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2026 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

Thinka Jun 2026 (V1) Cambridge International A Level-Style Mock — Chemistry (9701)

270 分465 分鐘2026
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

卷一 (選擇題)

Answer all 40 multiple-choice questions on the separate answer sheet using soft pencil. Each question carries 1 mark.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon \(\text{C}_x\text{H}_y\) was completely burnt in \(150\text{ cm}^3\) of oxygen (an excess).

After combustion, the total volume of gas remaining at room temperature and pressure was \(110\text{ cm}^3\).
Passing this remaining gas mixture through aqueous sodium hydroxide caused a decrease in volume of \(80\text{ cm}^3\).

All gas volumes were measured under the same conditions of temperature and pressure.

What is the molecular formula of the hydrocarbon?
  1. A.\(\text{C}_3\text{H}_8\)
  2. B.\(\text{C}_4\text{H}_8\)
  3. C.\(\text{C}_4\text{H}_{10}\)
  4. D.\(\text{C}_5\text{H}_{10}\)
查看答案詳解

解題

1. Volume of hydrocarbon \(\text{C}_x\text{H}_y = 20\text{ cm}^3\).
2. Sodium hydroxide absorbs carbon dioxide: volume of \(\text{CO}_2 = 80\text{ cm}^3\).
Therefore, \(x = \frac{80}{20} = 4\).
3. The unreacted \(\text{O}_2\) remaining \(= 110 - 80 = 30\text{ cm}^3\).
4. Volume of \(\text{O}_2\) reacted \(= 150 - 30 = 120\text{ cm}^3\).
5. From the combustion equation:
\[\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\]
\(x + \frac{y}{4} = \frac{120}{20} = 6\).
6. Substituting \(x = 4\):
\[4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\]
Thus, the molecular formula is \(\text{C}_4\text{H}_8\).

評分準則

[1 mark] for selecting B (\(\text{C}_4\text{H}_8\)).
題目 2 · multiple_choice
1 分
The first six successive ionisation energies of element \(X\), in \(\text{kJ mol}^{-1}\), are given:

789, 1577, 3232, 4356, 16091, 19785

Which statement about element \(X\) is correct?
  1. A.The chloride of \(X\) dissolves in water without reaction to form a neutral solution of pH 7.
  2. B.The oxide of \(X\) has a giant covalent structure and reacts with hot concentrated \(\text{NaOH(aq)}\).
  3. C.Element \(X\) forms a gaseous chloride with the empirical formula \(X\text{Cl}_3\).
  4. D.Element \(X\) is in Group 15 of the Periodic Table.
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解題

A large jump in ionisation energy occurs between the 4th and 5th ionisation energies (from 4356 to \(16091\text{ kJ mol}^{-1}\)), which indicates that element \(X\) has 4 valence electrons and belongs to Group 14 (silicon in Period 3).

- Group 14 oxide, \(\text{SiO}_2\), has a giant covalent lattice structure and reacts with hot concentrated alkali such as \(\text{NaOH(aq)}\) to form silicates, but does not react with \(\text{HCl(aq)}\).
- \(\text{SiCl}_4\) hydrolyses vigorously in water to produce an acidic solution of \(\text{pH } 1-2\), so option A is incorrect.
- Option C is incorrect as silicon forms \(\text{SiCl}_4\), not \(\text{SiCl}_3\).
- Option D is incorrect as \(X\) is in Group 14, not Group 15.

評分準則

[1 mark] for selecting B.
題目 3 · multiple_choice
1 分
Which pair of species both have the same shape and identical bond angles?
  1. A.\(\text{CO}_2\) and \(\text{SO}_2\)
  2. B.\(\text{NH}_4^+\) and \(\text{CH}_4\)
  3. C.\(\text{BF}_3\) and \(\text{PCl}_3\)
  4. D.\(\text{H}_2\text{O}\) and \(\text{BeCl}_2\)
查看答案詳解

解題

Both \(\text{NH}_4^+\) and \(\text{CH}_4\) have 4 bonding pairs and 0 lone pairs of electrons around the central atom. By VSEPR theory, both adopt a regular tetrahedral shape with identical bond angles of \(109.5^\circ\).

- \(\text{CO}_2\) is linear (\(180^\circ\)) while \(\text{SO}_2\) is non-linear/bent (\(\approx 119^\circ\)).
- \(\text{BF}_3\) is trigonal planar (\(120^\circ\)) while \(\text{PCl}_3\) is trigonal pyramidal (\(\approx 107^\circ\)).
- \(\text{H}_2\text{O}\) is non-linear/bent (\(104.5^\circ\)) while \(\text{BeCl}_2\) is linear (\(180^\circ\)).

評分準則

[1 mark] for selecting B.
題目 4 · multiple_choice
1 分
An anhydrous chloride of a Period 3 element, \(E\), is added to excess water.

The reaction is vigorous, releasing steamy acidic fumes, and leaves a white precipitate suspended in an acidic solution of \(\text{pH } \approx 2\).

What is the identity of chloride \(E\)?
  1. A.\(\text{MgCl}_2\)
  2. B.\(\text{AlCl}_3\)
  3. C.\(\text{SiCl}_4\)
  4. D.\(\text{PCl}_5\)
查看答案詳解

解題

\(\text{SiCl}_4\) is a liquid covalent chloride that hydrolyses violently upon addition to water:
\[\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}\]
This reaction forms an insoluble white precipitate of hydrated \(\text{SiO}_2\) and misty fumes of \(\text{HCl}\), leaving an acidic solution of \(\text{pH } 1-2\).

- \(\text{MgCl}_2\) dissolves to give a colourless solution with \(\text{pH } \approx 6.5\) and no precipitate.
- \(\text{AlCl}_3\) dissolves in excess water to give a clear acidic solution containing \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\) without forming a precipitate in excess water.
- \(\text{PCl}_5\) hydrolyses completely to form a clear solution containing \(\text{H}_3\text{PO}_4\) and \(\text{HCl}\) with no precipitate.

評分準則

[1 mark] for selecting C.
題目 5 · multiple_choice
1 分
Standard enthalpy changes of combustion, \(\Delta H^\circ_\text{c}\), are given in the table.

$$\begin{array}{|c|c|}\hline\text{Substance} & \Delta H^\circ_\text{c} / \text{kJ mol}^{-1} \\hline\text{C(graphite)} & -394 \\hline\text{H}_2\text{(g)} & -286 \\hline\text{CH}_3\text{OCH}_3\text{(g)} & -1460 \\hline\end{array}$$

What is the standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\), of methoxymethane, \(\text{CH}_3\text{OCH}_3\text{(g)}\)?
  1. A.\(-3106\text{ kJ mol}^{-1}\)
  2. B.\(-186\text{ kJ mol}^{-1}\)
  3. C.\(+186\text{ kJ mol}^{-1}\)
  4. D.\(+780\text{ kJ mol}^{-1}\)
查看答案詳解

解題

The equation for the formation of methoxymethane is:
\[2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OCH}_3\text{(g)}\]
Using Hess's Law with standard enthalpy changes of combustion:
\[\Delta H^\circ_\text{f} = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\]
\[\Delta H^\circ_\text{f} = [2(-394) + 3(-286)] - [-1460]\]
\[\Delta H^\circ_\text{f} = [-788 - 858] + 1460 = -1646 + 1460 = -186\text{ kJ mol}^{-1}\]

評分準則

[1 mark] for selecting B.
題目 6 · multiple_choice
1 分
Equal amounts of four different halogenoalkanes are added separately to warm ethanolic silver nitrate at \(50\ ^\circ\text{C}\).

Which halogenoalkane forms a precipitate in the shortest time?
  1. A.1-chlorobutane
  2. B.1-iodobutane
  3. C.2-bromo-2-methylpropane
  4. D.2-iodo-2-methylpropane
查看答案詳解

解題

The rate of hydrolysis of halogenoalkanes depends on two factors:
1. The mechanism and carbocation stability: Tertiary halogenoalkanes hydrolyse much faster via an \(\text{S}_\text{N}1\) pathway involving a stable tertiary carbocation intermediate than primary halogenoalkanes hydrolysing via \(\text{S}_\text{N}2\).
2. The carbon-halogen bond strength: The \(\text{C–I}\) bond is the weakest (lowest bond enthalpy) compared to \(\text{C–Br}\) and \(\text{C–Cl}\), and therefore cleaves most rapidly.

2-iodo-2-methylpropane is a tertiary iodoalkane, so it hydrolyses the fastest to precipitate \(\text{AgI(s)}\) in the shortest time.

評分準則

[1 mark] for selecting D.
題目 7 · multiple_choice
1 分
An unbranched hydrocarbon \(W\), \(\text{C}_6\text{H}_{10}\), contains two carbon–carbon double bonds.

When \(W\) is heated under reflux with hot concentrated acidified \(\text{KMnO}_4\), the only organic products obtained are propanoic acid and ethanedioic acid in a \(1:1\) molar ratio, alongside carbon dioxide and water.

What is the structural formula of \(W\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}=\text{CH}_2\)
  2. B.\(\text{CH}_3\text{CH}=\text{CHCH}=\text{CHCH}_3\)
  3. C.\(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}=\text{CH}_2\)
  4. D.\(\text{CH}_2=\text{CHCH}_2\text{CH}=\text{CHCH}_3\)
查看答案詳解

解題

Oxidative cleavage of carbon–carbon double bonds with hot concentrated acidified \(\text{KMnO}_4\) yields:
- \(\text{CH}_3\text{CH}_2\text{CH}=\) oxidises to propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
- \(=\text{CH–CH}=\) oxidises to ethanedioic acid, \(\text{HOOC–COOH}\).
- \(=\text{CH}_2\) oxidises to \(\text{CO}_2 + \text{H}_2\text{O}\).

Connecting these fragments gives hexa-1,3-diene:
\[\text{CH}_3\text{CH}_2\text{CH}=\text{CH–CH}=\text{CH}_2\]

Other options:
- \(\text{CH}_3\text{CH}=\text{CHCH}=\text{CHCH}_3\) gives ethanoic acid and ethanedioic acid (no \(\text{CO}_2\)).
- \(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}=\text{CH}_2\) gives \(\text{CO}_2\) and butanedioic acid.
- \(\text{CH}_2=\text{CHCH}_2\text{CH}=\text{CHCH}_3\) gives \(\text{CO}_2\), propanedioic acid, and ethanoic acid.

評分準則

[1 mark] for selecting A.
題目 8 · multiple_choice
1 分
Compound \(Q\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- \(Q\) gives an orange precipitate when treated with 2,4-dinitrophenylhydrazine.
- \(Q\) does not form a silver mirror when warmed with Tollens' reagent.
- \(Q\) forms a pale yellow precipitate when warmed with alkaline aqueous iodine.

What is the identity of compound \(Q\)?
  1. A.butanal
  2. B.butanone
  3. C.but-3-en-2-ol
  4. D.2-methylpropanal
查看答案詳解

解題

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine confirms \(Q\) is a carbonyl compound (aldehyde or ketone).
2. Negative Tollens' test confirms \(Q\) is a ketone, not an aldehyde (rules out butanal and 2-methylpropanal).
3. Formation of a yellow precipitate with alkaline aqueous iodine (tri-iodomethane test) confirms the presence of a \(\text{CH}_3\text{C}=\text{O}\) (methyl ketone) group.

With formula \(\text{C}_4\text{H}_8\text{O}\), the only methyl ketone is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

評分準則

[1 mark] for selecting B.
題目 9 · multiple_choice
1 分
The first eight successive ionisation energies of an element \(X\) in Period 3 of the Periodic Table are given in \(\text{kJ mol}^{-1}\):

1000, 2251, 3361, 4564, 7013, 8495, 27107, 31719

Which statement about element \(X\) is correct?
  1. A.It forms a basic oxide with formula \(X\text{O}_2\).
  2. B.It has six electrons in its valence shell and forms an ion \(X^{2-}\).
  3. C.Its outermost electrons occupy a \(4\text{p}\) subshell.
  4. D.The formula of its chloride in its highest oxidation state is \(X\text{Cl}_4\).
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解題

Looking at the successive ionisation energies, there is a large increase between the 6th (\(8495\text{ kJ mol}^{-1}\)) and 7th (\(27107\text{ kJ mol}^{-1}\)) ionisation energies. This indicates that the 7th electron is removed from an inner, fully-filled principal quantum shell, meaning element \(X\) has 6 valence electrons in Period 3 (sulfur, Group 16). Therefore, it readily gains two electrons to form the \(X^{2-}\) (sulfide) ion.

評分準則

[1] B is correct. Identifies the large jump between the 6th and 7th ionisation energies indicating 6 valence electrons (Group 16).
題目 10 · multiple_choice
1 分
A sample of an unknown gaseous hydrocarbon has a mass of \(0.387\text{ g}\). It occupies a volume of \(166.2\text{ cm}^3\) at a temperature of \(300\text{ K}\) and a pressure of \(1.00 \times 10^5\text{ Pa}\).

What is the molecular formula of this hydrocarbon?
  1. A.\(\text{C}_3\text{H}_6\)
  2. B.\(\text{C}_3\text{H}_8\)
  3. C.\(\text{C}_4\text{H}_{10}\)
  4. D.\(\text{C}_4\text{H}_8\)
查看答案詳解

解題

Using the ideal gas equation \(pV = nRT\):

\(V = 166.2\text{ cm}^3 = 1.662 \times 10^{-4}\text{ m}^3\)
\(n = \frac{pV}{RT} = \frac{1.00 \times 10^5 \times 1.662 \times 10^{-4}}{8.31 \times 300} = \frac{16.62}{2493} = 6.667 \times 10^{-3}\text{ mol}\)

\(M_r = \frac{m}{n} = \frac{0.387}{6.667 \times 10^{-3}} = 58.0\text{ g mol}^{-1}\)

Checking molar masses:
- \(\text{C}_3\text{H}_6 = (3 \times 12.0) + (6 \times 1.0) = 42.0\)
- \(\text{C}_3\text{H}_8 = (3 \times 12.0) + (8 \times 1.0) = 44.0\)
- \(\text{C}_4\text{H}_{10} = (4 \times 12.0) + (10 \times 1.0) = 58.0\)
- \(\text{C}_4\text{H}_8 = (4 \times 12.0) + (8 \times 1.0) = 56.0\)

Therefore, the hydrocarbon is \(\text{C}_4\text{H}_{10}\).

評分準則

[1] C is correct. Correct conversion of units and calculation of molar mass \(M_r = 58.0\text{ g mol}^{-1}\).
題目 11 · multiple_choice
1 分
Which species contains the smallest bond angle?
  1. A.\(\text{BF}_3\)
  2. B.\(\text{NH}_4^+\)
  3. C.\(\text{NH}_2^-\)
  4. D.\(\text{H}_3\text{O}^+\)
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解題

- \(\text{BF}_3\): 3 bonding pairs, 0 lone pairs; trigonal planar, bond angle is \(120^\circ\).
- \(\text{NH}_4^+\): 4 bonding pairs, 0 lone pairs; tetrahedral, bond angle is \(109.5^\circ\).
- \(\text{H}_3\text{O}^+\): 3 bonding pairs, 1 lone pair; trigonal pyramidal, bond angle is approximately \(107^\circ\).
- \(\text{NH}_2^-\): 2 bonding pairs, 2 lone pairs; non-linear / bent, bond angle is approximately \(104.5^\circ\).

Lone pair–lone pair and lone pair–bonding pair repulsions are greater than bonding pair–bonding pair repulsions, resulting in the smallest bond angle in \(\text{NH}_2^-\).

評分準則

[1] C is correct. Recognises that \(\text{NH}_2^-\text{ has 2 bonding pairs and 2 lone pairs, giving the greatest repulsion and smallest bond angle (}\approx 104.5^\circ\text{)}\).
題目 12 · multiple_choice
1 分
Standard enthalpy changes of combustion, \(\Delta H^\circ_c\), are provided:

- \(\Delta H^\circ_c(\text{C(s)}) = -394\text{ kJ mol}^{-1}\)
- \(\Delta H^\circ_c(\text{H}_2\text{(g)}) = -286\text{ kJ mol}^{-1}\)
- \(\Delta H^\circ_c(\text{CH}_3\text{OH(l)}) = -726\text{ kJ mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H^\circ_f\), of methanol, \(\text{CH}_3\text{OH(l)}\)?
  1. A.\(-240\text{ kJ mol}^{-1}\)
  2. B.\(+240\text{ kJ mol}^{-1}\)
  3. C.\(-46\text{ kJ mol}^{-1}\)
  4. D.\(-1692\text{ kJ mol}^{-1}\)
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解題

The equation for the formation of methanol is:
\(\text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\)

Using Hess's Law with enthalpies of combustion:
\(\Delta H^\circ_f = \sum \Delta H^\circ_c(\text{reactants}) - \sum \Delta H^\circ_c(\text{products})\)
\(\Delta H^\circ_f = [\Delta H^\circ_c(\text{C(s)}) + 2\Delta H^\circ_c(\text{H}_2\text{(g)})] - [\Delta H^\circ_c(\text{CH}_3\text{OH(l)})]\)
\(\Delta H^\circ_f = [(-394) + 2(-286)] - (-726)\)
\(\Delta H^\circ_f = [-394 - 572] + 726 = -966 + 726 = -240\text{ kJ mol}^{-1}\)

評分準則

[1] A is correct. Applies \(\Delta H^\circ_f = \sum \Delta H^\circ_c(\text{reactants}) - \sum \Delta H^\circ_c(\text{products}) = -240\text{ kJ mol}^{-1}\).
題目 13 · multiple_choice
1 分
A solid chloride of a Period 3 element reacts vigorously with water, releasing steamy white fumes and giving an acidic solution with a pH between 1 and 2.

When excess aqueous sodium hydroxide is added dropwise to this acidic solution, a white precipitate forms initially, which then redissolves to form a colourless solution.

Which element forms this chloride?
  1. A.Magnesium
  2. B.Aluminium
  3. C.Silicon
  4. D.Phosphorus
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解題

- Aluminium chloride (\(\text{AlCl}_3\)) hydrolyses vigorously in water to form \(\text{HCl}\) fumes (steamy white fumes) and an acidic solution containing \([\text{Al(H}_2\text{O)}_6]^{3+}\).
- Adding \(\text{NaOH(aq)}\) precipitates white \(\text{Al(OH)}_3\text{(s)}\).
- Because \(\text{Al(OH)}_3\) is amphoteric, it redissolves in excess \(\text{NaOH(aq)}\) to form the soluble tetrahydroxoaluminate ion, \([\text{Al(OH)}_4]^-\text{(aq)}\).
- Silicon tetrachloride (\(\text{SiCl}_4\)) and phosphorus pentachloride (\(\text{PCl}_5\)) do not form a precipitate that redissolves in this manner, while \(\text{Mg(OH)}_2\) does not dissolve in excess alkali.

評分準則

[1] B is correct. Identifies amphoteric behaviour of aluminium hydroxide and vigorous hydrolysis of aluminium chloride.
題目 14 · multiple_choice
1 分
Which row correctly describes the trends in properties of the Group 2 elements and their compounds on descending the group from magnesium to barium?
  1. A.Solubility of sulfates decreases; solubility of hydroxides increases; thermal stability of nitrates increases
  2. B.Solubility of sulfates decreases; solubility of hydroxides increases; thermal stability of nitrates decreases
  3. C.Solubility of sulfates increases; solubility of hydroxides decreases; thermal stability of nitrates increases
  4. D.Solubility of sulfates increases; solubility of hydroxides decreases; thermal stability of nitrates decreases
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解題

Descending Group 2 from \(\text{Mg}\) to \(\text{Ba}\):
1. Solubility of sulfates decreases: the lattice energy and enthalpy of hydration both decrease in magnitude, but the enthalpy of hydration decreases more rapidly because the sulfate ion is large.
2. Solubility of hydroxides increases: the lattice energy decreases more rapidly than the hydration enthalpy of the small hydroxide ion.
3. Thermal stability of nitrates increases: as the cation radius increases, charge density decreases, so the cation polarises the nitrate anion less effectively, making thermal decomposition require higher temperatures.

評分準則

[1] A is correct. Identifies that sulfate solubility decreases, hydroxide solubility increases, and nitrate thermal stability increases down Group 2.
題目 15 · multiple_choice
1 分
An alkene \(Y\) with molecular formula \(\text{C}_6\text{H}_{12}\) is heated under reflux with hot concentrated acidified potassium manganate(VII).

Propan-2-one, \(\text{CH}_3\text{COCH}_3\), is the only organic product obtained from this reaction.

What is the structure of alkene \(Y\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_2\text{CH}_3\)
  2. B.\((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\)
  3. C.\((\text{CH}_3)_2\text{CHCH}=\text{CHCH}_3\)
  4. D.\(\text{CH}_2=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\)
查看答案詳解

解題

Hot concentrated acidified \(\text{KMnO}_4\) causes oxidative cleavage of the alkene \(\text{C}=\text{C}\) double bond:
- A carbon with two alkyl groups (\(=\text{CR}_2\)) is oxidised to a ketone (\(\text{O}=\text{CR}_2\)).
- Since propan-2-one, \((\text{CH}_3)_2\text{C}=\text{O}\), is the only organic product, both ends of the double bond must be \(=C(\text{CH}_3)_2\).
- Combining these two fragments gives 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\), which has the formula \(\text{C}_6\text{H}_{12}\).

