📚 Year 12 OCR Chemistry: Mock Unit Test Walkthrough | Year 12 OCR 化学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test designed for Year 12 OCR Chemistry A, covering key concepts from Module 2 (Foundations in Chemistry) and Module 3 (Periodic Table and Energy). Each question is analysed with step-by-step explanations to reinforce your understanding and exam technique.
本文详细解析一套为 Year 12 OCR 化学 A 课程设计的单元测试模拟卷，涵盖模块 2（化学基础）与模块 3（元素周期表与能量）的核心概念。每道题目均配有逐步解析，帮助巩固理解并提升应试技巧。
1. Question 1: Electron Configuration and Ionisation Energy | 第 1 题：电子构型与电离能
Phosphorus (P) has atomic number 15. (a) Write the full electron configuration of a phosphorus atom. (b) State and explain whether the first ionisation energy of phosphorus is greater or smaller than that of sulfur.
磷（P）的原子序数为 15。（a）写出磷原子的完整电子构型。（b）判断并解释磷的第一电离能是高于还是低于硫。
The ground state electron configuration of phosphorus is 1s² 2s² 2p⁶ 3s² 3p³. This can also be written in shorthand as [Ne] 3s² 3p³, where the 3p subshell is exactly half-filled.
磷的基态电子构型为 1s² 2s² 2p⁶ 3s² 3p³，也可简写为 [Ne] 3s² 3p³，此时 3p 亚层恰好处于半充满状态。
Phosphorus has a first ionisation energy of approximately 1060 kJ mol⁻¹, which is higher than that of sulfur (around 1000 kJ mol⁻¹). The half-filled 3p³ configuration in phosphorus imparts extra stability, whereas sulfur (3p⁴) possesses one doubly occupied p orbital. The electron-electron repulsion in this paired orbital makes it easier to remove an electron, leading to a lower ionisation energy.
磷的第一电离能约为 1060 kJ·mol⁻¹，高于硫（约 1000 kJ·mol⁻¹）。磷的 3p³ 半充满构型具有额外的稳定性，而硫（3p⁴）有一个 p 轨道容纳了成对电子。该成对轨道中的电子-电子排斥使失去一个电子更容易，从而导致电离能降低。
Always consider the stability of half-filled and fully-filled subshells when comparing ionisation energies across a period; this often explains small anomalies in the general trend.
在比较同周期元素电离能时，务必考虑半充满和全充满亚层的稳定性；这往往能够解释总体趋势中的细微异常。
2. Question 2: Moles and Solution Concentration | 第 2 题：物质的量与溶液浓度
Calculate the concentration, in mol dm⁻³, of a solution prepared by dissolving 5.85 g of sodium chloride (NaCl, Mᵣ = 58.5) in water and making the solution up to 250.0 cm³.
计算将 5.85 g 氯化钠（NaCl，相对分子质量 58.5）溶于水并定容至 250.0 cm³ 后所得溶液的浓度，单位为 mol·dm⁻³。
First, determine the amount of NaCl in moles: n = mass / Mᵣ = 5.85 g / 58.5 g mol⁻¹ = 0.100 mol. The volume of the solution must be converted to dm³: 250.0 cm³ = 0.2500 dm³.
首先计算 NaCl 的物质的量：n = 质量 / 相对分子质量 = 5.85 g / 58.5 g·mol⁻¹ = 0.100 mol。然后将溶液体积转换为 dm³：250.0 cm³ = 0.2500 dm³。
Concentration c = n / V = 0.100 mol / 0.2500 dm³ = 0.400 mol dm⁻³. The final answer should be given to three significant figures, consistent with the data provided.
浓度 c = n / V = 0.100 mol / 0.2500 dm³ = 0.400 mol·dm⁻³。最终答案应与所给数据的有效数字保持一致，保留三位有效数字。
In OCR calculations, always show the conversion from cm³ to dm³ and pay attention to units; leaving the volume in cm³ would give a concentration 1000 times too large.
在 OCR 考试的计算中，务必展示 cm³ 到 dm³ 的换算并留意单位；若单位为 cm³，所得浓度将偏大 1000 倍。
3. Question 3: Hess’s Law and Enthalpy Change | 第 3 题：盖斯定律与焓变
Use the following standard enthalpy changes of combustion to determine the standard enthalpy change for the reaction C(s) + ½O₂(g) → CO(g). ΔH°c(C) = -394 kJ mol⁻¹ and ΔH°c(CO) = -283 kJ mol⁻¹.
利用下列标准燃烧焓变，计算反应 C(s) + ½O₂(g) → CO(g) 的标准焓变。已知 ΔH°c(C) = -394 kJ·mol⁻¹，ΔH°c(CO) = -283 kJ·mol⁻¹。
Construct a Hess cycle. The combustion of graphite to CO₂ is represented by: C(s) + O₂(g) → CO₂(g) ΔH₁ = -394 kJ mol⁻¹. The combustion of carbon monoxide is: CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = -283 kJ mol⁻¹.
