📚 AS AQA Chemistry Unit 1 June 2019 Mark Scheme: Key Points & Common Pitfalls | AS AQA 化学 Unit 1 2019年6月评分标准：关键得分点与常见误区
The June 2019 AQA AS Chemistry Unit 1 (7404/1) mark scheme gives a clear picture of how examiners award marks for definitions, calculations, equations, and practical skills. This article summarises the key marking points, common mistakes, and revision strategies that will help you turn mark-scheme language into full marks.
2019年6月AQA AS化学Unit 1（7404/1）评分标准清晰展示了考官如何对定义、计算、方程式和实验技能给分。本文总结关键得分点、常见错误和复习策略，帮助你把评分标准语言转化为满分答案。
1. Atomic Structure and Ionisation Energies | 原子结构与电离能
The mark scheme rewards precise definitions. For first ionisation energy, you must state gaseous atoms, one mole, and the removal of one electron. Missing 'gaseous' or 'one mole' loses a mark even if the rest of the definition is correct.
评分标准对定义要求精确。第一电离能必须说明气态原子、一摩尔以及失去一个电子。即使其余定义正确，缺少“气态”或“一摩尔”也会丢分。
The accepted equation for first ionisation energy is usually shown with state symbols. You must write X(g) on the left and X⁺(g) plus an electron on the right. The state symbol (g) is essential.
第一电离能的标准方程式通常需要写出状态符号。左边为 X(g)，右边为 X⁺(g) 加一个电子。状态符号 (g) 必不可少。
X(g) → X⁺(g) + e⁻
Successive ionisation energies show a large jump when an electron is removed from a new inner shell. For aluminium, the 4th ionisation energy jumps because the 4th electron comes from the 2p sub-shell, which is much closer to the nucleus and experiences less shielding than the 3p electron.
逐级电离能显示，当电子从新的内层被移除时会大幅跃升。对铝而言，第四电离能发生跳跃，因为第四个电子来自2p亚层，该亚层比3p电子更靠近原子核且屏蔽更少。
- Always include state symbols in ionisation energy equations.
- 电离能方程式中务必包含状态符号。
- Define first ionisation energy as one mole of gaseous atoms, not just 'an atom'.
- 第一电离能定义为一摩尔气态原子，而不是“一个原子”。
2. Amount of Substance and Titration Calculations | 物质的量与滴定计算
, so always show your working clearly. Even if the final answer is wrong, your method can still gain most of the marks.
2019年6月的摩尔计算占分很重。评分标准对使用 n = m/M 和 n = cV 给方法分，因此必须清晰展示计算步骤。即使最终答案错误，你的方法也可能拿到大部分分数。
n = m / M n = c × V pV = nRT
A common error in titration questions is forgetting to convert cm³ to dm³ by dividing by 1000. The June 2019 mark scheme penalised answers where the volume was left in cm³, but method marks were still available for using the correct mole ratio.
滴定题常见错误是忘记将cm³除以1000换算为dm³。2019年6月评分标准对体积未换算的答案扣分，但若摩尔比正确仍可获方法分。
When calculating percentage yield or atom economy, the mark scheme expects the formula and substitution. Percentage yield uses actual yield divided by theoretical yield, then multiplied by 100. Atom economy uses the mass of desired product divided by total mass of all products, multiplied by 100.
计算产率或原子经济性时，评分标准要求写出公式并代入数据。产率 = 实际产量 ÷ 理论产量 × 100。原子经济性 = 目标产物质量 ÷ 所有产物总质量 × 100。
| Formula | Units to check | Common error |
| n = m / M | Mass in g; molar mass in g mol⁻¹ | Using mass in kg |
| n = c × V | Volume in dm³ | Leaving volume in cm³ |
| pV = nRT | p in Pa; V in m³; T in K | Using °C or kPa without conversion |
3. Bonding and Shapes of Molecules | 化学键与分子形状
Shapes of molecules require both the shape name and the bond angle. The June 2019 mark scheme accepted names such as linear, trigonal planar, tetrahedral, trigonal pyramidal, bent, and octahedral. Missing the bond angle or giving the wrong shape name loses the mark.
分子形状需要写出形状名称和键角。2019年6月评分标准接受 linear、trigonal planar、tetrahedral、trigonal pyramidal、bent、octahedral 等名称。缺少键角或形状名称错误会失分。
For BF₃, the shape is trigonal planar with 120° bond angles because boron has three bonding pairs and no lone pairs. For NH₃, the shape is trigonal pyramidal with a 107° bond angle because the one lone pair exerts greater repulsion than bonding pairs.
以BF₃为例，形状为平面三角形，键角120°，因为硼周围有三个键对且无孤对。NH₃为三角锥形，键角107°，因为一对孤对的排斥力大于键对。
| Shape | Bond pairs | Lone pairs | Bond angle |
| Linear | 2 | 0 | 180° |
| Trigonal planar | 3 | 0 | 120° |
| Tetrahedral | 4 | 0 | 109.5° |
| Trigonal pyramidal | 3 | 1 | 107° |
| Bent | 2 | 2 | 104.5° |
| Octahedral | 6 | 0 | 90° |
When explaining polarity, the mark scheme expects a link between electronegativity difference and bond polarity. A molecule may have polar bonds but be non-polar overall if the dipoles cancel by symmetry, as in CO₂.
解释极性时，评分标准要求联系电负性差与键的极性。若偶极因对称性而抵消，如CO₂，则分子虽有极性键但整体非极性。
4. Energetics and Calorimetry | 能量学与量热法
00n depending on the units used. Many students lost marks by forgetting the negative sign for exothermic reactions or by using the mass of the solid rather than the mass of the solution.
2019年6月评分标准要求使用 Q = mcΔT，再用 ΔH = −Q/n 或 ΔH = −Q/1000n（取决于单位）。许多学生因放热反应漏掉负号，或误用固体质量而非溶液质量而丢分。
Q = mcΔT ΔH = −Q / n
Calorimetry assumptions appear repeatedly in mark schemes: the solution has the same specific heat capacity as water, there is no heat loss to the surroundings, and combustion or reaction is complete. State these assumptions exactly in practical questions to earn
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