CIE 化学
CIE Chemistry 9701 P4 Concept Map (A Level Structured)
34 个知识节点 · P4
1. Kinetics & rate laws
First order: half-life & average rate（一级反应）
📖 Definition一级反应：t_{1/2} 与初浓度无关；[A]=[A]_0/2^{t/t_{1/2}}；平均速率可用 \Delta[A]/\Delta t 估算。
🔑 Key教材 Ch.22：\text{rate}=k[X] 时 k=\ln 2/t_{1/2}。
🎯 Exam focus先判断总时间相当于几个半衰期，再求浓度变化。
🛠️ TechniqueShow that k 的步骤要写 \ln 2 与 t_{1/2} 的关系。
CIE 9701/42/O/N/24 Q1(a)(i)–(iii)
(i) Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t½, is 900 s at 20 °C. A solution of X with a concentration of 0.180 mol dm–3 is prepared at 20 °C. Calculate the average rate of reaction 1 over the first 1800 s. (ii) Complete the rate equation for reaction 1. (iii) Show that the rate constant, k, is 7.70 × 10–4 s–1 at 20 °C.
1800 s = 2×t½ \rightarrow [X] 减半两次；\Delta[X]/1800。\text{rate}=k[X]；k=\ln2/t_{1/2}。
[M] MS 9701_w24_ms_42 Q1(a) 链式给分。
CIE 9701/42/O/N/24 Q1(a)(iv)
(iv) Calculate the initial rate of reaction 1 when the concentration of X is 0.150 mol dm–3. Include units.
\text{rate}=k[X]；单位 \mathrm{mol\,dm^{-3}\,s^{-1}}。
[M] 9701_w24_qp_42。
Rate equation & units of \(k\)（速率常数单位）
📖 Definition总级数 n 决定 k 单位：k=\mathrm{rate}/([A]^a[B]^b\ldots) 量纲补齐。
🔑 Key二级（对总浓度为二）：k 常带 \mathrm{dm^3\,mol^{-1}\,s^{-1}}。
🎯 Exam focus题干给 \mathrm{rate} 与浓度时先代数字求 k，再单独写 units 格。
🛠️ Technique把各浓度幂次相加得总级数 → 反推 k。
CIE 9701/42/O/N/25 Q2(b)
The rate equation for the reaction between CH3CHO and NO2 is shown. rate = k [CH3CHO][NO2]. Under certain conditions, when the concentrations of both CH3CHO and NO2 are 0.200 mol dm–3, the rate of the reaction is 1.53 × 10–4 mol dm–3 s–1. Calculate the value of the rate constant, k, under these conditions. Give the units of k.
k=\text{rate}/([A][B])；总级数 2 \rightarrow k 单位 \mathrm{dm^3\,mol^{-1}\,s^{-1}}。
[M] 9701_w25_qp_42 Q2(b)。
CIE 9701/42/O/N/25 Q2(d)
The rate equation is shown. rate = k1[NO2]. Under certain conditions, the value of k1 is 0.0848 s–1. The reaction has a constant half-life under these conditions. Calculate the half-life in seconds.
一级：t_{1/2}=\ln2/k_1。
[M] 9701_w25_qp_42 Q2(d)。
2. Catalysis
Why transition metals catalyse（过渡金属催化）
📖 Definition可变氧化态、未充满 d 层、表面可吸附活化（异相）或形成中间体（均相）。
🔑 Key教材 Ch.22–24：用 **variable oxidation states / d-orbitals** 等关键词。
🎯 Exam focusExplain 题要因果齐全，不单写“有 d 电子”。
🛠️ Technique一句性质 + 一句如何降低 E_a / 提供表面。
CIE 9701/42/O/N/24 Q1(b)(i)
Platinum is a transition element. Explain why transition elements behave as catalysts.
variable oxidation state 与未充满 d 亚层 \rightarrow 易吸附/提供电子途径降低活化能。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/24 Q1(b)(ii)
Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst.
iron（或 acceptable 金属名）；固相催化剂与 \mathrm{N_2}、\mathrm{H_2} 气相反应物不同相。
[M] 9701_w24_qp_42 Q1(b)(ii)。
Heterogeneous surface action（多相催化机理）
📖 Definition吸附 \rightarrow 活化 \rightarrow 表面反应 \rightarrow 脱附；活性位点可再生。
🔑 Key对比 homogeneous：反应物与催化剂同相。
🎯 Exam focusDescribe mode of action 至少三句逻辑链。
🛠️ Technique可画简图：气体接触金属表面。
CIE 9701/42/O/N/24 Q1(b)(iii)
Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO2, from the exhaust gases of car engines. Describe the mode of action of a platinum catalyst in this process.
吸附 NO₂（及 CO 等）于表面；削弱键；反应；脱附产物。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/25 Q2(e)
NO2 is present in the exhaust gases of cars. It can react with carbon monoxide, CO, on the surface of a heterogeneous catalyst in the car’s catalytic converter. Describe the mode of action of this heterogeneous catalyst.
同结构答案：吸附—活化—反应—脱附。
[M] 9701_w25_qp_42 Q2(e)。
3. Lattice & cycles
Born–Haber style cycles（热化学循环）
📖 Definition\Delta H_\mathrm{latt} 常通过 Hess 闭合：溶解焓、水合焓、生成焓等组合。
🔑 Key教材 Ch.19：标状态符号；气相离子 \mathrm{Mg^{2+}(g)}、\mathrm{Cl^-(g)}。
🎯 Exam focusComplete line … 要画到正确物种与相态。
🛠️ Technique先抄齐已知箭头方向，再填缺失步骤名称。
CIE 9701/42/O/N/24 Q2(b)(i)–(ii)
Fig. 2.1 shows an incomplete energy cycle. (i) Complete line C on Fig. 2.1. Include state symbols. (ii) Use both words and symbols to identify change 2 on Fig. 2.1. Use changes 1 and 3 as examples of how this should be done.
line C 常为 \mathrm{Mg^{2+}(aq)}+2\mathrm{Cl^-}(aq)}；change 2 为 \Delta H_\mathrm{hyd} 或两步水合组合（按卷图）。
[M] 9701_w24_qp_42 Q2(b)。
CIE 9701/42/O/N/24 Q2(a)
Predict and explain the variation in enthalpy change of hydration for the ions Na+, Mg2+ and Al3+.
