A-Level OCR Chemistry: Detailed Walkthroughs of Typical Questions | A-Level OCR 化学：典型例题详解

📚 A-Level OCR Chemistry: Detailed Walkthroughs of Typical Questions | A-Level OCR 化学：典型例题详解

Mastering A-Level OCR Chemistry requires not only understanding theoretical concepts but also the ability to apply them to exam-style questions. This article presents a curated selection of typical questions across the specification, from atomic structure and bonding to energetics, kinetics, equilibrium, redox, transition metals, organic mechanisms, and spectroscopy. Each question is followed by a step-by-step reasoning and model answer, allowing you to see how marks are earned and how to structure your responses effectively.

掌握 A-Level OCR 化学不仅需要理解理论概念，还需要能够将其应用于考试题型。本文精选了涵盖原子结构、化学键合、能量学、动力学、平衡、氧化还原、过渡金属、有机机理和波谱分析等各章节的典型问题。每道题均配有分步推理和标准答案，让你明白如何得分以及如何有效组织答案。

1. Electron Configuration & Ionisation Energy | 电子构型与电离能

Question: Write the full electron configuration of a Cr atom (Z=24) and a Cr³⁺ ion. Explain why the first ionisation energy of chromium is lower than that of the preceding element, vanadium.

问题：写出 Cr 原子（Z=24）和 Cr³⁺ 离子的完整电子构型。解释为什么铬的第一电离能低于其前面的元素钒。

Solution: Cr atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹. Cr³⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ (the 4s electrons are removed first). The first ionisation energy of Cr is lower than that of V because Cr has an electron in the 4s orbital that is further from the nucleus and experiences greater shielding than the 3d electron removed from V. Additionally, the half-filled 3d subshell in Cr provides extra stability, but the electron removed is from the 4s, which requires less energy than removing a 3d electron from V (where configuration is [Ar] 3d³ 4s²).

解析：Cr 原子：1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹。Cr³⁺ 离子：1s² 2s² 2p⁶ 3s² 3p⁶ 3d³（先失去 4s 电子）。Cr 的第一电离能低于 V 是因为 Cr 失去的是 4s 电子，该电子离核更远、屏蔽效应更强，所需能量较 V 失去的 3d 电子少。V 的电子构型为 [Ar] 3d³ 4s²，失去的是 3d 电子，核吸引力更强。

2. Ionic and Covalent Bonding | 离子键与共价键

Question: Magnesium oxide has a much higher melting point than silicon dioxide. Explain this difference in terms of bonding and structure.

问题：氧化镁的熔点远高于二氧化硅。从键合和结构角度解释这一差异。

Solution: MgO is an ionic compound with a giant ionic lattice. The ions Mg²⁺ and O²⁻ are held together by strong electrostatic attractions. A large amount of energy is required to overcome these forces. SiO₂ is a covalent network solid (macromolecule) in which each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement. Although both are giant structures, the ionic bonds in MgO involve full charges (+2 and –2) leading to stronger attractions than the polar covalent bonds in SiO₂, resulting in a higher melting point for MgO (2852 °C vs 1713 °C).

解析：MgO 是离子化合物，形成巨型离子晶格，Mg²⁺ 和 O²⁻ 之间通过强大的静电引力结合，需要大量能量才能克服。SiO₂ 是共价网络固体（大分子），每个硅原子与四个氧原子以共价键形成四面体结构。尽管两者均为巨型结构，但 MgO 中的离子键涉及 ±2 的完整电荷，吸引力比 SiO₂ 中的极性共价键更强，因此 MgO 的熔点更高（2852 °C vs 1713 °C）。

3. Shapes of Molecules | 分子形状

Question: Predict the shape and bond angle of the PF₅ molecule and the ClF₃ molecule. Explain your reasoning using VSEPR theory.

问题：预测 PF₅ 分子和 ClF₃ 分子的形状与键角。用 VSEPR 理论解释你的推理。

Solution: PF₅: 5 bonding pairs, 0 lone pairs → trigonal bipyramidal shape. Bond angles: 90° (between axial and equatorial) and 120° (between equatorial groups). ClF₃: central chlorine has 5 electron pairs (3 bonding pairs to F and 2 lone pairs). The lone pairs occupy equatorial positions to minimise repulsion; shape is T-shaped. Bond angles slightly less than 90° due to lone pair–bonding pair repulsion.

