📚 Cambridge Pre-U Chemistry: Physical Chemistry Unit Test Mock Paper Analysis | 剑桥Pre-U化学：物理化学单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test covering core topics in Cambridge Pre-U Physical Chemistry. Each question is followed by model answers and step-by-step explanations that target common misconceptions and exam technique. Use this analysis to consolidate your understanding of electron configurations, bonding, energetics, kinetics, equilibrium, acid-base equilibria and electrochemistry.
本文为一份剑桥Pre-U物理化学单元模拟测试卷提供详尽的解析。每道题目均附有参考答案和逐步解释，直击常见误区与应试技巧。通过本文，你可以巩固电子构型、化学键、热力学、动力学、平衡、酸碱平衡和电化学等核心内容。
1. Question 1: Electron Configurations and Anomalies | 第1题：电子构型与异常
Write the ground‑ 24) and explain why the observed configuration differs from the simple aufbau prediction.
写出铬（Z = 24）的基态电子构型，并解释为什么实际构型与简单的构造原理预测不同。
The aufbau order would suggest [Ar] 4s² 3d⁴. However, the experimentally determined configuration is [Ar] 4s¹ 3d⁵.
按构造原理顺序应为 [Ar] 4s² 3d⁴。然而，实验测得的构型却是 [Ar] 4s¹ 3d⁵。
The anomaly arises because a half‑filled 3d subshell (d⁵) possesses extra stability. With one electron in the 4s orbital and five unpaired electrons in the 3d orbitals, all spins are parallel, maximising exchange energy. The energy gained by achieving the half‑filled d⁵ arrangement outweighs the energy cost of promoting an electron from 4s to 3d.
异常源于半充满的 3d 亚层（d⁵）具有额外的稳定性。4s 轨道上有一个电子，3d 轨道上有五个自旋平行的未成对电子，交换能达到最大。实现半充满 d⁵ 排布所获得的能量，超过了将一个电子从 4s 激发到 3d 所需的能量。
er than the expected [Ar] 4s² 3d⁹, because a completely filled d¹⁰ subshell is also exceptionally stable.
铜（Z = 29）也有类似异常，其构型为 [Ar] 4s¹ 3d¹⁰ 而非预期的 [Ar] 4s² 3d⁹，因为全充满的 d¹⁰ 亚层同样特别稳定。
2. Question 2: First Ionisation Energies of Period 2 | 第2题：第二周期的第一电离能
The first ionisation energies (kJ mol⁻¹) of Period 2 elements are: Li 520, Be 900, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081. Explain the general trend and the deviations observed between Be and B, and between N and O.
第二周期元素的第一电离能（kJ mol⁻¹）如下：Li 520, Be 900, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081。解释总趋势以及 Be 与 B 之间、N 与 O 之间的偏差。
Ionisation energy generally increases across a period because nuclear charge increases while shielding remains roughly constant, drawing the outer electrons closer and making them harder to remove.
第一电离能通常在同一周期中从左到右递增，因为核电荷增大而屏蔽效应基本不变，外层电子被更紧地束缚，更难移去。
The drop from Be (1s²2s²) to B (1s²2s²2p¹) occurs because the outermost electron in boron occupies a 2p orbital, which has a higher energy than the 2s orbital. The 2p electron is better shielded by the filled 2s subshell and is therefore easier to remove.
从 Be (1s²2s²) 到 B (1s²2s²2p¹) 出现下降，因为硼的最外层电子占据 2p 轨道，其能量高于 2s 轨道。2p 电子受满填的 2s 亚层屏蔽更强，因此更容易移去。
The decrease from N (1s²2s²2p³) to O (1s²2s²2p⁴) is due to electron‑pair repulsion in the doubly occupied 2p orbital of oxygen. In nitrogen, three unpaired electrons occupy different p orbitals with parallel spins, whereas oxygen must pair one set of electrons, causing extra repulsion that lowers the ionisation energy.
从 N (1s²2s²2p³) 到 O (1s²2s²2p⁴) 的下降是因为氧原子中双占据的 2p 轨道存在电子配对排斥。氮的三个未成对电子分占不同的 p 轨道且自旋平行，而氧必须有一对电子自旋相反共占同一轨道，由此产生的额外排斥降低了电离能。
3. Question 3: VSEPR Shapes and Bond Angles | 第3题：VSEPR 形状与键角
Use VSEPR theory to predict the shapes and approximate bond angles of PF₅, SF₄ and ClF₃. Indicate the positions of any lone pairs.
运用 VSEPR 理论预测 PF₅、SF₄ 和 ClF₃ 的形状和近似键角，并标明孤对电子的位置。
PF₅ has five bonding pairs and no lone pairs on phosphorus, giving a trigonal bipyramidal shape. Equatorial bonds are 120° apart, axial bonds are 90° to the equatorial plane.
PF₅ 中磷原子有五对成键电子且无孤对电子，呈三角双锥形。赤道键之间夹角为 120°，轴向键与赤道平面夹角为 90°。
SF₄ has four bonding pairs and one lone pair. The lone pair occupies an equatorial position to minimise repulsion, resulting in a see‑saw (distorted tetrahedral) shape. The equatorial bond angles are compressed below 120° and the axial bonds are bent slightly away from the lone pair, giving angles less than 180°.
SF₄ 有四对成键电子和一对孤对电子。孤对电子占据赤道位置以减少排斥，形成跷跷板形（变形四面体）。赤道键角被压缩至 120° 以下，轴向键因孤对电子排斥而偏离直线，夹角小于 180°。
ClF₃ has three bonding pairs and two lone pairs. The lone pairs occupy two equatorial positions of a trigonal bipyramid, producing a T‑shaped molecule. The bond angles are reduced to about 87.5° because of the repulsion from the two equatorial lone pairs.
