Cambridge IAL · Thinka 原創模擬試題

2023 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

Thinka Nov 2023 (V1) Cambridge International A Level-Style Mock — Chemistry (9701)

270 分465 分鐘2023

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (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. Each question has four options.

40 題目 · 40 分

題目 1 · 選擇題

1 分

A sample of an unknown anhydrous metal carbonate, \(M\text{CO}_3\), weighing \(1.52\text{ g}\), is completely dissolved in \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid. The excess hydrochloric acid required \(28.0\text{ cm}^3\) of \(0.500\text{ mol dm}^{-3}\) sodium hydroxide solution for complete neutralisation. What is the identity of metal \(M\)? (Use of the Periodic Table is required: \(M_r\text{ of }\text{CO}_3^{2-} = 60.0\))

A.Barium
B.Magnesium
C.Calcium
D.Strontium

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解題

Initial moles of \(\text{HCl} = 50.0 \times 10^{-3} \times 1.00 = 0.0500\text{ mol}\). Moles of \(\text{NaOH}\) used for neutralisation = \(28.0 \times 10^{-3} \times 0.500 = 0.0140\text{ mol}\). Since \(\text{HCl}\) and \(\text{NaOH}\) react in a 1:1 ratio, the excess \(\text{HCl}\) is \(0.0140\text{ mol}\). The moles of \(\text{HCl}\) that reacted with the carbonate = \(0.0500 - 0.0140 = 0.0360\text{ mol}\). The carbonate \(M\text{CO}_3\) reacts with \(\text{HCl}\) in a 1:2 ratio: \(M\text{CO}_3 + 2\text{HCl} \rightarrow M\text{Cl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). Thus, moles of \(M\text{CO}_3 = 0.0360 / 2 = 0.0180\text{ mol}\). Molar mass of \(M\text{CO}_3 = 1.52 / 0.0180 = 84.4\text{ g mol}^{-1}\). The relative atomic mass of \(M = 84.4 - 60.0 = 24.4\text{ g mol}^{-1}\), which corresponds to magnesium (\(\text{Mg}\)).

評分準則

1 mark for the correct calculation of moles of reacting acid, carbonate, molar mass of the metal, and identifying Magnesium as the correct option.

題目 2 · 選擇題

1 分

A mixture containing only copper(II) oxide (\(\text{CuO}\), \(M_{\text{r}} = 79.5\)) and copper(I) oxide (\(\text{Cu}_2\text{O}\), \(M_{\text{r}} = 143.0\)) has a total mass of \(4.25\text{ g}\). When this mixture is completely reduced by heating in a stream of hydrogen gas, \(3.56\text{ g}\) of pure copper metal (\(A_{\text{r}} = 63.5\)) is obtained. What is the mass percentage of \(\text{CuO}\) in the original mixture?

A.24.7%
B.43.5%
C.56.5%
D.75.3%

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解題

Let \(x\) be the mass of \(\text{CuO}\) and \(y\) be the mass of \(\text{Cu}_2\text{O}\). We have \(x + y = 4.25\). Total moles of copper produced = \(3.56 / 63.5 = 0.05606\text{ mol}\). From the reduction stoichiometry, 1 mole of \(\text{CuO}\) yields 1 mole of \(\text{Cu}\) and 1 mole of \(\text{Cu}_2\text{O}\) yields 2 moles of \(\text{Cu}\). Therefore, \(x / 79.5 + 2y / 143.0 = 0.05606\). Substituting \(y = 4.25 - x\) gives: \(x / 79.5 + 2(4.25 - x) / 143.0 = 0.05606\). Solving for \(x\) gives \(x = 2.40\text{ g}\). The mass percentage of \(\text{CuO}\) is \((2.40 / 4.25) \times 100\% = 56.5\%\).

評分準則

1 mark for setting up the simultaneous equations for the mixture, calculating the mass of CuO as 2.40 g, and finding the correct percentage of 56.5%.

題目 3 · 選擇題

1 分

When \(0.740\text{ g}\) of an organic compound containing only carbon, hydrogen, and oxygen is completely combusted in excess oxygen, \(1.320\text{ g}\) of carbon dioxide and \(0.540\text{ g}\) of water are formed. What is the empirical formula of the compound? (Relative atomic masses: \(\text{C} = 12.0\), \(\text{H} = 1.0\), \(\text{O} = 16.0\))

A.\(\text{C}_3\text{H}_4\text{O}_3\)
B.\(\text{C}_3\text{H}_6\text{O}_2\)
C.\(\text{C}_4\text{H}_8\text{O}\)
D.\(\text{C}_5\text{H}_{10}\text{O}_2\)

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解題

Mass of \(\text{C}\) in the compound = \((12.0 / 44.0) \times 1.320 = 0.360\text{ g}\). Moles of \(\text{C} = 0.360 / 12.0 = 0.030\text{ mol}\). Mass of \(\text{H}\) in the compound = \((2.0 / 18.0) \times 0.540 = 0.060\text{ g}\). Moles of \(\text{H} = 0.060 / 1.0 = 0.060\text{ mol}\). Mass of \(\text{O}\) in the compound = \(0.740 - (0.360 + 0.060) = 0.320\text{ g}\). Moles of \(\text{O} = 0.320 / 16.0 = 0.020\text{ mol}\). Dividing the mole values by the smallest number (0.020) gives a ratio of \(\text{C} : \text{H} : \text{O} = 1.5 : 3.0 : 1.0\). Multiplying by 2 to get whole numbers gives \(3 : 6 : 2\). Therefore, the empirical formula is \(\text{C}_3\text{H}_6\text{O}_2\).

評分準則

1 mark for calculating the mass and moles of carbon, hydrogen, and oxygen, obtaining the correct mole ratio, and selecting the correct empirical formula.

題目 4 · 選擇題

1 分

A standard solution of potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), is prepared by dissolving \(1.471\text{ g}\) of the solid in distilled water to make exactly \(250.0\text{ cm}^3\) of solution. A \(25.0\text{ cm}^3\) portion of this solution is pipetted into a flask, acidified, and titrated against an aqueous solution of iron(II) sulfate, \(\text{FeSO}_4\). The ionic equation for the reaction is: \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 6\text{Fe}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\). If the titration requires \(22.50\text{ cm}^3\) of the \(\text{FeSO}_4\) solution to reach the endpoint, what is the concentration of \(\text{Fe}^{2+}(\text{aq})\)? (\(M_{\text{r}} \text{ of } \text{K}_2\text{Cr}_2\text{O}_7 = 294.2\))

A.\(0.0222\text{ mol dm}^{-3}\)
B.\(0.133\text{ mol dm}^{-3}\)
C.\(0.267\text{ mol dm}^{-3}\)
D.\(1.33\text{ mol dm}^{-3}\)

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解題

First, find the moles of \(\text{K}_2\text{Cr}_2\text{O}_7\) in the total solution: \(n = 1.471 / 294.2 = 0.00500\text{ mol}\). The moles in the \(25.0\text{ cm}^3\) portion used in the titration = \(0.00500 \times (25.0 / 250.0) = 0.000500\text{ mol}\). From the balanced chemical equation, 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\) reacts with 6 moles of \(\text{Fe}^{2+}\). Therefore, moles of \(\text{Fe}^{2+}\) reacting = \(6 \times 0.000500 = 0.00300\text{ mol}\). The concentration of \(\text{Fe}^{2+}\) in the \(22.50\text{ cm}^3\) solution is: \(0.00300 / (22.50 \times 10^{-3}) = 0.133\text{ mol dm}^{-3}\).

評分準則

1 mark for scaling the moles of potassium dichromate(VI) to 25.0 cm3, applying the 1:6 stoichiometric ratio, and calculating the final concentration of Fe2+.

題目 5 · 選擇題

1 分

An alkene, \(X\), has the molecular formula \(\text{C}_6\text{H}_{12}\). When \(X\) is reacted with hot, concentrated, acidified potassium manganate(VII), two organic products, \(Y\) and \(Z\), are formed. Product \(Y\) reacts with sodium carbonate to produce carbon dioxide gas. Product \(Z\) is a ketone that does not undergo further oxidation. Which of the following is a possible structure of \(X\)?

A.3-methylpent-2-ene
B.2,3-dimethylbut-2-ene
C.hex-2-ene
D.2-methylpent-1-ene

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解題

Reaction of alkenes with hot, concentrated, acidified \(\text{KMnO}_4\) causes oxidative cleavage of the double bond. A double bond carbon with one hydrogen (\(\text{R-CH=}\)) oxidises to a carboxylic acid, and a double bond carbon with two alkyl substituents (\(\text{R}_1\text{R}_2\text{C=}\)) oxidises to a ketone. Since \(Y\) reacts with sodium carbonate, it must be a carboxylic acid. Since \(Z\) is a ketone, the alkene must have the structure \(\text{R-CH=C(R}_1\text{)R}_2\). Looking at the options, 3-methylpent-2-ene is \(\text{CH}_3\text{CH=C(CH}_3\text{)CH}_2\text{CH}_3\). On cleavage, it produces ethanoic acid (carboxylic acid \(Y\)) and butanone (ketone \(Z\)).

評分準則

1 mark for identifying that oxidative cleavage of 3-methylpent-2-ene gives a carboxylic acid and a ketone, matching the prompt.

題目 6 · 選擇題

1 分

When propene is reacted with cold, dilute, alkaline potassium manganate(VII), organic compound \(W\) is formed. When propene is reacted with bromine in the presence of concentrated aqueous sodium chloride, organic compound \(V\) is formed as one of the major products. What are the structures of \(W\) and \(V\)?

A.\(W\) is \(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\); \(V\) is \(\text{CH}_3\text{CH(Cl)CH}_2\text{Br}\)
B.\(W\) is \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\); \(V\) is \(\text{CH}_3\text{CH(Br)CH}_2\text{Cl}\)
C.\(W\) is \(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\); \(V\) is \(\text{CH}_3\text{CH(Br)CH}_2\text{Cl}\)
D.\(W\) is \(\text{CH}_3\text{COCH}_3\); \(V\) is \(\text{CH}_3\text{CH(Cl)CH}_2\text{Br}\)

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解題

Reaction of propene with cold, dilute, alkaline \(\text{KMnO}_4\) forms propane-1,2-diol via dihydroxylation, so \(W\) is \(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\). In the reaction of propene with \(\text{Br}_2\) in aqueous \(\text{NaCl}\), the electrophile \(\text{Br}^+\) adds to the carbon-carbon double bond to form the more stable secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+-\text{CH}_2\text{Br}\). The highly concentrated nucleophile \(\text{Cl}^-\)\text{ (from }\text{NaCl)} then attacks this carbocation, yielding \(\text{CH}_3\text{CH(Cl)CH}_2\text{Br}\) (1-bromo-2-chloropropane) as one of the major products. Therefore, \(V\) is \(\text{CH}_3\text{CH(Cl)CH}_2\text{Br}\).

評分準則

1 mark for identifying the correct diol product for cold KMnO4 reaction, and the correct regiochemistry for the mixed halogen addition product.

題目 7 · 選擇題

1 分

Which statement about the first-row transition elements or their compounds is correct?