离子半径减小、电荷增大 \rightarrow 水合焓更负（更放热）。
[M] 9701_w24_qp_42 Q2(a)。
\(\Delta H_\mathrm{latt}\) from data（由表求晶格能）
📖 Definition\Delta H_\mathrm{latt}=\Delta H_\mathrm{f}^\circ-(\sum\Delta H_\mathrm{hyd})-\cdots 按循环方向逐项加减。
🔑 Key注意化学式中离子数目（\mathrm{MgCl_2} 两个 \mathrm{Cl^-}）。
🎯 Exam focus数据表只选用题指定行。
🛠️ Technique列式前在草稿上画箭头方向，避免符号反。
CIE 9701/42/O/N/24 Q2(b)(iii)
Calculate a value for the lattice energy of magnesium chloride, ΔHlatt MgCl2(s), by selecting and using appropriate data from Table 2.1.
代入 \Delta H_\mathrm{sol}、\Delta H_\mathrm{hyd}(\mathrm{Mg^{2+}})、\Delta H_\mathrm{hyd}(\mathrm{Cl^-})、\Delta H_\mathrm{f} 等求 \Delta H_\mathrm{latt}。
[M] MS 9701_w24_ms_42 Q2(b)(iii)。
CIE 9701/42/O/N/24 Q2(b) Table 2.1
Table 2.1 lists enthalpy change of solution of magnesium chloride –155; enthalpy change of formation –642; first and second IE of Mg; electron affinity of chlorine; enthalpy change of hydration of Mg2+ and Cl–.
核对每项符号与单位 \mathrm{kJ\,mol^{-1}} 后再运算。
[M] 9701_w24_qp_42。
4. Hydration & entropy / Gibbs
\(\Delta H_\mathrm{hyd}\) trends（水合焓趋势）
📖 Definition同周期阳离子：电荷 \uparrow、半径 \downarrow \Rightarrow 水合焓更负。
🔑 Key用 ionic charge density 表述更得分。
🎯 Exam focusCompare Na⁺, Mg²⁺, Al³⁺ 等典型组。
🛠️ TechniqueExplain 题：趋势 + 原因（静电与水分子作用）。
CIE 9701/42/O/N/24 Q2(a)
Predict and explain the variation in enthalpy change of hydration for the ions Na+, Mg2+ and Al3+.
\mathrm{Al^{3+}} 最负 \rightarrow \mathrm{Na^+} 最不负；原因：电荷/半径。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/25 Q1(b)–(c)
MgO and SrO both react with dilute sulfuric acid. MgO forms a soluble salt, A. SrO forms an insoluble salt, B. (ii) Explain why A is more soluble than B.
离子半径与晶格能/水合焓差：\mathrm{Mg^{2+}} 更小电荷密度更大等（按 MS 可接受表述）。
[M] 9701_w25_qp_42 Q1。
\(S\), \(\Delta G\) & solubility（熵与吉布斯）
📖 Definition\Delta G=\Delta H-T\Delta S；\Delta G<0 常对应溶解自发倾向（需结合语境）。
🔑 Key熵定义题用 **disorder / number of ways particles are arranged**（教材 Ch.23）。
🎯 Exam focus单位：\Delta S 用 \mathrm{J\,K^{-1}\,mol^{-1}}，\Delta G 转 \mathrm{kJ\,mol^{-1}} 时除以 1000。
🛠️ Technique先换算 T\Delta S 与 \Delta H 同单位再加减。
CIE 9701/42/O/N/24 Q2(c)
Define entropy.
系统无序度/微观排列方式数目的度量（标准定义句）。
[M] 9701_w24_qp_42 Q2(c)。
CIE 9701/42/O/N/24 Q2(d)–(e)
(d) At 25 °C the enthalpy change of solution of compound Z is +26 kJ mol–1. The entropy change of solution of Z at the same temperature is +52 J K–1 mol–1. Calculate the value of the Gibbs free energy change, ΔG, for the solution of Z at 25 °C. (e) Use your answer to (d) to predict whether or not Z is soluble in water at 25 °C. Explain your answer. (ii) Predict whether Z becomes more or less soluble as the water is heated from 25 °C to 95 °C. Explain your answer.
\Delta G=\Delta H-T\Delta S；符号判自发；升温对 \Delta G 影响用 \Delta S 讨论。
[M] 9701_w24_qp_42 Q2(d)(e)。
5. Group 2 & hydroxides
Nitrate decomposition（硝酸盐热分解）
📖 Definition碱土金属硝酸盐分解生成金属氧化物、\mathrm{NO_2}、\mathrm{O_2}（配平方程式）。
🔑 Key教材 Group 2：自上而下热稳定性升高 \rightarrow 分解温度升高。
🎯 Exam focusExplain 用 ionic radius / polarisation of \mathrm{NO_3^-}。
🛠️ Technique方程式电子守恒：金属价态与气体计量。
CIE 9701/42/O/N/25 Q1(a)–(b)
(a) Write an equation for the thermal decomposition of Mg(NO3)2. (b) State which of Mg(NO3)2 or Sr(NO3)2 decomposes at a lower temperature. Explain your answer.
\mathrm{Mg(NO_3)_2} 更低温分解；半径小极化强。
[M] 9701_w25_qp_42 Q1。
CIE 9701/42/M/J/25 Q1(a)(i)–(ii)
(i) Write an equation for the decomposition of calcium nitrate. (ii) Describe the trend in the decomposition temperature of the Group 2 nitrates. Explain your answer.
