OCR A Level Chemistry Unit Test Mock Paper Analysis | OCR A水平化学单元测试模拟卷解析

📚 OCR A Level Chemistry Unit Test Mock Paper Analysis | OCR A水平化学单元测试模拟卷解析

This mock paper walkthrough is designed to help Year 13 OCR Chemistry students consolidate key concepts from AS and early A2 topics. Each question mirrors the style of OCR unit tests, with detailed bilingual explanations to reinforce understanding and exam technique.

本模拟卷解析旨在帮助OCR化学Year 13学生巩固AS及A2初期核心知识点。每道题目均仿照OCR单元测试风格编写，并提供详细的双语解析，以强化理解与应试技巧。

1. Multiple‑Choice: Electronic Configuration | 选择题：电子排布

Question: Which of the following is the ground‑state electronic configuration of a Cr³⁺ ion? A. [Ar] 3d³ B. [Ar] 3d⁴ 4s¹ C. [Ar] 3d⁵ D. [Ar] 3d² 4s¹

题目： 下列哪一个是Cr³⁺离子的基态电子排布？A. [Ar] 3d³ B. [Ar] 3d⁴ 4s¹ C. [Ar] 3d⁵ D. [Ar] 3d² 4s¹

A chromium atom (Z = 24) has the exceptional configuration [Ar] 3d⁵ 4s¹ due to the extra stability of a half‑filled d subshell. When forming Cr³⁺, the 4s and one 3d electron are removed first; the 4s orbital is higher in energy when occupied, so electrons are lost from 4s before 3d. This leaves [Ar] 3d³, choice A.

铬原子（原子序数24）由于半充满d亚层的额外稳定性，具有特殊的[Ar] 3d⁵ 4s¹排布。形成Cr³⁺时，首先失去4s电子和一个3d电子；由于4s轨道在填充后能量高于3d，先失去4s电子。最终留下[Ar] 3d³，答案选A。

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2. Structure & Bonding: Shapes of Molecules | 结构与键合：分子形状

Question: Predict the shape and bond angle of the PF₃ molecule. Explain your reasoning in terms of electron‑pair repulsion.

题目： 预测PF₃分子的形状和键角，并用电子对互斥理论加以解释。

Phosphorus has 5 valence electrons; in PF₃ it forms three single bonds to fluorine and retains one lone pair. That gives four electron pairs around the central atom: three bonding pairs and one lone pair. According to VSEPR theory, the electron pairs adopt a tetrahedral arrangement, but the molecular shape is determined by the positions of the bonding pairs only — trigonal pyramidal. The ideal tetrahedral angle is 109.5°, but the lone pair repels bonding pairs more strongly, compressing the bond angle to about 107°.

磷原子有5个价电子；在PF₃中它与三个氟原子形成三根单键，并保留一对孤对电子。中心原子周围共有四对电子：三对成键电子和一对孤对电子。根据VSEPR理论，电子对呈四面体排布，但分子形状仅由成键电子对位置决定——三角锥形。理想四面体键角为109.5°，但孤对电子对成键电子对有更大的排斥，将键角压缩至约107°。

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3. Moles & Stoichiometry: Empirical Formula | 摩尔与化学计量：经验式

Question: A 2.00 g sample of a hydrocarbon is burned completely in excess oxygen, producing 6.29 g of CO₂ and 2.57 g of H₂O. Determine the empirical formula of the hydrocarbon.

题目： 将2.00 g烃样品在过量氧气中完全燃烧，生成6.29 g CO₂和2.57 g H₂O。求该烃的经验式。

Mass of carbon in CO₂ = (12.0/44.0) × 6.29 g = 1.715 g. Mass of hydrogen in H₂O = (2.0/18.0) × 2.57 g = 0.2856 g. Moles of C = 1.715 g / 12.0 g mol⁻¹ = 0.1429 mol. Moles of H = 0.2856 g / 1.0 g mol⁻¹ = 0.2856 mol. Divide by the smallest number of moles (0.1429): C = 1, H = 2.00. The empirical formula is CH₂.