評分準則

[1] B is correct. Symmetrical cleavage of 2,3-dimethylbut-2-ene yields two molecules of propan-2-one.
題目 16 · multiple_choice
1 分
Compound \(Z\) undergoes the following reactions:

- It produces an orange precipitate with 2,4-dinitrophenylhydrazine.
- It does not produce a silver mirror when warmed with Tollens' reagent.
- It forms a pale yellow precipitate when warmed with alkaline aqueous iodine.

Which compound is \(Z\)?
  1. A.Propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\)
  2. B.Pentan-3-one, \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\)
  3. C.Butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\)
  4. D.Ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\)
查看答案詳解

解題

- Reaction with 2,4-dinitrophenylhydrazine gives an orange precipitate: \(Z\) is a carbonyl compound (aldehyde or ketone), which excludes ethanol.
- No reaction with Tollens' reagent: \(Z\) is not an aldehyde (it is a ketone), which excludes propanal.
- Reaction with alkaline aqueous iodine gives a yellow precipitate of \(\text{CHI}_3\): \(Z\) contains a methyl carbonyl group, \(\text{CH}_3\text{C}(=\text{O})-\).
- Butanone is \(\text{CH}_3\text{COCH}_2\text{CH}_3\), which contains a \(\text{CH}_3\text{C}(=\text{O})-\) group and therefore gives a positive tri-iodomethane test. Pentan-3-one (\(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\)) does not contain this group.

評分準則

[1] C is correct. Deduces that \(Z\) is a methyl ketone (butanone).
題目 17 · multiple_choice
1 分
The first seven successive ionisation energies of a Period 3 element, $E$, are shown in the table.

$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline\text{Ionisation energy} & 1^{\text{st}} & 2^{\text{nd}} & 3^{\text{rd}} & 4^{\text{th}} & 5^{\text{th}} & 6^{\text{th}} & 7^{\text{th}} \\hline\text{Value / kJ mol}^{-1} & 1012 & 1903 & 2912 & 4956 & 6274 & 22233 & 25397 \\hline\end{array}$$

Which formula represents the chloride formed when element $E$ is heated in an excess of chlorine gas?
  1. A.$\text{ECl}_2$
  2. B.$\text{ECl}_3$
  3. C.$\text{ECl}_4$
  4. D.$\text{ECl}_5$
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解題

The successive ionisation energies show a moderate increase from the $1^{\text{st}}$ to the $5^{\text{th}}$ electron removed ($1012$ to $6274\text{ kJ mol}^{-1}$), followed by a very large jump between the $5^{\text{th}}$ and $6^{\text{th}}$ ionisation energies ($6274\text{ kJ mol}^{-1}$ to $22233\text{ kJ mol}^{-1}$). This indicates that element $E$ has five valence electrons and belongs to Group 15 of the Periodic Table (phosphorus, $\text{P}$). When phosphorus is heated in an excess of chlorine, it forms phosphorus(V) chloride, $\text{PCl}_5$, corresponding to the formula $\text{ECl}_5$.

評分準則

[1] for selecting option D.
[0] for any other option.
題目 18 · multiple_choice
1 分
A $20\text{ cm}^3$ sample of a gaseous hydrocarbon, $\text{C}_x\text{H}_y$, was completely burned in $150\text{ cm}^3$ of oxygen gas (an excess).

After cooling to room temperature, the total volume of residual gas was $110\text{ cm}^3$.

When this residual gas was shaken with an excess of aqueous sodium hydroxide, the volume decreased by $80\text{ cm}^3$.

All gas volumes were measured under identical room conditions.

What is the molecular formula of the hydrocarbon?
  1. A.$\text{C}_3\text{H}_8$
  2. B.$\text{C}_4\text{H}_6$
  3. C.$\text{C}_4\text{H}_8$
  4. D.$\text{C}_4\text{H}_{10}$
查看答案詳解

解題

The decrease in volume upon treatment with aqueous $\text{NaOH}$ corresponds to the volume of carbon dioxide gas produced:
$$V(\text{CO}_2) = 80\text{ cm}^3$$
Since the original volume of hydrocarbon was $20\text{ cm}^3$, the number of carbon atoms $x$ is:
$$x = \frac{80}{20} = 4$$
The remaining gas after $\text{NaOH}$ absorption is unreacted excess oxygen:
$$V(\text{unreacted O}_2) = 110 - 80 = 30\text{ cm}^3$$
Volume of $\text{O}_2$ consumed in the reaction is:
$$V(\text{reacted O}_2) = 150 - 30 = 120\text{ cm}^3$$
From the general equation for the complete combustion of a hydrocarbon:
$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}(\text{l})$$
$$\frac{V(\text{O}_2)}{V(\text{hydrocarbon})} = x + \frac{y}{4} = \frac{120}{20} = 6$$
Substituting $x = 4$:
$$4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8$$
Therefore, the molecular formula is $\text{C}_4\text{H}_8$.

評分準則

[1] for selecting option C.
[0] for any other option.
題目 19 · multiple_choice
1 分
Which molecule contains exactly two lone pairs of electrons on the central atom and has a non-linear (bent) shape?
  1. A.$\text{CO}_2$
  2. B.$\text{H}_2\text{S}$
  3. C.$\text{SO}_2$
  4. D.$\text{NH}_3$
查看答案詳解

解題

In hydrogen sulfide ($\text{H}_2\text{S}$), sulfur is in Group 16 with 6 valence electrons. It forms two single covalent bonds to hydrogen atoms and retains two non-bonding lone pairs. The 4 electron pairs adopt a tetrahedral arrangement, giving a non-linear (bent) molecular geometry with a bond angle of approximately $104.5^\circ$.
- $\text{CO}_2$ has 0 lone pairs on the central carbon atom and is linear.
- $\text{SO}_2$ has 1 lone pair on the sulfur atom and is non-linear.
- $\text{NH}_3$ has 1 lone pair on the nitrogen atom and is trigonal pyramidal.

評分準則

[1] for selecting option B.
[0] for any other option.
題目 20 · multiple_choice
1 分
The standard enthalpy changes of combustion, $\Delta H^\circ_{\text{c}}$, are given:

$$\begin{aligned}
\text{C}(\text{graphite}) & = -394\text{ kJ mol}^{-1} \\
\text{H}_2(\text{g}) & = -286\text{ kJ mol}^{-1} \\
\text{CH}_3\text{C}\equiv\text{CH}(\text{g}) & = -1938\text{ kJ mol}^{-1}
\end{aligned}$$

What is the standard enthalpy change of formation, $\Delta H^\circ_{\text{f}}$, of propyne, $\text{CH}_3\text{C}\equiv\text{CH}(\text{g})$?
  1. A.$-184\text{ kJ mol}^{-1}$
  2. B.$+184\text{ kJ mol}^{-1}$
  3. C.$-1258\text{ kJ mol}^{-1}$
  4. D.$+1258\text{ kJ mol}^{-1}$
查看答案詳解

解題

The equation for the standard enthalpy of formation of propyne is:
$$3\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_4(\text{g})$$
Using Hess's Law with enthalpies of combustion:
$$\Delta H^\circ_{\text{f}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products})$$
$$\Delta H^\circ_{\text{f}} = [3(-394) + 2(-286)] - [-1938]$$
$$\Delta H^\circ_{\text{f}} = [-1182 - 572] + 1938 = -1754 + 1938 = +184\text{ kJ mol}^{-1}$$

評分準則

[1] for selecting option B.
[0] for any other option.
題目 21 · multiple_choice
1 分
An element $M$ in Period 3 forms an oxide that reacts with aqueous sodium hydroxide but does not react with dilute hydrochloric acid.

The chloride of $M$ reacts vigorously with water to form an acidic solution of $\text{pH } \approx 2$ and a white solid precipitate.

What is the identity of element $M$?
  1. A.Aluminium
  2. B.Silicon
  3. C.Phosphorus
  4. D.Sulfur
查看答案詳解

解題

Silicon dioxide ($\text{SiO}_2$) is an acidic oxide that reacts with hot concentrated sodium hydroxide to form silicates, but being a non-metal covalent network solid, it does not react with dilute hydrochloric acid.
Silicon tetrachloride ($\text{SiCl}_4$) undergoes complete hydrolysis when added to water according to:
$$\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})$$
This produces a white solid precipitate of hydrated silicon dioxide and misty fumes / an acidic solution of $\text{HCl}(\text{aq})$ ($\text{pH } 1-2$).
Therefore, element $M$ is silicon.

評分準則

[1] for selecting option B.
[0] for any other option.
題目 22 · multiple_choice
1 分
Two separate experiments are carried out using Group 2 compounds.

- Experiment 1: Equal molar amounts of solid $\text{Mg(OH)}_2$ and $\text{Ba(OH)}_2$ are added to equal volumes of pure water at $298\text{ K}$.
- Experiment 2: Equal molar amounts of solid $\text{MgCO}_3$ and $\text{BaCO}_3$ are heated strongly under identical conditions with a Bunsen burner.

Which row correctly identifies the compound that forms the solution with the higher pH in Experiment 1 and the compound that decomposes more readily in Experiment 2?

$$\begin{array}{|c|c|c|}\hline & \text{Higher pH (Experiment 1)} & \text{Decomposes more readily (Experiment 2)} \\hline\text{A} & \text{Mg(OH)}_2 & \text{BaCO}_3 \\text{B} & \text{Mg(OH)}_2 & \text{MgCO}_3 \\text{C} & \text{Ba(OH)}_2 & \text{BaCO}_3 \\text{D} & \text{Ba(OH)}_2 & \text{MgCO}_3 \\hline\end{array}$$
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
查看答案詳解

解題

In Experiment 1, the solubility of Group 2 hydroxides increases down the group. $\text{Ba(OH)}_2$ is far more soluble in water than $\text{Mg(OH)}_2$, liberating a much higher concentration of hydroxide ions, $\text{OH}^-$, and resulting in a higher pH.
In Experiment 2, the thermal stability of Group 2 carbonates increases down the group. The $\text{Mg}^{2+}$ cation has a smaller ionic radius and thus a higher charge density than $\text{Ba}^{2+}$, polarising the carbonate ion more strongly and weakening the $\text{C}-\text{O}$ bond. Consequently, $\text{MgCO}_3$ decomposes more readily (at a lower temperature) than $\text{BaCO}_3$.

評分準則

[1] for selecting option D.
[0] for any other option.
題目 23 · multiple_choice
1 分
An alkene $Q$ is heated under reflux with a hot, concentrated, acidified solution of potassium manganate(VII).

The organic products formed are propanone, $(\text{CH}_3)_2\text{CO}$, and propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$, in a $1:1$ molar ratio.

What is the IUPAC name of alkene $Q$?
  1. A.2-methylpent-1-ene
  2. B.2-methylpent-2-ene
  3. C.3-methylpent-2-ene
  4. D.4-methylpent-2-ene
查看答案詳解

解題

Hot concentrated acidified $\text{KMnO}_4$ cleaves the carbon-carbon double bond ($\text{C}=\text{C}$):
- A disubstituted carbon atom, $=\text{C}(\text{CH}_3)_2$, oxidises to a ketone: propanone, $(\text{CH}_3)_2\text{C}=\text{O}$.
- A monosubstituted carbon atom, $=\text{CH}-\text{CH}_2\text{CH}_3$, oxidises to a carboxylic acid: propanoic acid, $\text{HOOC}-\text{CH}_2\text{CH}_3$.
Recombining these two fragments across the original double bond gives:
$$(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_2\text{CH}_3$$
Numbering the continuous 5-carbon chain from the end closest to the double bond gives carbon-2 as the alkene position with a methyl substituent on carbon-2: 2-methylpent-2-ene.

評分準則

[1] for selecting option B.
[0] for any other option.
題目 24 · multiple_choice
1 分
Compound $W$ has the molecular formula $\text{C}_4\text{H}_8\text{O}$.

Compound $W$:
- forms an orange precipitate when reacted with 2,4-dinitrophenylhydrazine,
- produces no silver mirror when warmed with Tollens' reagent,
- forms a yellow precipitate when warmed with alkaline aqueous iodine.

Which structure represents compound $W$?
  1. A.$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
  2. B.$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$
  3. C.$\text{CH}_3\text{COCH}_2\text{CH}_3$
  4. D.$(\text{CH}_3)_2\text{CHCHO}$
查看答案詳解

解題

1. Forming an orange precipitate with 2,4-DNPH confirms the presence of a carbonyl group (aldehyde or ketone).
2. Giving a negative Tollens' test (no silver mirror) eliminates aldehydes, confirming that $W$ is a ketone.
3. Forming a yellow precipitate of tri-iodomethane ($\text{CHI}_3$) with alkaline aqueous iodine indicates the presence of a methyl carbonyl group, $\text{CH}_3\text{C}(=\text{O})-$.
For a compound with formula $\text{C}_4\text{H}_8\text{O}$, the only ketone is butanone, $\text{CH}_3\text{COCH}_2\text{CH}_3$, which contains the $\text{CH}_3\text{C}(=\text{O})-$ group.

評分準則

[1] for selecting option C.
[0] for any other option.
題目 25 · multiple_choice
1 分
An uncharged atom of an element in Period 3 has exactly three unpaired electrons in its ground state.

What is the formula of the chloride formed when this element reacts completely with an excess of chlorine gas?
  1. A.\(\text{AlCl}_3\)
  2. B.\(\text{SiCl}_4\)
  3. C.\(\text{PCl}_3\)
  4. D.\(\text{PCl}_5\)
查看答案詳解

解題

1. Determine the electronic configurations of Period 3 elements:
- \(\text{Na}\): \([\text{Ne}]\,3\text{s}^1\) (1 unpaired electron)
- \(\text{Mg}\): \([\text{Ne}]\,3\text{s}^2\) (0 unpaired electrons)
- \(\text{Al}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}^1\) (1 unpaired electron)
- \(\text{Si}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}^2\) (2 unpaired electrons)
- \(\text{P}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}_x^1\,3\text{p}_y^1\,3\text{p}_z^1\) (3 unpaired electrons by Hund's rule)
- \(\text{S}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}^4\) (2 unpaired electrons)
- \(\text{Cl}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}^5\) (1 unpaired electron)
- \(\text{Ar}\): \([\text{Ne}]\,3\text{s}^2\,3\text{p}^6\) (0 unpaired electrons)

2. The element with three unpaired electrons is phosphorus (\(\text{P}\)).

3. Phosphorus reacts with an excess of chlorine gas to form phosphorus(V) chloride, \(\text{PCl}_5\).

評分準則

D [1 mark]: Correctly identifies phosphorus from ground-state orbital occupancy and deduces the higher chloride formed in excess chlorine.
題目 26 · multiple_choice
1 分
A gaseous hydrocarbon \(Q\) was completely burned in an excess of oxygen.

\(10\,\text{cm}^3\) of \(Q\) required \(65\,\text{cm}^3\) of oxygen for complete combustion, producing \(40\,\text{cm}^3\) of carbon dioxide. All gas volumes were measured at the same temperature and pressure.

Which statement about compound \(Q\) is correct?
  1. A.It rapidly decolourises aqueous bromine in the dark.
  2. B.There are only two structural isomers with the molecular formula of \(Q\).
  3. C.Its empirical formula is \(\text{C}_2\text{H}_3\).
  4. D.It contains one carbon–carbon double bond per molecule.
查看答案詳解

解題

1. Write the general equation for the complete combustion of a hydrocarbon:
\[\text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\,\text{CO}_2\text{(g)} + \frac{y}{2}\,\text{H}_2\text{O}\text{(l)}\]

2. By Avogadro's law, mole ratios equal volume ratios at constant temperature and pressure:
- From \(\text{CO}_2\): \(x = \frac{40}{10} = 4\)
- From \(\text{O}_2\): \(x + \frac{y}{4} = \frac{65}{10} = 6.5\)
- Substitute \(x = 4\): \(4 + \frac{y}{4} = 6.5 \implies \frac{y}{4} = 2.5 \implies y = 10\)

3. The molecular formula of \(Q\) is \(\text{C}_4\text{H}_{10}\).

4. \(\text{C}_4\text{H}_{10}\) has exactly two structural isomers: butane and 2-methylpropane.

評分準則

B [1 mark]: Correctly applies reacting gas volume stoichiometry to deduce molecular formula C4H10 and identifies that it has two structural isomers.
題目 27 · multiple_choice
1 分
The boiling points of the Group 16 hydrides are shown.

- \(\text{H}_2\text{O}\): \(373\,\text{K}\)
- \(\text{H}_2\text{S}\): \(213\,\text{K}\)
- \(\text{H}_2\text{Se}\): \(232\,\text{K}\)
- \(\text{H}_2\text{Te}\): \(271\,\text{K}\)

Which statement explains why the boiling point of \(\text{H}_2\text{Te}\) is higher than that of \(\text{H}_2\text{S}\)?
  1. A.\(\text{H}_2\text{Te}\) forms stronger hydrogen bonds between molecules than \(\text{H}_2\text{S}\).
  2. B.The \(\text{H}-\text{Te}\) covalent bond is stronger than the \(\text{H}-\text{S}\) covalent bond.
  3. C.\(\text{H}_2\text{Te}\) has more electrons per molecule, resulting in stronger instantaneous dipole–induced dipole forces.
  4. D.\(\text{H}_2\text{Te}\) has a greater permanent dipole moment than \(\text{H}_2\text{S}\).
查看答案詳解

解題

1. Boiling point differences between molecular covalent substances depend on the strength of intermolecular forces, not intramolecular covalent bonds.
2. Neither \(\text{H}_2\text{S}\) nor \(\text{H}_2\text{Te}\) exhibits hydrogen bonding (hydrogen bonding requires \(\text{H}\) bonded to highly electronegative \(\text{N}\), \(\text{O}\), or \(\text{F}\)).
3. \(\text{H}_2\text{Te}\) has significantly more electrons per molecule than \(\text{H}_2\text{S}\), leading to greater polarisability and stronger instantaneous dipole–induced dipole (London dispersion) forces, requiring more thermal energy to overcome.

評分準則

C [1 mark]: Explains trend in terms of number of electrons and strength of instantaneous dipole–induced dipole forces.
題目 28 · multiple_choice
1 分
Standard enthalpy changes of combustion, \(\Delta H^\circ_c\), are given:

- \(\Delta H^\circ_c[\text{C}_3\text{H}_6\text{(g)}] = -2058\,\text{kJ}\,\text{mol}^{-1}\)
- \(\Delta H^\circ_c[\text{C}_3\text{H}_7\text{OH(l)}] = -2021\,\text{kJ}\,\text{mol}^{-1}\)

What is the standard enthalpy change, \(\Delta H^\circ_r\), for the hydration of propene?

\[\text{C}_3\text{H}_6\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\]
  1. A.\(-4079\,\text{kJ}\,\text{mol}^{-1}\)
  2. B.\(-37\,\text{kJ}\,\text{mol}^{-1}\)
  3. C.\(+37\,\text{kJ}\,\text{mol}^{-1}\)
  4. D.\(+4079\,\text{kJ}\,\text{mol}^{-1}\)
查看答案詳解

解題

1. Using Hess's law and standard enthalpy changes of combustion:
\[\Delta H^\circ_r = \sum \Delta H^\circ_c(\text{reactants}) - \sum \Delta H^\circ_c(\text{products})\]

2. Note that \(\text{H}_2\text{O(l)}\) is fully oxidised and cannot undergo combustion, so \(\Delta H^\circ_c[\text{H}_2\text{O(l)}] = 0\).

3. Calculate:
\[\Delta H^\circ_r = [\Delta H^\circ_c(\text{C}_3\text{H}_6\text{(g)}) + 0] - [\Delta H^\circ_c(\text{C}_3\text{H}_7\text{OH(l)})]\]
\[\Delta H^\circ_r = (-2058) - (-2021) = -37\,\text{kJ}\,\text{mol}^{-1}\]

評分準則

B [1 mark]: Correct application of Hess's Law cycle using combustion data: -2058 - (-2021) = -37 kJ mol^-1.
題目 29 · multiple_choice
1 分
Two Period 3 elements, \(X\) and \(Y\), react with chlorine to form chlorides \(X\text{Cl}_n\) and \(Y\text{Cl}_m\).

- Adding solid \(X\text{Cl}_n\) to water gives a neutral solution (\(\text{pH} = 7\)).
- Adding solid \(Y\text{Cl}_m\) to water causes a vigorous reaction, producing an acidic solution and steamy white fumes.
- The oxide of \(X\) is basic.
- The oxide of \(Y\) reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.

Which row correctly identifies elements \(X\) and \(Y\)?
  1. A.\(X = \text{magnesium};\, Y = \text{silicon}\)
  2. B.\(X = \text{sodium};\, Y = \text{aluminium}\)
  3. C.\(X = \text{sodium};\, Y = \text{phosphorus}\)
  4. D.\(X = \text{magnesium};\, Y = \text{aluminium}\)
查看答案詳解

解題

1. \(X\text{Cl}_n\) dissolves in water without hydrolysing to give \(\text{pH} = 7\), and its oxide is basic. This corresponds to sodium (\(\text{NaCl}\) forms neutral solution, \(\text{Na}_2\text{O}\) is basic).
2. \(Y\text{Cl}_m\) hydrolyses in water to give an acidic solution with misty fumes of \(\text{HCl}\), and its oxide reacts with both acids and alkalis (i.e. is amphoteric). This corresponds to aluminium (\(\text{AlCl}_3\) hydrolyses, \(\text{Al}_2\text{O}_3\) is amphoteric).
3. Silicon(IV) oxide (\(\text{SiO}_2\)) is acidic and does not react with dilute \(\text{HCl}\); magnesium oxide is basic only.

