is an unbranched alkene.
(a) In the mass spectrum of compound X, the molecular ion peak, $[\text{M}]^+$, occurs at $m/e = 70$ with a relative abundance of $28.4\%$. The $[\text{M}+1]^+$ peak occurs at $m/e = 71$ with a relative abundance of $1.56\%$.
(i) Use these data to show that a molecule of X contains 5 carbon atoms. Show your working. [2]
(ii) Deduce the molecular formula of X. [1]
(b) Compound X exhibits stereoisomerism.
(i) State the systematic IUPAC name of compound X. [1]
(ii) State the two structural features of X that allow it to show stereoisomerism. [2]
(iii) Draw the displayed formulas of the cis and trans stereoisomers of X. [2]
(c) Compound X reacts with hydrogen bromide, $\text{HBr}\text{(g)}$, to form a mixture of two structural isomers, Y and Z (2-bromopentane and 3-bromopentane).
(i) Outline the mechanism for the reaction of X with $\text{HBr}$ to form 2-bromopentane.
Include curly arrows, relevant dipoles, lone pairs, and full charges. [4]
(ii) Explain why 2-bromopentane and 3-bromopentane are formed in approximately equal quantities in this reaction. [1]
(d) Compound X can undergo addition polymerisation under suitable conditions.
(i) State what is meant by the term addition polymerisation. [1]
(ii) Draw the structure of the addition polymer formed from X, showing two repeat units. [1]
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解題
(a)(i) Using the formula for the number of carbon atoms: $$n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } [\text{M}]^+}$$ $$n = \frac{100 \times 1.56}{1.1 \times 28.4} = \frac{156}{31.24} = 4.994 \approx 5$$ Hence, the number of carbon atoms is 5.
(a)(ii) For an alkene with 5 carbon atoms: $$\text{Molecular formula} = \text{C}_5\text{H}_{10}$$ ($M_{\text{r}} = (5 \times 12.0) + (10 \times 1.0) = 70.0$, matching $m/e = 70$).
(b)(i) Since X is an unbranched alkene with 5 carbons that shows stereoisomerism, the double bond must be at carbon-2. Therefore, X is pent-2-ene. (Note: pent-1-ene has two hydrogen atoms on C1 and cannot show stereoisomerism).
(b)(ii) 1. Restricted/hindered rotation around the $\text{C=C}$ double bond (due to the presence of a $\pi$ bond). 2. Each of the two carbon atoms of the $\text{C=C}$ double bond is attached to two different groups/atoms (C2 is bonded to $-\text{H}$ and $-\text{CH}_3$; C3 is bonded to $-\text{H}$ and $-\text{CH}_2\text{CH}_3$).
(b)(iii) - cis-pent-2-ene / (Z)-pent-2-ene: High priority/alkyl groups ($-\text{CH}_3$ and $-\text{CH}_2\text{CH}_3$) on the same side of the double bond; both $-\text{H}$ atoms on the same side. - trans-pent-2-ene / (E)-pent-2-ene: Alkyl groups on opposite sides of the double bond; $-\text{H}$ atoms on opposite sides.
(c)(i) Mechanism of electrophilic addition: 1. First step: Dipole shown on $\text{H}^{\delta+}-\text{Br}^{\delta-}$. A curly arrow originates from the $\text{C=C}$ double bond pointing to the $\text{H}^{\delta+}$ atom, and a second curly arrow originates from the $\text{H}-\text{Br}$ single bond pointing to the $\text{Br}^{\delta-}$ atom. 2. Intermediate: Correct carbocation intermediate: $\text{CH}_3-\text{CH}^+-\text{CH}_2\text{CH}_2\text{CH}_3$ (secondary carbocation at C2) and bromide ion, $: \text{Br}^-$, with a lone pair and negative charge. 3. Second step: Curly arrow from the lone pair on $: \text{Br}^-$ pointing directly to the positively charged carbon atom ($\text{C}^+$) to form $\text{CH}_3\text{CH(Br)}\text{CH}_2\text{CH}_2\text{CH}_3$.