同族自上而下分解温度升高；阳离子半径增大极化减弱。
[M] 9701_s25_qp_42 Q1。
Strong base dilution & pH（强碱溶液 pH）
📖 Definition\mathrm{pOH}=-\lg[\mathrm{OH^-}]；\mathrm{pH}+\mathrm{pOH}=14（298 K）。
🔑 Key先由质量求 n(\mathrm{SrO})，再换算 n(\mathrm{OH^-}) 与稀释后体积。
🎯 Exam focusVolumetric flask 定容体积参与浓度。
🛠️ Technique题目要求 two d.p. 则 pH 亦保留两位小数。
CIE 9701/42/M/J/25 Q1(b)
A sample of 0.333 g of strontium oxide, SrO, is completely dissolved in distilled water to form a solution of strontium hydroxide, Sr(OH)2. The resulting solution is added to a volumetric flask and made up to 250.0 cm3 with distilled water. Calculate the pH of this solution at 298 K. Give your answer to two decimal places.
n(\mathrm{SrO})\rightarrown(\mathrm{OH^-})=2n\rightarrow[\mathrm{OH^-}]\rightarrowpH。
[M] 9701_s25_qp_42 Q1(b)。
CIE 9701/42/O/N/24 Q3(a)(i)
The pH of a saturated solution of calcium hydroxide is 12.35 at 298 K. (i) Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 mol dm–3 at 298 K.
由 pH 求 pOH 再求 [\mathrm{OH^-}]。
[M] 9701_w24_qp_42 Q3(a)(i)。
6. \(K_\mathrm{sp}\) & solubility
\(K_\mathrm{sp}\) from saturated pH（溶度积）
📖 Definition对 \mathrm{Ca(OH)_2}，K_\mathrm{sp}=[\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2；单位常为三因子幂次组合。
🔑 Key[\mathrm{Ca^{2+}}]=[\mathrm{OH^-}]/2（由溶解平衡）。
🎯 Exam focusUnits of K_\mathrm{sp} 单独一格。
🛠️ Technique写清表达式再代入数值。
CIE 9701/42/O/N/24 Q3(a)(ii)
Use data given in (i) to calculate the solubility product, Ksp, of calcium hydroxide at 298 K. Include the units of Ksp in your answer.
代入 [\mathrm{OH^-}] \rightarrow [\mathrm{Ca^{2+}}] \rightarrow K_\mathrm{sp}。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/24 Q3(a)(i)
The pH of a saturated solution of calcium hydroxide is 12.35 at 298 K. (i) Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 mol dm–3 at 298 K.
\mathrm{pOH}=14-12.35\rightarrow[\mathrm{OH^-}]=10^{-\mathrm{pOH}}=0.0224\ \mathrm{mol\,dm^{-3}}。
[M] 9701_w24_qp_42 Q3(a)(i)。
Common ion & sulfate solubility（硫酸盐溶解性对比）
📖 Definition\mathrm{CaSO_4} vs \mathrm{BaSO_4}：离子半径与晶格能差异。
🔑 Key教材 Ch.21：用 lattice energy / hydration enthalpy balance 解释。
🎯 Exam focus与 charge density 联系。
🛠️ Technique分写：阳离子半径 \rightarrow 晶格能 \rightarrow 溶解度趋势。
CIE 9701/42/O/N/24 Q3(a)(iv)
Explain why calcium sulfate is more soluble in water than barium sulfate.
\mathrm{Ca^{2+}} 半径较小…晶格能较高但水合能变化…（按 MS 采分点）。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/24 Q3(a)(iii)
A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves. Describe one other observation that would be made and give an estimated value of the pH of the solution obtained. Explain both your answers.
同离子 \mathrm{Ca^{2+}} 抑制溶解 \rightarrow 析出 \mathrm{Ca(OH)_2} / 浑浊；pH 下降（约 12 以下，按 MS 范围）。
[M] 9701_w24_qp_42 Q3(a)(iii)。
7. Buffers & weak acids
Conjugate pairs in mixture（共轭酸碱对）
📖 Definition相差一个 \mathrm{H^+} 的一对物种；弱酸与其共轭碱构成缓冲对。
🔑 Key有机共轭对常写 \mathrm{CH_3COOH}/\mathrm{CH_3COO^-}。
🎯 Exam focusPair 1 若要求 organic，勿写无机对。
🛠️ Technique标清 conjugate acid / conjugate base 两列。
CIE 9701/42/O/N/24 Q3(b)(ii)
Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D. Pair 1 should consist of organic species.
\mathrm{CH_3COOH}/\mathrm{CH_3COO^-}；另一对可为 \mathrm{H_3O^+}/H_2O} 等（视混合物）。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/24 Q3(b)(i)
Write an equation for the reaction of calcium with CH3COOH.
\mathrm{Ca}+2\mathrm{CH_3COOH}\rightarrow (\mathrm{CH_3COO})_2\mathrm{Ca}+\mathrm{H_2}。
[M] 9701_w24_qp_42。
pH of buffer + equations（缓冲 pH）
📖 Definition\mathrm{pH}=\mathrm{p}K_a+\lg([\mathrm{A^-}]/[\mathrm{HA}])；或从 K_a 定义严格解。
🔑 Key检查浓度是否可直接代入 Henderson。
🎯 Exam focus缓冲作用两式：加 \mathrm{H^+}、加 \mathrm{OH^-} 各写一条。
🛠️ Technique先算 \mathrm{p}K_a=-\lg K_a。
CIE 9701/42/O/N/24 Q3(b)(iv)–(v)
(iv) The concentration of calcium ethanoate, (CH3COO)2Ca, in mixture D is 0.394 mol dm–3. The concentration of CH3COOH in mixture D is 0.270 mol dm–3. The Ka of CH3COOH is 1.74 × 10–5 mol dm–3 at 298 K. Calculate the pH of mixture D. (v) Write two equations to show how mixture D can act as a buffer solution.