評分準則

B [1 mark]: Correctly identifies X as sodium and Y as aluminium from periodicity rules of Period 3 chlorides and oxides.
題目 30 · multiple_choice
1 分
An unlabelled white powder is known to be a Group 2 compound.

- The powder is insoluble in water.
- When dilute nitric acid is added to the powder, vigorous effervescence occurs, and a colourless solution is formed.
- When aqueous sodium sulfate is added to this resulting colourless solution, a dense white precipitate forms immediately.

What is the identity of the unlabelled white powder?
  1. A.\(\text{BaCO}_3\)
  2. B.\(\text{BaSO}_4\)
  3. C.\(\text{MgCO}_3\)
  4. D.\(\text{Mg(OH)}_2\)
查看答案詳解

解題

1. Vigorous effervescence with dilute nitric acid indicates the presence of carbonate ion, \(\text{CO}_3^{2-}\), releasing \(\text{CO}_2\text{(g)}\). This eliminates hydroxides and sulfates.
2. The reaction produces a solution containing the Group 2 metal nitrate: \(\text{MCO}_3\text{(s)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{M(NO}_3)_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\).
3. Addition of aqueous sodium sulfate gives a white precipitate of the insoluble Group 2 sulfate:
\[\text{M}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{MSO}_4\text{(s)}\]
4. Magnesium sulfate (\(\text{MgSO}_4\)) is soluble in water and would not precipitate. Barium sulfate (\(\text{BaSO}_4\)) is insoluble in water and forms a dense white precipitate.
5. Therefore, the compound is barium carbonate, \(\text{BaCO}_3\).

評分準則

A [1 mark]: Identifies carbonate from effervescence with acid and Ba2+ from sulfate precipitation.
題目 31 · multiple_choice
1 分
Four separate samples of halogenoalkanes are each mixed with an equal volume of ethanolic silver nitrate and placed in a water bath at \(50\,^\circ\text{C}\).

Which halogenoalkane produces a precipitate in the shortest time?
  1. A.1-chlorobutane
  2. B.1-iodobutane
  3. C.2-chloro-2-methylpropane
  4. D.2-iodo-2-methylpropane
查看答案詳解

解題

1. The rate of hydrolysis of halogenoalkanes depends on:
- The strength of the carbon–halogen bond: \(\text{C}-\text{I}\) (bond enthalpy \(240\,\text{kJ}\,\text{mol}^{-1}\)) is weaker than \(\text{C}-\text{Cl}\) (bond enthalpy \(340\,\text{kJ}\,\text{mol}^{-1}\)), so iodoalkanes react faster than chloroalkanes.
- The mechanism and structure of the alkyl group: tertiary halogenoalkanes hydrolyse much faster than primary halogenoalkanes because they react via the \(\text{S}_\text{N}1\) pathway involving a relatively stable tertiary carbocation intermediate.
2. Combining both factors, 2-iodo-2-methylpropane (a tertiary iodoalkane) undergoes hydrolysis fastest and forms a precipitate of \(\text{AgI}\) in the shortest time.

評分準則

D [1 mark]: Identifies that tertiary iodoalkanes have both the weakest C-X bond and fastest SN1 rate.
題目 32 · multiple_choice
1 分
Compound \(Z\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- \(Z\) forms an orange precipitate when reacted with 2,4-dinitrophenylhydrazine.
- \(Z\) does not produce a silver mirror when warmed with Tollens' reagent.
- \(Z\) forms a yellow precipitate when warmed with alkaline aqueous iodine.

What is the structure of \(Z\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
  2. B.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
  3. C.\((\text{CH}_3)_2\text{CHCHO}\)
  4. D.\(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{OH}\)
查看答案詳解

解題

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine indicates that \(Z\) contains a carbonyl group (aldehyde or ketone).
2. A negative Tollens' reagent test indicates that \(Z\) is a ketone, not an aldehyde (eliminating butanal and 2-methylpropanal).
3. Formation of a yellow precipitate of tri-iodomethane (\(\text{CHI}_3\)) with alkaline aqueous iodine indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C}=\text{O}\).
4. Butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), fits all three criteria.

評分準則

B [1 mark]: Correctly deduces butanone from functional group test responses.
題目 33 · multiple_choice
1 分
Element \(X\) is in Period 3 of the Periodic Table. The first seven successive ionisation energies of \(X\), in \(\text{kJ mol}^{-1}\), are shown:

1012, 1907, 2914, 4964, 6274, 21267, 25431

Which statement about the oxide of \(X\) in its highest oxidation state is correct?
  1. A.It is an ionic solid that dissolves in water to form a neutral solution.
  2. B.It reacts with water to form an acidic solution with a pH of approximately 1 to 2.
  3. C.It is an amphoteric oxide that dissolves readily in both dilute acids and dilute alkalis.
  4. D.It has a giant covalent lattice structure with a very high melting point.
查看答案詳解

解題

1. Identify the group of element \(X\):
Looking at the successive ionisation energies:
- 1st: 1012
- 2nd: 1907
- 3rd: 2914
- 4th: 4964
- 5th: 6274
- 6th: 21267
The largest jump occurs between the 5th and 6th ionisation energies (from 6274 to 21267 \(\text{kJ mol}^{-1}\)), which corresponds to removing an electron from an inner quantum shell. Therefore, \(X\) has 5 valence electrons and belongs to Group 15. In Period 3, element \(X\) is phosphorus (\(\text{P}\)).

2. Determine the highest oxide of phosphorus:
The highest oxidation state oxide of phosphorus is phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\).

3. Properties of \(\text{P}_4\text{O}_{10}\):
\(\text{P}_4\text{O}_{10}\) is a simple molecular acidic oxide. It reacts vigorously with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\):
\[\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\]
This produces a strongly acidic solution with a \(\text{pH}\) of approximately 1 to 2.

Therefore, statement B is correct.

評分準則

[1 mark] for selecting B.
- A is incorrect: \(\text{P}_4\text{O}_{10}\) is covalent, not ionic.
- C is incorrect: \(\text{P}_4\text{O}_{10}\) is purely acidic, not amphoteric (\(\text{Al}_2\text{O}_3\) is amphoteric).
- D is incorrect: \(\text{P}_4\text{O}_{10}\) has a simple molecular structure, not giant covalent (\(\text{SiO}_2\) is giant covalent).
題目 34 · multiple_choice
1 分
A \(0.174\text{ g}\) sample of an anhydrous Group 2 metal carbonate, \(M\text{CO}_3\), was added to \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}(\text{aq})\) (an excess).

The resulting unreacted acid required \(22.00\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide, \(\text{NaOH}(\text{aq})\), for complete neutralisation.

What is the relative atomic mass, \(A_r\), of metal \(M\)?
  1. A.24.3
  2. B.40.1
  3. C.64.3
  4. D.87.6
查看答案詳解

解題

1. Calculate initial moles of \(\text{HCl}\):
\[n(\text{HCl})_{\text{total}} = 50.0 \times 10^{-3}\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\]

2. Calculate moles of unreacted (excess) \(\text{HCl}\) using \(\text{NaOH}\):
\[n(\text{NaOH}) = 22.00 \times 10^{-3}\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.20 \times 10^{-3}\text{ mol}\]
Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), \(n(\text{HCl})_{\text{excess}} = 2.20 \times 10^{-3}\text{ mol}\).

3. Calculate moles of \(\text{HCl}\) that reacted with \(M\text{CO}_3\):
\[n(\text{HCl})_{\text{reacted}} = 5.00 \times 10^{-3} - 2.20 \times 10^{-3} = 2.80 \times 10^{-3}\text{ mol}\]

4. Calculate moles of \(M\text{CO}_3\):
\[M\text{CO}_3 + 2\text{HCl} \rightarrow M\text{Cl}_2 + \text{H}_2\text{O} + \text{CO}_2\]
\[n(M\text{CO}_3) = \frac{2.80 \times 10^{-3}}{2} = 1.40 \times 10^{-3}\text{ mol}\]

5. Determine \(M_r\) of \(M\text{CO}_3\) and \(A_r(M)\):
\[M_r(M\text{CO}_3) = \frac{0.174\text{ g}}{1.40 \times 10^{-3}\text{ mol}} = 124.3\text{ g mol}^{-1}\]
\[A_r(M) = 124.3 - [12.0 + (3 \times 16.0)] = 124.3 - 60.0 = 64.3\]

Therefore, \(A_r(M) = 64.3\) (Option C).

評分準則

[1 mark] for selecting C.
- A (24.3) corresponds to omitting the stoichiometry of 2:1 for HCl : MCO3.
- B (40.1) corresponds to arithmetic miscalculation.
- D (87.6) corresponds to taking total HCl instead of reacted HCl.
題目 35 · multiple_choice
1 分
Which list places the molecules in order of increasing bond angle, with the smallest bond angle first?
  1. A.\(\text{H}_2\text{O} < \text{NH}_3 < \text{CH}_4 < \text{BF}_3\)
  2. B.\(\text{NH}_3 < \text{H}_2\text{O} < \text{BF}_3 < \text{CH}_4\)
  3. C.\(\text{H}_2\text{O} < \text{CH}_4 < \text{NH}_3 < \text{BF}_3\)
  4. D.\(\text{BF}_3 < \text{CH}_4 < \text{NH}_3 < \text{H}_2\text{O}\)
查看答案詳解

解題

According to VSEPR theory:
- \(\text{H}_2\text{O}\): 2 bonding pairs and 2 lone pairs around oxygen (tetrahedral arrangement of pairs). Lone pair–lone pair repulsion reduces the bond angle to approximately \(104.5^\circ\).
- \(\text{NH}_3\): 3 bonding pairs and 1 lone pair around nitrogen (trigonal pyramidal). Lone pair–bonding pair repulsion reduces the bond angle to approximately \(107^\circ\).
- \(\text{CH}_4\): 4 bonding pairs and 0 lone pairs around carbon (regular tetrahedral). Bond angle is \(109.5^\circ\).
- \(\text{BF}_3\): 3 bonding pairs and 0 lone pairs around boron (trigonal planar). Bond angle is \(120^\circ\).

Ordering from smallest to largest:
\[\text{H}_2\text{O}\ (104.5^\circ) < \text{NH}_3\ (107^\circ) < \text{CH}_4\ (109.5^\circ) < \text{BF}_3\ (120^\circ)\]

Therefore, option A is correct.

評分準則

[1 mark] for selecting A.
題目 36 · multiple_choice
1 分
The standard enthalpy changes of combustion, \(\Delta H^\circ_c\), for carbon, hydrogen, and propene are given.

\(\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})\quad \Delta H^\circ_c = -394\text{ kJ mol}^{-1}\)

\(\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})\quad \Delta H^\circ_c = -286\text{ kJ mol}^{-1}\)

\(\text{C}_3\text{H}_6(\text{g}) + \frac{9}{2}\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\quad \Delta H^\circ_c = -2058\text{ kJ mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H^\circ_f\), of propene, \(\text{C}_3\text{H}_6(\text{g})\)?
  1. A.\(-2040\text{ kJ mol}^{-1}\)
  2. B.\(-18\text{ kJ mol}^{-1}\)
  3. C.\(+18\text{ kJ mol}^{-1}\)
  4. D.\(+2040\text{ kJ mol}^{-1}\)
查看答案詳解

解題

The target formation reaction for propene is:
\[3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_6(\text{g})\]

Using Hess's Law with standard enthalpy changes of combustion:
\[\Delta H^\circ_f(\text{C}_3\text{H}_6) = \sum \Delta H^\circ_c(\text{reactants}) - \sum \Delta H^\circ_c(\text{products})\]
\[\Delta H^\circ_f(\text{C}_3\text{H}_6) = [3 \times \Delta H^\circ_c(\text{C})] + [3 \times \Delta H^\circ_c(\text{H}_2)] - [\Delta H^\circ_c(\text{C}_3\text{H}_6)]\]
\[\Delta H^\circ_f = [3(-394)] + [3(-286)] - (-2058)\]
\[\Delta H^\circ_f = -1182 - 858 + 2058 = -2040 + 2058 = +18\text{ kJ mol}^{-1}\]

Therefore, option C is correct.

評分準則

[1 mark] for selecting C.
- A (-2040 kJ/mol) omits the combustion value of propene.
- B (-18 kJ/mol) arises from reversing the sign in the subtraction (products - reactants).
- D (+2040 kJ/mol) is the total combustion enthalpy of the elements with wrong sign.
題目 37 · multiple_choice
1 分
Two Period 3 chlorides, \(\text{SiCl}_4\) and \(\text{PCl}_5\), are each added separately to an excess of cold water.

Which row correctly describes the observations and the \(\text{pH}\) of the resulting solutions?
  1. A.\(\text{SiCl}_4\): white solid formed and pH 1–2 ; \(\text{PCl}_5\): colourless solution formed and pH 1–2
  2. B.\(\text{SiCl}_4\): colourless solution formed and pH 7 ; \(\text{PCl}_5\): white solid formed and pH 1–2
  3. C.\(\text{SiCl}_4\): white solid formed and pH 6–7 ; \(\text{PCl}_5\): colourless solution formed and pH 1–2
  4. D.\(\text{SiCl}_4\): colourless solution formed and pH 1–2 ; \(\text{PCl}_5\): colourless solution formed and pH 6–7
查看答案詳解

解題

1. Reaction of \(\text{SiCl}_4\) with water:
\[\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\]
Observations: Vigorous hydrolysis producing steamy white fumes of \(\text{HCl}\) and a white solid / precipitate of hydrated silicon dioxide (\(\text{SiO}_2\)). The solution is strongly acidic with \(\text{pH} \approx 1\text{--}2\).

2. Reaction of \(\text{PCl}_5\) with water:
\[\text{PCl}_5(\text{s}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{PO}_4(\text{aq}) + 5\text{HCl}(\text{aq})\]
Observations: Vigorous hydrolysis producing steamy white fumes of \(\text{HCl}\). The solid completely dissolves/reacts to form a clear, colourless solution containing phosphoric(V) acid and hydrochloric acid with \(\text{pH} \approx 1\text{--}2\).

Therefore, row A is correct.

評分準則

[1 mark] for selecting A.
題目 38 · multiple_choice
1 分
Separate samples of solid anhydrous magnesium nitrate, \(\text{Mg}(\text{NO}_3)_2\), and barium nitrate, \(\text{Ba}(\text{NO}_3)_2\), are heated strongly over a Bunsen flame.

Which statement correctly explains the difference in their thermal stability?
  1. A.Barium nitrate decomposes at a lower temperature because the \(\text{Ba}^{2+}\) ion has a higher charge density than \(\text{Mg}^{2+}\).
  2. B.Magnesium nitrate decomposes to form a brown gas, whereas barium nitrate decomposes to form only a colourless gas.
  3. C.Magnesium nitrate decomposes more readily because the \(\text{Mg}^{2+}\) ion polarises the nitrate anion more strongly than \(\text{Ba}^{2+}\).
  4. D.Both nitrates decompose completely to form the corresponding Group 2 metal and nitrogen dioxide gas.
查看答案詳解

解題

1. Trend in thermal stability of Group 2 nitrates:
Thermal stability increases down Group 2 from \(\text{Mg}(\text{NO}_3)_2\) to \(\text{Ba}(\text{NO}_3)_2\).

2. Explanation:
- The \(\text{Mg}^{2+}\) ion has a much smaller ionic radius than the \(\text{Ba}^{2+}\) ion while carrying the same \(+2\) charge.
- Therefore, \(\text{Mg}^{2+}\) has a significantly higher charge density and greater polarising power.
- It polarises and distorts the electron cloud of the nitrate (\(\text{NO}_3^-\)) anion more strongly, weakening the \(\text{N}-\text{O}\) bond within the nitrate ion and causing it to decompose at a lower temperature (more readily).

Both nitrates decompose to give the metal oxide, \(\text{NO}_2\) (brown gas), and \(\text{O}_2\):
\[2M(\text{NO}_3)_2(\text{s}) \rightarrow 2M\text{O}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\]

Therefore, statement C is correct.

評分準則

[1 mark] for selecting C.
題目 39 · multiple_choice
1 分
An alkene, \(Y\), has the molecular formula \(\text{C}_6\text{H}_{12}\).

When \(Y\) is heated under reflux with hot concentrated acidified potassium manganate(VII), the only organic products obtained are propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), and propanone, \((\text{CH}_3)_2\text{CO}\).

What is the systematic IUPAC name of alkene \(Y\)?
  1. A.2-methylpent-1-ene
  2. B.2-methylpent-2-ene
  3. C.3-methylpent-2-ene
  4. D.4-methylpent-2-ene
查看答案詳解

解題

1. Recall the oxidative cleavage rules for alkenes with hot concentrated acidified \(\text{KMnO}_4\):
- A terminal \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\).
- A \(=\text{CH}-R\) group is oxidised to a carboxylic acid, \(R-\text{COOH}\).
- A \(=\text{C}(R_1)(R_2)\) group is oxidised to a ketone, \(R_1-\text{C}(=\text{O})-R_2\).

2. Deduce the fragments of alkene \(Y\):
- The formation of propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) indicates the presence of a \(\text{CH}_3\text{CH}_2\text{CH}=\) fragment.
- The formation of propanone (\((\text{CH}_3)_2\text{C}=\text{O}\)) indicates the presence of a \(=\text{C}(\text{CH}_3)_2\) fragment.

3. Combine the fragments:
\[\text{CH}_3-\text{CH}_2-\text{CH}=\text{C}(\text{CH}_3)_2\]

4. Name the compound:
- Longest continuous carbon chain containing the double bond = 5 carbons (pent-2-ene).
- Number from the end closer to the double bond and substituents: C1 is on the right, giving the double bond between C2 and C3, and a methyl branch at C2.
- IUPAC name: 2-methylpent-2-ene.

Therefore, option B is correct.

評分準則

[1 mark] for selecting B.
- A (2-methylpent-1-ene) produces propan-2-one, CO2, and H2O.
- C (3-methylpent-2-ene) produces ethanoic acid and butanone.
- D (4-methylpent-2-ene) produces ethanoic acid and 2-methylpropanoic acid.
題目 40 · multiple_choice
1 分
Compound \(Z\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- \(Z\) gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent.
- \(Z\) shows no reaction when warmed with Tollens' reagent.
- \(Z\) forms a yellow precipitate when warmed with alkaline aqueous iodine.

What is the structural formula of \(Z\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
  2. B.\((\text{CH}_3)_2\text{CHCHO}\)
  3. C.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
  4. D.\(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{OH}\)
查看答案詳解

解題

1. Analyse each test observation:
- Orange precipitate with 2,4-DNPH: \(Z\) is a carbonyl compound (aldehyde or ketone).
- No reaction / no silver mirror with Tollens' reagent: \(Z\) is not an aldehyde, so it must be a ketone.
- Yellow precipitate of tri-iodomethane (\(\text{CHI}_3\)) with alkaline aqueous iodine (iodoform test): \(Z\) contains a methyl ketone group, \(\text{CH}_3\text{C}=\text{O}\).

2. Deduce the structure:
A 4-carbon ketone (\(\text{C}_4\text{H}_8\text{O}\)) with a \(\text{CH}_3\text{C}=\text{O}\) group is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\) and \((\text{CH}_3)_2\text{CHCHO}\) are aldehydes (they would react positively with Tollens' reagent).
- \(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{OH}\) is an unsaturated alcohol (it would not react with 2,4-DNPH).

Therefore, option C is correct.

評分準則

[1 mark] for selecting C.

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卷二 (AS Level 結構題)

Answer all questions in the spaces provided on the question paper. Show all working in calculations.
5 題目 · 60 分
題目 1 · structured
12 分
(a) Define relative atomic mass, \(A_r\). [2]

(b) A naturally occurring sample of element \(\text{M}\) contains four isotopes with the following percentage abundances:
- \(^{84}\text{M}\): 0.56%
- \(^{86}\text{M}\): 9.86%
- \(^{87}\text{M}\): 7.00%
- \(^{88}\text{M}\): 82.58%

Calculate the relative atomic mass, \(A_r\), of \(\text{M}\) to two decimal places. [2]

(c) A volatile organic compound \(\text{Y}\) contains carbon, hydrogen, and oxygen only. Elemental analysis shows that \(\text{Y}\) has a percentage composition of 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen by mass.

(i) Show that the empirical formula of \(\text{Y}\) is \(\text{C}_2\text{H}_4\text{O}\). [2]

(ii) In an experiment, a \(0.220\text{ g}\) sample of liquid \(\text{Y}\) was completely vaporised in a gas syringe at \(373\text{ K}\) and \(101\text{ kPa}\). The volume of vaporised gas collected was \(77.2\text{ cm}^3\).

Use the ideal gas equation, \(pV = nRT\), to calculate the relative molecular mass, \(M_r\), of \(\text{Y}\), and deduce its molecular formula. [\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)] [4]

(d) State two basic assumptions of the kinetic theory of ideal gases that are not valid for real gases at high pressure. [2]
查看答案詳解

解題

(a) Relative atomic mass is the weighted average mass of an atom of an element relative to 1/12th the mass of an atom of carbon-12 (\(^{12}\text{C}\)).