A.The electronic configuration of a \(\text{Cu}^{2+}\) ion is \([\text{Ar}] 3\text{d}^8 4\text{s}^1\).
B.When excess aqueous ammonia is added to aqueous copper(II) ions, the geometry of the complex changes from octahedral to tetrahedral.
C.\(\text{Fe}^{3+}\) has a more stable d-subshell configuration than \(\text{Fe}^{2+}\) because it has a half-filled d-subshell.
D.In the complex ion \([\text{Co}(\text{en})_3]^{3+}\), where 'en' represents ethane-1,2-diamine, the coordination number of cobalt is 3.

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解題

A: The electronic configuration of a \(\text{Cu}^{2+}\) ion is \([\text{Ar}] 3\text{d}^9\), not \([\text{Ar}] 3\text{d}^8 4\text{s}^1\). B: When excess aqueous ammonia is added to aqueous copper(II) ions, the geometry of the complex remains distorted octahedral, not tetrahedral. C: \(\text{Fe}^{3+}\) has the electronic configuration \([\text{Ar}] 3\text{d}^5\). A half-filled d-subshell has extra stability due to exchange energy and a symmetrical distribution of electron density, making it more stable than \(\text{Fe}^{2+}\) (\([\text{Ar}] 3\text{d}^6\)). D: In \([\text{Co}(\text{en})_3]^{3+}\), ethane-1,2-diamine is a bidentate ligand, so three ligands form six coordinate bonds, making the coordination number of cobalt 6, not 3.

評分準則

1 mark for identifying statement C as the correct description of the electronic configuration stability of Fe3+.

題目 8 · 選擇題

1 分

A transition metal complex has the formula \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\). Which type of isomerism is shown by this complex, and what is the coordination number of the cobalt ion?

A.Geometrical isomerism only; coordination number = 6
B.Optical isomerism only; coordination number = 6
C.Both geometrical and optical isomerism; coordination number = 6
D.Geometrical isomerism only; coordination number = 4

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解題

The complex \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) is octahedral with a coordination number of 6 because it has six monodentate ligands (four ammonia molecules and two chloride ions) coordinated to the central cobalt ion. It exhibits geometrical (cis-trans) isomerism. The cis-isomer has the two chloride ligands at a 90-degree angle, and the trans-isomer has them opposite at a 180-degree angle. Neither isomer is chiral, so it does not exhibit optical isomerism.

評分準則

1 mark for identifying that the complex shows geometrical isomerism only and has a coordination number of 6.

題目 9 · 選擇題

1 分

An excess of solid calcium carbonate, \(\text{CaCO}_3(s)\), is added to \(50.0\text{ cm}^3\) of \(2.00\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}(aq)\), at \(25^\circ\text{C}\).

The reaction goes to completion:

\(\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\)

What is the volume, in \(\text{dm}^3\), of carbon dioxide gas collected at \(25^\circ\text{C}\) and \(101\text{ kPa}\)?
(Assume 1 mole of gas occupies \(24.0\text{ dm}^3\) under these conditions.)

A.1.20
B.2.40
C.0.60
D.4.80

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解題

First, calculate the number of moles of \(\text{HCl}\) used:
\(n(\text{HCl}) = \frac{50.0}{1000} \times 2.00 = 0.100\text{ mol}\).

According to the stoichiometry of the equation, 2 moles of \(\text{HCl}\) produce 1 mole of \(\text{CO}_2\).
Therefore, \(n(\text{CO}_2) = \frac{0.100}{2} = 0.0500\text{ mol}\).

Finally, calculate the volume of \(\text{CO}_2\) gas:
\(V(\text{CO}_2) = 0.0500\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\).

評分準則

Award 1 mark for the correct calculation of \(1.20\text{ dm}^3\) (Option A).
- Method: Calculate moles of limiting reactant (\(\text{HCl}\)), use stoichiometry to find moles of gas, and multiply by molar gas volume.
- Incorrect options represent arithmetic errors or failure to use the 1:2 stoichiometric ratio.

題目 10 · 選擇題

1 分

A sample of copper consists of two isotopes, \(^{63}\text{Cu}\) and \(^{65}\text{Cu}\). The relative atomic mass of copper in this sample is \(63.55\).

What is the percentage abundance of the \(^{65}\text{Cu}\) isotope in this sample?

A.27.5%
B.35.5%
C.72.5%
D.63.5%

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解題

Let the relative abundance of \(^{65}\text{Cu}\) be \(x\) (as a fraction) and that of \(^{63}\text{Cu}\) be \(1 - x\).

Set up the relative atomic mass equation:
\(65x + 63(1 - x) = 63.55\)
\(65x + 63 - 63x = 63.55\)
\(2x = 0.55\)
\(x = 0.275\)

Converting the fraction to a percentage: \(0.275 \times 100 = 27.5\%\).

評分準則

Award 1 mark for the correct calculation of \(27.5\%\) (Option A).
- Method: Formulate the equation for weighted average mass and solve for the unknown percentage abundance of the heavier isotope.
- Reject other choices as they result from mathematical slips or calculating the abundance of the lighter isotope (72.5%).

題目 11 · 選擇題

1 分

An organic compound contains \(40.0\%\) carbon, \(6.7\%\) hydrogen, and \(53.3\%\) oxygen by mass. Its relative molecular mass is \(180\).

What is the molecular formula of the compound?

A.\(\text{C}_3\text{H}_6\text{O}_3\)
B.\(\text{C}_4\text{H}_8\text{O}_4\)
C.\(\text{C}_5\text{H}_{10}\text{O}_5\)
D.\(\text{C}_6\text{H}_{12}\text{O}_6\)

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解題

First, find the empirical formula by calculating the molar ratios of the elements:
- Moles of \(\text{C} = \frac{40.0}{12.0} = 3.33\text{ mol}\)
- Moles of \(\text{H} = \frac{6.7}{1.0} = 6.70\text{ mol}\)
- Moles of \(\text{O} = \frac{53.3}{16.0} = 3.33\text{ mol}\)

Divide each by the smallest value (3.33):
- \(\text{C} = 1\), \(\text{H} = 2\), \(\text{O} = 1\).
Thus, the empirical formula is \(\text{CH}_2\text{O}\).

The empirical formula mass is \(12.0 + 2(1.0) + 16.0 = 30.0\).

Divide the relative molecular mass by the empirical formula mass:
\(\frac{180}{30.0} = 6\).

Therefore, the molecular formula is \(6 \times (\text{CH}_2\text{O}) = \text{C}_6\text{H}_{12}\text{O}_6\).

評分準則

Award 1 mark for selecting the correct molecular formula \(\text{C}_6\text{H}_{12}\text{O}_6\) (Option D).
- Method: Calculate empirical formula first, then find the multiplier by dividing molecular mass by empirical formula mass.
- Other options are incorrect multiples of the empirical formula.

題目 12 · 選擇題

1 分

How many chloride ions, \(\text{Cl}^-\), are present in \(25.0\text{ cm}^3\) of a \(0.400\text{ mol dm}^{-3}\) solution of calcium chloride, \(\text{CaCl}_2\)?
(Take Avogadro's constant, \(L = 6.02 \times 10^{23}\text{ mol}^{-1}\))

A.\(6.02 \times 10^{21}\)
B.\(1.20 \times 10^{22}\)
C.\(2.41 \times 10^{22}\)
D.\(4.82 \times 10^{22}\)

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解題

1. Calculate the moles of \(\text{CaCl}_2\):
\(n(\text{CaCl}_2) = \frac{25.0}{1000} \times 0.400 = 0.0100\text{ mol}\).

2. Determine the moles of chloride ions, \(\text{Cl}^-\):
Each formula unit of \(\text{CaCl}_2\) contains 2 chloride ions.
\(n(\text{Cl}^-) = 0.0100 \times 2 = 0.0200\text{ mol}\).

3. Calculate the number of chloride ions:
\(\text{Number of ions} = 0.0200\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 1.204 \times 10^{22}\).

評分準則

Award 1 mark for selecting the correct number of ions, \(1.20 \times 10^{22}\) (Option B).
- Method: Multiply the volume and concentration to find moles of \(\text{CaCl}_2\), double this value to account for the stoichiometry of chloride ions, and multiply by Avogadro's constant.
- Distractor A fails to multiply by 2; distractor C is based on an incorrect calculation or wrong multiplier.

題目 13 · 選擇題

1 分

An alkene, \(X\), reacts with hot, concentrated, acidified potassium manganate(VII) to produce a single organic compound, propanone, \(\text{CH}_3\text{COCH}_3\), as the only organic product.

What is the structural formula of alkene \(X\)?

A.\(\text{CH}_3\text{CH}=\text{CHCH}_3\)
B.\((\text{CH}_3)_2\text{C}=\text{CHCH}_3\)
C.\((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\)
D.\(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\)

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解題

Oxidative cleavage of alkenes using hot, concentrated, acidified \(\text{KMnO}_4\) breaks the \(\text{C}=\text{C}\) double bond:
- A \(\text{R}_1\text{R}_2\text{C}=\text{C}\) group is oxidised to a ketone, \(\text{R}_1\text{CO}\text{R}_2\).
- A \(\text{R}\text{CH}=\text{C}\) group is oxidised to a carboxylic acid, \(\text{R}\text{COOH}\).
- A \(\text{CH}_2=\text{C}\) group is oxidised to carbon dioxide, \(\text{CO}_2\).

Since propanone (a ketone, \((\text{CH}_3)_2\text{C}=\text{O}\)) is the only product, both carbon atoms involved in the double bond must be attached to two methyl groups. Thus, the original alkene must be symmetrical with the structure \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\) (2,3-dimethylbut-2-ene).

評分準則

Award 1 mark for the correct structural formula (Option C).
- Method: Analyze the products of oxidative cleavage; identify that a ketone is formed from a fully substituted double-bonded carbon.
- Incorrect options produce different products: A gives ethanoic acid; B gives propanone and ethanoic acid; D gives propanoic acid and carbon dioxide.

題目 14 · 選擇題

1 分

When 2-methylbut-2-ene is reacted with cold, dilute, alkaline potassium manganate(VII), an organic product, \(Y\), is formed.

What is the structure of product \(Y\)?

A.\(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{OH}\)
B.\(\text{CH}_3\text{C(OH)}(\text{CH}_3)\text{CH(OH)}\text{CH}_3\)
C.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
D.\(\text{CH}_3\text{C(OH)}(\text{CH}_3)\text{CH}_2\text{CH}_3\)

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解題

Reaction of an alkene with cold, dilute, alkaline potassium manganate(VII) results in mild oxidation where a diol is formed. Two hydroxyl (\(-\text{OH}\)) groups are added across the double bond (dihydroxylation).

2-methylbut-2-ene has the structure \(\text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_3\).
Adding \(-\text{OH}\) groups to both carbons of the former double bond yields 2-methylbutane-2,3-diol, which has the structure \(\text{CH}_3\text{C(OH)}(\text{CH}_3)\text{CH(OH)}\text{CH}_3\).

評分準則

Award 1 mark for identifying the correct diol product (Option B).
- Method: Recall that cold, dilute, alkaline KMnO4 results in the addition of two OH groups across the double bond.
- Incorrect options: A is a primary alcohol; C is a ketone (from cleavage/oxidation); D is a tertiary alcohol lacking the second OH group.