CO₂中碳的质量 = (12.0/44.0) × 6.29 g = 1.715 g。H₂O中氢的质量 = (2.0/18.0) × 2.57 g = 0.2856 g。碳的物质的量 = 1.715 g / 12.0 g mol⁻¹ = 0.1429 mol。氢的物质的量 = 0.2856 g / 1.0 g mol⁻¹ = 0.2856 mol。除以最小物质的量(0.1429)：C = 1，H = 2.00。经验式为CH₂。

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4. Energetics: Enthalpy Change Calculation | 能量学：焓变计算

Question: Using the data below, calculate the standard enthalpy change for the reaction: 2CO(g) + O₂(g) → 2CO₂(g). ΔH°f (CO) = –110.5 kJ mol⁻¹, ΔH°f (CO₂) = –393.5 kJ mol⁻¹.

题目： 利用以下数据计算反应2CO(g) + O₂(g) → 2CO₂(g)的标准焓变。ΔH°f (CO) = –110.5 kJ mol⁻¹，ΔH°f (CO₂) = –393.5 kJ mol⁻¹。

ΔH° = ΣΔH°f (products) − ΣΔH°f (reactants). Products: 2 × (–393.5) = –787.0 kJ. Reactants: 2 × (–110.5) + 0 = –221.0 kJ. ΔH° = –787.0 – (–221.0) = –566.0 kJ. The value is –566 kJ per mole of reaction (or –283 kJ per mole of CO).

ΔH° = ΣΔH°f (生成物) − ΣΔH°f (反应物)。生成物：2 × (–393.5) = –787.0 kJ。反应物：2 × (–110.5) + 0 = –221.0 kJ。ΔH° = –787.0 – (–221.0) = –566.0 kJ。该值为每摩尔反应放热566 kJ（或每摩尔CO放热283 kJ）。

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5. Kinetics: Maxwell–Boltzmann Distribution | 动力学：麦克斯韦–玻尔兹曼分布

Question: Sketch the Maxwell–Boltzmann distribution curve for a gas at two different temperatures, T₁ and T₂ (T₂ > T₁). Explain how the shape changes and why the rate of reaction increases with temperature.

题目： 画出同一气体在两种不同温度T₁和T₂（T₂ > T₁）下的麦克斯韦–玻尔兹曼分布曲线。解释曲线形状的变化，以及为什么反应速率随温度升高而增加。

At the higher temperature T₂, the curve shifts to the right and flattens, with the peak lowering and moving towards higher energy. The area under both curves remains equal (same number of particles). Crucially, the fraction of particles with energy equal to or greater than the activation energy Ea increases significantly. Since only collisions with energy ≥ Ea lead to reaction, the rate increases — both because a greater proportion of collisions are successful and because collision frequency itself rises slightly.

在较高温度T₂下，曲线右移且变得平缓，峰值降低并向高能量方向移动。两条曲线下面积相等（粒子总数相同）。关键的是，能量大于或等于活化能Ea的粒子比例显著增加。由于只有能量≥ Ea的碰撞才能发生反应，反应速率因此提高——不仅因为成功碰撞比例增大，还因为碰撞频率本身也略有提升。

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6. Equilibrium: Kc Expression & Le Chatelier | 平衡：Kc表达式与勒夏特列原理

Question: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = –92 kJ mol⁻¹), write the Kc expression and predict the effect of increasing pressure on the equilibrium position.

题目： 对于反应N₂(g) + 3H₂(g) ⇌ 2NH₃(g)（ΔH = –92 kJ mol⁻¹），写出Kc表达式，并预测增大压力对平衡位置的影响。

Kc = [NH₃]² / ([N₂] [H₂]³)

The forward reaction converts 4 moles of gas (1 + 3) into 2 moles, reducing the number of particles. According to Le Chatelier’s principle, increasing pressure favours the side with fewer gaseous moles. Therefore the equilibrium shifts to the right, increasing the yield of ammonia. Kc itself is unaffected by pressure change; only temperature alters its value.

正反应将4摩尔气体（1+3）转化为2摩尔，粒子数减少。根据勒夏特列原理，增大压力会向气体分子总数减少的方向移动。因此平衡向右移动，提高氨的产率。Kc本身不受压力变化的影响；只有温度改变才会使Kc值变化。

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7. Acid–Base: pH of Strong Acid | 酸碱：强酸pH

Question: Calculate the pH of a 0.015 mol dm⁻³ solution of HNO₃, and the pH after a hundred‑fold dilution.