注意 [\mathrm{CH_3COO^-}]=2\times[(\mathrm{CH_3COO})_2\mathrm{Ca}] 等换算；缓冲方程写 \mathrm{CH_3COO^-} 与强酸强碱。
[M] 9701_w24_qp_42。
CIE 9701/42/O/N/24 Q6(c)
HOOCCOOH ionises as shown. HOOCCOOH ⇌ HOOCCOO– + H+. HOOCCOOH is a much stronger acid than methanoic acid, HCOOH. Suggest an explanation for this difference in acidity.
二羧酸第二个羧基拉电子/稳定共轭碱等（按 MS）。
[M] 9701_w24_qp_42 Q6(c)。
8. Partition & chromatography
\(K_\mathrm{pc}\) & iodine extraction（分配系数）
📖 Definition两不互溶溶剂达平衡时，溶质在两相浓度比为分配系数 K_\mathrm{pc}（定义式须与卷面一致）。
🔑 Key教材 Ch.23：先列 n(\mathrm{I_2}) 守恒或 K=[\mathrm{I_2}]_\mathrm{org}/[\mathrm{I_2}]_\mathrm{aq} 与体积联立。
🎯 Exam focus有机相提取后水相残留质量：不要漏体积比与 K_\mathrm{pc} 幂次。
🛠️ Technique设 x 为水相质量或浓度，列方程比死记公式更不易错。
CIE 9701/42/F/M/22 Q1(a)(ii)
15.0 cm3 of C6H12 is shaken with 20.0 cm3 of an aqueous solution containing I2 until no further change is seen. It is found that 0.390 g of I2 is extracted into the C6H12. The partition coefficient of I2 between C6H12 and water, Kpc, is 93.8. Calculate the mass of I2 that remains in the aqueous layer. Show your working.
用 K_\mathrm{pc} 与两相体积把已萃取量与总量关联，求水相剩余质量。
[M] 9701_m22_qp_42 Q1(a)(ii)。
CIE 9701/42/F/M/22 Q1(a)(iii)
Suggest how the value of Kpc of I2 between hexan-2-one, CH3(CH2)3COCH3, and water compares to the value given in (a)(ii). Explain your answer.
酮与水部分互溶/极性较环己烷大 \rightarrow 水相中 \mathrm{I_2} 溶解相对增加 \rightarrow K_\mathrm{pc} 通常低于非极性溶剂体系（按 MS 可接受推理）。
[M] 9701_m22_qp_42 Q1(a)(iii)。
TLC: \(R_\mathrm{f}\) & phases（薄层色谱）
📖 DefinitionR_\mathrm{f}=\dfrac{\text{spot 到原点距离}}{\text{溶剂前沿到原点距离}}，恒在 0–1；固定相常硅胶，流动相为展开溶剂。
🔑 Key极性差异决定相对迁移：与固定相氢键/吸附越强，R_\mathrm{f} 往往越小。
🎯 Exam focusIdentify spot：对照表中 R_\mathrm{f} 与图中各斑位置。
🛠️ TechniqueExplain R_\mathrm{f} 差别用 **relative affinity** for stationary vs mobile phase。
CIE 9701/42/F/M/22 Q6(e)(i)–(iii)
The purity of lidocaine can be checked using thin-layer chromatography. Ethyl ethanoate is used as a solvent. The Rf values of X and lidocaine are given in Table 6.1 (X 0.49, lidocaine 0.71). (i) Identify the substances used as the mobile and stationary phases in this thin-layer chromatography experiment. (ii) Describe how an Rf value can be calculated. (iii) Suggest why the Rf value for X is less than that for lidocaine.
(i) mobile = ethyl ethanoate（溶剂）；stationary = silica / alumina 等吸附剂。(ii) 比值定义。(iii) X 对硅胶吸附更强或极性相对更大故迁移较慢。
[M] 9701_m22_qp_42 Q6(e)。
CIE 9701/42/F/M/23 Q6(a)(i)–(ii)
(i) Suggest a compound that could be used as the stationary phase in this experiment. (ii) Table 6.1 shows the Rf values for different metal cations when separated by TLC using water as a solvent. Suggest the identity of the cation that causes the spot at M in Fig. 6.1. Explain your answer.
(i) 硅胶/氧化铝等。(ii) 将斑 M 的 R_\mathrm{f} 与表对照，最接近者为答案（如 \mathrm{Co^{2+}} 等，按图与表）。
[M] 9701_m23_qp_42 Q6(a)。
8. Electrochemistry & redox
\(E^\circ_\mathrm{cell}\) & oxidising agent（选氧化剂）
📖 DefinitionE^\circ_\mathrm{cell}=E^\circ_\mathrm{reduced}-E^\circ_\mathrm{oxidised}；须为正才自发按所写方向。
🔑 Key半反应方向统一：全写还原式或按卷面。
🎯 Exam focusSelect oxidising agent：找能使目标被氧化且 E^\circ_\mathrm{cell}>0 的物种。
🛠️ Technique写 overall equation 时配平 \mathrm{H^+}、\mathrm{H_2O}、电子。
CIE 9701/42/O/N/24 Q1(c)
SO2 dissolves in water, forming H2SO3. H2SO3 can be oxidised under acidic conditions. The relevant electrode reaction and its E o value are shown. SO42– + 4H+ + 2e– ⇌ H2SO3 + H2O E o = +0.17 V. Four more half-equations … are shown. Select the oxidising agent that could oxidise H2SO3 to SO42– ions under acidic conditions. Write an equation, and give the E o cell value, for the reaction that occurs.
比较 E^\circ：氧化剂对应还原电势更正；算 E^\circ_\mathrm{cell}。
[M] 9701_w24_qp_42 Q1(c)。
CIE 9701/42/M/J/25 Q2(c)
Use only the species listed in Table 2.2 to suggest: one reaction in which H2O2 acts as an oxidising agent and one reaction in which H2O2 acts as a reducing agent. Include the value of the standard cell potential, E o cell, and an overall equation for each reaction.