ClF₃ 有三对成键电子和两对孤对电子。孤对电子占据三角双锥的两个赤道位置，形成 T 形分子。受两个赤道孤对电子排斥，键角缩小至约 87.5°。
4. Question 4: Hybridisation and Bonding in Ethyne | 第4题：乙炔中的杂化与成键
Describe the hybridisation of carbon in ethyne (C₂H₂) and deduce the number of σ- and π‑bonds in the molecule.
描述乙炔 (C₂H₂) 中碳原子的杂化方式，并推断分子中 σ 键和 π 键的数目。
Each carbon atom in ethyne is sp hybridised. One 2s and one 2p orbital mix to form two sp hybrid orbitals oriented 180° apart, while the two remaining unhybridised 2p orbitals (2pᵧ and 2p₂) lie perpendicular to the molecular axis.
乙炔中每个碳原子均为 sp 杂化。一个 2s 轨道与一个 2p 轨道混合，形成两个呈 180° 分布的 sp 杂化轨道，剩下两个未杂化的 2p 轨道（2pᵧ 和 2p₂）垂直于分子轴。
The C–H σ‑bond arises from overlap of a carbon sp orbital with the hydrogen 1s orbital. The C–C σ‑bond results from end‑on overlap of two sp orbitals. Two π‑bonds are formed by side‑on overlap of the two sets of parallel p orbitals. Overall, the molecule contains three σ‑bonds (one C–C and two C–H) and two π‑bonds.
C–H σ 键由碳 sp 杂化轨道与氢 1s 轨道端向重叠形成。C–C σ 键由两个 sp 轨道端向重叠形成。两个 π 键则分别由两组平行的 p 轨道肩并肩重叠形成。总共有三个 σ 键（一个 C–C 和两个 C–H）和两个 π 键。
5. Question 5: Enthalpy Change Using Hess's Law | 第5题：用赫斯定律求焓变
Use the following data to calculate the standard enthalpy of formation of ethane, C₂H₆(g).
C(s) + O₂(g) → CO₂(g) ΔH° = –394 kJ mol⁻¹
H₂(g) + ½O₂(g) → H₂O(l) ΔH° = –286 kJ mol⁻¹
C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH° = –1560 kJ mol⁻¹
使用下列数据计算乙烷 C₂H₆(g) 的标准生成焓。
C(s) + O₂(g) → CO₂(g) ΔH° = –394 kJ mol⁻¹
H₂(g) + ½O₂(g) → H₂O(l) ΔH° = –286 kJ mol⁻¹
C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH° = –1560 kJ mol⁻¹
The formation reaction is: 2C(s) + 3H₂(g) → C₂H₆(g). According to Hess's Law:
生成反应为：2C(s) + 3H₂(g) → C₂H₆(g)。根据赫斯定律：
ΔH₆° = 2 × ΔH₆°(CO₂) + 3 × ΔH₆°(H₂O) – ΔH₆°(combustion of C₂H₆)
Substituting: ΔH₆° = 2(–394) + 3(–286) – (–1560) = –788 – 858 + 1560 = –86 kJ mol⁻¹.
代入：ΔH₆° = 2(–394) + 3(–286) – (–1560) = –788 – 858 + 1560 = –86 kJ mol⁻¹。
The standard enthalpy of formation of ethane is –86 kJ mol⁻¹. The negative sign indicates the formation is exothermic relative to the elements in their standard states.
乙烷的标准生成焓为 –86 kJ mol⁻¹。负号表示由标准状态下的元素生成乙烷为放热过程。
6. Question 6: Bond Enthalpy Calculations | 第6题：键焓计算
Using the bond enthalpy data (kJ mol⁻¹: N≡N 945, H–H 436, N– e for N₂(g) + 3H₂(g) → 2NH₃(g). Account for any discrepancy with the experimental value of –92.4 kJ mol⁻¹.
利用键焓数据（kJ mol⁻¹：N≡N 945, H–H 436, N–H 391），估算 N₂(g) + 3H₂(g) → 2NH₃(g) 的焓变，并解释与实验值 –92.4 kJ mol⁻¹ 的偏差。
Bonds broken: 1 N≡N (945) + 3 H–H (3 × 436 = 1308) = 2253 kJ mol⁻¹.
Bonds formed: 6 N–H (6 × 391 = 2346) kJ mol⁻¹.
ΔH ≈ Σ(bond enthalpies broken) – Σ(bond enthalpies formed) = 2253 – 2346 = –93 kJ mol⁻¹.
断裂的键：1 个 N≡N (945) + 3 个 H–H (3 × 436 = 1308) = 2253 kJ mol⁻¹。
形成的键：6 个 N–H (6 × 391 = 2346) kJ mol⁻¹。
ΔH ≈ Σ(断裂键焓) – Σ(形成键焓) = 2253 – 2346 = –93 kJ mol⁻¹。
The estimated value (–93 kJ mol⁻¹) is very close to the experimental –92.4 kJ mol⁻¹. The small difference arises because average bond enthalpies are used; actual N–H bond enthalpies in NH₃ are slightly different from the average value, and the calculation does not account for intermolecular forces or changes in physical state.
估算值（–93 kJ mol⁻¹）与实验值 –92.4 kJ mol⁻¹ 非常接近。微小差异来源于使用了平均键焓；NH₃ 中实际的 N–H 键焓与平均值略有不同，且计算未考虑分子间作用力或物态变化。
7. Question 7: Rate Equation and Rate Constant | 第7题：速率方程与速率常数
For the reaction 2A + B → C, the following initial rate data were collected at 298 K:
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
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