(c)(ii) Both isomeric products are formed via secondary carbocation intermediates ($\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_2\text{CH}_3$ and $\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_2\text{CH}_3$), which have the same number of alkyl groups and therefore have similar stabilities.
(d)(i) The formation of a long-chain polymer by the joining together of monomer molecules (containing $\text{C=C}$ double bonds) without the elimination / formation of any other product / by-product.
(d)(ii) Structure showing two repeat units with a continuous carbon-carbon backbone: $$-\left[\text{CH}(\text{CH}_3)-\text{CH}(\text{C}_2\text{H}_5)-\text{CH}(\text{CH}_3)-\text{CH}(\text{C}_2\text{H}_5)\right]-$$ Backbone consists of 4 carbon atoms with open extension bonds at both ends, having alternating $-\text{CH}_3$ and $-\text{CH}_2\text{CH}_3$ side groups (or consistent placement).
評分準則
(a)(i) [2 marks] - M1: Correct working shown using $[\text{M}+1]^+$ formula: $n = \frac{100 \times 1.56}{1.1 \times 28.4}$ [1] - M2: Final value $n = 5$ / $4.99$ deduced [1]
(b)(ii) [2 marks] - M1: Restricted rotation / lack of free rotation about the carbon-carbon double bond / $\text{C=C}$ bond [1] - M2: Each carbon atom of the $\text{C=C}$ bond is bonded to two different atoms / groups [1]
(b)(iii) [2 marks] - M1: Fully correct displayed formula of cis-pent-2-ene / (Z)-pent-2-ene with all bonds shown [1] - M2: Fully correct displayed formula of trans-pent-2-ene / (E)-pent-2-ene with all bonds shown [1] (Deduct 1 mark if hydrogens in ethyl/methyl groups are not fully displayed if displayed formula requested, allow skeletal if unambiguous with correct stereochemistry)
(c)(i) [4 marks] - M1: Correct dipole $\text{H}^{\delta+}-\text{Br}^{\delta-}$ AND curly arrow from $\text{C=C}$ bond to $\text{H}^{\delta+}$ [1] - M2: Curly arrow from $\text{H}-\text{Br}$ bond onto $\text{Br}$ atom [1] - M3: Correct secondary carbocation structure $\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_2\text{CH}_3$ [1] - M4: Curly arrow from lone pair on $: \text{Br}^-$ to the $\text{C}^+$ carbon atom [1]
(c)(ii) [1 mark] - Both carbocation intermediates are secondary / have similar stability [1]
(d)(i) [1 mark] - Addition polymerisation: Monomers join together to form a polymer as the sole/only product (or no small molecule eliminated) [1]
(d)(ii) [1 mark] - Correct structure with 4 carbon atoms in the continuous polymer backbone and open-ended bonds at both ends, with $-\text{CH}_3$ and $-\text{CH}_2\text{CH}_3$ substituents correctly positioned on adjacent carbons [1]
Paper 3 (Advanced Practical Skills)
Answer all practical tasks. Record all data to appropriate precision in formatted tables and carry out quantitative deductions.
3 題目 · 40 分
題目 1 · Qualitative Inorganic Analysis Experiment
15 分
Read through the entire question before starting any practical work.
FA 5 is an aqueous solution containing two different cations and one anion, all of which are listed in the Qualitative Analysis Notes.
FA 6 is an aqueous solution of a transition metal salt.
(a) Perform the following tests on FA 5 and record all your observations in the spaces provided.
(i) To a 1 cm\(^3\) portion of FA 5 in a boiling tube, add aqueous sodium hydroxide dropwise until in excess. Then warm the mixture gently and test any gas evolved with damp red litmus paper.
Observations with dropwise addition: \vspace{0.5cm}
Observations with excess: \vspace{0.5cm}
Observations on warming: \vspace{0.5cm}
[3]
(ii) To a 1 cm\(^3\) portion of FA 5 in a test-tube, add aqueous ammonia dropwise until in excess.