(b) \(A_r = \frac{(84 \times 0.56) + (86 \times 9.86) + (87 \times 7.00) + (88 \times 82.58)}{100} = \frac{47.04 + 847.96 + 609.00 + 7267.04}{100} = \frac{8771.04}{100} = 87.71\)

(c)(i) Moles of each element in 100 g:
- \(\text{C}: \frac{54.5}{12.0} = 4.542\text{ mol}\)
- \(\text{H}: \frac{9.1}{1.0} = 9.100\text{ mol}\)
- \(\text{O}: \frac{36.4}{16.0} = 2.275\text{ mol}\)

Dividing by the smallest value (2.275):
- \(\text{C}: \frac{4.542}{2.275} = 2.00\)
- \(\text{H}: \frac{9.100}{2.275} = 4.00\)
- \(\text{O}: \frac{2.275}{2.275} = 1.00\)

Empirical formula = \(\text{C}_2\text{H}_4\text{O}\).

(c)(ii) Convert units: \(p = 101\text{ kPa} = 101\,000\text{ Pa}\), \(V = 77.2\text{ cm}^3 = 7.72 \times 10^{-5}\text{ m}^3\), \(T = 373\text{ K}\).
\(n = \frac{pV}{RT} = \frac{101000 \times 7.72 \times 10^{-5}}{8.31 \times 373} = \frac{7.7972}{3099.63} = 2.5155 \times 10^{-3}\text{ mol}\)
\(M_r = \frac{m}{n} = \frac{0.220}{2.5155 \times 10^{-3}} = 87.46 \approx 88.0\)
Empirical formula mass of \(\text{C}_2\text{H}_4\text{O} = (2 \times 12.0) + (4 \times 1.0) + 16.0 = 44.0\).
\(\frac{M_r}{\text{empirical mass}} = \frac{88.0}{44.0} = 2\)
Molecular formula = \(\text{C}_4\text{H}_8\text{O}_2\).

(d) 1. Real gas particles have finite (non-negligible) volume.
2. There are attractive forces (intermolecular forces) between particles in a real gas.

評分準則

(a) [2 marks]
- Weighted average mass of an atom of an element [1]
- Compared to 1/12th of the mass of an atom of carbon-12 (or relative to \(^{12}\text{C} = 12\)) [1]

(b) [2 marks]
- Correct expression: \(\frac{(84 \times 0.56) + (86 \times 9.86) + (87 \times 7.00) + (88 \times 82.58)}{100}\) [1]
- Correct answer to 2 decimal places: 87.71 [1]

(c)(i) [2 marks]
- Correct calculation of mole ratio: \(\text{C} = 4.54\), \(\text{H} = 9.10\), \(\text{O} = 2.275\) [1]
- Simplest whole number ratio shown clearly: \(2 : 4 : 1\) giving \(\text{C}_2\text{H}_4\text{O}\) [1]

(c)(ii) [4 marks]
- Conversion of volume to \(\text{m}^3\) (\(7.72 \times 10^{-5}\)) and pressure to \(\text{Pa}\) (\(1.01 \times 10^5\)) [1]
- Correct calculation of amount in moles \(n = 2.52 \times 10^{-3}\text{ mol}\) [1]
- Calculation of \(M_r = 87.5\text{--}88.0\) [1]
- Molecular formula deduced as \(\text{C}_4\text{H}_8\text{O}_2\) [1]

(d) [2 marks]
- The volume occupied by gas particles/molecules is negligible compared to the volume of the container [1]
- There are no intermolecular forces of attraction between gas particles [1]
題目 2 · structured
12 分
This question is about Period 3 elements and their compounds.

(a) Write the full electronic configuration of the following ions:
(i) \(\text{Al}^{3+}\) [1]
(ii) \(\text{P}^{3-}\) [1]

(b) Silicon tetrachloride, \(\text{SiCl}_4\), and sodium chloride, \(\text{NaCl}\), are Period 3 chlorides.
(i) Describe the observations when \(\text{SiCl}_4\) is added to water and write a balanced equation for the reaction. [2]
(ii) State the approximate pH of the solution formed when \(\text{NaCl}\) is dissolved in water and when \(\text{AlCl}_3\) is dissolved in water. Explain the difference in pH between these two solutions. [3]

(c) Chlorine gas reacts with cold dilute aqueous sodium hydroxide.
(i) Construct a balanced chemical equation for this reaction. [1]
(ii) State the oxidation state of chlorine in each chlorine-containing product and explain why this reaction is classified as a disproportionation reaction. [2]

(d) Solid potassium bromide reacts with concentrated sulfuric acid.
(i) State two observations made during this reaction. [1]
(ii) Write a balanced chemical equation for the redox reaction in which bromide ions are oxidised by concentrated sulfuric acid. [1]
查看答案詳解

解題

(a)(i) \(\text{Al}^{3+}: 1s^2 2s^2 2p^6\)
(a)(ii) \(\text{P}^{3-}: 1s^2 2s^2 2p^6 3s^2 3p^6\)

(b)(i) Observations: Steamy/misty white fumes (of \(\text{HCl}\)) and a white solid/precipitate (\(\text{SiO}_2\)).
Equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) (or \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\))

(b)(ii) \(\text{NaCl}\): \(\text{pH} = 7\) (neutral).
\(\text{AlCl}_3\): \(\text{pH} = 2\text{--}3\) (acidic).
Explanation: \(\text{Na}^+\) ions do not polarise water molecules significantly because of low charge density. \(\text{Al}^{3+}\) has a small ionic radius and high charge (high charge density), which polarises the coordinated water molecules in \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\), weakening the \(\text{O}-\text{H}\) bond and releasing \(\text{H}^+\) ions (hydrolysis).

(c)(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)

(c)(ii) Oxidation states: in \(\text{NaCl}\), \(\text{Cl} = -1\); in \(\text{NaClO}\), \(\text{Cl} = +1\).
It is disproportionation because the same element (chlorine in oxidation state 0) is simultaneously oxidised (from 0 to +1) and reduced (from 0 to -1).

(d)(i) Observations: Red-brown/orange vapour/gas (\(\text{Br}_2\)) and steamy/white misty fumes (\(\text{HBr}\)) / choking gas (\(\text{SO}_2\)).
(d)(ii) \(2\text{KBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) (or ionic: \(2\text{Br}^- + 4\text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)).

評分準則

(a)(i) [1 mark] \(1s^2 2s^2 2p^6\)
(a)(ii) [1 mark] \(1s^2 2s^2 2p^6 3s^2 3p^6\)

(b)(i) [2 marks]
- White precipitate / white solid AND misty/steamy fumes [1]
- Balanced equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) [1]

(b)(ii) [3 marks]
- States \(\text{pH} = 7\) for \(\text{NaCl}\) and \(\text{pH} = 1\text{--}3\) for \(\text{AlCl}_3\) [1]
- \(\text{Al}^{3+}\) has high charge density / high charge-to-size ratio [1]
- Polarises/weakens \(\text{O}-\text{H}\) bond in coordinated water, releasing \(\text{H}^+\) [1]

(c)(i) [1 mark] \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (allow ionic equation)

(c)(ii) [2 marks]
- Oxidation states: \(-1\) in \(\text{NaCl}\) and \(+1\) in \(\text{NaClO}\) [1]
- Explanation: Chlorine is simultaneously oxidised and reduced [1]

(d)(i) [1 mark] Orange-brown vapour / reddish-brown gas AND steamy / misty fumes
(d)(ii) [1 mark] \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) or full balanced formula equation
題目 3 · structured
12 分
Sulfur trioxide is manufactured industrially by the oxidation of sulfur dioxide:

\[2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H^\circ = -196\text{ kJ mol}^{-1}\]

(a) Define standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\). [2]

(b) The standard enthalpy change of formation of \(\text{SO}_2(\text{g})\) is \(-296.8\text{ kJ mol}^{-1}\).
Calculate the standard enthalpy change of formation of \(\text{SO}_3(\text{g})\). [2]

(c) Use Le Chatelier's principle to predict and explain the effect of increasing the temperature on the equilibrium position and the value of \(K_\text{p}\). [2]

(d) In a closed vessel of fixed volume \(5.00\text{ dm}^3\) at a constant temperature \(T\), an initial mixture of \(2.00\text{ mol}\) of \(\text{SO}_2(\text{g})\) and \(1.00\text{ mol}\) of \(\text{O}_2(\text{g})\) was allowed to reach dynamic equilibrium. At equilibrium, \(1.40\text{ mol}\) of \(\text{SO}_3(\text{g})\) was present.

(i) Calculate the equilibrium amounts, in moles, of \(\text{SO}_2(\text{g})\) and \(\text{O}_2(\text{g})\). [2]

(ii) Write the expression for the equilibrium constant, \(K_\text{c}\), for this reaction. [1]

(iii) Calculate the value of \(K_\text{c}\) at temperature \(T\) and state its units. [3]
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解題

(a) The standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (\(298\text{ K}\) and \(101\text{ kPa}\)).

(b) \(\Delta H^\circ_\text{r} = \sum \Delta H^\circ_\text{f}(\text{products}) - \sum \Delta H^\circ_\text{f}(\text{reactants})\)
\(-196 = [2 \times \Delta H^\circ_\text{f}(\text{SO}_3)] - [2 \times (-296.8) + 0]\)
\(-196 = 2\Delta H^\circ_\text{f}(\text{SO}_3) + 593.6\)
\(2\Delta H^\circ_\text{f}(\text{SO}_3) = -196 - 593.6 = -789.6\text{ kJ}\)
\(\Delta H^\circ_\text{f}(\text{SO}_3) = \frac{-789.6}{2} = -394.8\text{ kJ mol}^{-1}\)

(c) The forward reaction is exothermic (\(\Delta H < 0\)). Increasing the temperature causes the equilibrium to shift in the endothermic direction (to the left) to absorb heat. Therefore, the yield of \(\text{SO}_3\) decreases and the value of \(K_\text{p}\) decreases.

(d)(i) Reaction stoichiometry: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\)
Since \(1.40\text{ mol}\) of \(\text{SO}_3\) is formed:
- Moles of \(\text{SO}_2\) reacted = \(1.40\text{ mol}\) \(\implies n(\text{SO}_2)_\text{eq} = 2.00 - 1.40 = 0.60\text{ mol}\)
- Moles of \(\text{O}_2\) reacted = \(\frac{1.40}{2} = 0.70\text{ mol}\) \(\implies n(\text{O}_2)_\text{eq} = 1.00 - 0.70 = 0.30\text{ mol}\)

(d)(ii) \(K_\text{c} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\)

(d)(iii) Equilibrium concentrations:
\([\text{SO}_3] = \frac{1.40}{5.00} = 0.280\text{ mol dm}^{-3}\)
\([\text{SO}_2] = \frac{0.60}{5.00} = 0.120\text{ mol dm}^{-3}\)
\([\text{O}_2] = \frac{0.30}{5.00} = 0.060\text{ mol dm}^{-3}\)

\(K_\text{c} = \frac{(0.280)^2}{(0.120)^2 \times (0.060)} = \frac{0.0784}{0.0144 \times 0.060} = \frac{0.0784}{0.000864} = 90.74 \approx 90.7\)
Units of \(K_\text{c} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 (\text{mol dm}^{-3})} = \text{mol}^{-1}\text{ dm}^3\) (or \(\text{dm}^3\text{ mol}^{-1}\)).

評分準則

(a) [2 marks]
- Enthalpy change when one mole of a compound is formed from its elements [1]
- Under standard conditions and in standard states [1]

(b) [2 marks]
- Correct cycle or expression: \(2\Delta H_\text{f}(\text{SO}_3) = -196 + 2(-296.8)\) [1]
- Correct calculation: \(-394.8\text{ kJ mol}^{-1}\) [1]

(c) [2 marks]
- Equilibrium shifts to the left / favours reverse reaction because forward reaction is exothermic [1]
- \(K_\text{p}\) decreases [1]

(d)(i) [2 marks]
- \(n(\text{SO}_2) = 0.60\text{ mol}\) [1]
- \(n(\text{O}_2) = 0.30\text{ mol}\) [1]

(d)(ii) [1 mark]
- \(K_\text{c} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\)

(d)(iii) [3 marks]
- Correct division of equilibrium moles by 5.00 to find concentrations [1]
- Correct calculation of \(K_\text{c} = 90.7\) (accept 90.74) [1]
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\) or \(\text{mol}^{-1}\text{ dm}^3\) [1]
題目 4 · structured
12 分
Halogenoalkanes undergo nucleophilic substitution and elimination reactions.

(a) 2-bromobutane, \(\text{CH}_3\text{CH(Br)CH}_2\text{CH}_3\), reacts with warm aqueous sodium hydroxide.
(i) Name the organic product formed and state the type of stereoisomerism it exhibits. [2]
(ii) Outline the mechanism for the \(\text{S}_\text{N}2\) hydrolysis of 2-bromobutane by hydroxide ions. Include relevant dipoles, curly arrows, and the structure of the transition state. [3]
(iii) Explain why tertiary halogenoalkanes, such as 2-bromo-2-methylpropane, react via the \(\text{S}_\text{N}1\) mechanism rather than the \(\text{S}_\text{N}2\) mechanism. [2]

(b) When 2-bromobutane is heated with sodium hydroxide dissolved in ethanol, an elimination reaction occurs to form a mixture of three isomeric alkenes.
(i) Draw the skeletal formulas of all three isomeric alkenes formed and name each one. [3]
(ii) State the type of isomerism shown between the stereoisomers formed in (b)(i). [1]
(iii) Describe a chemical test to distinguish an alkene from an alkane, including the reagent and the observation with the alkene. [1]
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解題

(a)(i) Organic product: butan-2-ol.
Type of stereoisomerism: optical isomerism (enantiomerism / presence of a chiral centre).

(a)(ii) \(\text{S}_\text{N}2\) Mechanism:
1. Nucleophilic attack: A curly arrow originates from a lone pair of electrons on the \(\text{OH}^-\) ion and points directly to the electron-deficient carbon atom (\(\text{C}^{\delta+}\)) bonded to bromine.
2. Heterolytic fission: A dipole is shown on the \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) bond, and a curly arrow originates from the \(\text{C}-\text{Br}\) single bond pointing towards the \(\text{Br}\) atom.
3. Transition state: Square brackets with a partial bond from \(\text{O}\) to \(\text{C}\) (\(\text{HO}\cdots\text{C}\)) and a partial bond from \(\text{C}\) to \(\text{Br}\) (\(\text{C}\cdots\text{Br}\)), carrying an overall negative charge (\(^{\delta-}\text{HO}\cdots\text{C}\cdots\text{Br}^{\delta-}\) inside \([\;]^-\)).

(a)(iii) Two main reasons:
1. Steric hindrance: The three bulky methyl groups prevent the nucleophile (\(\text{OH}^-\)) from attacking the central carbon from the back in an \(\text{S}_\text{N}2\) process.
2. Carbocation stability: In \(\text{S}_\text{N}1\), the tertiary carbocation formed is relatively stable due to the electron-donating (positive inductive) effect of the three alkyl (methyl) groups.

(b)(i) Three isomeric alkenes:
1. But-1-ene: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\)
2. \((E)\)-but-2-ene (trans-but-2-ene): \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) (opposite sides)
3. \((Z)\)-but-2-ene (cis-but-2-ene): \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) (same side)

(b)(ii) Geometric isomerism / cis-trans isomerism / \(E/Z\) isomerism.

(b)(iii) Reagent: Aqueous bromine / bromine water (\(\text{Br}_2(\text{aq})\)).
Observation: Decolourises (orange-brown to colourless).

評分準則

(a)(i) [2 marks]
- Name: butan-2-ol [1]
- Isomerism: optical isomerism / enantiomerism [1]

(a)(ii) [3 marks]
- Curly arrow from lone pair on \(\text{OH}^-\) to \(\text{C}^{\delta+}\) AND dipole \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) [1]
- Curly arrow from \(\text{C}-\text{Br}\) bond to \(\text{Br}\) [1]
- Correct transition state structure with partial dashed bonds and overall negative charge (or direct inverted product + \(\text{Br}^-\)) [1]

(a)(iii) [2 marks]
- Steric hindrance prevents backside attack in \(\text{S}_\text{N}2\) [1]
- Tertiary carbocation is stabilised by inductive electron donation of 3 alkyl groups [1]

(b)(i) [3 marks]
- Correct skeletal structure and name for but-1-ene [1]
- Correct skeletal structure and name for (E)-but-2-ene / trans-but-2-ene [1]
- Correct skeletal structure and name for (Z)-but-2-ene / cis-but-2-ene [1]

(b)(ii) [1 mark] \(cis\)-\(trans\) / \(E/Z\) / geometrical isomerism

(b)(iii) [1 mark] Bromine water decolourised / turns from orange to colourless
題目 5 · structured
12 分
Three structural isomers, \(\text{P}\), \(\text{Q}\), and \(\text{R}\), have the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- Isomer \(\text{P}\) reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH) to give an orange precipitate. It forms a silver mirror with Tollens' reagent, and has a branched carbon chain.
- Isomer \(\text{Q}\) also forms an orange precipitate with 2,4-DNPH. When \(\text{Q}\) is warmed with alkaline aqueous iodine, a yellow precipitate is formed.
- Isomer \(\text{R}\) does not react with 2,4-DNPH. It reacts with sodium metal to release hydrogen gas, and readily decolourises bromine water.

(a) Identify isomers \(\text{P}\) and \(\text{Q}\) by giving their systematic names and structural formulas. [4]

(b) (i) Identify the yellow precipitate formed when \(\text{Q}\) reacts with alkaline aqueous iodine. [1]
(ii) State the specific structural feature in \(\text{Q}\) responsible for the positive tri-iodomethane test. [1]

(c) Isomer \(\text{Q}\) can be reduced to an alcohol, \(\text{S}\), using sodium borohydride, \(\text{NaBH}_4\).
(i) Write an equation for the reduction of \(\text{Q}\) to form \(\text{S}\), using \([\text{H}]\) to represent the reducing agent. [1]
(ii) Describe how the infrared spectra of \(\text{Q}\) and \(\text{S}\) differ by stating the bonds and characteristic absorption wave numbers involved. [2]

(d) Draw one possible displayed formula for isomer \(\text{R}\). [1]

(e) Alcohol \(\text{S}\) is heated under reflux with ethanoic acid in the presence of a few drops of concentrated sulfuric acid.
(i) Draw the skeletal formula of the organic ester produced. [1]
(ii) State the role of concentrated sulfuric acid in this reaction. [1]
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解題

(a)
- Isomer \(\text{P}\): Reacts with 2,4-DNPH \(\implies\) carbonyl compound. Reacts with Tollens' \(\implies\) aldehyde. Branched carbon chain with 4 carbons \(\implies\) 2-methylpropanal, \((\text{CH}_3)_2\text{CHCHO}\).
- Isomer \(\text{Q}\): Reacts with 2,4-DNPH \(\implies\) carbonyl compound. Positive tri-iodomethane test \(\implies\) contains \(\text{CH}_3\text{C}=\text{O}\) group. Molecular formula \(\text{C}_4\text{H}_8\text{O} \implies\) butan-2-one, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

(b)(i) Tri-iodomethane / \(\text{CHI}_3\).
(b)(ii) The presence of a \(\text{CH}_3\text{C}=\text{O}\) (methyl ketone) group.

(c)(i) \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)
(c)(ii) In the IR spectrum of \(\text{Q}\), there is a strong, sharp peak at \(1670\text{--}1740\text{ cm}^{-1}\) due to the \(\text{C}=\text{O}\) bond, which is absent in \(\text{S}\). In \(\text{S}\), there is a broad absorption peak at \(3200\text{--}3650\text{ cm}^{-1}\) due to the alcohol \(\text{O}-\text{H}\) group.

(d) \(\text{R}\) does not react with 2,4-DNPH (not a carbonyl), reacts with \(\text{Na}\) (contains an \(\text{OH}\) alcohol group), and decolourises \(\text{Br}_2\) (contains \(\text{C}=\text{C}\) double bond). Possible displayed structures include but-2-en-1-ol (\(\text{HO}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3\)), but-3-en-1-ol (\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}=\text{CH}_2\)), but-3-en-2-ol (\(\text{CH}_3-\text{CH(OH)}-\text{CH}=\text{CH}_2\)), or 2-methylprop-2-en-1-ol (\(\text{HO}-\text{CH}_2-\text{C(CH}_3)=\text{CH}_2\)) drawn with all bonds shown.

(e)(i) Esterification between butan-2-ol and ethanoic acid gives butan-2-yl ethanoate: skeletal formula showing \(\text{CH}_3-\text{C}(=\text{O})-\text{O}-\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3\).
(e)(ii) Catalyst / speeds up the rate of reaction.