解析：PF₅：5 个键对，0 个孤对 → 三角双锥形。键角：90°（轴向与赤道之间）和 120°（赤道间）。ClF₃：中心氯原子有 5 个电子对（3 个键对连 F，2 个孤对）。孤对占据赤道位置以降低排斥，形状为 T 形。键角因孤对-键对排斥略小于 90°。

4. Enthalpy Changes | 焓变

Question: The combustion of 0.50 g of ethanol (C₂H₅OH) raised the temperature of 200 g of water by 13.2 °C. Calculate the enthalpy change of combustion of ethanol in kJ mol⁻¹. (Specific heat capacity of water = 4.18 J g⁻¹ K⁻¹)

问题：0.50 g 乙醇（C₂H₅OH）燃烧使 200 g 水的温度升高 13.2 °C。计算乙醇的燃烧焓变（kJ mol⁻¹）。（水的比热容 = 4.18 J g⁻¹ K⁻¹）

Solution: Heat absorbed by water, q = mcΔT = 200 g × 4.18 J g⁻¹ K⁻¹ × 13.2 K = 11035.2 J = 11.035 kJ. Moles of ethanol = mass / Mᵣ = 0.50 g / 46.0 g mol⁻¹ = 0.01087 mol. ΔcH = – q / n = – 11.035 kJ / 0.01087 mol = – 1015 kJ mol⁻¹ (approx – 1000 kJ mol⁻¹ to 3 sf). Negative sign indicates exothermic reaction.

解析：水吸收的热量 q = mcΔT = 200 × 4.18 × 13.2 = 11035.2 J = 11.035 kJ。乙醇物质的量 = 0.50 / 46.0 = 0.01087 mol。燃烧焓 ΔcH = – q / n = – 11.035 / 0.01087 ≈ –1015 kJ mol⁻¹（保留三位有效数字约 –1000 kJ mol⁻¹）。负号表示放热。

5. Born-Haber Cycle | 玻恩-哈伯循环

Question: Use the data below to construct a Born-Haber cycle and calculate the lattice enthalpy of calcium chloride, CaCl₂. Data (kJ mol⁻¹): ΔH⦵ₐₜₐ Ca(s) → Ca(g) = +178; 1st IE Ca = +590; 2nd IE Ca = +1145; bond energy Cl–Cl = +243; electron affinity of Cl = –348; enthalpy of formation of CaCl₂(s) = –795.

问题：利用以下数据构建 Born-Haber 循环，计算氯化钙 CaCl₂ 的晶格能。数据（kJ mol⁻¹）：Ca(s) → Ca(g) ΔH⦵ₐₜₐ = +178；第一电离能 Ca = +590；第二电离能 Ca = +1145；Cl–Cl 键能 = +243；Cl 的电子亲合能 = –348；CaCl₂(s) 的生成焓 = –795。

Solution: Born-Haber path: Ca(s) → Ca(g) = +178; Ca(g) → Ca²⁺(g) + 2e⁻ = +590+1145 = +1735; Cl₂(g) → 2Cl(g) = +243; 2Cl(g) + 2e⁻ → 2Cl⁻(g) = 2 × (–348) = –696. Sum of steps to form gaseous ions = +178 +1735 +243 –696 = +1460 kJ mol⁻¹. Lattice enthalpy ΔH⦵ₗₑ = ΔH⦵ₙ – sum = –795 – (+1460) = –2255 kJ mol⁻¹.

解析：路径：Ca(s) → Ca(g) = +178；Ca(g) → Ca²⁺(g) + 2e⁻ = +1735；Cl₂(g) → 2Cl(g) = +243；2Cl(g) + 2e⁻ → 2Cl⁻(g) = –696。气态离子生成步总计 = +1460 kJ mol⁻¹。晶格能 ΔH⦵ₗₑ = 生成焓 – 总计 = –795 – (+1460) = –2255 kJ mol⁻¹。

6. Rates of Reaction | 反应速率

Question: The reaction 2A + B → C was studied by measuring the initial rate. Results: Expt 1: [A]=0.10, [B]=0.10, rate = 4.0×10⁻⁴; Expt 2: [A]=0.20, [B]=0.10, rate = 1.6×10⁻³; Expt 3: [A]=0.10, [B]=0.20, rate = 8.0×10⁻⁴. Deduce the rate equation and calculate the rate constant with units.

问题：研究了反应 2A + B → C 的初速率。结果：实验1 [A]=0.10, [B]=0.10, 速率 4.0×10⁻⁴；实验2 [A]=0.20, [B]=0.10, 速率 1.6×10⁻³；实验3 [A]=0.10, [B]=0.20, 速率 8.0×10⁻⁴。推导速率方程并计算速率常数及单位。

Solution: Compare Expt 1 and 2: [A] doubles, [B] constant → rate ×4, so order wrt A = 2. Compare Expt 1 and 3: [B] doubles, [A] constant → rate ×2, so order wrt B = 1. Rate = k[A]²[B]. Using Expt 1: k = rate / ([A]²[B]) = 4.0×10⁻⁴ / (0.10² × 0.10) = 4.0×10⁻⁴ / 1.0×10⁻³ = 0.40 dm⁶ mol⁻² s⁻¹.

