📚 Year 12 Cambridge Chemistry Unit Test Mock Paper Walkthrough | 剑桥化学Year 12单元测试模拟卷解析
This article provides a step-by-step walkthrough of a mock unit test designed for Year 12 Cambridge AS Level Chemistry. The paper targets core topics such as atomic structure, bonding, stoichiometry, energetics, redox, equilibrium and an introduction to organic chemistry. Each section takes one representative question, unpacks the underlying concept, and demonstrates the logical pathway to the correct answer. The dual-language format helps you consolidate subject terminology and reasoning skills in both English and Chinese.
本文针对剑桥AS化学Year 12单元测试模拟卷进行逐题解析。试卷覆盖原子结构、化学键、化学计量、能量变化、氧化还原、化学平衡以及有机化学入门等核心主题。每一节选取一道典型题，梳理概念，展示解题思路。中英双语对照的讲解，有助于你同时巩固学科术语和逻辑推理能力。
1. Electron Configuration of Ions | 离子电子排布
The question asked for the full electron configuration of the sulfide ion, S²⁻. A neutral sulfur atom has 16 electrons. Gaining two electrons to form the anion increases the total count to 18. The configuration is therefore 1s² 2s² 2p⁶ 3s² 3p⁶, which is isoelectronic with argon. Students often forget to add the extra electrons, writing the neutral atom arrangement instead.
题目要求写出硫离子 S²⁻ 的完整电子排布。中性硫原子有16个电子，得到两个电子形成阴离子后，电子总数变为18。因此排布为 1s² 2s² 2p⁶ 3s² 3p⁶，与氩原子等电子。常见错误是忘记增加电子，仍按中性原子排布作答。
A quick check using the periodic table confirms that sulfur is in Group 16; gaining two electrons gives the stable noble gas configuration. Remember that for anions, the extra electrons occupy the lowest available energy sub-levels, so 3p fills to 3p⁶ without entering 3d.
借助周期表快速检验：硫位于第16族，得到两个电子即获得稳定的惰性气体电子构型。注意，对于阴离子，新增电子首先填入能量最低的空轨道，因此填满 3p 至 3p⁶，而不会进入 3d。
2. Mole Calculations and Empirical Formula | 摩尔计算与经验式
alculate moles of carbon from CO₂ (mass of CO₂ ÷ 44.0) and moles of hydrogen from H₂O (mass of H₂O ÷ 18.0 × 2). Then convert the mole ratio to the simplest whole-number ratio.
该题给出某碳氢化合物的质量及完全燃烧后生成的 CO₂ 和 H₂O 的质量。求经验式时，首先由 CO₂ 质量计算碳的物质的量（CO₂质量 ÷ 44.0），再由 H₂O 质量计算氢的物质的量（H₂O质量 ÷ 18.0 × 2）。然后将物质的量之比化为最简整数比。
Suppose the data gave 0.0400 mol C and 0.100 mol H. Dividing by the smaller value (0.0400) yields C₁H₂.₅. Multiplying by 2 gives the empirical formula C₂H₅. Common pitfalls include mishandling the hydrogen factor or rounding too early.
假设数据求得碳 0.0400 mol，氢 0.100 mol，除以较小值 0.0400 得 C₁H₂.₅，乘以 2 后得到经验式 C₂H₅。常见失误包括忘记氢原子数需乘以2，或过早取整导致比例错误。
3. First Ionisation Energy Trends Across Period 3 | 第三周期第一电离能趋势
The question presented a sketch of first ionisation energies for Period 3 elements and asked to explain the general increase, as well as the dips at aluminium and sulfur. The general rise from Na to Ar is driven by increasing nuclear charge while electrons are added to the same principal quantum shell, resulting in stronger attraction and a smaller atomic radius.
题目给出第三周期元素第一电离能的示意图，要求解释整体上升趋势以及铝和硫处出现的下降。从钠到氩，核电荷递增，而电子填入同一主量子壳层，吸引增强、原子半径减小，因此第一电离能总体呈上升趋势。
The drop at aluminium occurs because the 3p electron removed experiences slightly less effective nuclear charge and is at a higher energy level than the 3s electron removed in magnesium. Sulfur's dip arises from electron-electron repulsion in the doubly occupied 3p orbital; removing one electron from this paired arrangement requires less energy than expected from the general trend.
铝处下降的原因是移走的 3p 电子感受到的有效核电荷较镁的 3s 电子略低，且能级更高。硫处的下降则是由于 3p 轨道中出现电子成对，电子间斥力使得移走一个成对电子所需能量低于趋势预期。
4. Shapes of Molecules and Polarity | 分子形状与极性
The exam item asked for the shape, bond angle and polarity of NH₃ and BF₃. Ammonia has 3 bond pairs and 1 lone pair on nitrogen; electron-pair geometry is tetrahedral but molecular shape is trigonal pyramidal with a bond angle of approximately 107°. The lone pair compresses the H–N–H angle below the ideal tetrahedral 109.5°.
此题要求写出 NH₃ 和 BF₃ 的空间构型、键角与极性。氨分子中氮原子有3对成键电子和1对孤电子对，电子对排布为四面体，但分子形状为三角锥形，键角约 107°。孤电子对使 H–N–H 键角被压缩至四面体理想角 109.5° 以下。
Boron trifluoride has 3 bond pairs and no lone pairs on the central boron, giving a trigonal planar shape with 120° bond angles. The molecule is non-polar because the individual B–F dipoles cancel in the symmetrical planar arrangement. By contrast, NH₃ is polar; the dipole moment from the lone pair and N–H bonds does not cancel, resulting in a net dipole pointing towards the nitrogen.