評分準則

(a) [4 marks]
- \(\text{P}\) name: 2-methylpropanal [1]
- \(\text{P}\) formula: \((\text{CH}_3)_2\text{CHCHO}\) / \(\text{CH}_3\text{CH(CH}_3)\text{CHO}\) [1]
- \(\text{Q}\) name: butan-2-one (butanone) [1]
- \(\text{Q}\) formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) [1]

(b)(i) [1 mark] Tri-iodomethane / \(\text{CHI}_3\)
(b)(ii) [1 mark] \(\text{CH}_3\text{C}=\text{O}\) / methyl carbonyl / methyl ketone group

(c)(i) [1 mark] \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)

(c)(ii) [2 marks]
- Loss/disappearance of \(\text{C}=\text{O}\) absorption at \(1670\text{--}1740\text{ cm}^{-1}\) [1]
- Appearance of alcohol \(\text{O}-\text{H}\) absorption at \(3200\text{--}3650\text{ cm}^{-1}\) [1]

(d) [1 mark] Correct displayed formula of an unsaturated alcohol of formula \(\text{C}_4\text{H}_8\text{O}\) (all atoms and bonds shown)

(e)(i) [1 mark] Correct skeletal formula of butan-2-yl ethanoate
(e)(ii) [1 mark] (Acid) catalyst

Paper 3 (Advanced Practical Skills)

Complete all practical experiments, qualitative observations, and quantitative titrimetric/gravimetric recordings.
3 題目 · 39.99 分
題目 1 · Qualitative and quantitative laboratory tasks
13.33 分
FA 1 is a solution containing \(6.15\text{ g dm}^{-3}\) of an impure sample of hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}\).
FA 2 is aqueous potassium manganate(VII), \(\text{KMnO}_4\), of concentration \(0.0200\text{ mol dm}^{-3}\).
FA 3 is \(1.00\text{ mol dm}^{-3}\) sulfuric acid, \(\text{H}_2\text{SO}_4\).

You are to determine the percentage purity of the hydrated ethanedioic acid in the sample by titrating FA 1 with FA 2 in the presence of dilute sulfuric acid.

Method
1. Fill the burette with FA 2.
2. Pipette \(25.0\text{ cm}^3\) of FA 1 into a conical flask.
3. Using a measuring cylinder, add \(20.0\text{ cm}^3\) of FA 3 to the conical flask.
4. Heat the contents of the conical flask to approximately \(65^{\circ}\text{C}\).
5. Titrate the hot solution with FA 2 until a permanent pale pink colour is obtained.
6. Repeat the titration until concordant results are achieved.

(a) Record your titration results in an appropriate table in the space below.

| Titration number | 1 | 2 | 3 |
| :--- | :---: | :---: | :---: |
| Final burette reading / \(\text{cm}^3\) | 24.50 | 48.65 | 24.15 |
| Initial burette reading / \(\text{cm}^3\) | 0.00 | 24.50 | 0.00 |
| Titre / \(\text{cm}^3\) | 24.50 | 24.15 | 24.15 |

From your titration results, obtain a suitable value for the volume of FA 2 to be used in your calculations. Show clearly how this value is obtained.

(b)
(i) Calculate the amount, in \(\text{mol}\), of \(\text{MnO}_4^-\) present in the mean titre volume of FA 2.

(ii) The equation for the reaction is:
\[2\text{MnO}_4^-(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l)\]
Calculate the amount, in \(\text{mol}\), of \(\text{H}_2\text{C}_2\text{O}_4\) in \(25.0\text{ cm}^3\) of FA 1.

(iii) Calculate the concentration, in \(\text{mol dm}^{-3}\), of \(\text{H}_2\text{C}_2\text{O}_4\) in FA 1.

(iv) Calculate the mass of pure hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}\), present in \(1.00\text{ dm}^3\) of FA 1. \([M_r: \text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O} = 126.0]\)

(v) Calculate the percentage purity by mass of the hydrated ethanedioic acid sample in FA 1.

(c) The maximum error associated with each reading of the \(50.0\text{ cm}^3\) burette is \(\pm 0.05\text{ cm}^3\). Calculate the maximum percentage error in the titre volume for Titration 2.
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解題

(a) Concordant titres are Titrations 2 and 3 (within \(0.10\text{ cm}^3\)).
\[\text{Mean titre} = \frac{24.15 + 24.15}{2} = 24.15\text{ cm}^3\]

(b)
(i) Amount of \(\text{MnO}_4^-\) reacted:
\[n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{24.15}{1000}\text{ dm}^3 = 4.83 \times 10^{-4}\text{ mol}\]

(ii) From the mole ratio \(2\text{MnO}_4^- : 5\text{H}_2\text{C}_2\text{O}_4\):
\[n(\text{H}_2\text{C}_2\text{O}_4) = 4.83 \times 10^{-4} \times \frac{5}{2} = 1.2075 \times 10^{-3}\text{ mol} = 1.21 \times 10^{-3}\text{ mol}\]

(iii) Concentration of \(\text{H}_2\text{C}_2\text{O}_4\) in FA 1:
\[[\text{H}_2\text{C}_2\text{O}_4] = \frac{1.2075 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0483\text{ mol dm}^{-3}\]

(iv) Mass of \(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}\) per \(\text{dm}^3\):
\[\text{Mass} = 0.0483\text{ mol dm}^{-3} \times 126.0\text{ g mol}^{-1} = 6.0858\text{ g dm}^{-3} = 6.09\text{ g dm}^{-3}\]

(v) Percentage purity:
\[\text{Percentage purity} = \left(\frac{6.0858}{6.15}\right) \times 100\% = 98.96\% = 99.0\%\]

(c) Each titre consists of two burette readings (initial and final), so maximum absolute uncertainty = \(2 \times 0.05 = 0.10\text{ cm}^3\).
\[\text{Percentage error} = \frac{0.10}{24.15} \times 100\% = 0.414\%\]

評分準則

(a) [4 marks total]
- Table layout: unambiguous column/row headings with units (e.g. \(/\text{cm}^3\)) [1]
- All burette readings recorded to \(0.05\text{ cm}^3\) [1]
- Correct selection of concordant titres (within \(0.10\text{ cm}^3\)) with working clearly shown [1]
- Correct mean titre calculated: \(24.15\text{ cm}^3\) [1]

(b) [7 marks total]
- (i) \(n(\text{MnO}_4^-) = 0.0200 \times (\text{mean titre} / 1000) = 4.83 \times 10^{-4}\text{ mol}\) [1]
- (ii) Correct mole ratio applied (\(\times 5/2\)) giving \(1.21 \times 10^{-3}\text{ mol}\) (allow ecf) [2]
- (iii) Division by \(0.0250\text{ dm}^3\) to give \(0.0483\text{ mol dm}^{-3}\) [1]
- (iv) Multiplication of (iii) by \(126.0\) to give \(6.09\text{ g}\) [1]
- (v) Calculation of percentage purity: \((\text{mass} / 6.15) \times 100 = 99.0\%\) (quoted to 3 or 4 significant figures) [2]

(c) [2.33 marks total]
- Recognition of two readings contributing to uncertainty: \(2 \times 0.05 = 0.10\text{ cm}^3\) [1]
- Correct percentage calculation: \((0.10 / 24.15) \times 100 = 0.414\%\) [1.33]
題目 2 · Qualitative and quantitative laboratory tasks
13.33 分
FA 6, FA 7, and FA 8 are aqueous solutions. Each contains one cation and one anion from the Qualitative Analysis Notes.

Perform the tests described in the table below. Record your observations and deduce the identities of the ions present.

(a) Complete the observations table:

| Test | Observations with FA 6 | Observations with FA 7 | Observations with FA 8 |
| :--- | :--- | :--- | :--- |
| 1. To a \(1\text{ cm}\) depth of solution in a test-tube, add aqueous sodium hydroxide dropwise, then in excess. | Green precipitate formed, insoluble in excess; turns brown at the surface on standing. | Grey-green precipitate formed, dissolves in excess to give a dark green solution. | White precipitate formed, soluble in excess to give a colourless solution. |
| 2. To a \(1\text{ cm}\) depth of solution in a test-tube, add aqueous ammonia dropwise, then in excess. | Green precipitate formed, insoluble in excess; turns brown at the surface. | Grey-green precipitate formed, insoluble in excess. | White precipitate formed, insoluble in excess. |
| 3. To a \(1\text{ cm}\) depth of solution in a test-tube, add a few drops of aqueous barium nitrate followed by dilute nitric acid. | No precipitate / no change. | White precipitate formed, insoluble in dilute nitric acid. | White precipitate formed, dissolves with effervescence in dilute nitric acid; gas turns limewater milky. |
| 4. To a \(1\text{ cm}\) depth of solution in a test-tube, add acidified potassium manganate(VII), \(\text{KMnO}_4\), dropwise. | Purple solution is decolourised / turns pale yellow-brown. | No change / solution remains purple. | No change / solution remains purple. |

(b) Deduce the chemical formula for each solution:

(i) Cation in FA 6:
Anion in FA 6:
Formula of FA 6:

(ii) Cation in FA 7:
Anion in FA 7:
Formula of FA 7:

(iii) Cation in FA 8:
Anion in FA 8:
Formula of FA 8:

(c) Write an ionic equation for the reaction observed in Test 4 with FA 6.
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解題

(a) & (b)
- FA 6:
- In Test 1, \(\text{NaOH(aq)}\) gives a green precipitate insoluble in excess, turning brown on standing \(\rightarrow \text{Fe}^{2+}\).
- In Test 2, \(\text{NH}_3\text{(aq)}\) confirms \(\text{Fe}^{2+}\).
- In Test 3, no precipitate with \(\text{Ba(NO}_3)_2\) confirms absence of \(\text{SO}_4^{2-}\) / \(\text{SO}_3^{2-}\) / \(\text{CO}_3^{2-}\).
- In Test 4, acidified \(\text{KMnO}_4\) is decolourised as \(\text{Fe}^{2+}\) is oxidised to \(\text{Fe}^{3+}\). (Anion can be non-interfering e.g. \(\text{Cl}^-\) or \(\text{NO}_3^-\)). Cation = \(\text{Fe}^{2+}\).

- FA 7:
- In Test 1, \(\text{NaOH(aq)}\) gives a grey-green precipitate that dissolves in excess to a dark green solution \(\rightarrow \text{Cr}^{3+}\).
- In Test 2, \(\text{NH}_3\text{(aq)}\) gives a grey-green precipitate insoluble in excess \(\rightarrow \text{Cr}^{3+}\).
- In Test 3, a white precipitate with \(\text{Ba(NO}_3)_2\) insoluble in \(\text{HNO}_3\) confirms sulfate, \(\text{SO}_4^{2-}\).
- Identity of FA 7: \(\text{Cr}_2(\text{SO}_4)_3\).

- FA 8:
- In Test 1, \(\text{NaOH(aq)}\) gives a white precipitate soluble in excess giving a colourless solution \(\rightarrow \text{Al}^{3+}\) or \(\text{Zn}^{2+}\).
- In Test 2, \(\text{NH}_3\text{(aq)}\) gives a white precipitate insoluble in excess \(\rightarrow \text{Al}^{3+}\) (since \(\text{Zn}^{2+}\) would be soluble in excess \(\text{NH}_3\)).
- In Test 3, white precipitate dissolving with effervescence producing \(\text{CO}_2\) confirms carbonate, \(\text{CO}_3^{2-}\).
- Cation = \(\text{Al}^{3+}\), Anion = \(\text{CO}_3^{2-}\).

(c) Ionic equation for oxidation of \(\text{Fe}^{2+}\) by \(\text{MnO}_4^-\) in acidic solution:
\[5\text{Fe}^{2+}(aq) + \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow 5\text{Fe}^{3+}(aq) + \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)\]

評分準則

(a) [4 marks total]
- Accurate recording of observations consistent with qualitative test notes for \(\text{Fe}^{2+}\) [1]
- Accurate recording of observations for \(\text{Cr}^{3+}\) [1]
- Accurate recording of observations for \(\text{Al}^{3+}\) [1]
- Accurate recording of observations for \(\text{SO}_4^{2-}\) and \(\text{CO}_3^{2-}\) [1]

(b) [6 marks total]
- Cation in FA 6 identified as \(\text{Fe}^{2+}\) with valid justification [2]
- Cation in FA 7 identified as \(\text{Cr}^{3+}\) and anion as \(\text{SO}_4^{2-}\) / formula \(\text{Cr}_2(\text{SO}_4)_3\) [2]
- Cation in FA 8 identified as \(\text{Al}^{3+}\) and anion as \(\text{CO}_3^{2-}\) [2]

(c) [3.33 marks total]
- Correct species on reactants and products side: \(\text{Fe}^{2+} + \text{MnO}_4^- + \text{H}^+ \rightarrow \text{Fe}^{3+} + \text{Mn}^{2+} + \text{H}_2\text{O}\) [1.33]
- Fully balanced equation: \(5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}\) [2]
題目 3 · free-response
13.33 分
In this experiment, you will determine the concentration of potassium manganate(VII) in solution FA 1 by titrating it against a standard solution of hydrated iron(II) sulfate, \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\).

FA 1 is aqueous potassium manganate(VII), \(\text{KMnO}_4\).
FA 2 is dilute sulfuric acid, \(1.00\text{ mol dm}^{-3}\text{ H}_2\text{SO}_4\).
Solid \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) has a molar mass of \(278.0\text{ g mol}^{-1}\).

(a) A student weighs out \(6.950\text{ g}\) of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\), dissolves it in FA 2 and distilled water, and makes the solution up to exactly \(250.0\text{ cm}^3\) in a volumetric flask to prepare solution FA 3.

Calculate the concentration, in \(\text{mol dm}^{-3}\), of \(\text{Fe}^{2+}\text{(aq)}\) in solution FA 3.

(b) The student pipettes \(25.0\text{ cm}^3\) of FA 3 into a conical flask and adds \(20\text{ cm}^3\) of FA 2 using a measuring cylinder. FA 1 is placed in the burette. The titration results are recorded below:

- Rough Titre: Initial = \(0.00\text{ cm}^3\), Final = \(22.80\text{ cm}^3\), Titre = \(22.80\text{ cm}^3\)
- Titre 1: Initial = \(0.50\text{ cm}^3\), Final = \(22.90\text{ cm}^3\), Titre = \(22.40\text{ cm}^3\)
- Titre 2: Initial = \(22.90\text{ cm}^3\), Final = \(45.40\text{ cm}^3\), Titre = \(22.50\text{ cm}^3\)

(i) Calculate the suitable average titre of FA 1 to use in your calculations.
(ii) In acidic solution, manganate(VII) ions react with iron(II) ions according to the equation:
$$\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})$$
Calculate the concentration, in \(\text{mol dm}^{-3}\), of \(\text{KMnO}_4\) in FA 1. Give your answer to 3 significant figures.

(c) State the colour change observed at the end-point of this titration.

(d) The student used a balance with an uncertainty of \(\pm 0.005\text{ g}\) to weigh the \(6.950\text{ g}\) of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) by difference (two weighings). Calculate the percentage uncertainty in this mass.
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解題

(a) Amount of $\text{FeSO}_4 \cdot 7\text{H}_2\text{O} = \frac{6.950\text{ g}}{278.0\text{ g mol}^{-1}} = 0.02500\text{ mol}$.
Concentration of $\text{Fe}^{2+} = \frac{0.02500\text{ mol}}{0.2500\text{ dm}^3} = 0.100\text{ mol dm}^{-3}$.

(b)(i) Suitable concordant titres are Titre 1 ($22.40\text{ cm}^3$) and Titre 2 ($22.50\text{ cm}^3$).
Average titre $= \frac{22.40 + 22.50}{2} = 22.45\text{ cm}^3$.

(ii) Amount of $\text{Fe}^{2+}$ in $25.0\text{ cm}^3 = 0.100\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 2.500 \times 10^{-3}\text{ mol}$.
From stoichiometry ($5\text{Fe}^{2+} : 1\text{MnO}_4^-$):
Amount of $\text{MnO}_4^- = \frac{2.500 \times 10^{-3}}{5} = 5.000 \times 10^{-4}\text{ mol}$.
Concentration of $\text{KMnO}_4 = \frac{5.000 \times 10^{-4}\text{ mol}}{0.02245\text{ dm}^3} = 0.02227 \approx 0.0223\text{ mol dm}^{-3}$.

(c) The solution in the conical flask changes from colourless (or pale green/yellow) to the first permanent pale pink.

(d) Total uncertainty for two weighings $= 2 \times 0.005\text{ g} = 0.010\text{ g}$.
Percentage uncertainty $= \left(\frac{0.010}{6.950}\right) \times 100\% = 0.144\%\text{ (to 3 s.f.)}$.

評分準則

[13.33 marks total, distributed across sub-parts]
(a) [2 marks]
- Correct calculation of moles of hydrated salt: $0.0250\text{ mol}$ [1]
- Correct calculation of concentration: $0.100\text{ mol dm}^{-3}$ [1]

(b)(i) [2 marks]
- Selection of concordant titres ($22.40$ and $22.50\text{ cm}^3$) and mean correctly shown as $22.45\text{ cm}^3$ [2]

(b)(ii) [4 marks]
- Moles of $\text{Fe}^{2+}$ in $25.0\text{ cm}^3$ ($2.50 \times 10^{-3}\text{ mol}$) [1]
- Moles of $\text{MnO}_4^-$ using 1:5 ratio ($5.00 \times 10^{-4}\text{ mol}$) [1]
- Division of moles by candidate's mean titre in $\text{dm}^3$ [1]
- Final answer given as $0.0223\text{ mol dm}^{-3}$ to 3 significant figures [1]

(c) [2 marks]
- Colourless / pale green / pale yellow to (first permanent) pale pink [2]
(Reject: purple end-point or pink to colourless)

(d) [3.33 marks]
- Identifies 2 balance readings (total uncertainty $= 0.010\text{ g}$) [1.33]
- Correct calculation showing $(0.010 / 6.950) \times 100 = 0.144\%$ [2]

Paper 4 (A Level 結構題)

Answer all questions in the spaces provided. Calculator and Periodic Table provided. Show full mathematical working.
10 題目 · 110.99 分
題目 1 · structured
11.11 分
Cobalt forms a wide variety of coordination complexes in the +2 and +3 oxidation states.

(a) An aqueous solution of cobalt(II) chloride contains the hexaaquacobalt(II) ion, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), which is pink.
(i) State the electronic configuration of the \(\text{Co}^{2+}\) ion in terms of \(s\), \(p\), and \(d\) orbitals.
(ii) Explain why transition element complexes, such as \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), are coloured.

(b) When concentrated hydrochloric acid is added to an aqueous solution containing \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), a deep blue solution containing \([\text{CoCl}_4]^{2-}\) is formed.
(i) Write a balanced ionic equation for this ligand exchange reaction.
(ii) State the geometry and coordination number of the cobalt ion in \([\text{CoCl}_4]^{2-}\).

(c) The complex \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) exhibits stereoisomerism.
(i) Name the type of stereoisomerism shown by \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\).
(ii) Draw three-dimensional diagrams to show the structures of the stereoisomers of \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\).

(d) The stability constant, \(K_{\text{stab}}\), for the formation of the complex \([\text{Co}(\text{NH}_3)_6]^{2+}\) from \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) is \(7.7 \times 10^4\text{ dm}^{18}\text{ mol}^{-6}\), whereas \(K_{\text{stab}}\) for \([\text{Co}(\text{en})_3]^{2+}\) (where \(\text{en} = \text{ethane-1,2-diamine}\)) is \(3.8 \times 10^{14}\text{ dm}^9\text{ mol}^{-3}\).
Explain, with reference to entropy changes, why \([\text{Co}(\text{en})_3]^{2+}\) has a significantly larger stability constant than \([\text{Co}(\text{NH}_3)_6]^{2+}\).
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解題

(a)(i) Cobalt has atomic number 27 (\([\text{Ar}] 4s^2 3d^7\)). On forming \(\text{Co}^{2+}\), the two \(4s\) electrons are lost first: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\) or \([\text{Ar}] 3d^7\).
(ii) In an isolated gaseous transition metal ion, the five \(d\)-orbitals are degenerate. When ligands approach, repulsion between ligand lone pairs and metal \(d\)-electrons splits the \(d\)-orbitals into two distinct energy sets separated by an energy gap \(\Delta E\). An electron absorbs a photon of visible light (\(\Delta E = h\nu\)) and is promoted from a lower energy \(d\)-orbital to a higher energy \(d\)-orbital. The non-absorbed wavelengths are transmitted/reflected, producing the observed complementary colour.
(b)(i) \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)
(ii) Geometry: Tetrahedral; Coordination number: 4.
(c)(i) Cis-trans (or geometrical) isomerism.
(ii) In cis-\( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \), the two \(\text{Cl}^-\) ligands are adjacent at \(90^\circ\). In trans-\( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \), the two \(\text{Cl}^-\) ligands are opposite at \(180^\circ\).
(d) The substitution reaction \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightleftharpoons [\text{Co}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}\) converts 4 reactant particles into 7 product particles. This results in a significant increase in disorder and therefore a large positive entropy change of the system (\(\Delta S_{\text{sys}} > 0\)). In contrast, for \(\text{NH}_3\), 7 reactant species produce 7 product species (\(\Delta S \approx 0\)). Since \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), the large \(T\Delta S^\circ\) term makes \(\Delta G^\circ\) much more negative for \([\text{Co}(\text{en})_3]^{2+}\), giving a much greater \(K_{\text{stab}}\).