Observations with dropwise addition: \vspace{0.5cm}
Observations with excess: \vspace{0.5cm}
[2]
(iii) To a 1 cm\(^3\) portion of FA 5 in a test-tube, add 1 cm\(^3\) of dilute nitric acid, followed by a few drops of aqueous silver nitrate. Then add aqueous ammonia until no further change occurs.
Observations on adding dilute nitric acid and aqueous silver nitrate: \vspace{0.5cm}
Observations on adding aqueous ammonia: \vspace{0.5cm}
[2]
(iv) Identify the two cations and the anion present in FA 5. Support your deductions with evidence from your observations in (a).
Cations: .................................... and ....................................
(ii) To a 1 cm\(^3\) portion of FA 6 in a test-tube, add aqueous sodium hydroxide dropwise until in excess. Then add 5 drops of aqueous hydrogen peroxide, \(\text{H}_2\text{O}_2\).
*Observations with \(\text{NaOH(aq)}\):* ........................................................................................
*Observations on adding \(\text{H}_2\text{O}_2\text{(aq)}\):* .....................................................................................
[2]
(iii) Identify the cation present in FA 6.
Cation in FA 6: ....................................
[1]
(iv) Construct an ionic equation for the reaction occurring in (b)(i). State symbols are not required.
(a)(i) - Dropwise \(\text{NaOH(aq)}\): White precipitate formed (\(\text{Al(OH)}_3\)). - Excess \(\text{NaOH(aq)}\): White precipitate dissolves to give a colourless solution (forming \([\text{Al(OH)}_4]^-\)). - Warming: Colourless/pungent gas turns damp red litmus paper blue (\(\text{NH}_3\) gas released from \(\text{NH}_4^+\)).
(a)(ii) - Dropwise \(\text{NH}_3\text{(aq)}\): White precipitate formed. - Excess \(\text{NH}_3\text{(aq)}\): Precipitate remains insoluble (distinguishing \(\text{Al}^{3+}\) from \(\text{Zn}^{2+}\)).
(a)(iii) - Addition of \(\text{HNO}_3\) and \(\text{AgNO}_3\): White precipitate of \(\text{AgCl}\) formed. - Addition of \(\text{NH}_3\text{(aq)}\): Precipitate dissolves completely to form a colourless solution (\([\text{Ag(NH}_3)_2]^+\)).
(a)(iv) - Cations: \(\text{Al}^{3+}\) (or aluminium) and \(\text{NH}_4^+\) (or ammonium). - Anion: \(\text{Cl}^-\) (or chloride). - Evidence: White ppt soluble in excess \(\text{NaOH}\) but insoluble in excess \(\text{NH}_3\) confirms \(\text{Al}^{3+}\); gas turning red litmus blue upon heating with alkali confirms \(\text{NH}_4^+\); white ppt with \(\text{AgNO}_3\) soluble in \(\text{NH}_3\text{(aq)}\) confirms \(\text{Cl}^-\).
(b)(i) - Purple solution of \(\text{KMnO}_4\) is decolourised / turns colourless (or very pale yellow due to \(\text{Fe}^{3+}\)).
(b)(ii) - With \(\text{NaOH(aq)}\): Green precipitate insoluble in excess \(\text{NaOH}\). - With \(\text{H}_2\text{O}_2\text{(aq)}\): Green precipitate turns red-brown / darkens rapidly as \(\text{Fe(OH)}_2\) is oxidised to \(\text{Fe(OH)}_3\) (effervescence may also be observed).
(a)(i) [3 marks] - M1: White precipitate / ppt formed on adding \(\text{NaOH}\) dropwise. - M2: Precipitate dissolves / soluble in excess \(\text{NaOH}\) to form a colourless solution. - M3: Gas / ammonia evolved turns damp red litmus paper blue on warming.
(a)(ii) [2 marks] - M1: White precipitate formed on adding \(\text{NH}_3\) dropwise. - M2: Precipitate insoluble / remains in excess \(\text{NH}_3\).
(a)(iii) [2 marks] - M1: White precipitate formed with \(\text{HNO}_3\) and \(\text{AgNO}_3\). - M2: Precipitate dissolves / soluble in \(\text{NH}_3\text{(aq)}\).