构建盖斯循环。石墨燃烧生成 CO₂ 的热化学方程式为：C(s) + O₂(g) → CO₂(g) ΔH₁ = -394 kJ·mol⁻¹。一氧化碳的燃烧方程式为：CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = -283 kJ·mol⁻¹。
The target reaction can be obtained by subtracting the second equation from the first: C(s) + O₂(g) – [CO(g) + ½O₂(g)] → CO₂(g) – CO₂(g), which simplifies to C(s) + ½O₂(g) → CO(g). Therefore, ΔH = ΔH₁ – ΔH₂ = -394 – (-283) = -111 kJ mol⁻¹.
目标反应可由反应一减去反应二得到：C(s) + O₂(g) – [CO(g) + ½O₂(g)] → CO₂(g) – CO₂(g)，简化得 C(s) + ½O₂(g) → CO(g)。因此 ΔH = ΔH₁ – ΔH₂ = -394 – (-283) = -111 kJ·mol⁻¹。
The formation of carbon monoxide from its elements is exothermic, releasing 111 kJ per mole. Always ensure the arrows in your cycle match the direction of the known enthalpy changes.
单质生成一氧化碳的反应为放热反应，每摩尔释放 111 kJ 热量。绘制盖斯循环时，务必确保箭头方向与实际发生的焓变方向一致。
4. Question 4: Organic Nomenclature | 第 4 题：有机命名
Give the systematic name of the following compound: CH₃CH(CH₃)CH₂CH₃.
给出下列化合物的系统命名：CH₃CH(CH₃)CH₂CH₃。
Identify the longest continuous carbon chain. The parent chain contains four carbon atoms, so the base name is butane. The branching occurs at the second carbon, where a methyl substituent (-CH₃) is attached.
首先识别最长的连续碳链。该主链含有四个碳原子，因此母体名称为丁烷（butane）。支链位于第二个碳原子上，连接了一个甲基取代基（-CH₃）。
Number the chain from the end nearest the substituent, giving the methyl group the lowest possible locant. The correct IUPAC name is therefore 2-methylbutane. The name should be written without spaces, and hyphens separate numbers from words.
从最靠近取代基的一端开始编号，使甲基获得最小的位次编号。因此正确的 IUPAC 名称为 2-甲基丁烷（2-methylbutane）。书写时名称中不含空格，数字与文字之间用连字符分隔。
Common mistakes include identifying a shorter parent chain (e.g. propane) or numbering from the wrong end, which would give 3-methylbutane – an incorrect name for this structure.
常见错误包括误判较短的主链（如丙烷），或从错误的一端编号，从而得到 3-甲基丁烷，这对应该结构是错误的名称。
5. Question 5: Electronegativity and Molecular Shape | 第 5 题：电负性与分子形状
Explain, using VSEPR theory, why the CO₂ molecule is linear and non-polar, whereas SO₂ is bent and polar.
运用 VSEPR 理论解释：为何 CO₂ 分子呈直线形且非极性，而 SO₂ 则为弯曲形且具有极性。
In CO₂, the central carbon atom has two bonding pairs of electrons and no lone pairs. According to VSEPR theory, these two regions of electron density repel each other to positions 180° apart, resulting in a linear shape. The C=O bonds are polar, but the two identical bond dipoles cancel each other exactly due to symmetry, making the molecule non-polar.
在 CO₂ 中，中心碳原子有两个成键电子对且无孤对电子。根据 VSEPR 理论，这两个电子密度区域相互排斥至 180° 的位置，形成直线形结构。虽然 C=O 键是极性的，但由于分子的对称性，两个相同的键偶极恰好抵消，使分子整体无极性。
In SO₂, sulfur is the central atom. It forms two S=O bonding pairs and also possesses one lone pair of electrons. The three regions of electron density arrange themselves in a trigonal planar geometry, but the lone pair causes a bent molecular shape with a bond angle of approximately 119°. Because the molecule is bent, the individual S=O bond dipoles do not cancel; they produce a net dipole moment, making SO₂ polar.
在 SO₂ 中，硫为中心原子。它形成两个 S=O 成键对，同时还拥有一对孤对电子。三组电子密度区域在空间上呈平面三角形排布，但孤对电子的存在导致分子形状为弯曲形，键角约为 119°。由于分子并非直线，S=O 键的偶极无法抵消，产生了净偶极矩，因此 SO₂ 具有极性。
Lone pairs exert a greater repulsive force than bonding pairs, so the O-S-O bond angle is compressed slightly from the ideal 120°. This distinction between electron-pair geometry and molecular shape is essential for predicting molecular polarity.
孤对电子的排斥力大于成键电子对，因此 O-S-O 键角较理想的 120° 略微压缩。这种电子对几何构型与分子形状之间的区别对于预测分子极性至关重要。
6. Question 6: Equilibrium Constant Calculations | 第 6 题：平衡常数计算
For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g
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