題目 15 · 選擇題

1 分

The stability constant, \(K_{\text{stab}}\), for the formation of the complex ion \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) from \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) in aqueous ammonia is \(1.2 \times 10^{13}\text{ dm}^{12}\text{ mol}^{-4}\).

The reaction is represented by the equilibrium:

\([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{NH}_3(aq) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 4\text{H}_2\text{O}(l)\)

Which statement about this system is correct?

A.The value of \(K_{\text{stab}}\) indicates that ammonia is a weaker ligand than water.
B.Addition of a large excess of water shifts the equilibrium to the right.
C.The high value of \(K_{\text{stab}}\) indicates that the tetraamminediaquacopper(II) ion is thermodynamically more stable than the hexaaquacopper(II) ion.
D.The unit for this \(K_{\text{stab}}\) expression is \(\text{mol dm}^{-3}\).

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解題

A stability constant (\(K_{\text{stab}}\)) is the equilibrium constant for the formation of a complex ion in a solvent.
- A large value of \(K_{\text{stab}}\) (such as \(1.2 \times 10^{13}\)) indicates that the equilibrium lies heavily to the right, meaning the product complex ion is thermodynamically much more stable than the starting hexaaqua complex. This makes option C correct.
- Option A is incorrect because a large \(K_{\text{stab}}\) indicates ammonia is a stronger ligand than water.
- Option B is incorrect because adding excess water (solvent) would drive the equilibrium to the left.
- Option D is incorrect because the unit is indeed \(\text{dm}^{12}\text{ mol}^{-4}\) as stated in the question, not \(\text{mol dm}^{-3}\).

評分準則

Award 1 mark for the correct statement on thermodynamic stability (Option C).
- Method: Evaluate the significance of a very large stability constant value on the relative thermodynamic stability of coordination complexes.
- Reject other options based on chemical equilibria principles.

題目 16 · 選擇題

1 分

When \(10\text{ cm}^3\) of a gaseous hydrocarbon is completely combusted in \(80\text{ cm}^3\) of oxygen (which is in excess), the total volume of gas remaining after cooling to room temperature is \(70\text{ cm}^3\).

After shaking the remaining gas with aqueous sodium hydroxide (which absorbs carbon dioxide), the volume of gas decreases to \(50\text{ cm}^3\).

All gas volumes are measured at the same temperature and pressure.

What is the molecular formula of the hydrocarbon?

A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)

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解題

Let the hydrocarbon formula be \(\text{C}_x\text{H}_y\).
1. The volume of hydrocarbon reacting is \(10\text{ cm}^3\).
2. Upon cooling to room temperature, water condenses to liquid, so its volume is negligible.
3. The gas remaining after cooling (\(70\text{ cm}^3\)) consists of produced \(\text{CO}_2\) and unreacted excess \(\text{O}_2\).
4. After treatment with \(\text{NaOH}\), the carbon dioxide is absorbed, leaving only the unreacted \(\text{O}_2\), which is \(50\text{ cm}^3\).
5. Therefore, the volume of \(\text{CO}_2\) produced is \(70 - 50 = 20\text{ cm}^3\).
6. The volume of oxygen used in the reaction is \(80\text{ cm}^3\text{ (initial)} - 50\text{ cm}^3\text{ (leftover)} = 30\text{ cm}^3\).

Using Avogadro's hypothesis (gas volume ratio is equal to mole ratio):
- Ratio of hydrocarbon to \(\text{CO}_2 = 10 : 20 = 1 : 2\). Therefore, \(x = 2\).
- Ratio of hydrocarbon to reacting \(\text{O}_2 = 10 : 30 = 1 : 3\).

The general combustion equation is:
\(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\)

Since \(x = 2\) and the coefficient for \(\text{O}_2\) is 3:
\(2 + \frac{y}{4} = 3 \implies \frac{y}{4} = 1 \implies y = 4\).

Thus, the molecular formula is \(\text{C}_2\text{H}_4\).

評分準則

Award 1 mark for identifying \(\text{C}_2\text{H}_4\) as the correct molecular formula (Option B).
- Method: Determine the volume of CO2 produced and O2 reacted using differences in gaseous volumes before and after NaOH absorption. Apply Avogadro's law to solve for the coefficients of the combustion equation.
- Reject other options because they do not match the volume relationships (e.g., CH4 would produce 10 cm3 CO2 and use 20 cm3 O2; C2H6 would use 35 cm3 O2).

題目 17 · 選擇題

1 分

A 1.38 g sample of a Group 1 metal carbonate, \(X_2\text{CO}_3\), is reacted completely with excess hydrochloric acid. This reaction produces 240 \(\text{cm}^3\) of carbon dioxide gas, measured at room temperature and pressure. What is the identity of metal \(X\)? [Molar volume of gas at r.t.p. is 24.0 \(\text{dm}^3\,\text{mol}^{-1}\)]

A.Lithium
B.Sodium
C.Potassium
D.Rubidium

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解題

1. Calculate the moles of carbon dioxide: \(n(\text{CO}_2) = \frac{240\,\text{cm}^3}{24000\,\text{cm}^3\,\text{mol}^{-1}} = 0.0100\,\text{mol}\). 2. Since 1 mole of \(X_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\), \(n(X_2\text{CO}_3) = 0.0100\,\text{mol}\). 3. Calculate the molar mass of \(X_2\text{CO}_3\): \(M_r = \frac{1.38\,\text{g}}{0.0100\,\text{mol}} = 138\,\text{g\,mol}^{-1}\). 4. Use the molar mass of the carbonate ion to find \(A_r(X)\): \(2 \times A_r(X) + 12.0 + (3 \times 16.0) = 138 \Rightarrow 2 \times A_r(X) + 60.0 = 138 \Rightarrow 2 \times A_r(X) = 78.0 \Rightarrow A_r(X) = 39.0\). This corresponds to Potassium (K, \(A_r = 39.1\)).

評分準則

Award 1 mark for the correct choice C. Method: Calculate moles of gas, relate to moles of carbonate, calculate Mr, and identify the Group 1 metal.

題目 18 · 選擇題

1 分

An organic compound \(Y\) with molecular formula \(\text{C}_6\text{H}_{12}\) reacts with cold, dilute, acidified potassium manganate(VII) to form a diol. When \(Y\) is reacted with hot, concentrated, acidified potassium manganate(VII), the only organic product obtained is propanone. What is the systematic IUPAC name of \(Y\)?

A.2,3-dimethylbut-2-ene
B.hex-3-ene
C.2-methylpent-2-ene
D.3-methylpent-2-ene

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解題

1. Symmetrical cleavage of \(Y\) by hot, concentrated \(\text{KMnO}_4\) yields only one organic product, propanone, \(\text{CH}_3\text{COCH}_3\). 2. This indicates that the starting alkene must be symmetrical and composed of two \((\text{CH}_3)_2\text{C}=\) fragments joined together: \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\). 3. The IUPAC name of this alkene is 2,3-dimethylbut-2-ene.

評分準則

Award 1 mark for the correct choice A. Method: Identify the structure of the alkene based on the cleavage product and determine its systematic IUPAC name.

題目 19 · 選擇題

1 分

Which of the following transition metal species in its ground state has the same number of unpaired d-electrons as a \(\text{Fe}^{3+}\) ion?

A.\(\text{Cr}^{3+}\)
B.\(\text{Mn}^{2+}\)
C.\(\text{Co}^{2+}\)
D.\(\text{Ni}^{2+}\)

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解題

1. The electronic configuration of a neutral Fe atom is \([\text{Ar}] 3d^6 4s^2\). For \(\text{Fe}^{3+}\), it is \([\text{Ar}] 3d^5\). In a d5 configuration, there are 5 unpaired d-electrons. 2. The electronic configuration of a neutral Mn atom is \([\text{Ar}] 3d^5 4s^2\). For \(\text{Mn}^{2+}\), it is \([\text{Ar}] 3d^5\), which also contains 5 unpaired d-electrons. 3. Other options: \(\text{Cr}^{3+}\) is \([\text{Ar}] 3d^3\) (3 unpaired), \(\text{Co}^{2+}\) is \([\text{Ar}] 3d^7\) (3 unpaired), and \(\text{Ni}^{2+}\) is \([\text{Ar}] 3d^8\) (2 unpaired).

評分準則

Award 1 mark for the correct choice B. Method: Determine the electronic configurations of the transition metal ions and count the number of unpaired electrons in the d-orbitals.

題目 20 · 選擇題

1 分

An organic compound consists of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. A 0.150 g gaseous sample of this compound occupies a volume of 76.7 \(\text{cm}^3\) at a temperature of 373 K and a pressure of 101 kPa. What is the molecular formula of this compound? [\(R = 8.31\,\text{J\,K}^{-1}\,\text{mol}^{-1}\)]

A.\(\text{CH}_2\text{O}\)
B.\(\text{C}_2\text{H}_4\text{O}_2\)
C.\(\text{C}_3\text{H}_6\text{O}_3\)
D.\(\text{C}_4\text{H}_8\text{O}_4\)

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解題

1. Find empirical formula: Moles of C = \(40.0/12.0 = 3.33\); H = \(6.7/1.0 = 6.7\); O = \(53.3/16.0 = 3.33\). The ratio is \(1:2:1\), giving the empirical formula \(\text{CH}_2\text{O}\) (empirical formula mass = 30.0). 2. Use the ideal gas equation \(pV = nRT = \frac{m}{M_r}RT\) to find the molecular mass: \(M_r = \frac{mRT}{pV} = \frac{0.150\,\text{g} \times 8.31\,\text{J\,K}^{-1}\,\text{mol}^{-1} \times 373\,\text{K}}{101 \times 10^3\,\text{Pa} \times 76.7 \times 10^{-6}\,\text{m}^3} = 60.0\,\text{g\,mol}^{-1}\). 3. Since \(60.0 / 30.0 = 2\), the molecular formula is twice the empirical formula: \(\text{C}_2\text{H}_4\text{O}_2\).

評分準則

Award 1 mark for the correct choice B. Method: Determine the empirical formula, calculate the relative molecular mass from the ideal gas equation, and scale the empirical formula to the molecular formula.

題目 21 · 選擇題

1 分

When 2-methylbut-2-ene reacts with cold concentrated hydrogen bromide (HBr), 2-bromo-2-methylbutane is the major product. Which statement correctly explains this observation?

A.The reaction proceeds via a secondary carbocation intermediate which is more stable than a tertiary carbocation.
B.The reaction proceeds via a tertiary carbocation intermediate which is more stable than a secondary carbocation.
C.The bromine atom is highly electronegative and preferentially attacks the less hindered carbon.
D.The reaction is a free-radical addition that favors the formation of a more substituted product.

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解題

The addition of a proton to the double bond of 2-methylbut-2-ene can form either a secondary or a tertiary carbocation. The reaction proceeds preferentially via the tertiary carbocation intermediate, \((\text{CH}_3)_2\text{C}^+-\text{CH}_2-\text{CH}_3\), because it is more stable than the alternative secondary carbocation, due to the greater positive inductive effect of three alkyl groups compared to two. Nucleophilic attack by bromide then yields 2-bromo-2-methylbutane as the major product.

評分準則

Award 1 mark for the correct choice B. Method: Apply Markownikoff's rule and transition state stability concepts based on carbocation intermediates.