\mathrm{H_2O_2} 既可被氧化又可被还原：选对半反应组合。
[M] 9701_s25_qp_42 Q2(c)。
Homogeneous catalytic cycle（均相催化两方程）
📖 Definition催化剂在反应开始与结束化学式相同；中间生成再消耗。
🔑 Key\mathrm{NO_2} 催化 \mathrm{SO_2} 氧化为典型例子。
🎯 Exam focus两式各体现 \mathrm{NO_2} 的消耗与再生。
🛠️ Technique不要写总反应代替两步。
CIE 9701/42/O/N/24 Q1(c)(iv)
NO2 acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO2. Write equations for the two reactions that occur.
第一步 \mathrm{NO_2} 被还原（如生成 \mathrm{NO}）；第二步 \mathrm{NO} 再被 \mathrm{O_2} 氧化回 \mathrm{NO_2}（与 MS 所给可接受方程式一致）。
[M] 9701_w24_ms_42 Q1(c)(iv)。
9. Co complexes & ligands
\(\mathrm{NaOH}\) precipitation（钴配合物）
📖 Definition[\mathrm{Co(H_2O)_6}]^{2+} 加碱生成 \mathrm{Co(OH)_2} 沉淀；同时脱质子配位。
🔑 Key教材 Ch.24：颜色变化 + ionic equation。
🎯 Exam focusPrecipitation 亦可答 acid–base with ligands。
🛠️ Technique配平方程：\mathrm{OH^-} 计量与电荷。
CIE 9701/42/O/N/24 Q4(a)(i)–(iii)
When NaOH(aq) is added to an aqueous solution containing [Co(H2O)6]2+ a precipitation reaction occurs accompanied by a colour change. (i) State the colour change seen in this precipitation reaction. (ii) Complete the ionic equation for this precipitation reaction. [Co(H2O)6]2+ + … (iii) This precipitation reaction can also be described as a different type of reaction. Name this type of reaction.
粉红→蓝/蓝绿沉淀等（按卷面 MS）；方程式生成 \mathrm{Co(OH)_2(H_2O)_4} 等；类型 acid-base。
[M] 9701_w24_qp_42 Q4(a)。
CIE 9701/42/M/J/25 Q2(a)(i)–(ii)
Complete Table 2.1 to show the formula and colour of each of the cobalt-containing species present in A, B and C. Identify the type of reaction forming each of B and C. (CoSO4 scheme with NH3 and NaOH.)
\mathrm{[Co(H_2O)_6]^{2+}}、\mathrm{Co(OH)_2}、\mathrm{[Co(NH_3)_6]^{2+}} 等颜色与反应类型。
[M] 9701_s25_qp_42 Q2(a)。
Tridentate L & \(d\)-splitting（三齿配体与 d 轨道）
📖 Definition八面体场：t_{2g} 与 e_g 分裂；高/低自旋与电子数有关。
🔑 Keynon-degenerate：能量不同的 d 轨道集合。
🎯 Exam focus氧化态由配体电荷与配离子总电荷推算。
🛠️ Technique数清：较高能级 d 轨道条数（八面体常为 2 个 e_g）。
CIE 9701/42/O/N/24 Q4(b)(i)–(iv)
L is an uncharged tridentate ligand. Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge. (i) Give the formula of E. (ii) Identify the oxidation state of cobalt in E. (iii) State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level. (iv) Define the term non-degenerate d-orbitals.
\mathrm{[CoL_2]^{2+}}；+2；2 与 3；定义句。
[M] 9701_w24_qp_42 Q4(b)。
CIE 9701/42/M/J/25 Q2(b)(i)
Complete Fig. 2.2 to show the relative energies of the 3d orbitals in an isolated Co2+ ion and in Co2+ in a tetrahedral complex.
四面体分裂：两组轨道能量相反于八面体标记习惯（按卷图完成）。
[M] 9701_s25_qp_42 Q2(b)。
10. Redox titration chains
\(\mathrm{MnO_4^-}\) / \(\mathrm{Fe^{2+}}\)（滴定反应式）
📖 Definition酸性介质：\mathrm{MnO_4^-}+5\mathrm{Fe^{2+}}+8\mathrm{H^+}\rightarrow\mathrm{Mn^{2+}}+5\mathrm{Fe^{3+}}+4\mathrm{H_2O}。
🔑 Key电子数：\mathrm{Mn} +5，\mathrm{Fe} +1。
🎯 Exam focus介质酸化：卷面若写 \mathrm{H^+} 必须配平。
🛠️ Technique先半反应再加总。
CIE 9701/42/O/N/24 Q4(c)(i)
Write an equation for the reaction of Fe2+ ions with MnO4– ions in acid solution.
上式或等价配平（按 MS 接受项）。
[M] 9701_w24_qp_42 Q4(c)(i)。
CIE 9701/42/O/N/24 Q4(c)(ii)
Calculate the number of moles of Fe2+ ions in 25.0 cm3 of solution F.
n(\mathrm{MnO_4^-})=cV\times5=n(\mathrm{Fe^{2+}})。
[M] 9701_w24_qp_42。
\(M_r\) & stoichiometric \(n\)（化学式推断）
📖 Definition由滴定求 \mathrm{Fe^{2+}} 物质的量 \rightarrow 整试样 \rightarrow M_r(\mathrm{FeCr_nO_4})\rightarrow 整数 n。
🔑 Key\mathrm{Fe} 为 +2，\mathrm{Cr} 氧化态用总电荷平衡。
🎯 Exam focus体积 250 cm³ 与取样 25.0 cm³ 比例。
🛠️ Technique最后 n 取最近整数并自检电荷。
CIE 9701/42/O/N/24 Q4(c)(iii)
Calculate the Mr of FeCrnO4 and use your answer to deduce the value of n. One MnO4– ion reacts with five Fe2+ ions.
