IB Chemistry: Common Mistake Questions Explained | IB 化学：易错题精讲

📚 IB Chemistry: Common Mistake Questions Explained | IB 化学：易错题精讲

In IB Chemistry, exam questions are carefully designed to probe understanding, not just recall. Common misconception traps appear in every topic – from the misuse of molar volume at non‑standard conditions to the incorrect writing of equilibrium expressions. This article analyses the most frequent errors encountered in past papers and internal assessments, explaining the underlying principles and how to reach the correct answer.

在 IB 化学考试中，题目精心设计，以考查理解而非死记硬背。从非标准状况下摩尔体积的误用，到平衡表达式书写错误，常见的理解陷阱贯穿各个主题。本文分析历年真题和内部评估中最常见的错误，阐释背后的原理，并说明如何得出正确答案。

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1. Mole Calculations and Gas Volumes | 摩尔计算与气体体积

A classic trap is to assume that one mole of any gas always occupies 22.7 dm³. The IB Data Booklet gives the molar volume as 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa). When a question states conditions of 298 K and 100 kPa, many students still multiply moles by 22.7, leading to an incorrect volume. At non‑STP conditions, the ideal gas equation pV = nRT must be used. Remember also to convert cm³ to dm³ by dividing by 1000, and to match the units of pressure with the value of R (e.g. 8.31 J K⁻¹ mol⁻¹ when p is in kPa and V in dm³).

一个经典陷阱是认为任何条件下 1 摩尔气体的体积总是 22.7 dm³。IB 数据手册给出的摩尔体积是标准状况（273 K, 100 kPa）下的 22.7 dm³ mol⁻¹。当题设条件为 298 K、100 kPa 时，许多学生仍直接用摩尔数乘以 22.7，导致体积算错。在非标准状况下，必须使用理想气体方程 pV = nRT。还要注意将 cm³ 转换成 dm³（除以 1000），并确保压强单位与所用 R 值匹配（如 p 用 kPa、V 用 dm³ 时，R = 8.31 J K⁻¹ mol⁻¹）。

Another common slip is omitting the conversion between mass and moles. Always calculate n = m / M first; only then apply the stoichiometric ratio or gas law. For example, a question might give the mass of calcium carbonate and ask for the volume of CO₂ evolved at 25 °C and 1 atm. Those who skip the step of finding moles of CaCO₃ will end up with a nonsense answer.

另一个常见错误是忽略质量与摩尔数之间的转换。务必先用 n = m / M 计算摩尔量，然后再应用化学计量比或气体定律。例如，题目给出碳酸钙的质量，要求计算在 25 °C、1 atm 下产生的 CO₂ 体积。如果跳过求 CaCO₃ 摩尔量这一步，就会得出荒谬的结果。

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2. Limiting Reactant Pitfalls | 限量试剂陷阱

Students often pick one reactant and use its amount to calculate the theoretical yield of the product without checking whether it is the limiting reagent. In a question where 2.0 mol of A reacts with 3.0 mol of B according to A + 2B → C, the mole ratio demands 2 mol B per 1 mol A. Here B is the limiting reactant (3.0 mol B requires only 1.5 mol A, leaving A in excess). Yet many candidates simply use A as the basis and obtain an inflated yield. Always divide the available moles by the stoichiometric coefficient; the smaller value indicates the limiting reactant.

学生常常随意选择一个反应物，用其物质的量直接计算产物的理论产量，而并未检查其是否为限量试剂。在 2.0 mol A 与 3.0 mol B 按 A + 2B → C 反应的题目中，化学计量比要求每摩尔 A 消耗 2 mol B。此处 B 是限量试剂（3.0 mol B 只要求 1.5 mol A，A 过量）。但许多考生直接用 A 计算，得到偏高的产量。正确的做法是：将每种反应物的现有摩尔量除以其化学计量系数，所得值较小者即为限量试剂。

Sometimes masses are given instead of moles, requiring a double step: convert both masses to moles, then perform the limiting‑reactant check. Without this, students might wrongly assume the reactant with the larger mass is always in excess, ignoring the molar masses. This is especially deceptive when the reactant with a higher molar mass appears to be “more” in mass but is actually the limiting one.

有时题目给出质量而非摩尔量，这就需要两步操作：先将质量换算为摩尔量，再进行限量试剂判断。否则，学生可能错误地认为质量较大的反应物一定过量，而忽略了摩尔质量的影响。当某反应物摩尔质量很大，其质量显得很“多”，实际上却可能是限量试剂，这种情况特别具有迷惑性。

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3. Percentage Yield and Experimental Error | 产率计算与实验误差

The formula for percentage yield is (actual yield / theoretical yield) × 100%. A common mistake is to invert the fraction, dividing theoretical by actual. Another error is to treat the “actual yield” as the mass obtained directly from a limiting‑reactant calculation, forgetting that the limiting reactant determines the theoretical yield. In a typical exam problem, you first identify the limiting reactant, calculate the theoretical mass of product, and then apply the actual mass provided to find the percentage yield. Confusing these steps leads to swapping numbers.

