📚 IB & WJEC Chemistry: Quick Tricks for Multiple Choice Questions | IB WJEC 化学：选择题秒杀技巧
Multiple choice questions in IB and WJEC chemistry exams can be tackled swiftly with strategic approaches, saving precious time and boosting accuracy. Whether you face Paper 1 in IB or the multiple choice section in WJEC, mastering these 'quick kill' techniques will sharpen your edge.
在 IB 和 WJEC 化学考试中，选择题可以用策略性方法快速解决，节省宝贵时间并提高准确率。无论你面对 IB 试卷一还是 WJEC 的选择题部分，掌握这些“秒杀”技巧都能让你更胜一筹。
1. Elimination & Common Sense | 排除法与常识判断
Start by eliminating obviously wrong options. If a question asks for a colourless aqueous ion, any transition metal ion like Cu²⁺ (blue) or Fe²⁺ (pale green) can be crossed out instantly.
首先排除明显错误的选项。如果题目问无色水合离子，任何过渡金属离子如 Cu²⁺（蓝）或 Fe²⁺（浅绿）可立刻划掉。
Use common sense: ionic compounds are mostly solid at room temperature, non‑polar molecules like CCl₄ have low boiling points, and noble gases are monatomic. Such knowledge helps to ditch absurd options.
利用常识：离子化合物室温下多为固体，非极性分子如 CCl₄ 沸点低，稀有气体是单原子的。这类知识能帮你丢掉荒谬选项。
When two options are direct opposites (e.g., 'increases' vs. 'decreases'), one of them is often correct. Cross-check with fundamental trends to confirm.
当两个选项直接相反（如“增大”与“减小”），其中往往有一个正确。对照基本趋势验证即可。
Look out for options with extreme language like 'always' or 'never'; chemistry rarely works in absolutes, so these are prime candidates for elimination.
警惕带有“总是”“从不”等绝对化措辞的选项；化学中极少有绝对情况，这类选项是首选排除对象。
2. Unit & Dimensional Analysis | 量纲分析法
]ⁿ and plug in typical units: mol dm⁻³ s⁻¹ divided by (mol dm⁻³)ⁿ gives the unit directly.
如果题目问速率常数的单位，不必死记公式，直接利用速率方程：速率 = k[A]ⁿ。变形为 k = 速率 / [A]ⁿ，代入常用单位 mol dm⁻³ s⁻¹ 除以 (mol dm⁻³)ⁿ 即可得出单位。
For second order overall: k units = mol⁻¹ dm³ s⁻¹
对于总级数为二级：k 的单位 = mol⁻¹ dm³ s⁻¹
Dimensional analysis also works for equilibrium constants. Kc has units (mol dm⁻³)Δⁿ, where Δn is the change in gaseous moles. Spot the powers by balancing concentration units.
量纲分析同样适用平衡常数。Kc 的单位为 (mol dm⁻³)Δⁿ，其中 Δn 为气体分子数的变化。通过匹配浓度单位次方可快速确定。
If the options for a physical quantity contain mismatched units, eliminate them immediately. For example, entropy change must be J K⁻¹ mol⁻¹, not kJ mol⁻¹.
若某个物理量的选项出现单位不符，立即排除。例如熵变必须是 J K⁻¹ mol⁻¹，而不是 kJ mol⁻¹。
3. Extreme Case Reasoning | 极值假设法
er the extremes: tiny Mg²⁺ vs huge Ba²⁺. Smaller cation → stronger electrostatic attraction → more exothermic lattice energy.
比较两种相似化合物时，设想极端情形。例如判断 MgO 与 BaO 谁的晶格焓更大，考虑极端：极小的 Mg²⁺ 与极大的 Ba²⁺。阳离子越小 → 静电引力越强 → 晶格能越放热。
For acid strength of hydrogen halides, think of extremes: HF has strong H‑F bond, so it is a weak acid; HI has a weak bond, making it strong. The trend is clear without calculating pKa.