由 M_r 与原子量表反推 n。
[M] MS 9701_w24_ms_42 Q4(c)(iii)。
CIE 9701/42/O/N/24 Q4(c) stem
The mineral chromite contains a compound which has the formula FeCrnO4. The oxidation state of iron in FeCrnO4 is +2. A sample of 4.18 g of FeCrnO4 is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm3. This is solution F. All the Fe2+ ions in 25.0 cm3 of solution F are oxidised to Fe3+ ions by exactly 18.7 cm3 of 0.0200 mol dm–3 KMnO4.
先建立 Fe 物质的量链，再求摩尔质量。
[M] 9701_w24_qp_42 Q4(c)。
11. \(K_\mathrm{stab}\) & isomerism
\(K_\mathrm{stab}\) & ligand exchange（稳定常数）
📖 DefinitionK_\mathrm{stab} 越大，配合物越稳定；多齿配体通常显著增大 K_\mathrm{stab}。
🔑 Key表达式：生成常数 = 产物浓度幂之积 / 反应物浓度幂之积。
🎯 Exam focusen 与 \mathrm{NH_3} 竞争时看 K_\mathrm{stab} 数量级。
🛠️ Technique配平 ligand exchange：\mathrm{H_2O} 与 en 交换计量。
CIE 9701/42/O/N/24 Q5(a)–(b)(ii)
Complete the expression for the Kstab of [Ni(en)3]2+. (b)(i) Predict which complex ion, [Ni(NH3)6]2+ or [Ni(en)3]2+, is present in the resulting mixture in the highest concentration. Explain your answer. (ii) Complete the equation for the ligand exchange reaction occurring in (i).
K_\mathrm{stab} 写法；选 [\mathrm{Ni(en)_3}]^{2+} 因常数大得多；方程式。
[M] 9701_w24_qp_42 Q5。
CIE 9701/42/O/N/24 Q5(b)(i) context
The numerical values of two stability constants, Kstab, are given in Table 5.1. [Ni(NH3)6]2+ 4.8 × 107. [Ni(en)3]2+ 2.0 × 1018. Explain why [Ni(en)3]2+ is present at higher concentration than [Ni(NH3)6]2+ when both NH3 and en are available.
K_\mathrm{stab} 大得多 \Rightarrow en 为螯合配体置换水/氨后配合物更稳定。
[M] 9701_w24_qp_42 Q5(b)(i) 思路。
Optical isomers of \([\mathrm{Ni(en)_3}]^{2+}\)（光学异构）
📖 Definition八面体与三双齿配体可形成一对非重叠镜像异构体（propeller）。
🔑 Key教材 Ch.24：类型名称 **optical isomerism**。
🎯 Exam focus题目要求用 N–N 表示 en 时勿画成全结构。
🛠️ Technique两图呈镜像且不可叠合。
CIE 9701/42/O/N/24 Q5(c)
Complete Fig. 5.1 to show the three-dimensional structures of the two isomers of [Ni(en)3]2+. Use N—N to represent the en ligand. Name the type of isomerism shown.
画两个镜像 propeller；optical isomerism。
[M] 9701_w24_qp_42 Q5(c)。
CIE 9701/42/M/J/25 Q2(a)(iii)
Suggest a suitable reagent for the formation of [CoCl4]2– from solution A.
浓 \mathrm{HCl} 或 \mathrm{NH_4Cl} 等增加 [\mathrm{Cl^-}] 使平衡移向四氯合钴(II)酸根（按 MS）。
[M] 9701_s25_qp_42 Q2(a)(iii)。
12. Acyl chlorides & polymers
Oxalic acid / \(\mathrm{SOCl_2}\) / repeat unit（草酰氯与聚酯）
📖 Definition二元酸 + 过量 \mathrm{SOCl_2}\rightarrow 双酰氯；与二醇缩聚成聚酯键。
🔑 Key教材 Ch.26：显示 **ester linkage** fully displayed。
🎯 Exam focusRepeat unit 只画一个重复单元且键线完整。
🛠️ Technique标清 linkage 两端连接羰基与氧。
CIE 9701/42/O/N/24 Q6(a)–(b)
Fig. 6.1 shows two reactions of ethanedioic acid, HOOCCOOH. (a)(i) Draw the organic product G in the box in Fig. 6.1. (ii) … SOCl2 … Identify a different reagent that also reacts with HOOCCOOH to produce G. (b) Identify two different reagents that oxidise HOOCCOOH to form carbon dioxide and water.
G 为 (\mathrm{COCl})_2；另一试剂 \mathrm{PCl_5} 等；氧化剂 \mathrm{KMnO_4}、\mathrm{MnO_2} 热等（按 MS）。
[M] 9701_w24_qp_42 Q6(a)(b)。
CIE 9701/42/O/N/24 Q6(d)(iii)–(iv)
(iii) Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, HOCH2CH2OH. The linkage formed between the monomers should be shown fully displayed. (iv) State the type of polymerisation … and name the linkage …
缩聚；酯键 ester linkage。
[M] 9701_w24_qp_42 Q6(d)。
Benzene-1,4-dicarboxylic route（对苯二甲酸合成）
📖 Definition苯环上引入两个羧基：常经甲基氧化或酰基路线（卷面给中间体 J）。
🔑 Key教材 Ch.25–26：氧化用 \mathrm{KMnO_4} 热、碱性或酸性条件。
🎯 Exam focus两步 reagents and conditions 分开写。
🛠️ TechniqueJ 常为对二甲苯或对二取代烷基苯。
CIE 9701/42/O/N/24 Q6(d)(i)–(ii)
Benzene-1,4-dicarboxylic acid, HOOC—C6H4—COOH, can be made from benzene, C6H6, in two steps as shown in Fig. 6.2. (i) Suggest the identity of J by drawing its structure in the box in Fig. 6.2. (ii) Identify the reagents and conditions for step 1 and step 2.