产率计算公式是 （实际产量 / 理论产量）× 100%。常见错误是分子分母颠倒，用理论产量除以实际产量。另一错误是将从限量试剂算出的产量直接当作“实际产量”，忘记了限量试剂决定的是理论产量。在典型考题中，应先确定限量试剂，算出产物的理论质量，再根据题目所给的实际质量求出百分比产率。混淆上述步骤就会造成数据代入错误。

Questions sometimes ask to comment on a yield below 100%, expecting reasons such as incomplete reaction, side reactions, loss during purification, or the reaction being reversible. A yield above 100% usually indicates an impurity or a damp product. Students losing marks here tend to give vague answers like “some product was lost”, without linking to the specific experimental context described in the question.

题目有时要求对低于 100% 的产率进行评论，期望的答案包括反应不完全、发生副反应、纯化过程中损失或反应为可逆反应。产率高于 100% 则通常意味着含有杂质或产物未干燥。失分的学生往往给出模糊回答，如“一些产物损失了”，却未与题目所描述的具体实验背景相联系。

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4. Electron Configurations of Transition Metals | 过渡金属电子排布

Chromium and copper atoms show anomalous electron configurations: Cr is [Ar] 4s¹ 3d⁵, not [Ar] 4s² 3d⁴; Cu is [Ar] 4s¹ 3d¹⁰, not [Ar] 4s² 3d⁹. This arises from the extra stability of a half‑filled or fully filled d‑subshell. In examinations, candidates are often asked to write the configuration of these atoms or their ions. A frequent mistake is to keep the 4s² arrangement for Cr and Cu. Furthermore, when forming cations, electrons are removed from the 4s orbital before the 3d. Thus Fe²⁺ is [Ar] 3d⁶, not [Ar] 4s² 3d⁴. Writing Fe³⁺ as [Ar] 3d⁵ is correct, but many students wrongly remove two 4s electrons and then one 3d electron for Fe²⁺, ending up with [Ar] 4s⁰ 3d⁶? Actually that is correct in shorthand as [Ar] 3d⁶, but they might erroneously write [Ar] 4s² 3d⁴ for Fe²⁺. Being explicit about the order of removal is crucial.

铬和铜原子表现出反常的电子排布：Cr 为 [Ar] 4s¹ 3d⁵，而非 [Ar] 4s² 3d⁴；Cu 为 [Ar] 4s¹ 3d¹⁰，而非 [Ar] 4s² 3d⁹。这是由于半充满和全充满 d 亚层具有额外稳定性。考试中常要求书写这些原子或离子的电子排布，常见错误是让 Cr 和 Cu 仍保持 4s² 的排列。此外，形成阳离子时，电子优先从 4s 轨道失去，然后才失去 3d 电子。因此 Fe²⁺ 为 [Ar] 3d⁶，而非 [Ar] 4s² 3d⁴；Fe³⁺ 为 [Ar] 3d⁵。许多学生错误地写出 Fe²⁺：[Ar] 4s² 3d⁴，这便是失分点。

For ions like Zn²⁺, the configuration is [Ar] 3d¹⁰, as the 4s electrons are lost. A good way to remember is: “first in, first out” – the 4s fills before 3d but also empties before 3d. Practice writing configurations for Sc³⁺, Mn²⁺ and Cu⁺ helps solidify the pattern.

对于 Zn²⁺ 等离子，其排布为 [Ar] 3d¹⁰，因为 4s 电子已失去。一个有效的记忆方法是：“先进先出”——4s 轨道先于 3d 填充，也先于 3d 失去。通过练习书写 Sc³⁺、Mn²⁺ 和 Cu⁺ 等的电子排布，可以巩固这一规律。

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5. Molecular Geometry and Polarity | 分子形状与极性

When predicting shape using VSEPR theory, students often forget to count the lone pairs on the central atom. For ammonia, NH₃, the central nitrogen has one lone pair and three bonding pairs, leading to a trigonal pyramidal shape, not trigonal planar. Water, H₂O, has two lone pairs: the shape is bent or V‑shaped, not linear. In exam questions, a Lewis structure must be drawn first to tally the total electron domains; only then can the molecular geometry be assigned. The mistake of ignoring lone pairs is widespread in questions involving species like SF₄ (seesaw) or ClF₃ (T‑shaped).