題目 22 · 選擇題

1 分

Which factor is NOT required for a transition metal complex to exhibit colour due to a d–d electron transition?

A.The presence of a partially filled d-orbital subshell.
B.The splitting of d-orbitals into different energy levels by ligands.
C.The absorption of a photon of light in the visible region of the spectrum to promote an electron.
D.The emission of a photon of light in the visible region when an electron falls back to the lower d-orbital level.

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解題

Color in transition metal complexes arises from the promotion of an electron from a lower energy split d-orbital to a higher energy split d-orbital by the absorption of a photon of light in the visible region (d-d transition). The remaining unabsorbed wavelengths are transmitted/reflected to give the complementary color that is observed. Emission of a photon of light is not involved in the visible color of standard coordination complexes.

評分準則

Award 1 mark for the correct choice D. Method: Recall the physical chemistry principles explaining transition metal complex color, highlighting absorption rather than emission.

題目 23 · 選擇題

1 分

What is the total number of ions present in 7.41 g of anhydrous calcium hydroxide, \(\text{Ca(OH)}_2\)? [Take the Avogadro constant, \(L = 6.02 \times 10^{23}\,\text{mol}^{-1}\); \(A_r\): \(\text{Ca} = 40.1\), \(\text{O} = 16.0\), \(\text{H} = 1.0\)]

A.\(6.02 \times 10^{22}\)
B.\(1.20 \times 10^{23}\)
C.\(1.81 \times 10^{23}\)
D.\(5.42 \times 10^{23}\)

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解題

1. Calculate the molar mass of \(\text{Ca(OH)}_2\): \(M_r = 40.1 + 2 \times (16.0 + 1.0) = 74.1\,M_r\,\text{g\,mol}^{-1}\). 2. Find the moles of \(\text{Ca(OH)}_2\): \(n = 7.41 / 74.1 = 0.100\,\text{mol}\). 3. Each formula unit of \(\text{Ca(OH)}_2\) contains 3 ions (one \(\text{Ca}^{2+}\) and two \(\text{OH}^-\)). 4. Total moles of ions = \(0.100 \times 3 = 0.300\,\text{mol}\). 5. Total number of ions = \(0.300 \times 6.02 \times 10^{23} = 1.81 \times 10^{23}\).

評分準則

Award 1 mark for the correct choice C. Method: Calculate moles of the ionic compound, determine the number of constituent ions per formula unit, and multiply by the Avogadro constant.

題目 24 · 選擇題

1 分

The mass spectrum of a sample of element \(Z\) shows the following peaks: \(m/e = 24\) (relative abundance = 79.0%), \(m/e = 25\) (relative abundance = 10.0%), and \(m/e = 26\) (relative abundance = 11.0%). What is the relative atomic mass of this sample of element \(Z\) to three significant figures?

A.24.0
B.24.3
C.24.5
D.25.0

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解題

1. Calculate the weighted average mass of the isotopes: \(A_r(Z) = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}\). 2. Sum the products: \(1896.0 + 250.0 + 286.0 = 2432.0\). 3. Divide by 100 to get the relative atomic mass: \(24.32\). Rounded to three significant figures, this is \(24.3\).

評分準則

Award 1 mark for the correct choice B. Method: Determine relative atomic mass from mass spectrometry isotopic abundance data by computing a weighted average.

題目 25 · 選擇題

1 分

A 1.50 g sample of an organic compound containing only carbon, hydrogen, and oxygen was burned completely in excess oxygen. This produced 3.30 g of carbon dioxide and 1.80 g of water. What is the empirical formula of the compound?

A.\(\text{C}_3\text{H}_8\text{O}\)
B.\(\text{C}_3\text{H}_6\text{O}_2\)
C.\(\text{C}_4\text{H}_{10}\text{O}\)
D.\(\text{C}_3\text{H}_5\text{O}\)

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解題

First, calculate the mass of carbon in the \(\text{CO}_2\):
\(\text{Mass of C} = 3.30 \times \frac{12.0}{44.0} = 0.90\,\text{g}\)
\(\text{Moles of C} = \frac{0.90}{12.0} = 0.075\,\text{mol}\)

Next, calculate the mass of hydrogen in the \(\text{H}_2\text{O}\):
\(\text{Mass of H} = 1.80 \times \frac{2.0}{18.0} = 0.20\,\text{g}\)
\(\text{Moles of H} = \frac{0.20}{1.0} = 0.20\,\text{mol}\)

Calculate the mass of oxygen by subtraction:
\(\text{Mass of O} = 1.50 - 0.90 - 0.20 = 0.40\,\text{g}\)
\(\text{Moles of O} = \frac{0.40}{16.0} = 0.025\,\text{mol}\)

Determine the simplest mole ratio:
\(\text{C} : \text{H} : \text{O} = \frac{0.075}{0.025} : \frac{0.20}{0.025} : \frac{0.025}{0.025} = 3 : 8 : 1\)

Therefore, the empirical formula is \(\text{C}_3\text{H}_8\text{O}\).

評分準則

1 mark for the correct option A. Method: Calculate moles of C, H, and O from combustion data, then determine the simplest whole-number ratio.

題目 26 · 選擇題

1 分

A gaseous hydrocarbon of volume 10 cm\(^3\) was exploded with 100 cm\(^3\) of oxygen (an excess). After cooling to room temperature and pressure, the remaining gas volume was 85 cm\(^3\). After passing this gas mixture through a concentrated aqueous solution of potassium hydroxide to absorb carbon dioxide, the remaining gas volume was 55 cm\(^3\). What is the molecular formula of the hydrocarbon?

A.\(\text{C}_3\text{H}_4\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)

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解題

Let the hydrocarbon be \(\text{C}_x\text{H}_y\).
1. The volume of \(\text{CO}_2\) produced is the change in volume when passed through \(\text{KOH}\):
\(\text{Volume of CO}_2 = 85 - 55 = 30\,\text{cm}^3\).
Since 10 cm\(^3\) of hydrocarbon was used:
\(x = \frac{30}{10} = 3\).

2. The volume of remaining oxygen is 55 cm\(^3\), so the volume of oxygen reacted is:
\(\text{Volume of O}_2\text{ reacted} = 100 - 55 = 45\,\text{cm}^3\).

3. The combustion equation is:
\(\text{C}_3\text{H}_y + (3 + \frac{y}{4})\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\)
Since 10 cm\(^3\) of hydrocarbon reacted with 45 cm\(^3\) of oxygen:
Ratio = \(1 : 4.5\)
\(3 + \frac{y}{4} = 4.5 \implies \frac{y}{4} = 1.5 \implies y = 6\).
Thus, the molecular formula is \(\text{C}_3\text{H}_6\).

評分準則

1 mark for the correct option B. Method: Determine carbon content from the volume of absorbed CO2 and hydrogen content from the volume of reacted oxygen.

題目 27 · 選擇題

1 分

A 1.50 g sample of impure limestone (mainly calcium carbonate, \(\text{CaCO}_3\)) was dissolved in 50.0 cm\(^3\) of 0.500 mol dm\(^{-3}\) hydrochloric acid. The resulting mixture was filtered and the filtrate made up to 100.0 cm\(^3\) with distilled water in a volumetric flask. A 25.0 cm\(^3\) aliquot of this solution required 12.5 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium hydroxide solution for neutralisation. What is the percentage by mass of \(\text{CaCO}_3\) in the limestone? [Take \(M_r\) of \(\text{CaCO}_3\) as 100.1]

A.33.3%
B.50.0%
C.66.7%
D.83.3%

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解題

1. Calculate initial moles of \(\text{HCl}\) added:
\(\text{moles of HCl} = 0.0500\,\text{dm}^3 \times 0.500\,\text{mol}\,\text{dm}^{-3} = 0.0250\,\text{mol}\).

2. Calculate moles of \(\text{NaOH}\) used to neutralise excess \(\text{HCl}\) in a 25.0 cm\(^3\) aliquot:
\(\text{moles of NaOH} = 0.0125\,\text{dm}^3 \times 0.100\,\text{mol}\,\text{dm}^{-3} = 0.00125\,\text{mol}\).
Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the moles of excess \(\text{HCl}\) in 25.0 cm\(^3\) is 0.00125 mol.

3. Calculate total excess \(\text{HCl}\) in the 100.0 cm\(^3\) flask:
\(\text{total excess HCl} = 0.00125\,\text{mol} \times \frac{100.0}{25.0} = 0.00500\,\text{mol}\).

4. Calculate moles of \(\text{HCl}\) that reacted with \(\text{CaCO}_3\):
\(\text{moles of HCl reacted} = 0.0250 - 0.00500 = 0.0200\,\text{mol}\).

5. From the reaction \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\):
\(\text{moles of CaCO}_3 = \frac{0.0200}{2} = 0.0100\,\text{mol}\).

6. Calculate the mass and percentage of \(\text{CaCO}_3\):
\(\text{mass of CaCO}_3 = 0.0100\,\text{mol} \times 100.1\,\text{g}\,\text{mol}^{-1} = 1.001\,\text{g}\).
\(\text{percentage by mass} = \left(\frac{1.001}{1.50}\right) \times 100\% = 66.7\%\).

評分準則

1 mark for the correct option C. Method: Apply back-titration concepts: find total HCl, find unreacted HCl, find reacted HCl, and use the 1:2 stoichiometry to find the percentage purity.

題目 28 · 選擇題

1 分

A hypothetical element Y has only two naturally occurring isotopes: \(^{63}\text{Y}\) and \(^{65}\text{Y}\). The relative atomic mass of Y is 63.60. What is the percentage abundance of the heavier isotope?

A.20%
B.30%
C.60%
D.70%

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解題

Let the fractional abundance of \(^{65}\text{Y}\) be \(x\). Then the fractional abundance of \(^{63}\text{Y}\) is \(1 - x\).
The relative atomic mass equation is:
\(63.60 = 65(x) + 63(1 - x)\)
\(63.60 = 65x + 63 - 63x\)
\(2x = 0.60\)
\(x = 0.30\)
Therefore, the percentage abundance of the heavier isotope (\(^{65}\text{Y}\)) is 30%.

評分準則

1 mark for the correct option B. Method: Express relative atomic mass as a weighted average and solve for the unknown abundance.

題目 29 · 選擇題

1 分

How many total ions are present in 14.2 g of anhydrous sodium sulfate, \(\text{Na}_2\text{SO}_4\)? [Take the Avogadro constant, \(L\), as \(6.02 \times 10^{23}\,\text{mol}^{-1}\); \(M_r\) of \(\text{Na}_2\text{SO}_4 = 142.1\)]

A.\(6.02 \times 10^{22}\)
B.\(1.20 \times 10^{23}\)
C.\(1.81 \times 10^{23}\)
D.\(2.41 \times 10^{23}\)

查看答案詳解

解題

1. Calculate the number of moles of \(\text{Na}_2\text{SO}_4\):
\(\text{moles of Na}_2\text{SO}_4 = \frac{14.2}{142.1} \approx 0.100\,\text{mol}\).

2. Determine the number of ions per formula unit:
Each formula unit of \(\text{Na}_2\text{SO}_4\) dissociates into 3 ions: 2 \(\text{Na}^+\) ions and 1 \(\text{SO}_4^{2-}\) ion.