(a)(iv) [3 marks] - M1: Cation 1: \(\text{Al}^{3+}\) (allow aluminium) AND Cation 2: \(\text{NH}_4^+\) (allow ammonium). - M2: Anion: \(\text{Cl}^-\) (allow chloride). - M3: Valid supporting evidence linking observations to the identified ions: (e.g. white ppt soluble in excess \(\text{NaOH}\) and insoluble in excess \(\text{NH}_3\) for \(\text{Al}^{3+}\); ammonia evolved on warming with alkali for \(\text{NH}_4^+\); white ppt with \(\text{Ag}^+\) soluble in \(\text{NH}_3\) for \(\text{Cl}^-\)).
(b)(ii) [2 marks] - M1: Green precipitate formed with \(\text{NaOH}\) (insoluble in excess). - M2: Precipitate turns red-brown / brown / darkens on addition of \(\text{H}_2\text{O}_2\).
Hydrated zinc sulfate has the formula \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\), where \(x\) is an integer.
A student carries out a gravimetric experiment to determine the value of \(x\) by thermally decomposing a sample of hydrated zinc sulfate in a crucible.
$$\begin{array}{|l|c|} \hline \text{Mass of empty crucible} / \text{g} & 18.420 \\ \hline \text{Mass of crucible} + \text{ZnSO}_4 \cdot x\text{H}_2\text{O} / \text{g} & 21.295 \\ \hline \text{Mass of crucible} + \text{residue after 1st heating} / \text{g} & 20.065 \\ \hline \text{Mass of crucible} + \text{residue after 2nd heating} / \text{g} & 20.035 \\ \hline \text{Mass of crucible} + \text{residue after 3rd heating} / \text{g} & 20.035 \\ \hline \end{array}$$
(a) (i) Use the data in the table to determine: - the mass of \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\) used - the mass of anhydrous \(\text{ZnSO}_4\) formed - the mass of \(\text{H}_2\text{O}\) removed. [1]
(ii) Calculate the amount, in \(\text{mol}\), of anhydrous \(\text{ZnSO}_4\) formed. [\(A_r\): \(\text{Zn} = 65.4\), \(\text{S} = 32.1\), \(\text{O} = 16.0\)] [1]
(iii) Calculate the amount, in \(\text{mol}\), of \(\text{H}_2\text{O}\) lost. [\(A_r\): \(\text{H} = 1.0\), \(\text{O} = 16.0\)] [1]
(iv) Deduce the value of \(x\) in the formula \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\). [1]
(b) (i) Explain why the student heated the crucible to constant mass. [1]
(ii) During heating, some of the solid spat out of the open crucible. State and explain the effect this would have on the calculated value of \(x\). [2]
(c) (i) The balance used has an uncertainty of \(\pm 0.005\text{ g}\) for each reading. Calculate the maximum percentage uncertainty in the mass of \(\text{H}_2\text{O}\) removed. Show your working. [2]
(ii) Suggest one improvement to the experimental procedure, other than using a more precise balance, that would reduce the percentage uncertainty in the mass of \(\text{H}_2\text{O}\) removed. [1]
(b) (i) Heating to constant mass ensures that all water of crystallisation has been driven off / the reaction has gone to completion.
(ii) Loss of solid from spitting causes the final measured mass of the crucible and residue to be lower than expected. This increases the calculated mass (and moles) of water lost, leading to an overestimation / higher calculated value of \(x\).
(c) (i) Determining the mass of water lost involves two balance readings (mass before heating and final mass after heating): $$\text{Total absolute uncertainty} = 2 \times 0.005\text{ g} = 0.010\text{ g}$$ $$\text{Percentage uncertainty} = \frac{0.010\text{ g}}{1.260\text{ g}} \times 100 = 0.794\%$$ (or \(0.79\%\))
(ii) Use a larger mass of hydrated zinc sulfate (which increases the mass of water lost and reduces relative error).
評分準則
(a)(i) [1 mark] - Correctly determines all three masses: mass of salt = 2.875 g, mass of residue = 1.615 g, mass of water = 1.260 g.