对于氢卤酸的强弱，也采用极端思维：HF 的 H‑F 键极强，因此是弱酸；HI 键弱，为强酸。趋势清楚，无需计算 pKa。
In transition metal colour questions, imagine full vs. empty d orbitals: Sc³⁺ and Cu⁺ are colourless, because d⁰ and d¹⁰ configurations forbid d‑d transitions. This quickly eliminates coloured options.
在过渡金属颜色题中，设想全空或全满 d 轨道：Sc³⁺ 和 Cu⁺ 无色，因为 d⁰ 和 d¹⁰ 组态禁阻 d‑d 跃迁。这能快速排除有色选项。
4. Conservation Shortcuts | 守恒法速解
Mass and charge conservation can solve redox and stoichiometry questions in seconds. In a half‑equation, first balance atoms other than O and H; then add H₂O, H⁺ (or OH⁻), and electrons so that total charge is equal on both sides.
质量与电荷守恒能秒解氧化还原与化学计量题。在配平半反应时，先平衡非 O、H 原子，再通过 H₂O、H⁺（或 OH⁻）与电子使两侧总电荷相等。
For compound identification, if a metal chloride has 65.5% mass of Cl (35.5 g mol⁻¹) and simple ratio: mass of metal = 100 and mass of Cl₂ not needed—work per mole of Cl atoms.
对化合物鉴定，若某金属氯化物含氯 65.5%（质量分数），用 Cl 摩尔质量 35.5 g mol⁻¹ 和简单比例快速运算，不必陷入复杂流程。
When a question gives several reactions, sum them up to see overall change, making sure all intermediates cancel. This avoids stepwise calculations.
当题目给出多步反应时，将它们相加得到总反应，确保中间产物相消，免去逐步计算的麻烦。
5. Stoichiometry Quick Calculations | 化学计量速算
For mole‑ratio questions, pivot on the species with the known mass. Convert to moles and apply the simplest whole‑number ratio from the balanced equation. Avoid calculating all reactants.
涉及物质的量之比的问题，以已知质量物质为基准，换算成物质的量，再用配平方程的最简整数比推算，无需算遍所有反应物。
→ 2AlCl₃. If 5.4 g Al (0.2 mol) is used, moles of Cl₂ = 0.3 mol, mass = 0.3 × 71 = 21.3 g.
示例：2Al + 3Cl₂ → 2AlCl₃。若用 5.4 g Al（0.2 mol），则需 Cl₂ 0.3 mol，质量为 0.3 × 71 = 21.3 g。
Use the 'scaling factor' hen for X moles of A, B produced = X × (3/2). No need to write full proportions each time.
使用“倍率法”：若 2 mol A 生成 3 mol B，则 X mol A 生成 B = X × (3/2)，不必每次都列全比例式。
simply apply C₁V₁ = C₂V₂, but watch units: cm³ must be converted to dm³ if using mol dm⁻³.
稀释计算直接套用 C₁V₁ = C₂V₂，但注意单位：若用 mol dm⁻³，体积 cm³ 需转换为 dm³。
6. Interpreting Graphs & Diagrams | 图表快速解读
Boltzmann distribution curves: a catalyst lowers activation energy, so the peak shifts to a lower energy and a greater proportion of molecules exceed the new Ea. On the graph, the area beyond Ea increases sharply.
玻尔兹曼分布曲线：催化剂降低活化能，峰向低能量移动，超过新 Ea 的分子比例大增。在图上，Ea 右侧的面积显著扩大。
Rate vs. concentration graphs: zero‑order gives a horizontal line, first‑order a straight line through origin. Identify the order by the shape, not by plotting.
速率-浓度图：零级反应呈水平线，一级反应为过原点直线。根据形状直接判断反应级数，无需描点。
In electromotive force diagrams, a more negative E⁰ means a stronger reducing agent. Quickly locate the species on the left of the half‑equation with the most negative E⁰ value.
在电动序图中，E⁰ 越负，还原性越强。迅速找到半反应左侧 E⁰ 最负的物种即可确定最强还原剂。
7. Organic Reactions & Reagents Tricks | 有机反应推断技巧
Functional group interconversions can be recognised by the gain or loss of oxygen/hydrogen. Oxidation adds O or removes H; reduction does the opposite. This helps match reagents like K₂Cr₂O₇/H⁺ or NaBH₄.