J 为对二甲苯等；step1 \mathrm{CH_3COCl}/\mathrm{AlCl_3} 或甲基化路线；step2 热 \mathrm{KMnO_4} 氧化侧链。
[M] 9701_w24_ms_42 Q6(d)。
CIE 9701/42/M/J/25 Q6(a)(ii)
(ii) Complete the mechanism in Fig. 6.1 for the nitration of methylbenzene to form 1-methyl-2-nitrobenzene. Include all relevant curly arrows and charges.
邻位进攻 \mathrm{NO_2^+}；中间体带正电；去质子恢复苯环。
[M] 9701_s25_qp_42 Q6(a)(ii)（定位取代与卷面 Fig. 6.1）。
13. Arenes, ES & NMR
\(\mathrm{Cl^+}\) ES curly arrows（亲电取代弯箭头）
📖 Definition苯环 \pi 电子进攻 \mathrm{Cl^+} 形成 σ 络合物；去质子恢复芳香性。
🔑 Key指清 x、y 两对电子在 step 前后所在位置（卷面图注）。
🎯 Exam focusCurly arrow 从供电子区指向受电子原子。
🛠️ Technique用题目 diagram 编号作答，勿另画无关结构。
CIE 9701/42/O/N/24 Q7(b)(i)–(ii)
The mechanism for this reaction is shown. (i) The movement of a pair of electrons is represented by x in diagram 1. State where this pair of electrons is before step 1 takes place. State where this pair of electrons is after step 1 has taken place. (ii) The movement of another pair of electrons is represented by y in diagram 2. State where this pair of electrons is before step 2 takes place. State where this pair of electrons is after step 2 has taken place.
x：环上 \pi 键 \rightarrow \mathrm{Cl} 或环碳；y：\mathrm{C–H} 键电子给回环等（按图）。
[M] 9701_w24_qp_42 Q7(b)。
CIE 9701/42/O/N/24 Q7(a)
Give the name or formula of a catalyst that can be used for this reaction.
\mathrm{AlCl_3} / \mathrm{FeCl_3} 等 Lewis acid。
[M] 9701_w24_qp_42 Q7(a)。
CIE 9701/42/O/N/24 Q7(d)–(e)
(d) Complete the equation for this reaction between benzene and chlorine. (e) The mechanism for this reaction is electrophilic substitution. Complete the following sentence. During this reaction, the electrophile is … and a … atom in benzene is substituted by a … atom.
\mathrm{C_6H_6}+\mathrm{Cl_2}\rightarrow\mathrm{C_6H_5Cl}+\mathrm{HCl}（催化剂条件略）；electrophile \mathrm{Cl^+}；H 被 Cl 取代。
[M] 9701_w24_qp_42 Q7(d)(e)。
Wheland hybridisation & ¹H NMR（杂化与氢谱）
📖 Definitionσ 络合物：环上多数碳仍为 sp^2，与亲电碳相连的四面体碳为 sp^3。^1\mathrm{H} NMR：化学环境、裂分与相邻不等价氢数目相关（Ch.30）。
🔑 Key\mathrm{D_2O} 交换掉活泼 \mathrm{O–H}、\mathrm{N–H} 峰，简化谱图。
🎯 Exam focussplitting explanation 用 n+1 规则并说明为何等价/不等价。
🛠️ Technique表格题按列：化学位移组、裂分名称、解释。
CIE 9701/42/O/N/24 Q7(c)
There are six carbon atoms in diagram 2. State how many of these carbon atoms are sp hybridised, sp2 hybridised, and sp3 hybridised.
典型答案：0, 5, 1（以 MS 为准）。
[M] 9701_w24_ms_42 Q7(c)。
CIE 9701/42/O/N/24 Q8(f)
Complete Table 8.1 to describe the peaks seen in the proton (1H) NMR spectrum of HOCH2CH(NH2)COOH dissolved in D2O. Use as many rows in Table 8.1 as you need to, leaving the other rows blank. (group responsible for peak; name of splitting pattern; explanation for splitting pattern)
\mathrm{-CH_2OH}、\mathrm{-CH-}（与 \mathrm{NH_2}、\mathrm{COOH} 已交换）等环境；裂分如 doublet, triplet, doublet of doublets 等（按 MS）。
[M] 9701_w24_qp_42 Q8(f)（教材 Ch.30）。
14. Radical / 2nd order & exam craft
Radical step + \(k\) units（自由基与速率）
📖 Definition均裂 homolytic fission 产生自由基；总级数决定 k。
🔑 KeyVSEPR：\mathrm{HNO_2} 中心 N 键角约 non-linear。
🎯 Exam focusdot-and-cross 只画外层电子。
🛠️ Technique自由基用单点标示未成对电子。
CIE 9701/42/O/N/25 Q2(a)(i)
Use two words to complete the sentence. This reaction involves … … of the single covalent bond between a hydrogen atom and a carbon atom in CH3CHO.
homolytic fission。
[M] 9701_w25_qp_42 Q2(a)(i)。
CIE 9701/42/O/N/25 Q2(a)(ii)–(iii)
Draw dot-and-cross diagrams of NO2 and HNO2 in the boxes. Show outer shell electrons only. (iii) Use VSEPR theory to predict the bond angle at the nitrogen atom in an HNO2 molecule.
\mathrm{NO_2} 弯折；\mathrm{HNO_2} 非直线；键角约 120° 或略小（按 MS）。
[M] 9701_w25_qp_42 Q2(a)。
CIE 9701/42/O/N/25 Q2(c)
The reaction mixture described in (b) is monitored over a period of time. Predict whether the graph of [NO2] against time shows a constant half-life. Explain your answer.
总级数为 2 非一级 \rightarrow 半衰期不恒定。
[M] 9701_w25_qp_42 Q2(c)。
Working, units & command words（应试规范）
📖 DefinitionP4 全卷 **show all your working**；末位常查单位与有效数字。
🔑 Keydefine / describe / explain / suggest 按大纲定义书写（Coursebook margin）。
🎯 Exam focusexplain 必须因果链；suggest 允许合理推断但须自洽。
🛠️ Technique长计算分步写中间结果，便于 ecf。
CIE 9701/42 (generic paper instructions)
You should show all your working and use appropriate units.