(a)(ii) [1 mark] - Correct calculation of moles of ZnSO4: 1.615 / 161.5 = 0.0100 mol (minimum 2 significant figures; allow ECF from (a)(i)).
(a)(iii) [1 mark] - Correct calculation of moles of H2O: 1.260 / 18.0 = 0.0700 mol (minimum 2 significant figures; allow ECF from (a)(i)).
(a)(iv) [1 mark] - Correct deduction of x = 7 (must be an integer; allow ECF from (a)(ii) and (a)(iii)).
(b)(i) [1 mark] - States that heating to constant mass ensures all water of crystallisation has been removed / reaction is complete.
(b)(ii) [2 marks] - M1: Value of x would be higher / overestimated / larger [1]. - M2: Because the mass of residue is lower than it should be, so the calculated mass of water lost is greater than actual [1].
(c)(i) [2 marks] - M1: Identifies 2 readings used (total absolute error = 2 x 0.005 = 0.010 g) [1]. - M2: Correct calculation: (0.010 / 1.260) x 100 = 0.794% (allow 0.79% or 0.8%) [1]. (Award 1 mark max if candidate uses 1 reading: (0.005 / 1.260) x 100 = 0.397%).
(c)(ii) [1 mark] - Suggests using a larger mass / sample of hydrated zinc sulfate.
題目 3 · theory
15 分
In this experiment, you will determine the value of \(x\) in the formula of hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}\), by volumetric titration with aqueous sodium hydroxide.
FA 1 is \(0.125\text{ mol dm}^{-3}\text{ NaOH(aq)}\). FA 2 is solid hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}\). phenolphthalein indicator
Method 1. A student accurately weighed \(1.890\text{ g}\) of FA 2 using a 3-decimal-place balance. 2. The solid was dissolved in approximately \(100\text{ cm}^3\) of distilled water in a beaker and quantitatively transferred to a \(250.0\text{ cm}^3\) volumetric flask. 3. Distilled water was added to make the solution up to the \(250.0\text{ cm}^3\) mark, and the flask was inverted thoroughly to mix. This solution is FA 3. 4. A \(25.0\text{ cm}^3\) volumetric pipette was used to transfer \(25.0\text{ cm}^3\) of FA 3 into a clean conical flask. 5. Three drops of phenolphthalein indicator were added to the conical flask. 6. FA 1 was placed into a burette and titrated into the conical flask until the end-point was reached. 7. The titration was repeated until concordant results were obtained.
(a) The results of the student's titrations are shown in the table below:
(b)Calculations Show appropriate working in all parts. Give all numerical answers to 3 significant figures.
(i) Calculate the amount, in \(\text{mol}\), of \(\text{NaOH}\) present in the mean titre volume of FA 1.
$$\text{amount of } \text{NaOH} = \text{............................. } \text{mol}$$ [1]
(ii) Write the balanced chemical equation for the reaction between aqueous ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\), and aqueous sodium hydroxide, \(\text{NaOH(aq)}\). Hence, calculate the amount, in \(\text{mol}\), of \(\text{H}_2\text{C}_2\text{O}_4\) present in \(25.0\text{ cm}^3\) of FA 3.
$$\text{equation: } \text{......................................................................................................}$$ $$\text{amount of } \text{H}_2\text{C}_2\text{O}_4 \text{ in } 25.0\text{ cm}^3 = \text{............................. } \text{mol}$$ [2]
(iii) Calculate the amount, in \(\text{mol}\), of \(\text{H}_2\text{C}_2\text{O}_4\) present in the total \(250.0\text{ cm}^3\) of solution FA 3.
$$\text{amount of } \text{H}_2\text{C}_2\text{O}_4 \text{ in } 250.0\text{ cm}^3 = \text{............................. } \text{mol}$$ [1]
(iv) Using your answer to (b)(iii) and the mass of FA 2 weighed (\(1.890\text{ g}\)), calculate the relative formula mass, \(M_r\), of hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}\).
$$M_r \text{ of } \text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} = \text{.............................}$$ [2]
(v) Determine the value of \(x\) in the formula \(\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}\).
$$x = \text{.............................}$$ [2]
(c)Apparatus and Error Analysis
(i) State the colour change observed at the end-point in the conical flask.
$$\text{from .......................................... to ..........................................}$$ [1]
(ii) The maximum uncertainty in each individual burette reading is \(\pm 0.05\text{ cm}^3\). Calculate the maximum percentage uncertainty in the titre obtained in Titration 3 (\(24.00\text{ cm}^3\)). Show your working.