3. Calculate the total moles of ions:
\(\text{total moles of ions} = 0.100\,\text{mol} \times 3 = 0.300\,\text{mol}\).

4. Calculate the total number of ions:
\(\text{number of ions} = 0.300 \times 6.02 \times 10^{23} = 1.81 \times 10^{23}\).

評分準則

1 mark for the correct option C. Method: Compute moles of compound, multiply by the number of ions per formula unit (3), then multiply by the Avogadro constant.

題目 30 · 選擇題

1 分

An alkene with molecular formula \(\text{C}_6\text{H}_{12}\) is heated with a hot, concentrated, acidified solution of potassium manganate(VII). A ketone is the only organic product obtained from this reaction. What is the IUPAC name of the starting alkene?

A.2,3-dimethylbut-2-ene
B.hex-3-ene
C.2-methylpent-2-ene
D.3-methylpent-2-ene

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解題

When alkenes are heated with hot, concentrated, acidified \(\text{KMnO}_4\), the \(\text{C}=\text{C}\) double bond is cleaved.
- A \(=\text{CH}_2\) group oxidises to \(\text{CO}_2\).
- A \(=\text{CHR}\) group oxidises to a carboxylic acid.
- A \(=\text{CR}_2\) group oxidises to a ketone.
Since a ketone is the ONLY organic product obtained, the alkene must be completely symmetrical and have two alkyl substituents on each of the double-bonded carbon atoms (i.e. no hydrogen atoms attached directly to the double-bonded carbons).
Evaluating 2,3-dimethylbut-2-ene: \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\).
Cleavage of this \(\text{C}=\text{C}\) double bond yields two molecules of propanone (\(\text{CH}_3\text{COCH}_3\)), which is a ketone.
Therefore, the starting alkene is 2,3-dimethylbut-2-ene.

評分準則

1 mark for the correct option A. Method: Identify that a symmetrical tetrasubstituted alkene is required to produce a single ketone product upon strong oxidation.

題目 31 · 選擇題

1 分

Which transition metal complex can exist as a pair of optical isomers? [Let 'en' represent the bidentate ligand ethane-1,2-diamine]

A.\(\text{[Co(en)}_3\text{]}^{3+}\)
B.\(\textit{trans-}\text{[Co(en)}_2\text{Cl}_2\text{]}^+\)
C.\(\text{[Cu(NH}_3)_4\text{(H}_2\text{O)}_2\text{]}^{2+}\)
D.\(\text{[Pt(NH}_3)_2\text{Cl}_2\text{]}\)

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解題

Optical isomerism in octahedral complexes occurs when the complex lacks a plane of symmetry.
- \(\text{[Co(en)}_3\text{]}^{3+}\) is an octahedral complex with three bidentate ligands, which form a propeller-like structure. It has non-superimposable mirror images and thus exists as a pair of optical isomers.
- \(\textit{trans-}\text{[Co(en)}_2\text{Cl}_2\text{]}^+\) has a plane of symmetry (the Cl-Co-Cl axis and the equatorial plane containing the nitrogen atoms) and is therefore achiral.
- \(\text{[Cu(NH}_3)_4\text{(H}_2\text{O)}_2\text{]}^{2+}\) has a plane of symmetry and does not contain bidentate ligands that can cause optical isomerism in this arrangement.
- \(\text{[Pt(NH}_3)_2\text{Cl}_2\text{]}\) is square planar and has a plane of symmetry (the molecular plane).

評分準則

1 mark for the correct option A. Method: Identify that octahedral complexes containing three bidentate ligands lack a plane of symmetry and are chiral.

題目 32 · 選擇題

1 分

Which statement correctly describes the origin of color in transition metal complexes?

A.Light is emitted when electrons fall from higher-energy d-orbitals to lower-energy d-orbitals.
B.The d-orbitals split into two energy levels, and light of a specific frequency is absorbed to promote an electron to a higher d-orbital.
C.All five d-orbitals absorb light of different wavelengths, exciting electrons to the empty s-orbital.
D.Ligands lose electrons to the transition metal ion, which releases energy in the visible spectrum.

查看答案詳解

解題

In transition metal complexes, the presence of ligands causes the five d-orbitals to split into two groups of different energy levels. When light in the visible region passes through the complex, an electron from a lower d-orbital absorbs a photon of a specific frequency/energy and is promoted (excited) to a higher-energy d-orbital (d-d transition). The light that is transmitted (not absorbed) is seen as the complementary color.

評分準則

1 mark for the correct option B. Method: Distinguish between the absorption of a specific photon's energy (for electron promotion) and emission.

題目 33 · 選擇題

1 分

A gaseous hydrocarbon \( X \) has a volume of \( 10\text{ cm}^3 \). It is completely combusted in \( 100\text{ cm}^3 \) of oxygen (an excess). After cooling to room temperature, the remaining gas volume is \( 80\text{ cm}^3 \). On shaking this gas mixture with excess aqueous sodium hydroxide, the volume decreases to \( 50\text{ cm}^3 \).

What is the molecular formula of hydrocarbon \( X \)?

A.\(\text{C}_3\text{H}_4\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)

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解題

1. The reduction in volume on shaking with NaOH represents the volume of \(\text{CO}_2\) gas, which is absorbed: \( 80\text{ cm}^3 - 50\text{ cm}^3 = 30\text{ cm}^3 \).
2. Since \( 10\text{ cm}^3 \) of \(\text{C}_x\text{H}_y\) produces \( 30\text{ cm}^3 \) of \(\text{CO}_2\), the ratio is \( 1 : 3 \), which means \( x = 3 \).
3. The remaining \( 50\text{ cm}^3 \) of gas after NaOH treatment is unreacted \(\text{O}_2\).
4. The volume of \(\text{O}_2\) reacted is therefore \( 100\text{ cm}^3 - 50\text{ cm}^3 = 50\text{ cm}^3 \).
5. From the general combustion equation:
\(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\)
6. Ratio of hydrocarbon to reacted \(\text{O}_2\) is \( 10 : 50 = 1 : 5 \).
7. Therefore, \( x + \frac{y}{4} = 5 \).
8. Substituting \( x = 3 \) gives \( 3 + \frac{y}{4} = 5 \Rightarrow \frac{y}{4} = 2 \Rightarrow y = 8 \).
9. The molecular formula of \( X \) is \(\text{C}_3\text{H}_8\).

評分準則

Award 1 mark for selecting C.
Reject other options:
- A, B, and D do not satisfy the quantitative volumes of carbon dioxide produced and oxygen consumed during the reaction.

題目 34 · 選擇題

1 分

An oxide of chlorine contains \( 52.6\% \) chlorine by mass.

What is the empirical formula of this oxide?

\([A_r: \text{Cl} = 35.5, \text{O} = 16.0]\)

A.\(\text{Cl}_2\text{O}\)
B.\(\text{ClO}_2\)
C.\(\text{Cl}_2\text{O}_3\)
D.\(\text{Cl}_2\text{O}_7\)

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解題

1. Assume a \( 100\text{ g} \) sample of the oxide.
- Mass of \(\text{Cl} = 52.6\text{ g}\)
- Mass of \(\text{O} = 100\text{ g} - 52.6\text{ g} = 47.4\text{ g}\)
2. Calculate the moles of each element:
- Moles of \(\text{Cl} = \frac{52.6}{35.5} = 1.482\text{ mol}\)
- Moles of \(\text{O} = \frac{47.4}{16.0} = 2.963\text{ mol}\)
3. Determine the simplest whole-number ratio:
- \(\text{Cl} = \frac{1.482}{1.482} = 1\)
- \(\text{O} = \frac{2.963}{1.482} \approx 2\)
4. The empirical formula is \(\text{ClO}_2\).

評分準則

Award 1 mark for selecting B.
Reject A, C, and D as they represent different mass percentages of chlorine.

題目 35 · 選擇題

1 分

3-methylpent-2-ene reacts with hydrogen bromide, \(\text{HBr}\), to form a major organic product, \( Y \).

Which statement about \( Y \) is correct?

A.Y contains a chiral carbon atom and is 3-bromo-3-methylpentane.
B.Y does not contain a chiral carbon atom and is 3-bromo-3-methylpentane.
C.Y contains a chiral carbon atom and is 2-bromo-3-methylpentane.
D.Y does not contain a chiral carbon atom and is 2-bromo-3-methylpentane.

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解題

1. In the addition of \(\text{HBr}\) to 3-methylpent-2-ene (\(\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3\)), the reaction follows Markovnikov's rule.
2. The hydrogen atom adds to C2 (which has more hydrogen atoms attached) to form the more stable tertiary carbocation intermediate on C3.
3. The bromide ion then attacks the carbocation at C3 to yield 3-bromo-3-methylpentane as the major product.
4. Analyzing 3-bromo-3-methylpentane (\(\text{CH}_3\text{CH}_2\text{C}(\text{CH}_3)(\text{Br})\text{CH}_2\text{CH}_3\)), we see that the central carbon atom is bonded to a methyl group, a bromine atom, and two identical ethyl groups. Since it does not contain four different groups, it does not have a chiral carbon atom.

評分準則

Award 1 mark for selecting B.
Reject A, C, and D because the product is 3-bromo-3-methylpentane, which does not contain a chiral carbon atom due to the presence of two identical ethyl groups.

題目 36 · 選擇題

1 分

Which transition metal ion in the gaseous state has the same number of unpaired d-electrons as a gaseous \(\text{Fe}^{2+}\) ion?

A.\(\text{Cr}^{2+}\)
B.\(\text{Mn}^{2+}\)
C.\(\text{Co}^{2+}\)
D.\(\text{Cu}^{2+}\)

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解題

1. The electronic configuration of a gaseous iron atom is \([\text{Ar}]3\text{d}^6 4\text{s}^2\).
2. For the gaseous \(\text{Fe}^{2+}\) ion, two electrons are removed from the 4s orbital, leaving \([\text{Ar}]3\text{d}^6\).
3. In a gaseous ion, the five d-orbitals are degenerate. According to Hund's rule, 5 electrons occupy the five d-orbitals singly, and the 6th electron pairs up in one orbital, leaving 4 unpaired d-electrons.
4. Let us determine the number of unpaired electrons in the options:
- \(\text{Cr}^{2+}\): \([\text{Ar}]3\text{d}^4\) has 4 unpaired d-electrons.
- \(\text{Mn}^{2+}\): \([\text{Ar}]3\text{d}^5\) has 5 unpaired d-electrons.
- \(\text{Co}^{2+}\): \([\text{Ar}]3\text{d}^7\) has 3 unpaired d-electrons.
- \(\text{Cu}^{2+}\): \([\text{Ar}]3\text{d}^9\) has 1 unpaired d-electron.
5. Therefore, \(\text{Cr}^{2+}\) has the same number of unpaired d-electrons (4) as \(\text{Fe}^{2+}\).

評分準則

Award 1 mark for selecting A.
Reject B, C, and D because their d-orbital configuration results in 5, 3, and 1 unpaired electrons respectively, rather than 4.

題目 37 · 選擇題

1 分

What is the total number of ions present in \( 25.0\text{ cm}^3 \) of \( 0.0400\text{ mol dm}^{-3} \) aqueous aluminium sulfate, \(\text{Al}_2(\text{SO}_4)_3\)?