解析：比较实验1和2：[A] 加倍，[B] 不变 → 速率 ×4，对 A 为二级。实验1和3：[B] 加倍，[A] 不变 → 速率 ×2，对 B 为一级。速率方程：速率 = k[A]²[B]。代入实验1：k = 4.0×10⁻⁴ / (0.10² × 0.10) = 0.40 dm⁶ mol⁻² s⁻¹。

7. Equilibrium Constant Kc | 平衡常数 Kc

Question: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g) at 700 K, 0.50 mol of H₂ and 0.50 mol of I₂ were mixed in a 2.0 dm³ container. At equilibrium, 0.80 mol of HI was formed. Calculate Kc.

问题：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，在 700 K 下，将 0.50 mol H₂ 和 0.50 mol I₂ 混合于 2.0 dm³ 容器中。平衡时生成 0.80 mol HI。计算 Kc。

Solution: Moles at eq: HI = 0.80; H₂ reacted = 0.80/2 = 0.40; so H₂ eq = 0.50 – 0.40 = 0.10 mol; I₂ eq = 0.10 mol. Concentrations: [HI] = 0.80/2.0 = 0.40 mol dm⁻³; [H₂] = [I₂] = 0.10/2.0 = 0.050 mol dm⁻³. Kc = [HI]² / ([H₂][I₂]) = (0.40)² / (0.050 × 0.050) = 0.16 / 0.0025 = 64 (no units, as Δn = 0).

解析：平衡物质的量：HI = 0.80；反应的 H₂ = 0.80/2 = 0.40，故 H₂ 剩余 = 0.10 mol，I₂ 同。浓度：[HI] = 0.40 mol dm⁻³；[H₂]=[I₂]=0.050 mol dm⁻³。Kc = [HI]²/([H₂][I₂]) = 0.16/0.0025 = 64（无单位，Δn=0）。

8. Acid-Base Equilibria | 酸碱平衡

Question: Calculate the pH of a buffer solution made by mixing 50 cm³ of 0.10 mol dm⁻³ ethanoic acid with 100 cm³ of 0.050 mol dm⁻³ sodium ethanoate. Ka for ethanoic acid = 1.8×10⁻⁵.

问题：计算由 50 cm³ 0.10 mol dm⁻³ 乙酸与 100 cm³ 0.050 mol dm⁻³ 乙酸钠混合制成的缓冲溶液的 pH。Ka（乙酸）= 1.8×10⁻⁵。

Solution: Moles acid = 0.050 × 0.10 = 0.0050 mol; moles salt = 0.100 × 0.050 = 0.0050 mol. Total volume = 150 cm³ = 0.150 dm³. [HA] = 0.0050/0.150 = 0.0333 mol dm⁻³; [A⁻] = 0.0333 mol dm⁻³. [H⁺] = Ka × [HA]/[A⁻] = 1.8×10⁻⁵ × 1 = 1.8×10⁻⁵ mol dm⁻³. pH = –log(1.8×10⁻⁵) = 4.74.

解析：酸物质的量 = 0.050 × 0.10 = 0.0050 mol；盐物质的量 = 0.100 × 0.050 = 0.0050 mol。总体积 0.150 dm³。[HA] = 0.0050/0.150 = 0.0333 mol dm⁻³；[A⁻] 相同。[H⁺] = Ka × [HA]/[A⁻] = 1.8×10⁻⁵。pH = –log(1.8×10⁻⁵) = 4.74。

9. Electrode Potentials | 电极电势

Question: Using the standard electrode potentials: Fe³⁺/Fe²⁺ = +0.77 V; I₂/I⁻ = +0.54 V. Predict whether Fe³⁺ will oxidise I⁻ to I₂ under standard conditions. Write the cell reaction and calculate E⦵cell.

问题：利用标准电极电势：Fe³⁺/Fe²⁺ = +0.77 V；I₂/I⁻ = +0.54 V。预测在标准条件下 Fe³⁺ 能否将 I⁻ 氧化为 I₂。写出电池反应并计算 E⦵cell。

Solution: Fe³⁺/Fe²⁺ has higher potential, so Fe³⁺ is the oxidising agent, I⁻ the reducing agent. Cell reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. E⦵cell = E⦵(right) – E⦵(left) = +0.77 – (+0.54) = +0.23 V. Positive E⦵cell → reaction is feasible.