题目： 计算0.015 mol dm⁻³ HNO₃溶液的pH值，以及稀释100倍后的pH值。

Nitric acid is a strong monoprotic acid, so [H⁺] = 0.015 mol dm⁻³. pH = –log₁₀(0.015) = 1.82. After a 100‑fold dilution, the new concentration is 0.00015 mol dm⁻³. pH = –log₁₀(0.00015) = 3.82. Water autoprotolysis can be ignored here because the H⁺ concentration remains well above 10⁻⁷ mol dm⁻³.

硝酸是一种强一元酸，因此[H⁺] = 0.015 mol dm⁻³。pH = –log₁₀(0.015) = 1.82。稀释100倍后，新浓度为0.00015 mol dm⁻³。pH = –log₁₀(0.00015) = 3.82。此处可忽略水的自偶电离，因为H⁺浓度仍远高于10⁻⁷ mol dm⁻³。

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8. Redox: Oxidation Numbers | 氧化还原：氧化数

Question: Determine the oxidation number of manganese in KMnO₄ and of sulfur in SO₃²⁻. Hence state whether KMnO₄ can act as an oxidising or reducing agent.

题目： 确定KMnO₄中锰的氧化数和SO₃²⁻中硫的氧化数，并说明KMnO₄可以作为氧化剂还是还原剂。

In KMnO₄: K = +1, O = –2 × 4 = –8, sum must be 0 → Mn = +7. In SO₃²⁻: O = –2 × 3 = –6, overall charge –2 → S = +4. Manganese in the +7 state is in its highest common oxidation state, so it can only gain electrons (be reduced). Therefore KMnO₄ acts as a strong oxidising agent.

KMnO₄中：K = +1，O = –2 × 4 = –8，总和为0，故Mn = +7。SO₃²⁻中：O = –2 × 3 = –6，整体电荷 –2，故S = +4。锰的+7态是其常见最高氧化态，只能得到电子（被还原）。因此KMnO₄可作为强氧化剂。

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9. Organic Chemistry: Naming Alkanes | 有机化学：烷烃命名

Question: Give the IUPAC name of the following compound: (CH₃)₂CHCH(CH₃)CH₂CH₃.

题目： 给出下列化合物的IUPAC名称：(CH₃)₂CHCH(CH₃)CH₂CH₃。

Identify the longest continuous carbon chain: the backbone has 5 carbons (pentane). Number the chain from the end nearest the first substituent. The substituents are a methyl group on C‑2 and another methyl on C‑3. Alphabetical order gives 2,3‑dimethylpentane. The condensed structural formula corresponds to a chain with branching at positions 2 and 3.

先确定最长的连续碳链：主链有5个碳（戊烷）。从离第一个取代基最近的一端开始编号。取代基为C‑2上一个甲基和C‑3上另一个甲基。按字母顺序命名为2,3‑二甲基戊烷。该缩合结构式对应在2、3位有支链的碳链。

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10. Organic Reaction Mechanisms: Free Radical Substitution | 有机反应机理：自由基取代

Question: Methane reacts with chlorine in the presence of UV light to form chloromethane. Write the three steps of the mechanism, using curly arrows and showing all species.

题目： 甲烷与氯气在紫外光下反应生成一氯甲烷。写出该机理的三个步骤，用弯箭头标出所有物种。

1. Initiation: Cl₂ → 2Cl• (homolytic fission caused by UV). 2. Propagation: CH₄ + Cl• → •CH₃ + HCl; then •CH₃ + Cl₂ → CH₃Cl + Cl•. The Cl• radical is regenerated, allowing the chain to continue. 3. Termination: any two radicals combine, e.g. Cl• + Cl• → Cl₂, or •CH₃ + Cl• → CH₃Cl. Only a tiny amount of termination product forms under normal conditions because radical concentrations stay low.

1. 引发步：Cl₂ → 2Cl•（紫外光引起均裂）。2. 增长步：CH₄ + Cl• → •CH₃ + HCl；接着 •CH₃ + Cl₂ → CH₃Cl + Cl•。Cl•自由基得以再生，使链反应持续进行。3. 终止步：任意两个自由基结合，例如 Cl• + Cl• → Cl₂，或 •CH₃ + Cl• → CH₃Cl。在通常条件下，自由基浓度很低，终止产物生成量极少。

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