[Avogadro constant, \(L = 6.02 \times 10^{23}\text{ mol}^{-1}\)]

A.\( 6.02 \times 10^{20} \)
B.\( 1.20 \times 10^{21} \)
C.\( 1.81 \times 10^{21} \)
D.\( 3.01 \times 10^{21} \)

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解題

1. Calculate the number of moles of \(\text{Al}_2(\text{SO}_4)_3\):
\( n = c \times V = 0.0400\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.00100\text{ mol} \)
2. One formula unit of \(\text{Al}_2(\text{SO}_4)_3\) dissociates completely in water into 5 ions:
\(\text{Al}_2(\text{SO}_4)_3(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq})\)
3. Total moles of ions = \( 5 \times 0.00100\text{ mol} = 0.00500\text{ mol} \)
4. Total number of ions = \( 0.00500\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 3.01 \times 10^{21} \).

評分準則

Award 1 mark for selecting D.
Reject A: Calculated with only 1 ion per formula unit.
Reject B: Calculated with only 2 ions per formula unit.
Reject C: Calculated with only 3 ions per formula unit.

題目 38 · 選擇題

1 分

Anhydrous copper(II) nitrate decomposes on heating according to the equation shown.

\(2\text{Cu}(\text{NO}_3)_2(\text{s}) \rightarrow 2\text{CuO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\)

A \( 9.38\text{ g} \) sample of anhydrous copper(II) nitrate (\(M_r = 187.6\)) is heated until it is completely decomposed.

What is the total volume of gas produced, measured at room temperature and pressure (rtp)?

[Assume 1 mole of gas occupies \( 24.0\text{ dm}^3 \) at rtp.]

A.\( 1.20\text{ dm}^3 \)
B.\( 2.40\text{ dm}^3 \)
C.\( 3.00\text{ dm}^3 \)
D.\( 6.00\text{ dm}^3 \)

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解題

1. Find the moles of \(\text{Cu}(\text{NO}_3)_2\) heated:
\( n = \frac{9.38\text{ g}}{187.6\text{ g mol}^{-1}} = 0.0500\text{ mol} \)
2. Use the stoichiometric ratio from the equation:
2 moles of \(\text{Cu}(\text{NO}_3)_2\) produce \( 4\text{ mol of } \text{NO}_2\) and \( 1\text{ mol of } \text{O}_2\), making a total of 5 moles of gas.
3. Moles of gas produced = \( 0.0500\text{ mol} \times \frac{5}{2} = 0.125\text{ mol} \)
4. Calculate the gas volume at rtp:
\( V = 0.125\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 3.00\text{ dm}^3 \).

評分準則

Award 1 mark for selecting C.
Reject A: Obtained by incorrectly using a 2:1 ratio (only oxygen gas calculated).
Reject B: Obtained by incorrectly using a 2:4 ratio (only nitrogen dioxide gas calculated).
Reject D: Obtained by incorrectly multiplying the 2:5 ratio.

題目 39 · 選擇題

1 分

An alkene \( W \) reacts with bromine in the dark to form a single organic compound containing two chiral carbon atoms.

Which alkene could be \( W \)?

A.propene
B.but-1-ene
C.but-2-ene
D.2-methylbut-2-ene

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解題

1. Bromine addition to an alkene converts the double bond into a vicinal dibromoalkane.
2. Let's analyze the products of the addition of \(\text{Br}_2\) to each option:
- Propene + \(\text{Br}_2 \rightarrow\) 1,2-dibromopropane. Only C2 is chiral (1 chiral carbon).
- But-1-ene + \(\text{Br}_2 \rightarrow\) 1,2-dibromobutane. Only C2 is chiral (1 chiral carbon).
- But-2-ene + \(\text{Br}_2 \rightarrow\) 2,3-dibromobutane. Both C2 and C3 are chiral carbons (they each have four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{Br}\), and \(-\text{CH(Br)CH}_3\)). Thus, it has two chiral carbon atoms.
- 2-methylbut-2-ene + \(\text{Br}_2 \rightarrow\) 2,3-dibromo-2-methylbutane. C2 has two methyl groups (not chiral); only C3 is chiral (1 chiral carbon).

評分準則

Award 1 mark for selecting C.
Reject A, B, and D because their brominated products contain only a single chiral center.

題目 40 · 選擇題

1 分

An aqueous solution of cobalt(II) chloride is pink because it contains the octahedral complex ion \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\). When concentrated hydrochloric acid is added to this solution, the colour changes to blue due to the formation of a new complex ion.

What is the formula and the geometry of the blue complex ion formed?

A.\([\text{CoCl}_4]^{2-}\), tetrahedral
B.\([\text{CoCl}_4]^{2-}\), square planar
C.\([\text{CoCl}_6]^{4-}\), octahedral
D.\([\text{CoCl}_4]^{4-}\), tetrahedral

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解題

1. The ligand substitution of hexaaquacobalt(II) by excess chloride ions can be written as:
\([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)
2. The tetrachorocobaltate(II) ion, \([\text{CoCl}_4]^{2-}\), is formed.
3. Because chloride ligands are larger and carry a negative charge, a maximum of 4 chloride ligands can fit around the cobalt(II) ion to avoid steric hindrance and electrostatic repulsion. The resulting coordination number is 4, which adopts a tetrahedral geometry.

評分準則

Award 1 mark for selecting A.
Reject B: Square planar geometry is incorrect for this complex.
Reject C: Octahedral hexachlorocobaltate does not form under these conditions.
Reject D: The charge of the complex is \(2-\), not \(4-\).

卷二 (AS 結構題)

Answer all questions. Show your working and use appropriate units.

(a) (i) [2 marks] 1 mark for Co: 1s2 2s2 2p6 3s2 3p6 3d7 4s2; 1 mark for Co2+: 1s2 2s2 2p6 3s2 3p6 3d7. (ii) [1 mark] 1 mark for stating that it is a d-block element forming at least one stable ion with an incomplete d-subshell. (b) (i) [1 mark] 1 mark for [Co(H2O)6]2+. (ii) [2 marks] 1 mark for correct 3D octahedral representation (using wedges and dashes); 1 mark for coordinate arrows pointing from O of water to Co and showing the 2+ charge. (c) (i) [2 marks] 1 mark for formula [CoCl4]2-; 1 mark for balanced equation. (ii) [2 marks] 1 mark for identifying that chloride ions are larger than water molecules; 1 mark for explaining that steric hindrance / crowding prevents six chloride ions from coordinating. (d) [5 marks] 1 mark for stating that ligands cause d-orbitals to split into two different energy levels; 1 mark for explaining that electrons absorb light in the visible spectrum; 1 mark for stating that electrons are promoted from lower to higher energy d-orbitals (d-d transition); 1 mark for referencing the equation dE = h * v or dE = hc / lambda; 1 mark for explaining that the complementary colour (pink) is transmitted/seen.

Paper 3 (Advanced Practical Skills)

Perform both quantitative and qualitative tasks as described.

2 題目 · 40 分

題目 1 · practical

20 分

FA 1 is a solution prepared by dissolving \( 6.95\text{ g} \) of hydrated iron(II) sulfate, \( \text{FeSO}_4 \cdot x\text{H}_2\text{O} \), in dilute sulfuric acid and making up to \( 250.0\text{ cm}^3 \) with distilled water in a volumetric flask.

FA 2 is \( 0.0200\text{ mol dm}^{-3} \) potassium manganate(VII), \( \text{KMnO}_4 \).

In a series of titrations, \( 25.0\text{ cm}^3 \) portions of FA 1 were transferred into a conical flask using a pipette and titrated against FA 2.

The following results were obtained:
- Rough titration: Initial reading = \( 0.00\text{ cm}^3 \), Final reading = \( 25.60\text{ cm}^3 \)
- Titration 1: Initial reading = \( 0.00\text{ cm}^3 \), Final reading = \( 25.05\text{ cm}^3 \)
- Titration 2: Initial reading = \( 25.05\text{ cm}^3 \), Final reading = \( 50.00\text{ cm}^3 \)

(a) Complete the titration results table by calculating the titre volumes, select the appropriate titres to calculate the mean titre, and state your mean titre volume to 2 decimal places. (3 marks)

(b) Write a balanced ionic equation for the reaction between \( \text{MnO}_4^- \) and \( \text{Fe}^{2+} \) ions in acidic solution. (2 marks)

(c) Calculate the number of moles of manganate(VII) ions that reacted with the pipette volume of FA 1. (2 marks)

(d) Calculate the number of moles of \( \text{Fe}^{2+} \) ions present in \( 250.0\text{ cm}^3 \) of FA 1. (3 marks)

(e) Calculate the relative formula mass (\( M_r \)) of the hydrated iron(II) sulfate and find the integer value of \( x \). [\( A_r \): \( \text{Fe} = 55.8 \), \( \text{S} = 32.1 \), \( \text{O} = 16.0 \), \( \text{H} = 1.0 \)] (4 marks)

(f) The pipette used has an uncertainty of \( \pm 0.06\text{ cm}^3 \) and each burette reading has an uncertainty of \( \pm 0.05\text{ cm}^3 \).
(i) Calculate the percentage uncertainty in the pipette volume.
(ii) Calculate the percentage uncertainty in the titre volume obtained in Titration 1. (4 marks)

(g) Explain why excess dilute sulfuric acid is added to the flask before starting the titration, and why hydrochloric acid cannot be used instead. (2 marks)

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解題

(a)
- Rough titre: \( 25.60 - 0.00 = 25.60\text{ cm}^3 \)
- Titre 1: \( 25.05 - 0.00 = 25.05\text{ cm}^3 \)
- Titre 2: \( 50.00 - 25.05 = 24.95\text{ cm}^3 \)
Using Titres 1 and 2 (concordant within \( \pm 0.10\text{ cm}^3 \)):
Mean titre = \( \frac{25.05 + 24.95}{2} = 25.00\text{ cm}^3 \).

(b) Reaction equation:
\( \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)

(c) Moles of \( \text{MnO}_4^- \) in mean titre:
\( n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{25.00\text{ cm}^3}{1000} = 5.00 \times 10^{-4}\text{ mol} \)

(d)
Using the stoichiometry ratio from (b), \( n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-) \):
\( n(\text{Fe}^{2+}) \text{ in } 25.0\text{ cm}^3 = 5 \times (5.00 \times 10^{-4}\text{ mol}) = 2.50 \times 10^{-3}\text{ mol} \)
\( n(\text{Fe}^{2+}) \text{ in } 250.0\text{ cm}^3 = 2.50 \times 10^{-3}\text{ mol} \times 10 = 0.0250\text{ mol} \)

(e)
\( M_r \text{ of } \text{FeSO}_4 \cdot x\text{H}_2\text{O} = \frac{\text{mass}}{\text{moles}} = \frac{6.95\text{ g}}{0.0250\text{ mol}} = 278.0\text{ g mol}^{-1} \)
\( M_r \text{ of anhydrous } \text{FeSO}_4 = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1} \)
Mass of water of crystallisation = \( 278.0 - 151.9 = 126.1\text{ g mol}^{-1} \)
\( x = \frac{126.1}{18.0} = 7.01 \approx 7 \)

(f)
(i) Pipette percentage uncertainty = \( \frac{0.06}{25.0} \times 100\% = 0.24\% \)
(ii) A titre requires two burette readings, so the combined uncertainty is \( 2 \times 0.05 = \pm 0.10\text{ cm}^3 \).
Titre 1 percentage uncertainty = \( \frac{0.10}{25.05} \times 100\% = 0.40\% \)

(g) Dilute sulfuric acid is added to prevent the hydrolysis/oxidation of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) by air, and to provide the \( \text{H}^+ \) ions required for the reduction of manganate(VII). Hydrochloric acid cannot be used because chloride ions (\( \text{Cl}^- \)) are oxidized to chlorine gas (\( \text{Cl}_2 \)) by the strong oxidizing agent potassium manganate(VII), leading to an inaccurately high titre.