三氟化硼中心硼原子有3对成键电子，无孤电子对，构型为平面三角形，键角 120°。分子对称，B–F 键的极性相互抵消，故为非极性分子。而 NH₃ 中孤电子对与 N–H 键的极性无法抵消，净偶极矩指向氮原子，因此为极性分子。
5. Enthalpy Change Using Hess's Law | 用赫斯定律求焓变
In this part, standard enthalpies of formation were given: ΔHf° [CO₂(g)] = –394 kJ mol⁻¹, ΔHf° [H₂O(l)] = –286 kJ mol⁻¹ and ΔHf° [C₂H₅OH(l)] = – ¹. The task was to calculate the standard enthalpy of combustion of ethanol. Applying Hess's Law: ΔHc° = Σ ΔHf°(products) – Σ ΔHf°(reactants), taking care with stoichiometric coefficients.
该题给出了标准生成焓数据：ΔHf°[CO₂(g)] = –394 kJ mol⁻¹，ΔHf°[H₂O(l)] = –286 kJ mol⁻¹，ΔHf°[C₂H₅OH(l)] = –278 kJ mol⁻¹。要求计算乙醇的标准燃烧焓。运用赫斯定律：ΔHc° = Σ ΔHf°(生成物) – Σ ΔHf°(反应物)，注意配平系数。
Combustion equation: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). Therefore ΔHc° = [2(–394) + 3(–286)] – [–278 + 3×0] = –788 – 858 + 278 = – combustion. A frequent mistake is forgetting that the enthalpy of formation of O₂ is zero.
燃烧方程式：C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)。代入 ΔHc° = [2×(–394) + 3×(–286)] – [–278 + 3×0] = –788 – 858 + 278 = –1368 kJ mol⁻¹。数值为负且较大，反映了醇类燃烧的放热特点。常见错误是忽略 O₂ 的标准生成焓为零。
6. Balancing Redox Equations Under Acidic Conditions | 酸性条件下氧化还原方程式的配平
duction of manganate(VII) is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; oxidation of iron(II) is Fe²⁺ → Fe³⁺ + e⁻.
题干给出骨架方程式 MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺，要求在酸性条件下配平。首先写出半反应：高锰酸根还原为 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O；铁(II)氧化为 Fe²⁺ → Fe³⁺ + e⁻。
obtain the final balanced equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. The state symbols (aq) are expected for all ions in solution.
将氧化半反应乘以5以平衡电子，然后相加：MnO₄⁻ + 8H⁺ + 5e⁻ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ + 5e⁻。消去电子即得配平方程式：MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺。所有溶液中的离子均需标注 (aq) 状态符号。
7. Equilibrium Constant Kc Calculation | 平衡常数 Kc 计算
A typical equilibrium problem: 0.50 mol of ethanoic acid and 0.50 mol of ethanol were mixed, reaching equilibrium with 0.30 mol of ethyl ethanoate. The volume was 1.0 dm³. From the equation CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, the change table shows equilibrium amounts: acid 0.20 mol, alcohol 0.20 mol, ester 0.30 mol, water 0.30 mol.
一个典型的平衡计算题：0.50 mol 乙酸与 0.50 mol 乙醇混合，达到平衡时生成 0.30 mol 乙酸乙酯，容器体积为 1.0 dm³。根据方程式 CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O，通过变化表得到平衡时各物质的量：酸 0.20 mol、醇 0.20 mol、酯 0.30 mol、水 0.30 mol。
Substituting: (0.30 × 0.30) / (0.20 × 0.20) = 0.090 / 0.040 = 2.25. The units cancel, so Kc is dimensionless here. Always check whether the number of moles on each side affects the unit.
因体积为 1.0 dm³，平衡浓度数值与物质的量相等。Kc 表达式为 [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]。代入得 (0.30 × 0.30) / (0.20 × 0.20) = 0.090 / 0.040 = 2.25。反应前后分子总数相等，单位约去，因此 Kc 无量纲。每次均需检验方程式两侧摩尔数变化对单位的影响。
8. Free Radical Substitution Mechanism of Methane Chlorination | 甲烷氯代自由基取代机理
t. The process proceeds via three stages: initiation, propagation and termination. Initiation: Cl₂ undergoes homolytic fission, Cl–Cl bond breaks to form two chlorine radicals, Cl•.
最后一题要求描述紫外线照射下 CH₄ + Cl₂ → CH₃Cl + HCl 的反应机理。过程分为链引发、链增长和链终止三个阶段。引发：Cl₂ 发生均裂，Cl–Cl 键断裂生成两个氯自由基 Cl•。
Propagation: Cl• abstracts a hydrogen atom from CH₄ to produce HCl and a methyl radical •CH₃. The methyl radical then attacks a Cl₂ molecule, forming CH₃Cl and regenerating a Cl• radical. These two steps repeat many times, sustaining the chain reaction.
链增长：Cl• 从 CH₄ 中夺取一个氢原子，生成 HCl 和甲基自由基 •CH₃。随后 •CH₃ 进攻 Cl₂ 分子，生成 CH₃Cl 并再生一个 Cl• 自由基。这两步重复循环，使链反应持续进行。
Termination occurs when two radicals collide, for example Cl• + Cl• → Cl₂, •CH₃ + Cl• → CH₃Cl, or •CH₃ + • ) to show single-electron movement and for writing the correct structural representations.
链终止发生在两个自由基碰撞结合时，例如 Cl• + Cl• → Cl₂，•CH₃ + Cl• → CH₃Cl，或 •CH₃ + •CH₃ → C₂H₆。痕量乙烷的存在正是自由基机理的证据。考试中需要用弯箭头表示单电子转移，并写出正确结构式，才能获得完整分数。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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