評分準則

[1] (a)(i) \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\) or \([\text{Ar}] 3d^7\)
[2] (a)(ii) Splitting of \(d\)-orbitals into two non-degenerate energy levels by ligands [1]; promotion of an electron from lower to higher \(d\)-level by absorbing a specific wavelength/frequency of visible light, with complementary light transmitted [1]
[1] (b)(i) Correct balanced equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)
[1] (b)(ii) Geometry = tetrahedral AND coordination number = 4
[1] (c)(i) Cis-trans / geometrical isomerism
[2] (c)(ii) Correct 3D octahedral representation of cis-isomer (two Cl at \(90^\circ\)) [1] AND trans-isomer (two Cl at \(180^\circ\)) [1] with proper wedge/dash conventions and charge \(+\)
[3] (d) Replacement of 6 monodentate \(\text{H}_2\text{O}\) ligands by 3 bidentate \(\text{en}\) ligands produces more particles in solution (4 moles of reactants produce 7 moles of products) [1]; leading to a positive entropy change (\(\Delta S_{\text{system}} > 0\)) [1]; making \(\Delta G^\circ\) much more negative via \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), which increases \(K_{\text{stab}}\) (\(\ln K_{\text{stab}} = -\Delta G^\circ/RT\)) [1]
題目 2 · structured
11.11 分
Lactic acid, \(\text{CH}_3\text{CH(OH)COOH}\), is a weak monobasic acid that plays an essential role in biological processes. It can be represented as \(\text{HLac}\).

(a) (i) Write the expression for the acid dissociation constant, \(K_a\), of lactic acid.
(ii) A \(0.120\text{ mol dm}^{-3}\) aqueous solution of lactic acid has a \(\text{pH}\) of \(2.39\) at \(298\text{ K}\).
Calculate the value of \(K_a\) for lactic acid at \(298\text{ K}\). Include units in your answer.

(b) A buffer solution is prepared by mixing \(60.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\text{ HLac}\) with \(40.0\text{ cm}^3\) of \(0.0800\text{ mol dm}^{-3}\text{ NaOH}\).
(i) Define what is meant by a buffer solution.
(ii) Calculate the \(\text{pH}\) of this buffer solution at \(298\text{ K}\). Use your value of \(K_a\) from (a)(ii).

(c) Explain, with the aid of an ionic equation, how this buffer solution resists a change in \(\text{pH}\) when a small quantity of aqueous hydrochloric acid is added.

(d) Solid calcium lactate, \(\text{Ca(Lac)}_2\), is sparingly soluble in water. At \(298\text{ K}\), its solubility product, \(K_{\text{sp}}\), is \(2.40 \times 10^{-6}\text{ mol}^3\text{ dm}^{-9}\).
Calculate the solubility of calcium lactate in water at \(298\text{ K}\) in \(\text{g dm}^{-3}\).
[\(M_r(\text{Ca(Lac)}_2) = 218.2\)]
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解題

(a)(i) \(K_a = \frac{[\text{H}^+][\text{Lac}^-]}{[\text{HLac}]}\)
(ii) \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.39} = 4.074 \times 10^{-3}\text{ mol dm}^{-3}\).
\(K_a = \frac{[\text{H}^+]^2}{[\text{HLac}] - [\text{H}^+]} = \frac{(4.074 \times 10^{-3})^2}{0.120 - 4.074 \times 10^{-3}} = \frac{1.660 \times 10^{-5}}{0.1159} = 1.43 \times 10^{-4}\text{ mol dm}^{-3}\)
(Or using \(K_a = \frac{[\text{H}^+]^2}{c} = \frac{(4.074 \times 10^{-3})^2}{0.120} = 1.38 \times 10^{-4}\text{ mol dm}^{-3}\)).
(b)(i) A buffer solution is one which resists changes in \(\text{pH}\) when small amounts of an acid or a base are added.
(ii) Initial moles of \(\text{HLac} = 0.0600 \times 0.120 = 7.20 \times 10^{-3}\text{ mol}\).
Initial moles of \(\text{OH}^- = 0.0400 \times 0.0800 = 3.20 \times 10^{-3}\text{ mol}\).
\(\text{HLac} + \text{OH}^- \rightarrow \text{Lac}^- + \text{H}_2\text{O}\).
Moles of \(\text{Lac}^-\) formed \(= 3.20 \times 10^{-3}\text{ mol}\).
Remaining moles of \(\text{HLac} = 7.20 \times 10^{-3} - 3.20 \times 10^{-3} = 4.00 \times 10^{-3}\text{ mol}\).
\(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{Lac}^-]}{[\text{HLac}]}\right) = -\log(1.383 \times 10^{-4}) + \log\left(\frac{3.20 \times 10^{-3}}{4.00 \times 10^{-3}}\right) = 3.86 - 0.0969 = 3.76\).
(c) When \(\text{H}^+\) ions are added, the conjugate base \(\text{Lac}^-\) removes them according to:
\(\text{Lac}^- + \text{H}^+ \rightarrow \text{HLac}\).
Since \([\text{Lac}^-]\) is in large supply, the added \(\text{H}^+\) is consumed and the \([\text{H}^+]\) remains virtually constant.
(d) \(\text{Ca(Lac)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{Lac}^-(\text{aq})\).
Let solubility be \(s\text{ mol dm}^{-3}\).
\([\text{Ca}^{2+}] = s\), \([\text{Lac}^-] = 2s\).
\(K_{\text{sp}} = [\text{Ca}^{2+}][\text{Lac}^-]^2 = s(2s)^2 = 4s^3\).
\(4s^3 = 2.40 \times 10^{-6} \implies s^3 = 6.00 \times 10^{-7} \implies s = 8.434 \times 10^{-3}\text{ mol dm}^{-3}\).
Solubility in \(\text{g dm}^{-3} = s \times M_r = 8.434 \times 10^{-3} \times 218.2 = 1.84\text{ g dm}^{-3}\).

評分準則

[1] (a)(i) \(K_a = [\text{H}^+][\text{Lac}^-] / [\text{HLac}]\)
[2] (a)(ii) \([\text{H}^+] = 10^{-2.39} = 4.07 \times 10^{-3}\text{ mol dm}^{-3}\) [1]; \(K_a = 1.38 \times 10^{-4}\text{ mol dm}^{-3}\) (or \(1.43 \times 10^{-4}\text{ mol dm}^{-3}\)) with correct units [1]
[1] (b)(i) Definition: A solution that resists changes in pH when small quantities of acid or alkali are added
[3] (b)(ii) Moles of HLac remaining (\(4.00 \times 10^{-3}\)) and moles of Lac- formed (\(3.20 \times 10^{-3}\)) [1]; Henderson-Hasselbalch or \([\text{H}^+] = K_a \times (\text{acid}/\text{salt})\) expression set up correctly [1]; \(\text{pH} = 3.76\) (ecf on \(K_a\)) [1]
[1] (c) Equation \(\text{Lac}^- + \text{H}^+ \rightarrow \text{HLac}\) and statement that added \(\text{H}^+\) is removed by high concentration of lactate ions
[3] (d) \(K_{\text{sp}} = 4s^3\) relationship deduced [1]; \(s = 8.43 \times 10^{-3}\text{ mol dm}^{-3}\) [1]; solubility in \(\text{g dm}^{-3} = 1.84\text{ g dm}^{-3}\) (allow 1.83-1.85) [1]
題目 3 · structured
11.11 分
This question is about lattice energy, enthalpy changes, and Gibbs free energy.

(a) Define the term standard lattice energy (\(\Delta H^\circ_{\text{latt}}\)).

(b) Use the following data to calculate the standard lattice energy of strontium chloride, \(\text{SrCl}_2(\text{s})\).
- Standard enthalpy change of formation of \(\text{SrCl}_2(\text{s}) = -829\text{ kJ mol}^{-1}\)
- Standard enthalpy change of atomisation of \(\text{Sr}(\text{s}) = +164\text{ kJ mol}^{-1}\)
- First ionisation energy of \(\text{Sr}(\text{g}) = +550\text{ kJ mol}^{-1}\)
- Second ionisation energy of \(\text{Sr}(\text{g}) = +1064\text{ kJ mol}^{-1}\)
- Enthalpy change of atomisation of chlorine, \(\Delta H^\circ_{\text{at}}(\text{Cl}_2) = +122\text{ kJ mol}^{-1}\)
- Electron affinity of chlorine, \(\text{EA}_1(\text{Cl}) = -349\text{ kJ mol}^{-1}\)

(c) The lattice energy of barium chloride, \(\text{BaCl}_2\), is \(-2056\text{ kJ mol}^{-1}\).
Explain why the lattice energy of \(\text{SrCl}_2\) is more exothermic than that of \(\text{BaCl}_2\).

(d) The thermal decomposition of strontium carbonate is given by:
\[\text{SrCO}_3(\text{s}) \rightarrow \text{SrO}(\text{s}) + \text{CO}_2(\text{g}) \quad \Delta H^\circ = +235\text{ kJ mol}^{-1}\]
Standard molar entropies, \(S^\circ\), are given in the table:
| Substance | \(S^\circ / \text{J K}^{-1}\text{ mol}^{-1}\) |
|---|---|
| \(\text{SrCO}_3(\text{s})\) | 97.0 |
| \(\text{SrO}(\text{s})\) | 54.4 |
| \(\text{CO}_2(\text{g})\) | 213.6 |

(i) Calculate the standard entropy change of the system, \(\Delta S^\circ_{\text{sys}}\), for this reaction. Include sign and units.
(ii) Calculate the minimum temperature, in \(\text{K}\), at which this decomposition becomes thermodynamically feasible.
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解題

(a) Standard lattice energy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions (298 K, 101 kPa).
(b) By Hess's Law / Born-Haber cycle:
\(\Delta H^\circ_f[\text{SrCl}_2(\text{s})] = \Delta H^\circ_{\text{at}}[\text{Sr}] + \text{IE}_1[\text{Sr}] + \text{IE}_2[\text{Sr}] + 2\Delta H^\circ_{\text{at}}[\text{Cl}_2] + 2\text{EA}_1[\text{Cl}] + \Delta H^\circ_{\text{latt}}[\text{SrCl}_2]\)
\(-829 = 164 + 550 + 1064 + 2(122) + 2(-349) + \Delta H^\circ_{\text{latt}}\)
\(-829 = 1778 + 244 - 698 + \Delta H^\circ_{\text{latt}}\)
\(-829 = 1324 + \Delta H^\circ_{\text{latt}}\)
\(\Delta H^\circ_{\text{latt}} = -829 - 1324 = -2153\text{ kJ mol}^{-1}\).
(c) Both \(\text{Sr}^{2+}\) and \(\text{Ba}^{2+}\) have a \(+2\) charge, but the \(\text{Sr}^{2+}\) ion has a smaller ionic radius than \(\text{Ba}^{2+}\). Therefore, \(\text{Sr}^{2+}\) has a greater charge density and forms stronger electrostatic attractions with \(\text{Cl}^-\) ions, resulting in a more exothermic lattice energy.
(d)(i) \(\Delta S^\circ_{\text{sys}} = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}) = [S^\circ(\text{SrO}) + S^\circ(\text{CO}_2)] - [S^\circ(\text{SrCO}_3)]\)
\(\Delta S^\circ_{\text{sys}} = (54.4 + 213.6) - (97.0) = 268.0 - 97.0 = +171.0\text{ J K}^{-1}\text{ mol}^{-1}\).
(ii) For feasibility, \(\Delta G^\circ \le 0\).
\(\Delta H^\circ - T\Delta S^\circ = 0 \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}\)
\(T = \frac{235 \times 10^3\text{ J mol}^{-1}}{171.0\text{ J K}^{-1}\text{ mol}^{-1}} = 1374.3\text{ K} \approx 1374\text{ K}\) (or \(1101\,^\circ\text{C}\)).

評分準則

[2] (a) Enthalpy change when one mole of an ionic crystalline solid is formed [1]; from its separate gaseous ions under standard conditions [1]
[3] (b) Cycle expression with correct multiplier of 2 for Cl atomisation and EA [1]; correct substitution of signs and values [1]; \(\Delta H^\circ_{\text{latt}} = -2153\text{ kJ mol}^{-1}\) (with negative sign and units) [1]
[2] (c) \(\text{Sr}^{2+}\) has a smaller ionic radius / higher charge density than \(\text{Ba}^{2+}\) [1]; stronger ionic bonding / stronger electrostatic attraction between \(\text{Sr}^{2+}\) and \(\text{Cl}^-\) [1]
[2] (d)(i) Correct calculation: \(268.0 - 97.0 = +171.0\text{ J K}^{-1}\text{ mol}^{-1}\) (value + sign + units) [2]
[2] (d)(ii) Use of \(T = \Delta H / \Delta S\) with correct unit conversion (\(\times 10^3\)) [1]; \(T = 1374\text{ K}\) [1]
題目 4 · structured
11.11 分
Electrochemical cells provide valuable insights into redox thermodynamic tendencies.

(a) An electrochemical cell is set up under standard conditions consisting of a \(\text{Cr}^{3+}(\text{aq})/\text{Cr}(\text{s})\) half-cell and an \(\text{Fe}^{3+}(\text{aq})/\text{Fe}^{2+}(\text{aq})\) half-cell.
Standard electrode potentials are given:
\[\text{Cr}^{3+}(\text{aq}) + 3e^- \rightleftharpoons \text{Cr}(\text{s}) \quad E^\circ = -0.74\text{ V}\]
\[\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\circ = +0.77\text{ V}\]
(i) State the material used as the electrode in the \(\text{Fe}^{3+}(\text{aq})/\text{Fe}^{2+}(\text{aq})\) half-cell.
(ii) Calculate the standard cell potential, \(E^\circ_{\text{cell}}\), for this electrochemical cell.
(iii) Write the overall balanced ionic equation for the cell reaction that occurs spontaneously.

(b) The Nernst equation for a metal/metal-ion electrode can be expressed as:
\[E = E^\circ + \frac{0.059}{z} \log_{10}[\text{M}^{z+}]\]
(i) Calculate the electrode potential, \(E\), of the \(\text{Cr}^{3+}/\text{Cr}\) half-cell when \([\text{Cr}^{3+}] = 0.0200\text{ mol dm}^{-3}\) at \(298\text{ K}\).
(ii) Predict and explain how the cell potential, \(E_{\text{cell}}\), changes when the concentration of \(\text{Cr}^{3+}\) is decreased from \(1.00\text{ mol dm}^{-3}\) to \(0.0200\text{ mol dm}^{-3}\), keeping all other concentrations standard.

(c) An aqueous solution of chromium(III) sulfate, \(\text{Cr}_2(\text{SO}_4)_3\), is electrolysed using inert electrodes for \(45.0\text{ minutes}\) with a constant current of \(2.50\text{ A}\).
Calculate the mass of chromium deposited at the cathode.
[\(F = 9.65 \times 10^4\text{ C mol}^{-1}\); \(A_r(\text{Cr}) = 52.0\)]
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解題

(a)(i) Platinum (Pt) or graphite (carbon).
(ii) \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.77 - (-0.74) = +1.51\text{ V}\).
(iii) The half-cell with the more positive \(E^\circ\) undergoes reduction (\(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}\)), while the more negative \(E^\circ\) undergoes oxidation (\(\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-\)).
Balanced equation: \(\text{Cr}(\text{s}) + 3\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{Fe}^{2+}(\text{aq})\).
(b)(i) \(z = 3\). \(E = -0.74 + \frac{0.059}{3} \log_{10}(0.0200) = -0.74 + 0.01967 \times (-1.6990) = -0.74 - 0.0334 = -0.773\text{ V}\).
(ii) When \([\text{Cr}^{3+}]\) decreases, the position of equilibrium \(\text{Cr}^{3+}(\text{aq}) + 3e^- \rightleftharpoons \text{Cr}(\text{s})\) shifts to the left, making \(E(\text{Cr}^{3+}/\text{Cr})\) more negative. Since \(E_{\text{cell}} = E(\text{Fe}^{3+}/\text{Fe}^{2+}) - E(\text{Cr}^{3+}/\text{Cr})\), decreasing \(E(\text{Cr}^{3+}/\text{Cr})\) causes \(E_{\text{cell}}\) to increase.
(c) \(Q = I \times t = 2.50\text{ A} \times (45.0 \times 60\text{ s}) = 2.50 \times 2700 = 6750\text{ C}\).
Moles of electrons \(n(e^-) = \frac{Q}{F} = \frac{6750}{96500} = 0.06995\text{ mol}\).
Reduction reaction: \(\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}\).
Moles of \(\text{Cr} = \frac{n(e^-)}{3} = \frac{0.06995}{3} = 0.02332\text{ mol}\).
Mass of \(\text{Cr} = 0.02332 \times 52.0 = 1.21\text{ g}\).

評分準則

[1] (a)(i) Platinum / Pt / Graphite
[1] (a)(ii) \(E^\circ_{\text{cell}} = +1.51\text{ V}\) (sign required)
[1] (a)(iii) \(\text{Cr}(\text{s}) + 3\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{Fe}^{2+}(\text{aq})\)
[2] (b)(i) Correct substitution into Nernst equation [1]; \(E = -0.773\text{ V}\) (allow -0.77 V) [1]
[2] (b)(ii) \(E_{\text{cell}}\) increases [1]; because the electrode potential of \(\text{Cr}^{3+}/\text{Cr}\) becomes more negative / shifts left to replace lost ions [1]
[4] (c) \(Q = 2.50 \times 2700 = 6750\text{ C}\) [1]; moles of electrons \(= 6750 / 96500 = 0.06995\text{ mol}\) [1]; moles of \(\text{Cr} = 0.06995 / 3 = 0.02332\text{ mol}\) [1]; mass \(= 1.21\text{ g}\) (3 s.f.) [1]
題目 5 · structured
11.11 分
The reaction between persulfate ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\), in aqueous solution is represented by:
\[\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})\]

(a) Initial rate experiments were carried out at \(298\text{ K}\) to determine the kinetics of the reaction. The results obtained are shown:
| Experiment | \([\text{S}_2\text{O}_8^{2-}] / \text{mol dm}^{-3}\) | \([\text{I}^-] / \text{mol dm}^{-3}\) | Initial rate / \(\text{mol dm}^{-3}\text{ s}^{-1}\) |
|---|---|---|---|
| 1 | 0.040 | 0.030 | \(4.80 \times 10^{-5}\) |
| 2 | 0.080 | 0.030 | \(9.60 \times 10^{-5}\) |
| 3 | 0.040 | 0.060 | \(9.60 \times 10^{-5}\) |

(i) Deduce the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) and with respect to \(\text{I}^-\). Explain your reasoning.
(ii) Write the rate equation for this reaction.
(iii) Calculate the value of the rate constant, \(k\), at \(298\text{ K}\). Include the units of \(k\).

(b) This reaction is catalysed by aqueous iron(III) ions, \(\text{Fe}^{3+}(\text{aq})\).
(i) Explain why uncatalysed reaction between \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) has a high activation energy.
(ii) Construct two balanced equations to show how \(\text{Fe}^{3+}(\text{aq})\) acts as a homogeneous catalyst for this reaction.

(c) In a separate experiment, the decomposition of hydrogen peroxide is first order with a half-life of \(420\text{ s}\) at \(298\text{ K}\).
Calculate the time taken, in seconds, for the concentration of \(\text{H}_2\text{O}_2\) to decrease from \(0.800\text{ mol dm}^{-3}\) to \(0.0500\text{ mol dm}^{-3}\).
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解題

(a)(i) Comparing Experiment 1 and 2: \([\text{I}^-]\) is constant at \(0.030\text{ mol dm}^{-3}\). When \([\text{S}_2\text{O}_8^{2-}]\) doubles (from 0.040 to 0.080), the initial rate doubles (from \(4.80 \times 10^{-5}\) to \(9.60 \times 10^{-5}\)). Therefore, the reaction is 1st order with respect to \(\text{S}_2\text{O}_8^{2-}\).
Comparing Experiment 1 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant at \(0.040\text{ mol dm}^{-3}\). When \([\text{I}^-]\) doubles (from 0.030 to 0.060), the initial rate doubles (from \(4.80 \times 10^{-5}\) to \(9.60 \times 10^{-5}\)). Therefore, the reaction is 1st order with respect to \(\text{I}^-\).
(ii) \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).
(iii) Using Experiment 1:
\(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{4.80 \times 10^{-5}}{0.040 \times 0.030} = \frac{4.80 \times 10^{-5}}{1.20 \times 10^{-3}} = 0.0400\).
Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
\(k = 0.040\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
(b)(i) Both \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) are negatively charged ions. There is strong electrostatic repulsion between like charges, requiring high collision energy to bring them together, which results in a high activation energy.
(ii) Step 1: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)
Step 2: \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)
(Each step involves oppositely charged ions attracting each other, with lower \(E_a\)).
(c) For a 1st order reaction, the half-life is constant (\(t_{1/2} = 420\text{ s}\)).
Number of half-lives to go from \(0.800\) to \(0.0500\text{ mol dm}^{-3}\):
\(0.800 \xrightarrow{1} 0.400 \xrightarrow{2} 0.200 \xrightarrow{3} 0.100 \xrightarrow{4} 0.0500\).
This requires 4 half-lives.
Total time \(= 4 \times 420\text{ s} = 1680\text{ s}\).

評分準則

[2] (a)(i) Order 1 for \(\text{S}_2\text{O}_8^{2-}\) with valid deduction [1]; Order 1 for \(\text{I}^-\) with valid deduction [1]
[1] (a)(ii) Rate \(= k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
[2] (a)(iii) Correct numerical value \(k = 0.040\) [1]; correct units \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) [1]
[1] (b)(i) High activation energy due to electrostatic repulsion between two negatively charged ions / anions
[2] (b)(ii) Equation 1: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) [1]; Equation 2: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\) [1]
[3] (c) Identification of 4 half-lives (or use of \(k = \ln 2 / 420 = 1.65 \times 10^{-3}\text{ s}^{-1}\)) [1]; correct working \([\text{A}] = [\text{A}]_0 e^{-kt}\) or \(4 \times 420\) [1]; answer \(= 1680\text{ s}\) (or \(28.0\text{ min}\)) [1]
題目 6 · structured
11.11 分
Benzene and its derivatives undergo characteristic electrophilic substitution reactions.