評分準則

- (a) 1 mark for calculating all three titres correctly.
- 1 mark for choosing concordant titres (1 and 2) and calculating the correct mean value.
- 1 mark for showing the mean titre to 2 decimal places with units.
- (b) 1 mark for correct species and oxidation states.
- 1 mark for fully balancing the equation including water and protons.
- (c) 1 mark for correct formula substitution.
- 1 mark for calculating \( 5.00 \times 10^{-4}\text{ mol} \) to 3 significant figures.
- (d) 1 mark for 1:5 ratio calculation.
- 1 mark for scaling up to \( 250.0\text{ cm}^3 \) (multiplying by 10).
- 1 mark for stating final answer to 3 significant figures.
- (e) 1 mark for calculating the formula mass as \( 278.0 \).
- 1 mark for calculating the formula mass of anhydrous \( \text{FeSO}_4 \) as \( 151.9 \).
- 1 mark for setting up the relation for the water content.
- 1 mark for identifying \( x = 7 \).
- (f) 1 mark for pipette uncertainty of \( 0.24\% \).
- 1 mark for recognising two readings are needed for a titre volume, so uncertainty = \( \pm 0.10\text{ cm}^3 \).
- 2 marks for calculating burette uncertainty of \( 0.40\% \).
- (g) 1 mark for stating acid prevents oxidation/hydrolysis of \( \text{Fe}^{2+} \) or provides necessary hydrogen ions.
- 1 mark for stating hydrochloric acid is oxidized to chlorine by potassium manganate(VII).

題目 2 · practical

20 分

You are provided with three aqueous solutions of transition metal salts: FA 3, FA 4, and FA 5. Perform the following tests systematically and complete the tasks.

**Test 1**: Add aqueous sodium hydroxide, \( \text{NaOH(aq)} \), dropwise and then in excess to separate portions of FA 3, FA 4, and FA 5 in test-tubes.
**Test 2**: Add aqueous ammonia, \( \text{NH}_3\text{(aq)} \), dropwise and then in excess to separate portions of FA 3, FA 4, and FA 5 in test-tubes.
**Test 3**: Add concentrated hydrochloric acid, \( \text{HCl(aq)} \), dropwise and then in excess to separate portions of FA 3, FA 4, and FA 5.
**Test 4**: Add aqueous barium nitrate, \( \text{Ba(NO}_3)_2\text{(aq)} \), followed by dilute nitric acid to separate portions of FA 4 and FA 5.

**Observations:**
- FA 3 is a pink solution. With dropwise \( \text{NaOH} \), a blue precipitate forms, which turns pink/brown on standing and is insoluble in excess. With dropwise \( \text{NH}_3 \), a blue precipitate forms, which dissolves in excess to give a yellow-brown solution. With concentrated \( \text{HCl} \), the pink solution turns deep blue.
- FA 4 is a pale blue solution. With dropwise \( \text{NaOH} \), a pale blue precipitate forms, which is insoluble in excess. With dropwise \( \text{NH}_3 \), a pale blue precipitate forms, which dissolves in excess to form a deep blue solution. With concentrated \( \text{HCl} \), the solution turns yellow-green. With \( \text{Ba(NO}_3)_2 \) followed by dilute nitric acid, a dense white precipitate forms.
- FA 5 is a yellow-brown solution. With both \( \text{NaOH} \) and \( \text{NH}_3 \), a red-brown precipitate forms, which is insoluble in excess. With \( \text{Ba(NO}_3)_2 \) followed by dilute nitric acid, a dense white precipitate forms.

(a) Identify the transition metal cation present in each solution based on the observations. Provide brief supporting evidence for each identification. (6 marks)

(b) Identify the anion present in solutions FA 4 and FA 5, and write the ionic equation (including state symbols) for the precipitation reaction observed in Test 4. (3 marks)

(c) Write balanced ionic equations for the precipitation reactions of:
(i) FA 4 with aqueous sodium hydroxide.
(ii) FA 5 with aqueous ammonia. (4 marks)

(d) Write the chemical formula of the transition metal species responsible for the color change of FA 3 when concentrated hydrochloric acid is added in excess, and name the type of reaction occurring. (3 marks)

(e) Write a balanced equation for the reaction that occurs when excess aqueous ammonia is added to FA 4, explaining the chemical change in terms of ligand exchange. (4 marks)

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解題

(a)
- **FA 3** contains the cobalt(II) ion, \( \text{Co}^{2+} \). Supporting evidence: The pink aqueous solution (containing \( [\text{Co(H}_2\text{O)}_6]^{2+} \)) forms a blue precipitate of cobalt(II) hydroxide with alkali, which turns pink/brown on standing. It dissolves in excess ammonia to form a straw-yellow/brown amine complex and turns blue in concentrated \( \text{HCl} \) as the tetrahedral \( [\text{CoCl}_4]^{2-} \) is formed.
- **FA 4** contains the copper(II) ion, \( \text{Cu}^{2+} \). Supporting evidence: The pale blue solution forms a pale blue precipitate with both NaOH and ammonia. The precipitate dissolves in excess ammonia to give a highly characteristic deep blue solution containing \( [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \).
- **FA 5** contains the iron(III) ion, \( \text{Fe}^{3+} \). Supporting evidence: The yellow-brown solution forms a red-brown precipitate of \( \text{Fe(OH)}_3 \) with both NaOH and ammonia, which is completely insoluble in excess.

(b)
- The anion present in both FA 4 and FA 5 is the sulfate ion, \( \text{SO}_4^{2-} \).
- The dense white precipitate formed is barium sulfate, which is insoluble in dilute nitric acid.
- Ionic equation: \( \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} \)

(c)
(i) Precipitation of FA 4 (\( \text{Cu}^{2+} \)) with NaOH:
\( \text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Cu(OH)}_2\text{(s)} \)
(ii) Precipitation of FA 5 (\( \text{Fe}^{3+} \)) with aqueous ammonia:
\( \text{Fe}^{3+}\text{(aq)} + 3\text{NH}_3\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow \text{Fe(OH)}_3\text{(s)} + 3\text{NH}_4^+\text{(aq)} \)

(d)
- The species responsible for the deep blue color is the tetrachorocobaltate(II) ion, \( [\text{CoCl}_4]^{2-} \).
- The reaction type is ligand exchange (or ligand substitution).

(e)
- Reaction equation:
\( [\text{Cu(H}_2\text{O)}_6]^{2+}\text{(aq)} + 4\text{NH}_3\text{(aq)} \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)} \)
- Explanation: Four of the neutral monodentate water ligands in the octahedral hexaaquacopper(II) complex are replaced by four neutral monodentate ammonia ligands because ammonia forms a more stable complex due to a higher stability constant (ammonia is a stronger coordinate ligand than water).

評分準則

- (a) 1 mark for each correct identity (Co2+, Cu2+, Fe3+) and 1 mark for each corresponding valid, correct supporting observation (3 + 3 = 6 marks total).
- (b) 1 mark for identifying sulfate, \( \text{SO}_4^{2-} \).
- 2 marks for the correct, balanced ionic equation including state symbols: \( \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} \).
- (c) (i) 2 marks for balanced precipitation equation of copper(II) hydroxide (including states).
- (ii) 2 marks for balanced precipitation equation of iron(III) hydroxide using aqueous ammonia (including states).
- (d) 2 marks for stating \( [\text{CoCl}_4]^{2-} \).
- 1 mark for identifying the reaction as ligand exchange/substitution.
- (e) 2 marks for the fully balanced equation: \( [\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \).
- 2 marks for explaining that four water ligands are substituted/replaced by four ammonia ligands because ammonia is a stronger ligand/forms a more stable complex.

Paper 4 (A Level 結構題)

Answer all questions. Write your answers in the spaces provided.

**(a)(i)** [1 mark]
- Correctly balanced equation with correct formulas and state symbols:
\([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\)

**(a)(ii)** [1 mark]
- Pale blue to deep blue / dark blue (reject blue to dark blue; must specify pale blue starting color).

**(b)** [2 marks]
- Chloride ion is larger / has larger ionic radius than water molecule [1]
- Greater steric hindrance / repulsion prevents 6 ligands fitting (or coordination number of 6) [1]

**(c)(i)** [1 mark]
- Equilibrium constant for the formation of a complex ion from its constituent ions or molecules in a solvent.

**(c)(ii)** [1 mark]
- Correct expression: \(K_{stab} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4}\) (reject if \([H_2O]\) is included in the expression).

**(d)(i)** [3 marks]
- Correct ratio expression linking both stability constants [1]:
\(\text{Ratio} = \frac{K_{stab1} \times [NH_3]^4}{K_{stab2} \times [Cl^-]^4}\)
- Correct substitution of values [1]:
\(\text{Ratio} = \frac{1.0 \times 10^{13} \times (0.050)^4}{4.0 \times 10^{5} \times (0.20)^4}\)
- Correct calculated final value of \(9.8 \times 10^4\) (or \(9.77 \times 10^4\)) [1]

**(d)(ii)** [2 marks]
- \([Cu(NH_3)_4(H_2O)_2]^{2+}\) is the more stable complex [1]
- Because its \(K_{stab}\) value is much larger, meaning the equilibrium position lies much further to the right / product side [1]

Paper 5 (Planning, Analysis and Evaluation)

Answer both planning and data analysis questions.

2 題目 · 30 分

題目 1 · planning-structured

15 分

A student wants to determine the formula of a colored complex formed between iron(III) ions, \(\text{Fe}^{3+}\), and a monodentate ligand, \(\text{L}^-\), using the method of continuous variation (Job's method). In this method, the total concentration of the metal ion plus the ligand is kept constant, but their relative proportions (mole fractions) are varied. The absorbance of each mixture is measured using a colorimeter.

The student is provided with:
- \(0.050\text{ mol dm}^{-3}\) aqueous iron(III) nitrate (acidified with dilute nitric acid)
- \(0.050\text{ mol dm}^{-3}\) aqueous sodium salt of the ligand, \(\text{NaL}\)

(a) State the independent variable and the dependent variable in this continuous variation experiment. [2]

(b) Describe a procedure to prepare ten different mixtures of \(\text{Fe}^{3+}\) and \(\text{L}^-\), each with a total volume of \(10.0\text{ cm}^3\), covering the entire range of mole fractions from 0.1 to 0.9. Name the specific volumetric apparatus you would use to measure the volumes. [3]

(c) Explain why:
(i) the total volume of each mixture must be kept constant at \(10.0\text{ cm}^3\). [1]
(ii) the concentrations of the two stock solutions must be identical (both \(0.050\text{ mol dm}^{-3}\)). [1]

(d) Explain how the student would use a colorimeter to measure the absorbance. State how the correct color filter is selected for the colorimeter. [2]

(e) The student obtains the following data from the colorimeter:

- Mixture 4: Volume of \(0.050\text{ mol dm}^{-3}\) \(\text{Fe}^{3+}\) = \(3.3\text{ cm}^3\); Volume of \(0.050\text{ mol dm}^{-3}\) \(\text{L}^-\) = \(6.7\text{ cm}^3\); Absorbance = 0.50.