(iii) A student washed the conical flask with distilled water between titrations but did not dry the flask before pipetting the next \(25.0\text{ cm}^3\) sample of FA 3 into it.
State and explain what effect, if any, this leftover distilled water in the conical flask would have on the calculated value of \(x\).
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解題
(a) Concordant titres are those within \(0.10\text{ cm}^3\) of each other: Titrations 1, 2, and 3 (\(24.05\text{ cm}^3\), \(23.95\text{ cm}^3\), and \(24.00\text{ cm}^3\)).
(c)(i) From colourless to pale pink (or permanent pink).
(c)(ii) Two burette readings are taken to determine a titre (initial and final), so total maximum uncertainty = \(2 \times 0.05\text{ cm}^3 = 0.10\text{ cm}^3\).
(c)(iii) No effect. The amount (number of moles) of \(\text{H}_2\text{C}_2\text{O}_4\) transferred into the conical flask is fixed by the \(25.0\text{ cm}^3\) pipette. Extra water dilutes the solution but does not change the amount of acid present, so the titre volume and the calculated \(M_r\) and \(x\) remain unaffected.
評分準則
(a) [2 marks] - M1: Selects Titrations 1, 2, and 3 (or at least two concordant titres within \(0.10\text{ cm}^3\)) and shows clear calculation of the mean. [1] - M2: Correct calculation giving \(24.00\text{ cm}^3\) (quoted to 2 decimal places). [1]
(b)(ii) [2 marks] - M1: Correct balanced equation: \(\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}\) (allow ionic equation: \(\text{H}_2\text{C}_2\text{O}_4 + 2\text{OH}^- \rightarrow \text{C}_2\text{O}_4^{2-} + 2\text{H}_2\text{O}\)). [1] - M2: Divides moles of \(\text{NaOH}\) by 2 to give \(1.50 \times 10^{-3}\text{ mol}\) (allow ECF from (b)(i)). [1]
(b)(iii) [1 mark] - M1: Multiplies answer to (b)(ii) by 10 to give \(0.0150\text{ mol}\) (allow ECF). [1]
(b)(iv) [2 marks] - M1: Uses \(M_r = \frac{1.890}{\text{moles in 250 cm}^3}\). [1] - M2: Correct evaluation to give \(M_r = 126\) (allow 126.0, must be 3 sig. figs. or integer). [1]
(b)(v) [2 marks] - M1: Correct method to determine mass of water of crystallisation: \(x \times 18.0 = M_r - 90.0\) (allow ECF from (b)(iv)). [1] - M2: Deduces \(x = 2\) (must be given as an integer). [1]
(c)(i) [1 mark] - M1: Colourless to (pale) pink / permanent pink. [1] (Reject 'clear' instead of 'colourless'; reject 'red' or 'purple')
(c)(ii) [2 marks] - M1: Total absolute uncertainty = \(2 \times 0.05 = 0.10\text{ cm}^3\) used in calculation. [1] - M2: Correct percentage uncertainty \(= \frac{0.10}{24.00} \times 100 = 0.417\%\) (or \(0.42\%\)). [1] *(If only single reading uncertainty \(0.05\text{ cm}^3\) is used giving \(0.208\%\), award 1 mark max)*
(c)(iii) [2 marks] - M1: States no effect (on \(x\) / calculated values). [1] - M2: Explains that the number of moles / amount of acid is determined solely by the pipette volume transferred (adding water does not change the number of moles of \(\text{H}_2\text{C}_2\text{O}_4\) reacting). [1]
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