(a) Methylbenzene can be converted into 4-nitromethylbenzene.
(i) State the reagents and conditions needed to convert methylbenzene into 4-nitromethylbenzene.
(ii) Outline the mechanism for the mononitration of methylbenzene. Include all relevant curly arrows, dipoles, charges, and the generation of the electrophile.

(b) 4-Nitromethylbenzene can be converted into 4-aminobenzoic acid via a two-stage synthetic route:
- Stage 1: Oxidation of the methyl group to a carboxylic acid using hot alkaline \(\text{KMnO}_4\) followed by acidification.
- Stage 2: Reduction of the nitro group to an amine group.
(i) State the reagents and conditions required for Stage 2.
(ii) 4-Aminobenzoic acid can exist as a zwitterion in aqueous solution. Draw the displayed formula of the zwitterion formed by 4-aminobenzoic acid.

(c) Phenylamine reacts with nitrous acid, \(\text{HNO}_2\), at \(5\,^\circ\text{C}\) to form a diazonium salt, which is subsequently coupled with naphthalen-2-ol in alkaline solution.
(i) State how nitrous acid is prepared in situ.
(ii) Explain why the temperature of the reaction mixture must be kept below \(10\,^\circ\text{C}\).
(iii) Draw the structure of the azo dye formed when the benzenediazonium ion reacts with naphthalen-2-ol.
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解題

(a)(i) Concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\) (catalyst), warm at \(30\text{--}55\,^\circ\text{C}\).
(ii) Electrophile generation: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)).
Mechanism: A curly arrow from the \(\pi\) electron cloud of the benzene ring (para to the \(\text{CH}_3\) group) to the \(\text{N}\) atom of \(\text{NO}_2^+\). This forms an arenium intermediate with a horseshoe \(\pi\)-system spanning the remaining 5 carbon atoms and a \(+\) charge inside the horseshoe. A curly arrow from the \(\text{C}-\text{H}\) bond at the substituted carbon back into the ring system restores aromaticity, releasing \(\text{H}^+\).
(b)(i) Tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\), heated under reflux, followed by adding aqueous \(\text{NaOH}\).
(ii) Zwitterion structure: \(^+\text{H}_3\text{N}-\text{C}_6\text{H}_4-\text{COO}^-\), showing the protonated amino group \(-\text{NH}_3^+\) and deprotonated carboxylate group \(-\text{COO}^-\).
(c)(i) Sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)).
(ii) Benzenediazonium chloride is unstable at temperatures above \(10\,^\circ\text{C}\) and hydrolyses rapidly to form phenol, releasing nitrogen gas (\(\text{N}_2\)).
(iii) The benzenediazonium ion couples at position 1 of naphthalen-2-ol: \(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_{10}\text{H}_6\text{OH}\) with the azo linkage \(-\text{N}=\text{N}-\) bonded to carbon 1 adjacent to the \(-\text{OH}\) at carbon 2.

評分準則

[1] (a)(i) Concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\) at \(30\text{--}55\,^\circ\text{C}\)
[3] (a)(ii) Generation of \(\text{NO}_2^+\) equation [1]; curly arrow from ring to \(\text{NO}_2^+\) and correctly drawn carbocation intermediate with \(+\) charge [1]; curly arrow from \(\text{C}-\text{H}\) bond into ring restoring aromaticity [1]
[2] (b)(i) \(\text{Sn}\) and concentrated \(\text{HCl}\), reflux [1]; followed by \(\text{NaOH}(\text{aq})\) [1]
[1] (b)(ii) Correct zwitterion formula \(^+\text{H}_3\text{N}-\text{C}_6\text{H}_4-\text{COO}^-\)
[1] (c)(i) \(\text{NaNO}_2 + \text{HCl}\) (or \(\text{HNO}_2\) generated from sodium nitrite and acid)
[1] (c)(ii) Diazonium salt is unstable / decomposes to phenol and \(\text{N}_2\) at temperatures above \(10\,^\circ\text{C}\)
[2] (c)(iii) Correct azo group \(-\text{N}=\text{N}-\) [1]; correct coupling to position 1 of naphthalen-2-ol [1]
題目 7 · structured
11.11 分
Phenol, \(\text{C}_6\text{H}_5\text{OH}\), and ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\), both possess a hydroxyl group but show distinctly different chemical properties.

(a) By considering the electronic structures of the conjugate bases, explain why phenol is significantly more acidic than ethanol.

(b) Describe what would be observed, and write a balanced equation, when aqueous bromine is added to an aqueous solution of phenol at room temperature.

(c) Phenol reacts with ethanoyl chloride, \(\text{CH}_3\text{COCl}\), in the presence of a base to form an ester, \(\mathbf{E}\).
(i) Name the type of reaction occurring and identify ester \(\mathbf{E}\).
(ii) Contrast the reactivity of ethanoyl chloride, chloroethane, and chlorobenzene towards hydrolysis by water at room temperature. Explain these differences in reactivity in terms of their structures and bonding.

(d) When phenol is treated with dilute nitric acid at room temperature, a mixture of two mononitrated structural isomers is formed.
(i) Draw the structures of these two isomers.
(ii) Explain why phenol reacts with dilute nitric acid at room temperature, whereas benzene requires concentrated nitric acid and concentrated sulfuric acid with heating to undergo nitration.
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解題

(a) When phenol loses a proton, the resulting phenoxide ion (\(\text{C}_6\text{H}_5\text{O}^-\)) is stabilised because the lone pair on the oxygen atom overlaps and delocalises into the \(\pi\)-electron system of the aromatic ring. This disperses the negative charge. In the ethoxide ion (\(\text{CH}_3\text{CH}_2\text{O}^-\)), the alkyl ethyl group has a positive inductive effect (electron-donating), which concentrates negative charge on the oxygen atom, destabilising the ethoxide ion. Therefore, phenol dissociates more readily.
(b) Observation: The reddish-brown / orange bromine water is decolourised, and a white precipitate (of 2,4,6-tribromophenol) is formed.
Equation: \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\).
(c)(i) Type of reaction: Nucleophilic addition-elimination (or acylation / condensation / esterification). Ester \(\mathbf{E}\): Phenyl ethanoate, \(\text{CH}_3\text{COOC}_6\text{H}_5\).
(ii) Reactivity order: ethanoyl chloride > chloroethane > chlorobenzene.
- In \(\text{CH}_3\text{COCl}\), the carbonyl carbon is attached to both an electronegative oxygen and chlorine atom, making it strongly electron-deficient (\(\delta+\)), and the reaction proceeds readily via addition-elimination.
- In \(\text{CH}_3\text{CH}_2\text{Cl}\), the \(\text{C}-\text{Cl}\) bond is polar, allowing nucleophilic substitution at a moderate rate (requires heating with aqueous alkali).
- In \(\text{C}_6\text{H}_5\text{Cl}\), a lone pair of electrons on the chlorine atom overlaps with the delocalised \(\pi\)-system of the benzene ring, giving the \(\text{C}-\text{Cl}\) bond partial double-bond character, making it much stronger and resistant to nucleophilic attack.
(d)(i) 2-nitrophenol (ortho) and 4-nitrophenol (para).
(ii) The lone pair of electrons on the oxygen atom of the \(-\text{OH}\) group delocalises into the \(\pi\) ring system of benzene. This significantly increases the electron density of the aromatic ring, making it much more susceptible to electrophilic attack (activated) compared to benzene.

評分準則

[2] (a) Delocalisation of negative charge/lone pair into aromatic \(\pi\) system in phenoxide stabilizes the conjugate base [1]; inductive electron-donating effect of ethyl group intensifies charge on oxygen in ethoxide [1]
[2] (b) Observations: decolourises bromine water AND white precipitate [1]; balanced equation forming 2,4,6-tribromophenol and 3 HBr [1]
[3] (c)(i) Reaction: addition-elimination / acylation AND product: phenyl ethanoate [1]
(c)(ii) Correct reactivity order: ethanoyl chloride > chloroethane > chlorobenzene [1]; explanation: carbonyl carbon strongly \(\delta+\) in acyl chloride vs partial double bond character due to p-orbital overlap of Cl lone pair with \(\pi\) ring in chlorobenzene [1]
[2] (d)(i) Correct structures for 2-nitrophenol and 4-nitrophenol [1]
[2] (d)(ii) Lone pair of electrons on the oxygen atom delocalises into the benzene ring [1]; increasing electron density of the ring and activating it towards electrophiles [1]
題目 8 · structured
11.11 分
Condensation polymers are widely used in synthetic fibres and biomaterials.

(a) Kevlar is a tough polyamide prepared from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine.
(i) Draw the structure of the repeat unit of Kevlar, showing all bonds in the amide link.
(ii) State the type of intermolecular forces between adjacent polymer chains of Kevlar that account for its exceptional tensile strength.
(iii) Name the small molecule eliminated during the industrial polymerization reaction.

(b) Poly(lactic-co-glycolic acid), PLGA, is a biodegradable polyester synthesized from 2-hydroxypropanoic acid (lactic acid) and 2-hydroxyethanoic acid (glycolic acid).
(i) Draw the repeat unit of the polymer formed when one molecule of lactic acid alternates with one molecule of glycolic acid.
(ii) Explain why polyesters such as PLGA are biodegradable, whereas addition polymers like poly(propene) are non-biodegradable.

(c) A tripeptide, \(\mathbf{T}\), was hydrolysed using concentrated hydrochloric acid under reflux, yielding three amino acids:
- Glycine, \(\text{H}_2\text{NCH}_2\text{COOH}\)
- Alanine, \(\text{H}_2\text{NCH(CH}_3)\text{COOH}\)
- Cysteine, \(\text{H}_2\text{NCH(CH}_2\text{SH})\text{COOH}\)
(i) State how thin-layer chromatography (TLC) can be used to separate and identify these amino acids in the hydrolysate.
(ii) Draw the displayed structure of one possible tripeptide \(\mathbf{T}\) containing these three amino acids in a 1:1:1 ratio.
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解題

(a)(i) The repeat unit is: \(-[\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}]_n-\).
(ii) Extensive hydrogen bonding between \(\text{C}=\text{O}\) groups on one chain and \(\text{N}-\text{H}\) groups on an adjacent parallel chain, together with \(\pi-\pi\) stacking of benzene rings.
(iii) Water (\(\text{H}_2\text{O}\)) [or \(\text{HCl}\) if diacyl chloride is used].
(b)(i) Lactic acid contributes \(-\text{O}-\text{CH}(\text{CH}_3)-\text{CO}-\) and glycolic acid contributes \(-\text{O}-\text{CH}_2-\text{CO}-\).
Repeat unit: \(-[\text{O}-\text{CH}(\text{CH}_3)-\text{CO}-\text{O}-\text{CH}_2-\text{CO}]_n-\).
(ii) The ester linkages (\(-\text{COO}-\)) in polyesters are polar and contain a \(\delta+\) carbonyl carbon susceptible to nucleophilic attack by water (hydrolysis), which can be catalyzed by enzymes/microorganisms. Addition polymers such as poly(propene) contain only non-polar, chemically inert \(\text{C}-\text{C}\) single bonds that are not susceptible to hydrolysis.
(c)(i) Apply a small spot of the hydrolysate alongside reference samples of pure glycine, alanine, and cysteine onto a TLC plate (silica/alumina stationary phase). Place the plate vertically in a closed beaker with a suitable mobile phase (solvent). After the solvent front reaches near the top, remove and dry the plate, spray with ninhydrin (and warm) or view under UV light to visualize the purple/brown spots. Measure distance travelled by spot and solvent to determine \(R_f\) values (\(R_f = \text{distance by spot}/\text{distance by solvent}\)), and compare with known reference \(R_f\) values.
(ii) Tripeptide \(\mathbf{T}\) structure: e.g., Gly-Ala-Cys: \(\text{H}_2\text{N}-\text{CH}_2-\text{CONH}-\text{CH}(\text{CH}_3)-\text{CONH}-\text{CH}(\text{CH}_2\text{SH})-\text{COOH}\).

評分準則

[2] (a)(i) Correct repeat unit showing both amide links \(-\text{CONH}-\) and 1,4-disubstituted aromatic rings [1]; open valencies at both ends [1]
[1] (a)(ii) Hydrogen bonding (between \(\text{C}=\text{O}\) and \(\text{N}-\text{H}\))
[1] (a)(iii) Water / \(\text{H}_2\text{O}\) (allow \(\text{HCl}\))
[2] (b)(i) Correct alternating repeat unit with both ester links shown [2]
[2] (b)(ii) Ester link is polar / susceptible to nucleophilic attack/hydrolysis by enzymes/water [1]; addition polymers have strong, non-polar \(\text{C}-\text{C}\) bonds that resist hydrolysis [1]
[2] (c)(i) Correct TLC procedure: spotting plate, running solvent, visualizing with ninhydrin / UV [1]; comparing \(R_f\) values to reference values [1]
[1] (c)(ii) Correct tripeptide displayed/structural formula linking all 3 amino acids with two peptide bonds \(-\text{CONH}-\)
題目 9 · structured
11.11 分
Compound \(\mathbf{X}\) is an aromatic organic compound with the molecular formula \(\text{C}_9\text{H}_{10}\text{O}_2\).

(a) Compound \(\mathbf{X}\) gives the following analytical results:
- It effervesces with aqueous sodium carbonate, \(\text{Na}_2\text{CO}_3\), releasing a gas that turns limewater cloudy.
- When reacted with alkaline aqueous iodine, no yellow precipitate is formed.
- The \(^{13}\text{C}\) NMR spectrum of \(\mathbf{X}\) exhibits 7 distinct peaks.
- The \(^1\text{H}\) NMR spectrum of \(\mathbf{X}\) recorded in \(\text{CDCl}_3\) is summarized in the table:

| Chemical shift, \(\delta / \text{ppm}\) | Splitting pattern | Relative peak area |
|---|---|---|
| 1.25 | triplet | 3 |
| 2.68 | quartet | 2 |
| 7.30 | doublet | 2 |
| 8.02 | doublet | 2 |
| 11.80 | singlet | 1 |

(i) Identify the functional group responsible for the effervescence with \(\text{Na}_2\text{CO}_3\).
(ii) State what happens to the signal at \(\delta = 11.80\text{ ppm}\) when a drop of \(\text{D}_2\text{O}\) is added to the NMR tube. Write an equation to support your answer.
(iii) Deduce the structural fragment responsible for the peaks at \(\delta = 1.25\) and \(\delta = 2.68\text{ ppm}\). Explain the splitting patterns using the \(n+1\) rule.
(iv) Deduce the substitution pattern of the benzene ring from the signals at \(\delta = 7.30\) and \(\delta = 8.02\text{ ppm}\).
(v) Deduce the identity of compound \(\mathbf{X}\) and draw its full structural formula.

(b) An isomer of \(\mathbf{X}\), compound \(\mathbf{Y}\), does not react with \(\text{Na}_2\text{CO}_3\) but reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate. \(\mathbf{Y}\) gives a positive tri-iodomethane test and its \(^1\text{H}\) NMR spectrum shows a singlet corresponding to 3 protons at \(\delta = 2.15\text{ ppm}\).
Suggest a structure for compound \(\mathbf{Y}\).
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解題

(a)(i) Effervescence with \(\text{Na}_2\text{CO}_3\) releasing \(\text{CO}_2\) confirms the presence of a carboxylic acid functional group, \(-\text{COOH}\).
(ii) When \(\text{D}_2\text{O}\) is added, the carboxylic acid proton undergoes rapid deuterium exchange: \(\text{R}-\text{COOH} + \text{D}_2\text{O} \rightleftharpoons \text{R}-\text{COOD} + \text{HOD}\). Since deuterium (\(^2\text{H}\)) does not resonate at the proton frequency, the singlet at \(\delta = 11.80\text{ ppm}\) disappears from the spectrum.
(iii) The triplet (\(3\text{H}\)) at \(\delta = 1.25\) and quartet (\(2\text{H}\)) at \(\delta = 2.68\) are characteristic of an ethyl group, \(-\text{CH}_2\text{CH}_3\). By the \(n+1\) rule, the \(-\text{CH}_3\) protons have \(n=2\) neighbouring protons, giving \(2+1=3\) (triplet). The \(-\text{CH}_2-\) protons have \(n=3\) neighbouring protons, giving \(3+1=4\) (quartet). The chemical shift of \(\delta = 2.68\) indicates the \(-\text{CH}_2-\) is attached directly to the aromatic ring.
(iv) The two doublets of area 2 at \(\delta = 7.30\) and \(\delta = 8.02\) indicate a pair of mutually coupled aromatic protons in a 1,4-disubstituted (para-substituted) benzene ring with plane symmetry.
(v) Combining the fragments: \(-\text{COOH}\), \(-\text{C}_6\text{H}_4-\) (para), and \(-\text{CH}_2\text{CH}_3\) gives \(\text{C}_9\text{H}_{10}\text{O}_2\). The compound is 4-ethylbenzoic acid: \(\text{CH}_3\text{CH}_2-\text{C}_6\text{H}_4-\text{COOH}\).
Checking \(^{13}\text{C}\) environments: 1 (\(\text{CH}_3\)) + 1 (\(\text{CH}_2\)) + 4 (aromatic carbons due to symmetry) + 1 (\(\text{C}=\text{O}\)) = 7 peaks, consistent with the data.
(b) Compound \(\mathbf{Y}\) forms a 2,4-DNP derivative (contains \(\text{C}=\text{O}\)) and gives a positive tri-iodomethane test (contains a \(\text{CH}_3\text{C}=\text{O}\) methyl ketone group). The singlet at \(\delta = 2.15\) (\(3\text{H}\)) confirms \(\text{CH}_3\text{C}=\text{O}\). Formula is \(\text{C}_9\text{H}_{10}\text{O}_2\). Subtracting \(\text{CH}_3\text{CO}-\) (\(\text{C}_2\text{H}_3\text{O}\)) and \(-\text{C}_6\text{H}_4-\) (\(\text{C}_6\text{H}_4\)) leaves \(-\text{OCH}_3\). Therefore, \(\mathbf{Y}\) is 1-(4-methoxyphenyl)ethan-1-one (or 4-methoxyacetophenone), \(\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{COCH}_3\).

評分準則

[1] (a)(i) Carboxylic acid / \(-\text{COOH}\)
[2] (a)(ii) Signal disappears / vanishes [1]; exchange equation \(\text{RCOOH} + \text{D}_2\text{O} \rightleftharpoons \text{RCOOD} + \text{HOD}\) [1]
[2] (a)(iii) Identification of ethyl group \(-\text{CH}_2\text{CH}_3\) [1]; explanation using \(n+1\) rule: \(-\text{CH}_3\) split by 2 adjacent protons into triplet AND \(-\text{CH}_2-\) split by 3 adjacent protons into quartet [1]
[1] (a)(iv) 1,4-disubstitution / para-substituted benzene ring (due to symmetrical pair of 2H doublets)
[2] (a)(v) Correct name / identification: 4-ethylbenzoic acid [1]; clear structural formula \(\text{CH}_3\text{CH}_2\text{C}_6\text{H}_4\text{COOH}\) [1]
[3] (b) Deduces presence of \(\text{CH}_3\text{CO}-\) group from iodoform test and \(^1\text{H}\) NMR singlet [1]; deduces methoxy group \(-\text{OCH}_3\) on aromatic ring [1]; correct structure of 4-methoxyacetophenone (or structural isomer) [1]
題目 10 · structured
11 分
Cobalt is a transition element that forms a wide variety of coordination complexes with different coordination numbers, geometries, and colours.

(a) (i) Complete the electronic configuration of the \(\text{Co}^{2+}\) ion.

\(1\text{s}^2 2\text{s}^2 2\text{p}^6 \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\) [1]

(ii) Explain why aqueous transition metal complex ions are typically coloured. [3]

(b) An aqueous solution of cobalt(II) chloride contains the pink complex ion \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\).
When concentrated hydrochloric acid is added to this solution, a reversible reaction occurs to form a deep blue solution containing \([\text{CoCl}_4]^{2-}\).

(i) Construct an equation for this equilibrium reaction. [1]

(ii) State the shape and the coordination number of the complex ion present in the deep blue solution.

Shape: \(\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\)
Coordination number: \(\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\) [1]

(c) When 1,2-diaminoethane, \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\) (abbreviated as \(\text{en}\)), is added to an aqueous solution of cobalt(II) ions followed by reaction with a suitable oxidising agent and dilute hydrochloric acid, the octahedral complex \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) is formed.