解析：Fe³⁺/Fe²⁺ 电势更高，Fe³⁺ 作氧化剂，I⁻ 作还原剂。电池反应：2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂。E⦵cell = 0.77 – 0.54 = +0.23 V。电势为正，反应可自发进行。

10. Transition Metal Complexes | 过渡金属配合物

Question: [Cu(H₂O)₆]²⁺ appears blue in aqueous solution, while [CuCl₄]²⁻ is yellow-green. Explain the difference in colour and write the equation for the ligand substitution reaction.

问题：[Cu(H₂O)₆]²⁺ 在水溶液中呈蓝色，而 [CuCl₄]²⁻ 呈黄绿色。解释颜色差异，并写出配体取代反应的方程式。

Solution: In [Cu(H₂O)₆]²⁺, the Cu²⁺ (d⁹) ion is surrounded by six water ligands, which produce a smaller d-orbital splitting (Δoct) absorbing in the red region, transmitting blue. In [CuCl₄]²⁻, chloride ligands cause a smaller splitting (Δtet) due to tetrahedral geometry and weaker field ligand, shifting absorption to higher energy (violet), hence complementary yellow-green colour. Equation: [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O.

解析：[Cu(H₂O)₆]²⁺ 中 Cu²⁺ (d⁹) 被六个水配体包围，d 轨道分裂能较小，吸收红光，呈现蓝色。在 [CuCl₄]²⁻ 中，氯配体为弱场配体且形成四面体结构，分裂能更小，吸收紫光，因而呈现互补的黄绿色。反应：六水合铜(II) 离子在过量氯离子下发生配体取代。

11. Organic Reaction Mechanisms | 有机反应机理

Question: Show the mechanism for the electrophilic addition of HBr to propene. Explain why the major product is 2-bromopropane rather than 1-bromopropane.

问题：画出 HBr 与丙烯的亲电加成机理。解释为何主要产物为 2-溴丙烷而非 1-溴丙烷。

Solution: Step 1: Electrophilic attack by H⁺ on the C=C double bond. The secondary carbocation (CH₃–⁺CH–CH₃) is more stable than primary (CH₃–CH₂–⁺CH₂) due to greater inductive effect of two alkyl groups. Step 2: Br⁻ attacks the carbocation to form 2-bromopropane. Major product arises from the more stable carbocation intermediate (Markovnikov’s rule).

解析：第一步：H⁺ 进攻双键，形成碳正离子。二级碳正离子（CH₃–⁺CH–CH₃）比一级更稳定，因两个烷基的诱导效应更强。第二步：Br⁻ 进攻该碳正离子，得到 2-溴丙烷。主要产物遵循马氏规则，源于更稳定的中间体。

12. NMR Spectroscopy | 核磁共振波谱

Question: A compound with molecular formula C₄H₈O₂ gave the following ¹H NMR data: δ 1.3 (3H, triplet), δ 2.3 (2H, quartet), δ 3.7 (3H, singlet). Identify the compound and account for the splitting patterns.

问题：分子式为 C₄H₈O₂ 的化合物给出如下 ¹H NMR 数据：δ 1.3 (3H，三重峰)，δ 2.3 (2H，四重峰)，δ 3.7 (3H，单峰)。鉴定该化合物并解释裂分图案。

Solution: The signal at δ 3.7 (3H, singlet) indicates a –OCH₃ group isolated from neighbouring protons. The δ 1.3 triplet and δ 2.3 quartet (integrating 3:2) suggest an ethyl group –CH₂CH₃ adjacent to a carbonyl. Combined with formula C₄H₈O₂, the compound is methyl propanoate, CH₃CH₂COOCH₃. The triplet arises from CH₃ split by adjacent CH₂; the quartet from CH₂ split by CH₃. The methoxy singlet has no adjacent protons.

解析：δ 3.7 单峰（3H）表示 –OCH₃ 基团，无邻质子。δ 1.3 三重峰和 δ 2.3 四重峰（积分比 3:2）表明乙基 –CH₂CH₃ 连接在羰基旁。结合分子式，化合物为丙酸甲酯 CH₃CH₂COOCH₃。三重峰由 CH₃ 被邻位 CH₂ 裂分产生；四重峰由 CH₂ 被 CH₃ 裂分。甲氧基无邻质子，故为单峰。

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A-Level OCR Chemistry: Detailed Walkthroughs of Typical Questions | A-Level OCR 化学：典型例题详解