官能团转化可通过氧、氢的增减来识别。氧化增加 O 或减少 H；还原则相反。由此快速匹配试剂，如 K₂Cr₂O₇/H⁺ 或 NaBH₄。
For reaction mechanisms, if a carbocation intermediate is implied, the substrate is likely tertiary or benzylic. SN1 is favoured; no inversion of configuration.
若涉及碳正离子中间体，底物通常是叔碳或苄位，倾向于 SN1 机理，不发生构型翻转。
Use the 'Bond‑Breaking‑Bond‑Making' check: count the number of σ and π bonds broken vs. formed to ensure the mechanistic steps conserve total bond order in each elementary step.
用“断键-成键”检查法：计算每步基元反应中断裂和新生成的 σ 键与 π 键数目，确保总键级守恒。
8. Equilibrium & Rate Constant Hacks | 平衡与速率秒杀
When an equilibrium system is disturbed, Le Châtelier’s principle gives the shift. For temperature: endothermic reactions shift right upon heating. Check ΔH sign quickly.
平衡体系受扰动时，用勒夏特列原理判断移动方向。温度变化：吸热反应升温时平衡右移。快速查看 ΔH 符号。
Kc and temperature: if a reaction is exothermic, Kc decreases with increasing temperature. Look at the relative magnitude of Kc values at two temperatures to infer ΔH.
Kc 与温度：若反应放热，Kc 随温度升高而减小。根据两个温度下的 Kc 大小反推 ΔH 的符号。
Rate constant k depends only on temperature and activation energy (Arrhenius equation). In multiple choice, options with 'k changes with concentration' are always wrong.
速率常数 k 仅取决于温度和活化能（阿伦尼乌斯方程）。选择题中，声称“k 随浓度变化”的选项永远错误。
9. Electrochemistry Quick Wins | 电化学快速检查
anode. The more negative standard electrode potential is the anode (oxidation). Simply scan the list and subtract the smaller number from the larger.
原电池 E⁰池 = E⁰阴 − E⁰阳。标准电极电势较负的作为阳极（氧化）。只需扫描数据表，用大数减小数即可。
In electrolysis of aqueous solutions, check the standard potentials of water reduction/oxidation vs. the ions present. If a metal is more reactive than hydrogen, it will not be discharged at the cathode.
电解水溶液时，对比水的还原/氧化势与离子势。若金属比氢活泼，阴极不会析出该金属。
es of metal = (It)/(nF). Often you can use proportionality without calculating full value.
法拉第定律：电子物质的量 = It/F。金属析出的物质的量 = (It)/(nF)。多数题目用比例关系即可作答，无需全算。
10. Data‑Based & Unfamiliar Contexts | 数据与陌生情景应对
IB and WJEC often present data tables on unfamiliar reactions. Don't panic. Identify the pattern: if doubling [A] doubles initial rate, it is first order in A. Use the method of initial rates by inspection.
IB 与 WJEC 常给出陌生反应的数据表。不要慌张。找出规律：若 [A] 加倍，初始速率亦加倍，则是 A 的一级反应。直接视察法确定级数。
For thermodynamic cycles, use Hess’s Law: the sum of enthalpy changes around a closed loop is zero. Fill in the missing arrow with the difference of the other known values.
遇到热力学循环，用赫斯定律：闭合循环的焓变之和为零。用已知值之差补全未知箭头。
When a new ion or compound is introduced, relate it to a familiar trend. E.g., astatine (At₂) as a halogen: low melting point, insoluble in water but soluble in hexane, just like iodine but even weaker oxidising ability.
引入陌生离子或化合物时，联系已知规律。例如砹（At₂）作为卤素：熔点低，难溶于水而溶于己烷，类似碘但氧化能力更弱。
For graphical extrapolation, straight lines mean constant gradient; use the slope to extract activation energy or order. Simply count rise over run from given points.
在外推图中，直线意味着恒定斜率；利用斜率求活化能或级数，直接从给定点计算 Δy/Δx 即可。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图