(i) State the type of ligand represented by 1,2-diaminoethane. [1]

(ii) The complex \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) exhibits stereoisomerism.
- State the two types of stereoisomerism shown by \([\text{Co}(\text{en})_2\text{Cl}_2]^+\).
- Draw three-dimensional diagrams to show the two optical isomers of \([\text{Co}(\text{en})_2\text{Cl}_2]^+\). You may use the abbreviation \(\overset{\frown}{\text{N}\phantom{-}\text{N}}\) to represent 1,2-diaminoethane. [3]

(d) The stability constant, \(K_{\text{stab}}\), for the equilibrium between \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) and 1,2-diaminoethane is given by the reaction:

\[[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightleftharpoons [\text{Co}(\text{en})_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\]

Write an expression for \(K_{\text{stab}}\) for this equilibrium and deduce its units. [1]
查看答案詳解

解題

(a) (i) Co has atomic number 27 (configuration: \([\text{Ar}] 3\text{d}^7 4\text{s}^2\)). In forming \(\text{Co}^{2+}\), the two \(4\text{s}\) electrons are removed first, leaving \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7\).

(ii) In an isolated gaseous transition metal atom/ion, all five \(3\text{d}\) orbitals are degenerate. When ligands approach to form coordinate bonds, repulsion between ligand lone pairs and d electrons splits the \(3\text{d}\) subshell into two groups of non-degenerate orbitals with an energy difference \(\Delta E\). An electron absorbs a photon of visible light (
\(\Delta E = h\nu = hc/\lambda\)) and is promoted from a lower energy d-orbital to a higher energy d-orbital. The unabsorbed wavelengths of visible light are transmitted or reflected, producing the complementary colour seen.

(b) (i) Ligand exchange of six neutral water ligands by four monodentate chloride ligands:
\[[\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\]

(ii) \([\text{CoCl}_4]^{2-}\) has 4 chloride ligands around the central cobalt ion, giving a coordination number of 4 and a tetrahedral shape.

(c) (i) 1,2-diaminoethane has two lone pairs (one on each nitrogen atom) that can form two coordinate bonds to a metal ion, so it is a bidentate ligand.

(ii) \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) has two geometric isomers: trans (two \(\text{Cl}^-\) ligands at \(180^\circ\)) and cis (two \(\text{Cl}^-\) ligands at \(90^\circ\)). The cis isomer is chiral and has no plane of symmetry, thereby showing optical isomerism (enantiomers). The optical isomers are drawn as non-superimposable 3D mirror images of the cis form with two adjacent \(\text{Cl}\) groups and two \(\overset{\frown}{\text{N}\phantom{-}\text{N}}\) chelating rings.

(d) \(K_{\text{stab}} = \frac{[\text{Co}(\text{en})_3^{2+}]}{[\text{Co}(\text{H}_2\text{O})_6^{2+}][\text{en}]^3}\)
Units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = \frac{1}{(\text{mol dm}^{-3})^3} = \text{mol}^{-3} \text{dm}^9\).

評分準則

(a)(i) [1 mark]
- \(3\text{s}^2 3\text{p}^6 3\text{d}^7\) (or \([\text{Ne}] 3\text{s}^2 3\text{p}^6 3\text{d}^7\))

(a)(ii) [3 marks]
- 1 mark: d-orbitals split / lose degeneracy / split into two sets of different energy levels (by ligands)
- 1 mark: electron absorbs light / photon / energy (in the visible spectrum) AND is promoted from a lower to a higher d-orbital / d-d transition
- 1 mark: transmitted light / colour seen is complementary to the absorbed colour / remaining frequencies transmitted

(b)(i) [1 mark]
- \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\) (must show equilibrium arrow or state symbols not essential unless specified)

(b)(ii) [1 mark]
- Shape: tetrahedral AND Coordination number: 4 (both required for 1 mark)

(c)(i) [1 mark]
- Bidentate

(c)(ii) [3 marks]
- 1 mark: Geometric(al) / cis-trans AND optical isomerism (both needed)
- 1 mark: Octahedral 3D representation of cis-[Co(en)₂Cl₂]⁺ showing two Cl ligands in adjacent (cis) positions and wedges/dashes (or clear 3D axes)
- 1 mark: Non-superimposable mirror image accurately drawn showing corresponding reflection

(d) [1 mark]
- Expression: \(K_{\text{stab}} = \frac{[\text{Co}(\text{en})_3^{2+}]}{[\text{Co}(\text{H}_2\text{O})_6^{2+}][\text{en}]^3}\) AND units: \(\text{mol}^{-3}\text{dm}^9\) (both required for 1 mark; [H₂O] must be omitted)

Paper 5 (Planning, Analysis and Evaluation)

Answer all planning, data analysis, graph plotting, and evaluation questions in the spaces provided.
2 題目 · 30 分
題目 1 · Experimental planning, analysis and graphical evaluation
15 分
A student plans an experiment to determine the stoichiometric formula of a complex formed between aqueous nickel(II) ions, \(\text{Ni}^{2+}(\text{aq})\), and a bidentate ligand, 1,2-diaminoethane (abbreviated as \(\text{en}\)), using the method of continuous variation (Job's method) and a visible spectrophotometer (colorimeter).

The reaction can be represented by the general equation:
\[\text{Ni}^{2+}(\text{aq}) + n\,\text{en}(\text{aq}) \rightarrow [\text{Ni}(\text{en})_n]^{2+}(\text{aq})\]

Neither \(\text{Ni}^{2+}(\text{aq})\) nor \(\text{en}(\text{aq})\) absorbs significantly at the wavelength \(\lambda_{\text{max}} = 545\text{ nm}\), whereas the complex \([\text{Ni}(\text{en})_n]^{2+}\) exhibits strong absorbance at this wavelength.

(a) The student is provided with solid nickel(II) sulfate hexahydrate, \(\text{NiSO}_4\cdot 6\text{H}_2\text{O}\) (\(M_{\text{r}} = 262.8\)), and an aqueous solution of 1,2-diaminoethane of concentration \(0.100\text{ mol}\,\text{dm}^{-3}\).

(i) Calculate the mass of \(\text{NiSO}_4\cdot 6\text{H}_2\text{O}\) required to prepare \(250.0\text{ cm}^3\) of a \(0.100\text{ mol}\,\text{dm}^{-3}\) standard solution of \(\text{Ni}^{2+}(\text{aq})\). [1]

(ii) Describe step-by-step how the student should prepare this \(250.0\text{ cm}^3\) standard solution using standard laboratory volumetric apparatus. [3]

(b) In the continuous variation method, mixtures of the \(0.100\text{ mol}\,\text{dm}^{-3}\) \(\text{Ni}^{2+}(\text{aq})\) solution and \(0.100\text{ mol}\,\text{dm}^{-3}\) \(\text{en}(\text{aq})\) solution are prepared such that the total volume of each mixture is constantly maintained at \(20.00\text{ cm}^3\).

(i) Name a suitable piece of apparatus to measure the volumes of both solutions accurately. [1]

(ii) State the independent variable and the dependent variable in this investigation. [2]

(iii) Before measuring the absorbance of each mixture in the colorimeter, describe what the student must do to calibrate the colorimeter at \(545\text{ nm}\). [1]

(c) The results obtained by the student are shown in the table:

\[\begin{array}{|c|c|c|c|} \hline \text{Mixture} & \text{Volume of } 0.100\text{ M Ni}^{2+}\text{ / cm}^3 & \text{Volume of } 0.100\text{ M en / cm}^3 & \text{Absorbance at } 545\text{ nm} \\ \hline 1 & 18.00 & 2.00 & 0.09 \\ 2 & 15.00 & 5.00 & 0.23 \\ 3 & 12.00 & 8.00 & 0.36 \\ 4 & 10.00 & 10.00 & 0.45 \\ 5 & 7.00 & 13.00 & 0.59 \\ 6 & 5.00 & 15.00 & 0.68 \\ 7 & 4.00 & 16.00 & 0.54 \\ 8 & 2.00 & 18.00 & 0.27 \\ \hline \end{array}\]

(i) Explain why the maximum absorbance corresponds to the stoichiometric ratio of reactants in the complex. [1]

(ii) Determine the mole fraction of \(\text{en}\), \(x_{\text{en}} = \frac{V_{\text{en}}}{V_{\text{Ni}^{2+}} + V_{\text{en}}}\), at maximum absorbance. Hence, deduce the integer value of \(n\) in the formula \([\text{Ni}(\text{en})_n]^{2+}\). [2]

(iii) Draw the 3D stereochemical structure of one optical isomer of \([\text{Ni}(\text{en})_n]^{2+}\), clearly showing the coordination geometry around the nickel ion (represent the bidentate ligand as \(\text{N}\!\!\frown\!\!\text{N}\) or \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\)). [2]

(d) 1,2-diaminoethane is corrosive and harmful by inhalation. State one specific safety precaution, other than wearing eye protection and gloves, that the student must take when handling concentrated 1,2-diaminoethane. [1]

(e) Suggest why the experiment must be carried out at a constant temperature. [1]
查看答案詳解

解題

(a)(i) Moles of \(\text{NiSO}_4\cdot 6\text{H}_2\text{O} = c \times V = 0.100\text{ mol}\,\text{dm}^{-3} \times 0.2500\text{ dm}^3 = 0.0250\text{ mol}\).
Mass required \(= 0.0250\text{ mol} \times 262.8\text{ g}\,\text{mol}^{-1} = 6.57\text{ g}\).

(ii) 1. Weigh out accurately \(6.57\text{ g}\) of \(\text{NiSO}_4\cdot 6\text{H}_2\text{O}\) using a balance and dissolve it completely in approximately \(100\text{ cm}^3\) of distilled/deionised water in a beaker.
2. Transfer the solution quantitatively into a \(250.0\text{ cm}^3\) volumetric flask, rinsing the beaker and glass rod several times with distilled water and transferring all washings into the flask.
3. Add distilled water until the bottom of the meniscus is aligned with the graduation mark (using a dropping pipette for the final drops), stopper the flask, and invert several times to ensure thorough mixing.

(b)(i) A \(50\text{ cm}^3\) burette (or a \(10.0\text{ cm}^3 / 25.0\text{ cm}^3\) graduated/volumetric pipette).
(ii) Independent variable: Volume / mole fraction of 1,2-diaminoethane (or ratio of \(V_{\text{en}} : V_{\text{Ni}^{2+}}\)).
Dependent variable: Absorbance (at \(545\text{ nm}\)).
(iii) Place a cuvette containing pure distilled water (solvent blank) into the colorimeter and set the absorbance reading to zero (or \(100\%\) transmittance).

(c)(i) The concentration of the complex reaches a maximum when reactants are mixed in their exact stoichiometric reacting ratio. Since absorbance is directly proportional to complex concentration (Beer-Lambert law), the absorbance peaks at this point.
(ii) Maximum absorbance is observed at Mixture 6, where \(V_{\text{en}} = 15.00\text{ cm}^3\) and \(V_{\text{Ni}^{2+}} = 5.00\text{ cm}^3\).
\[x_{\text{en}} = \frac{15.00}{5.00 + 15.00} = \frac{15.00}{20.00} = 0.75\]
Ratio of \(\text{en} : \text{Ni}^{2+} = 15.00 : 5.00 = 3 : 1\).
Therefore, \(n = 3\) and the complex is \([\text{Ni}(\text{en})_3]^{2+}\).
(iii) Nickel is 6-coordinate with octahedral geometry. Three bidentate \(\text{en}\) ligands span adjacent (cis) positions on the octahedron forming a chiral propeller structure (\(\Delta\) or \(\Lambda\) enantiomer).

(d) Handle concentrated 1,2-diaminoethane inside a well-ventilated fume cupboard / fume hood to avoid inhaling harmful vapours.

(e) The complex formation reaction is in dynamic equilibrium; the stability constant \(K_{\text{stab}}\) (and thus the equilibrium position / extent of complex formation) changes with temperature because the reaction has an enthalpy change (\(\Delta H\)).

評分準則

[Total: 15 marks]

(a)(i) [1 mark]
- \(6.57\text{ g}\) (must be given to 3 significant figures).

(a)(ii) [3 marks]
- Dissolve solid in distilled/deionised water in a beaker [1]
- Quantitative transfer to \(250\text{ cm}^3\) volumetric flask including rinsings/washings of beaker and rod [1]
- Make up to mark with distilled water (bottom of meniscus on line) and invert/shake to homogenise [1]

(b)(i) [1 mark]
- Burette / graduated pipette. (Reject: measuring cylinder, beaker).

(b)(ii) [2 marks]
- Independent variable: Volume / mole fraction / proportion of 1,2-diaminoethane (or ratio of \(\text{en}\) to \(\text{Ni}^{2+}\)) [1]
- Dependent variable: Absorbance [1]

(b)(iii) [1 mark]
- Fill a cuvette with distilled water / solvent blank and set absorbance to zero / \(100\%\) transmittance [1]

(c)(i) [1 mark]
- Absorbance is proportional to concentration of complex, which reaches a maximum when reactants are present in the exact stoichiometric ratio [1]

(c)(ii) [2 marks]
- Calculation of mole fraction: \(x_{\text{en}} = 0.75\) (or volume ratio \(15.00 : 5.00 = 3 : 1\)) [1]
- Deducing \(n = 3\) / formula \([\text{Ni}(\text{en})_3]^{2+}\) [1]

(c)(iii) [2 marks]
- Octahedral shape with 6 coordinate bonds shown with wedge-dash convention [1]
- Three bidentate ligands correctly linking adjacent cis-coordination sites [1]

(d) [1 mark]
- Work in a fume cupboard / fume hood [1]

(e) [1 mark]
- Stability constant / position of equilibrium is temperature-dependent / complexation reaction has a non-zero enthalpy change [1]
題目 2 · Experimental planning, analysis and graphical evaluation
15 分
A student investigates the kinetics of the reaction between persulfate ions, \(\text{S}_2\text{O}_8^{2-}(\text{aq})\), and iodide ions, \(\text{I}^-(\text{aq})\), at different temperatures:
\[\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})\]

The rate constant \(k\) varies with absolute temperature \(T\) according to the Arrhenius equation:
\[\ln k = -\frac{E_{\text{a}}}{R}\left(\frac{1}{T}\right) + \ln A\]
where \(R = 8.31\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1}\), \(E_{\text{a}}\) is the activation energy in \(\text{J}\,\text{mol}^{-1}\), and \(A\) is the pre-exponential factor.

(a) The student measures the rate constant \(k\) across a range of temperatures. The experimental data obtained are given in the table below.

(i) Complete the table by calculating the values of \(\frac{1}{T} \times 10^3\) and \(\ln k\). Record \(\frac{1}{T} \times 10^3\) to 3 decimal places and \(\ln k\) to 2 decimal places. [2]

\[\begin{array}{|c|c|c|c|c|} \hline \text{Experiment} & \text{Temperature, } \theta\text{ / }^\circ\text{C} & T\text{ / K} & \frac{1}{T} \times 10^3\text{ / K}^{-1} & k\text{ / }\text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1} & \ln k \\ \hline 1 & 15.0 & 288.1 & 3.471 & 1.10 \times 10^{-3} & -6.81 \\ 2 & 22.5 & 295.6 & 3.383 & 1.85 \times 10^{-3} & -6.29 \\ 3 & 30.0 & 303.1 & 3.299 & 3.15 \times 10^{-3} & \text{...............} \\ 4 & 37.0 & 310.1 & \text{...............} & 5.12 \times 10^{-3} & -5.27 \\ 5 & 45.0 & 318.1 & 3.144 & 8.75 \times 10^{-3} & -4.74 \\ 6 & 52.5 & 325.6 & 3.071 & 1.18 \times 10^{-2} & -4.44 \\ 7 & 60.0 & 333.1 & 3.002 & 2.12 \times 10^{-2} & -3.85 \\ \hline \end{array}\]

(ii) Plotting \(\ln k\) on the \(y\)-axis against \(\frac{1}{T} \times 10^3\text{ / K}^{-1}\) on the \(x\)-axis yields a straight line.

Identify the experiment that contains an anomalous result. Suggest one possible experimental error that could have led to this specific anomaly. [2]

(iii) Using the non-anomalous points from the table, two coordinates on the best-fit line are \((3.000, -3.82)\) and \((3.450, -6.65)\).

Calculate the gradient of the line of best fit. State the units of your gradient. [2]

(b) Use your gradient from (a)(iii) and the Arrhenius equation to calculate a value for the activation energy, \(E_{\text{a}}\), in \(\text{kJ}\,\text{mol}^{-1}\). Include the appropriate sign. [2]

(c) The reaction is monitored using an iodine clock method, in which a small, fixed amount of sodium thiosulfate and starch indicator are added to the initial mixture.

(i) State the observation that indicates the end-point of the clock reaction. [1]

(ii) Explain the role of the thiosulfate ions in this clock reaction. [2]

(iii) Explain why the amount of sodium thiosulfate added must be very small relative to the amounts of persulfate and iodide ions. [1]

(d) In a further experiment, a student adds a few drops of aqueous iron(II) sulfate, \(\text{FeSO}_4(\text{aq})\), to the mixture.

(i) State the effect of adding \(\text{FeSO}_4(\text{aq})\) on the gradient of the Arrhenius plot of \(\ln k\) against \(\frac{1}{T}\). [1]

(ii) Explain your answer to (d)(i) in terms of transition metal chemistry and reaction pathways. [2]
查看答案詳解

解題

(a)(i) For Experiment 3: \(\ln(3.15 \times 10^{-3}) = -5.76\).
For Experiment 4: \(T = 310.1\text{ K}\), so \(\frac{1}{310.1} \times 10^3 = 3.225\text{ K}^{-1}\).

(ii) Experiment 6: \(k = 1.18 \times 10^{-2}\text{ dm}^3\,\text{mol}^{-1}\,\text{s}^{-1}\), \(\ln k = -4.44\). For \(3.071\), the expected \(\ln k\) from the trend is approximately \(-4.27\). Thus, \(k\) is too small. A possible reason is that the temperature of the reaction mixture was lower than recorded (e.g. water bath cooled down), or a higher volume of thiosulfate was accidentally added.

(iii) \(\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{-6.65 - (-3.82)}{(3.450 - 3.000) \times 10^{-3}\text{ K}^{-1}} = \frac{-2.83}{0.450 \times 10^{-3}\text{ K}^{-1}} = -6.29 \times 10^3\text{ K}\) (or \(-6290\text{ K}\)).

(b) Since \(\text{gradient} = -\frac{E_{\text{a}}}{R}\):
\[E_{\text{a}} = -\text{gradient} \times R = -(-6288.9\text{ K}) \times 8.31\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1} = +52261\text{ J}\,\text{mol}^{-1} = +52.3\text{ kJ}\,\text{mol}^{-1}\]

(c)(i) The mixture turns suddenly from colourless to blue-black.
(ii) \(\text{S}_2\text{O}_3^{2-}\) rapidly reduces the iodine (\(\text{I}_2\)) produced back to iodide (\(\text{I}^-\)):
\[2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-\]
This prevents free \(\text{I}_2\) from forming a complex with starch until all \(\text{S}_2\text{O}_3^{2-}\) has reacted.
(iii) The clock method approximates the initial rate; if a large amount of thiosulfate were used, a significant fraction of reactants would be consumed and the rate would decrease during the timing interval.

(d)(i) The gradient becomes less steep (less negative / smaller magnitude).
(ii) \(\text{Fe}^{2+}\) acts as a homogeneous catalyst because of its variable oxidation state (\(\text{Fe}^{2+}/\text{Fe}^{3+}\)), providing an alternative pathway with lower activation energy (\(E_{\text{a}}\)) by reacting with oppositely charged species in two steps rather than two negatively charged ions repelling each other. Since \(|\text{gradient}| = \frac{E_{\text{a}}}{R}\), a lower \(E_{\text{a}}\) reduces the steepness of the gradient.

評分準則

[Total: 15 marks]

(a)(i) [2 marks]
- \(\ln k = -5.76\) for Exp 3 [1]
- \(\frac{1}{T} \times 10^3 = 3.225\) for Exp 4 [1]

(a)(ii) [2 marks]
- Identifies Experiment 6 as anomalous [1]
- Valid explanation (e.g. actual reaction temperature lower than measured / water bath cooled / excess thiosulfate added / late stopping of timer) [1]

(a)(iii) [2 marks]
- Correct calculation of gradient: value in range \(-6.20 \times 10^3\) to \(-6.35 \times 10^3\) [1]
- Correct unit: \(\text{K}\) (kelvin) [1]

(b) [2 marks]
- Correct calculation of \(E_{\text{a}} = -\text{gradient} \times 8.31\) [1]
- Final value with positive sign and converted to \(\text{kJ}\,\text{mol}^{-1}\) (e.g. \(+52.3\text{ kJ}\,\text{mol}^{-1}\), accept \(51.5\) to \(53.0\)) to 3 sig figs [1]

(c)(i) [1 mark]
- Mixture turns blue-black / blue [1]

(c)(ii) [2 marks]
- Thiosulfate reduces \(\text{I}_2\) back to \(\text{I}^-\) / removes iodine as soon as it is formed [1]
- Colour only appears once all thiosulfate has been consumed, allowing free \(\text{I}_2\) to react with starch [1]

(c)(iii) [1 mark]
- To ensure rate measured is the initial rate / concentration of persulfate and iodide remains effectively constant [1]

(d)(i) [1 mark]
- Gradient becomes less negative / less steep / smaller in magnitude [1]

(d)(ii) [2 marks]
- \(\text{Fe}^{2+}\) acts as a catalyst by providing an alternative reaction route with lower \(E_{\text{a}}\) [1]
- Explains that \(\text{Fe}^{2+}\) undergoes variable oxidation state redox reactions avoiding high-barrier repulsion between two negative anions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) [1]

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