使用 VSEPR 理论预测分子形状时，学生经常忘记计算中心原子的孤电子对。对于 NH₃，中心氮原子有一对孤对电子和三对成键电子，分子形状为三角锥形，而非平面三角形。水分子 H₂O 有两对孤电子对，形状是 V 形（弯曲形），而非直线形。解题时必须先画出路易斯结构，统计总电子域数，然后才能确定分子几何形状。忽视孤电子对的错误广泛出现在涉及 SF₄（跷跷板形）或 ClF₃（T 形）等物种的题目中。

Polarity is the next pitfall. Many learners assume that all symmetrical molecules are non‑polar, but this fails when lone pairs break the symmetry. For example, SF₄ has a seesaw shape with a net dipole moment. Similarly, molecules with polar bonds can be non‑polar if the dipoles cancel, as in CO₂ (linear) or CCl₄ (tetrahedral). Students frequently predict CH₂Cl₂ to be non‑polar because they recognise the tetrahedral geometry, but its asymmetry makes it polar. Always draw the molecule in 3D and evaluate whether bond dipoles sum to a net vector.

极性的判断是下一个陷阱。许多学生认为所有对称分子均非极性，但若孤电子对破坏了对称性，该结论便不成立。例如，SF₄ 呈跷跷板形，具有净偶极矩。另一方面，含有极性键的分子若键矩相互抵消，则可能为非极性，如 CO₂（直线形）或 CCl₄（正四面体形）。学生常因识别出四面体构型而错误地判断 CH₂Cl₂ 为非极性，但其不对称性使其成为极性分子。解题时务必画出分子的三维结构，并评估键矩的矢量总和。

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6. Enthalpy Changes and Hess’s Law | 焓变与赫斯定律

Sign errors are extremely common in enthalpy calculations. Using ΔH = ΣΔHf(products) − ΣΔHf(reactants), students may mistakenly put reactants first or fail to account for the coefficient of each substance. When combustion data are used, the formula changes to ΔH = ΣΔHc(reactants) − ΣΔHc(products), and swapping the terms gives the opposite sign. In Hess’s law cycles, drawing an arrow in the wrong direction inverts the ΔH value. Always check whether the question asks for the enthalpy change of a specific reaction; adding an equation means adding the ΔH values with the correct sign.

焓变计算中符号错误极为常见。使用 ΔH = ΣΔHf(产物) − ΣΔHf(反应物) 时，学生可能将反应物与产物的位置颠倒，或忘记乘以各物质的化学计量系数。若题目提供燃烧数据，则应改用 ΔH = ΣΔHc(反应物) − ΣΔHc(产物)，互换顺序同样会导致正负号反转。在赫斯定律循环图中，箭头方向画反会使 ΔH 数值取相反符号。务必看清题目问的是哪个具体反应的焓变；叠加方程式时必须按正确的符号叠加相应 ΔH 值。

A typical exam question gives ΔHf values for several compounds and asks for the enthalpy change of a reaction that is not a simple formation. Students frequently miss negative signs for exothermic formation values and end up with a positive ΔH for an obviously exothermic reaction. Taking an extra moment to check whether the final sign makes intuitive sense (e.g., combustion should be exothermic) can save marks.

典型的考题会给出若干化合物的 ΔHf 值，求某非直接生成反应的焓变。学生常因漏掉放热生成的负号，使一个明显放热的反应计算出正的 ΔH。多花片刻检验最终符号是否符合直觉（如燃烧应为放热），便能避免失分。

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7. Rate Equations and Mechanisms | 速率方程与机理

The rate equation cannot be deduced from the stoichiometric equation; it must be determined experimentally. Many students look at the reaction 2A + B → C and immediately write rate = k[A]²[B], which may be incorrect. If the reaction proceeds via a multi‑step mechanism, the slow step (rate‑determining step) governs the rate law. In an IB context, a common task is to propose a rate equation from a given mechanism. When the slow step involves an intermediate that is produced in a fast equilibrium, the concentration of that intermediate must be expressed in terms of the reactants using the equilibrium constant. For instance, if a fast equilibrium 2X ⇌ X₂ precedes the slow step X₂ + Y → products, rate = k[X]²[Y] becomes the overall rate law, but students often incorrectly leave the intermediate X₂ in the expression.

速率方程不可由化学计量方程式直接推导，必须通过实验确定。许多学生看到反应 2A + B → C，就立刻写下 rate = k[A]²[B]，这可能是错误的。如果反应经历多步机理，则慢步骤（决速步）决定速率方程。在 IB 试题中，常见任务是根据给定机理提出速率方程。若慢步骤涉及一个由快速平衡产生的中间体，则必须利用平衡常数将该中间体的浓度用反应物浓度表示。例如，快平衡 2X ⇌ X₂ 在上一步中

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