CCEA Year 12 Chemistry Unit Test Mock Exam Analysis | CCEA 12年级化学单元测试模拟卷解析

📚 CCEA Year 12 Chemistry Unit Test Mock Exam Analysis | CCEA 12年级化学单元测试模拟卷解析

Welcome to this detailed analysis of a CCEA Year 12 Chemistry unit test mock paper. In this article, we work through a series of representative questions, unpack key concepts, and highlight common pitfalls. By studying these solutions, you will strengthen your understanding and refine your exam technique.

欢迎阅读这份针对 CCEA 12 年级化学单元测试模拟卷的详细解析。在本文中，我们将分析一系列典型题目，阐释核心概念，并指出常见错误。通过学习这些解答，你将加深理解并优化考试技巧。

1. Atomic Structure and Relative Atomic Mass | 原子结构与相对原子质量

A typical multiple-choice or short-answer question asks: ‘A sample of chlorine contains two isotopes, ³⁵Cl with an abundance of 75% and ³⁷Cl with 25%. Calculate the relative atomic mass of chlorine.’

一道典型选择题或简答题问道：“一个氯样品含有两种同位素，³⁵Cl 丰度为 75%，³⁷Cl 为 25%。计算氯的相对原子质量。”

Step 1: Multiply the mass number of each isotope by its percentage abundance. For ³⁵Cl: 35 × 75 = 2625. For ³⁷Cl: 37 × 25 = 925.

第一步：将每种同位素的质量数乘以其丰度百分比。³⁵Cl: 35 × 75 = 2625；³⁷Cl: 37 × 25 = 925。

Step 2: Add these contributions and divide by 100 to obtain the weighted average. (2625 + 925) / 100 = 3550 / 100 = 35.5.

第二步：将贡献值相加后除以 100，得到加权平均值。(2625 + 925) ÷ 100 = 3550 ÷ 100 = 35.5。

The weighted average is expressed as:

Ar(Cl) = (35 × 75 + 37 × 25) / 100 = 35.5

加权平均值计算式为：Ar(Cl) = (35×75 + 37×25) / 100 = 35.5。答案正是元素周期表上氯的相对原子质量。注意丰度必须使用百分比，而不是小数，除非先将百分比除以 100。

2. Electronic Configuration and Ionisation Energy | 电子排布与电离能

Question: ‘Write the full electronic configuration of a phosphorus atom (atomic number 15). Explain the general trend in first ionisation energy across Period 3, referring to any anomalies.’

题目：“写出磷原子（原子序数 15）的完整电子排布。解释第三周期第一电离能的总体趋势，并指出异常情况。”

The electronic configuration of phosphorus should be written using subshell notation: 1s² 2s² 2p⁶ 3s² 3p³.

磷的电子排布应采用亚层符号表示：1s² 2s² 2p⁶ 3s² 3p³。

Across Period 3, first ionisation energy generally increases. This is because nuclear charge increases from Na to Ar while shielding remains similar, so the outer electrons are held more tightly.

在第三周期中，第一电离能总体上从左到右增加。这是因为核电荷从钠到氩递增，而屏蔽作用相似，因此外层电子被更牢固地束缚。

There are two drops: between Mg and Al, Al’s outer electron is in a 3p orbital of higher energy, so it is easier to remove than Mg’s 3s electron. Between P and S, S has a pair of electrons in one 3p orbital, leading to electron-electron repulsion that makes the electron easier to remove.

存在两处下降：镁到铝之间，铝的外层电子位于能量更高的 3p 轨道，因此比镁的 3s 电子更容易移除。磷到硫之间，硫在一个 3p 轨道中有一对电子，电子间排斥使得电子更容易移除。

3. Chemical Bonding and Shapes of Molecules | 化学键与分子形状

Question: ‘Use VSEPR theory to predict the shape and bond angle of a molecule of BF₃ and a molecule of NH₃. Explain the difference.’

题目：“运用 VSEPR 理论预测 BF₃ 和 NH₃ 分子的形状和键角。解释其差异。”

In BF₃, boron has three bonding pairs and no lone pairs around the central atom. The electron pairs repel to positions of minimum repulsion, giving a trigonal planar shape with bond angles of 120°.

在 BF₃ 中，硼原子周围有三对键合电子对，无孤电子对。电子对互相排斥至最小排斥位置，形成平面三角形，键角为 120°。

In NH

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