(i) Calculate the mole fraction of \(\text{Fe}^{3+}\), \(x_{\text{Fe}}\), for Mixture 4. Show your working. [2]

(ii) Plotting a graph of absorbance against \(x_{\text{Fe}}\) yields two intersecting straight lines. The intersection of these lines represents the maximum absorbance, which occurs at \(x_{\text{Fe}} = 0.25\). Deduce the value of \(n\) in the complex \(\text{FeL}_n^{(3-n)+}\). [2]

(iii) Under what circumstances would this method show a significant deviation from two straight lines near the maximum absorbance? Suggest how the maximum absorbance would still be determined in this case. [2]

查看答案詳解

解題

(a) The independent variable is the mole fraction of \(\text{Fe}^{3+}\) (or the relative ratio of the volumes). The dependent variable is the absorbance.

(b) Use a burette or graduated pipette to measure the volumes. Set up ten test tubes and add systematic volumes: for tube 1, add \(1.0\text{ cm}^3\) of \(\text{Fe}^{3+}\) and \(9.0\text{ cm}^3\) of \(\text{L}^-\); for tube 2, \(2.0\text{ cm}^3\) of \(\text{Fe}^{3+}\) and \(8.0\text{ cm}^3\) of \(\text{L}^-\), continuing this pattern up to tube 9 containing \(9.0\text{ cm}^3\) of \(\text{Fe}^{3+}\) and \(1.0\text{ cm}^3\) of \(\text{L}^-\). This ensures that the total volume of each mixture remains constant at \(10.0\text{ cm}^3\).

(c) (i) Keeping the total volume constant ensures that the total concentration of metal ions plus ligands remains constant, so any change in absorbance is solely due to the changing ratio of reactants. (ii) Identical stock concentrations ensure that the mole fraction is directly proportional to the volume fraction, i.e., \(x_{\text{Fe}} = V_{\text{Fe}} / (V_{\text{Fe}} + V_{\text{L}})\).

(d) The colorimeter is first calibrated by placing a cuvette containing a blank (distilled water or stock \(\text{Fe}^{3+}\) solution) and setting the absorbance to 0.00. The filter chosen should be complementary to the color of the complex to ensure maximum sensitivity.

(e) (i) \(x_{\text{Fe}} = \frac{3.3}{3.3 + 6.7} = 0.33\).
(ii) At maximum absorbance, the stoichiometry of the complex matches the ratio of the reactants. Since \(x_{\text{Fe}} = 0.25\), the mole fraction of the ligand is \(1.00 - 0.25 = 0.75\). The ratio of \(\text{Fe}^{3+}\) to \(\text{L}^-\) is \(0.25 : 0.75 = 1 : 3\). Thus, \(n = 3\).
(iii) Deviation occurs if the complex has a low stability constant (\(K_{\text{stab}}\)), causing significant dissociation at equilibrium. The true maximum is determined by drawing best-fit straight lines from the outer points (where dissociation is suppressed by excess reagent) and extrapolating them to find the intersection point.

評分準則

1 mark: Correctly identifying independent variable (mole fraction of metal/volume ratio) AND dependent variable (absorbance).
1 mark: Use of a burette or graduated pipette for measuring volumes.
1 mark: Systematically varying volumes of \(\text{Fe}^{3+}\) and \(\text{L}^-\) to cover the range (e.g., 1.0 to 9.0 \(\text{cm}^3\)).
1 mark: Specifying that the total volume must sum to \(10.0\text{ cm}^3\) for all mixtures.
1 mark: Explaining that constant total volume maintains constant total concentration of reactants.
1 mark: Explaining that identical stock concentrations make the mole fraction directly proportional to the volume ratio.
1 mark: Describing calibration of colorimeter using a reference blank (e.g., distilled water).
1 mark: Selecting a complementary color filter to maximize absorbance/sensitivity.
1 mark: Calculation of mole fraction \(x_{\text{Fe}} = 0.33\) for Mixture 4.
1 mark: Showing working for \(x_{\text{Fe}}\).
1 mark: Explaining that at max absorbance, the mole ratio of metal to ligand is equal to the stoichiometry of the complex.
1 mark: Correctly deducing \(n = 3\).
1 mark: Linking the deviation to a low stability constant / significant dissociation at equilibrium.
1 mark: Describing extrapolation of the linear portions of the graph to find the intersection.

題目 2 · planning-structured

15 分

An unknown liquid alkene, \(X\), has the molecular formula \(\text{C}_n\text{H}_{2n}\). A student plans to determine the relative molecular mass, \(M_r\), of \(X\) by reacting a known mass of \(X\) with an excess of bromine dissolved in cyclohexane. The unreacted bromine is then determined by a back titration involving potassium iodide and standard sodium thiosulfate solution.

(a) Describe how the student would prepare exactly \(100\text{ cm}^3\) of a solution containing approximately \(1.5\text{ g}\) of alkene \(X\) in cyclohexane. Your answer should include:
- the name of the volumetric vessel used to prepare the final solution [1]
- a description of how the mass of the liquid alkene is measured accurately [2]

(b) A \(10.0\text{ cm}^3\) sample of the alkene solution is transferred to a conical flask, and \(25.0\text{ cm}^3\) of \(0.0500\text{ mol dm}^{-3}\) bromine in cyclohexane (an excess) is added.
(i) Explain why the reaction flask must be stoppered and kept in the dark until the reaction is complete. [2]
(ii) State the change that would be observed in the color of the solution in the flask as the reaction proceeds. [1]

(c) Once the reaction is complete, excess aqueous potassium iodide, \(\text{KI}(\text{aq})\), is added to the flask. This reacts with the remaining unreacted bromine to liberate iodine, \(\text{I}_2\).
(i) Write ionic equations for the reaction of \(\text{Br}_2\) with \(\text{I}^-\), and the subsequent titration of \(\text{I}_2\) with thiosulfate ions, \(\text{S}_2\text{O}_3^{2-}\). [2]
(ii) Name a suitable indicator for the titration and state the color change at the end-point. [2]

(d) In a practical trial, \(1.42\text{ g}\) of alkene \(X\) was dissolved to make \(100.0\text{ cm}^3\) of solution.
- \(10.0\text{ cm}^3\) of this solution was reacted with \(25.0\text{ cm}^3\) of \(0.0500\text{ mol dm}^{-3}\) bromine.
- The unreacted bromine required \(18.40\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\) for complete reaction with the liberated iodine.

(i) Calculate the number of moles of \(\text{Br}_2\) initially added to the \(10.0\text{ cm}^3\) sample of the alkene. [1]
(ii) Calculate the number of moles of unreacted \(\text{Br}_2\) present in the mixture after the reaction. [2]
(iii) Calculate the relative molecular mass, \(M_r\), of the alkene \(X\). Show all steps in your working. [2]

查看答案詳解

解題

(a) The volumetric vessel used to prepare the solution is a \(100\text{ cm}^3\) volumetric flask. To weigh the alkene accurately, use a weighing bottle on a 2-decimal or 3-decimal place balance. Weigh the bottle with the liquid alkene, pour the liquid into a beaker containing some cyclohexane, and then reweigh the empty bottle to determine the mass transferred by difference. Transfer the mixture to the volumetric flask, wash the beaker with cyclohexane, transfer the washings, and make up to the line with cyclohexane, followed by thorough mixing.

(b) (i) The flask must be stoppered to prevent the evaporation of volatile bromine and cyclohexane. It must be kept in the dark to prevent light-catalysed free radical substitution of cyclohexane (the solvent) or the alkyl chains of the alkene by bromine. (ii) The dark orange/brown color of bromine becomes paler (lighter orange or yellow).

(c) (i) \(\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2\)
\(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\)
(ii) Starch indicator is used. The color change at the end-point is from blue-black to colorless.

(d) (i) Moles of \(\text{Br}_2 = 0.0250\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 1.25 \times 10^{-3}\text{ mol}\).
(ii) Moles of \(\text{S}_2\text{O}_3^{2-} = 0.01840\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.84 \times 10^{-3}\text{ mol}\).
Since \(1\text{ mol of Br}_2\) produces \(1\text{ mol of I}_2\), which reacts with \(2\text{ mol of S}_2\text{O}_3^{2-}\), the mole ratio of unreacted \(\text{Br}_2\) to \(\text{S}_2\text{O}_3^{2-}\) is 1:2.
Moles of unreacted \(\text{Br}_2 = \frac{1}{2} \times 1.84 \times 10^{-3} = 9.20 \times 10^{-4}\text{ mol}\).
(iii) Moles of \(\text{Br}_2\) reacted with alkene = \(1.25 \times 10^{-3} - 9.20 \times 10^{-4} = 3.30 \times 10^{-4}\text{ mol}\).
Since 1 mole of alkene reacts with 1 mole of \(\text{Br}_2\), moles of alkene in \(10.0\text{ cm}^3\) of solution = \(3.30 \times 10^{-4}\text{ mol}\).
Moles of alkene in \(100.0\text{ cm}^3\) of stock solution = \(3.30 \times 10^{-3}\text{ mol}\).
\(M_r\) of alkene \(X = \frac{\text{mass}}{\text{moles}} = \frac{1.42\text{ g}}{3.30 \times 10^{-3}\text{ mol}} = 430\) (or 430.3).

評分準則

1 mark: Name of vessel: \(100\text{ cm}^3\) volumetric flask.
1 mark: Weighing the liquid alkene by difference using a weighing bottle on a 2/3 decimal place balance.
1 mark: Quantitative transfer: rinse beaker/funnel with cyclohexane and transfer washings to the volumetric flask before filling to the mark and mixing.
1 mark: Stoppering to prevent loss of volatile reactants/solvent (bromine/cyclohexane).
1 mark: Keeping in the dark to prevent photochemical/light-induced free-radical substitution reactions.
1 mark: Color change: Orange/brown to yellow/paler orange.
1 mark: Correct equation: \(\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2\).
1 mark: Correct equation: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\).
1 mark: Indicator: Starch.
1 mark: End-point color change: blue-black to colorless (reject: clear).
1 mark: Initial moles of \(\text{Br}_2 = 1.25 \times 10^{-3}\text{ mol}\).
1 mark: Deducing 1:2 stoichiometry between unreacted \(\text{Br}_2\) and thiosulfate.
1 mark: Correctly calculating unreacted \(\text{Br}_2 = 9.20 \times 10^{-4}\text{ mol}\).
1 mark: Calculating moles of reacted \(\text{Br}_2\) (and thus alkene in \(10.0\text{ cm}^3\)) as \(3.30 \times 10^{-4}\text{ mol}\) AND scaling to \(100\text{ cm}^3\) (\(3.30 \times 10^{-3}\text{ mol}\)).
1 mark: Correctly calculating the final \(M_r = 430\) (or 430.3) to 3 significant figures.

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