每题检查：公式 → 代入 → 结果+单位；与 MS 有效数字规则一致。
[M] 各卷首页 Instructions。
CIE Coursebook A2 + syllabus-map
Assessment intent: reward chemical reasoning chains; penalise missing state symbols / units.
状态符号、条件（acid, heat）、单位与方向性正确同样重要。
[M] 内部教研 syllabus-map.md。
15. Amino acids & peptides
Glutamic acid: pH & zwitterion（谷氨酸离子形态）
📖 Definition氨基酸含酸、碱基团；在某 pH 为 **zwitterion**（净电荷零）；低于/高于 pI 时净正/负电荷占优。谷氨酸侧链羧基使 pI 偏酸。
🔑 Key教材 Ch.28：按 pKa 顺序逐步去质子；画结构时标清 \mathrm{-NH_3^+}、\mathrm{-COO^-} 等。
🎯 Exam focusw24 Q8(h) 四格：pH 1、3、9、14 对应不同去质子阶段。
🛠️ Technique先列各可电离基团，再对照 pH 相对 pKa 决定主要物种。
CIE 9701/42/O/N/24 Q8(h)
The isoelectric point of glutamic acid is pH 3. A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values (at pH 1, 3, 9, 14).
pH1：两羧基质子化、\mathrm{-NH_3^+}；pH3：zwitterion为主；pH9：侧链去质子；pH14：两羧基与氨基全去质子（具体式按 MS）。
[M] 9701_w24_qp_42 Q8(h)。
CIE 9701/42/O/N/24 Q8(a)
Isomer P and isomer Q have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light. State the other property by which they differ.
与手性中心相连基团在三维空间排列不同 \rightarrow 与另一手性试剂反应速率/产物分布不同等（**different reaction with a chiral reagent** 等标准表述）。
[M] 9701_w24_qp_42 Q8(a)。
Dipeptide, pI & electrophoresis（肽与电泳）
📖 Definition**Isoelectric point (pI)**：氨基酸净电荷为零时的 pH；此时在电场中基本不迁移。
🔑 KeyDipeptide：酰胺键（肽键）fully displayed；水解/保护策略见教材 Ch.28。
🎯 Exam focus电泳缓冲 pH 与各物种 pI 比较决定向阳极或阴极移动。
🛠️ Techniques25 Q9：pH 5.7 = serine pI \Rightarrow ser 净电荷 0；lys 带正；肽端基团决定 ser–lys 净电荷。同卷 Q9(c) 常考 **多步合成**（腈水解/还原成胺等），按 Fig. 9.3 写 reagents + conditions。
CIE 9701/42/M/J/25 Q9(a)
Draw the structure for the dipeptide, ser–lys, with molecular formula C9H19N3O4. The peptide functional group formed should be displayed.
Ser 羧基与 Lys 氨基成肽键；侧链 \mathrm{-CH_2OH}、\mathrm{-CH_2CH_2CH_2CH_2NH_2} 保留；C 端、N 端端基正确。
[M] 9701_s25_qp_42 Q9(a)。
CIE 9701/42/M/J/25 Q9(b)
The isoelectric point of serine is 5.7 and of lysine is 9.7. (i) State what is meant by isoelectric point. (ii) A mixture of serine, lysine and ser–lys is analysed by electrophoresis using a buffer at pH 5.7. Draw and label three spots on Fig. 9.2 to indicate the predicted position of each of these three species, after electrophoresis. Explain your answer.
(i) pH at which species has no net charge / does not migrate in electric field。(ii) Ser 在 pI 不移；Lys 带正移向阴极；肽的净电荷由端基与侧链决定（按 MS 位置与解释）。
[M] 9701_s25_qp_42 Q9(b)。
16. \(^{13}\mathrm{C}\) / MS & haloarenes
Aryl vs alkyl halide + \(^{13}\mathrm{C}\)（卤代烃对比）
📖 Definition氯乙烷：亲核取代（\mathrm{S_N1/S_N2}）易；氯苯：C–Cl 键因苯环离域而增强，需苛刻条件，常温 \mathrm{NaOH(aq)} 不反应。
🔑 KeyExplain 题写清 **bond strength / overlap with ring \pi electrons / lone pair on Cl delocalised** 等采分点。
🎯 Exam focus与 w24 Q7 亲电取代机理题同卷联动：芳环上 C–Cl 与脂肪 C–Cl 对比常成对考查。
🛠️ TechniqueQ7(f)：mechanism name + major product 各一分；解释两段：键能 + 轨道/离域。
CIE 9701/42/O/N/24 Q7(f)
Chloroethane reacts with NaOH(aq). Chlorobenzene does not. (i) Name the mechanism of the reaction that chloroethane undergoes with NaOH(aq), and identify the major organic product that is formed. (ii) Explain the difference in reactivity of chloroethane and chlorobenzene when treated with NaOH(aq).
(i) nucleophilic substitution（\mathrm{S_N2} 为主语境）；产物 ethanol。(ii) 氯苯 C–Cl 键短、键能高；氯孤对离域进环 \rightarrow 难亲核取代。
[M] 9701_w24_qp_42 Q7(f)。
CIE 9701/42/O/N/24 Q8(g)
Proline is a naturally occurring amino acid. The skeletal formula of proline is shown. State the number of peaks in the carbon-13 (13C) NMR spectrum of proline.
数清非等价碳环境（环对称性）：通常为 **5**（以 MS 为准）。
[M] 9701_w24_qp_42 Q8(g)。
\(^{13}\mathrm{C}\) / \(^{1}\mathrm{H}\) ketones & MS（碳谱氢谱与质谱）
📖 Definition^{13}\mathrm{C} NMR：数 **chemically distinct** 碳环境；对称性减少峰数。MS：**molecular ion** M^+ 对应分子离子峰 m/z=M_r（整数值取最近）。
🔑 Key鉴别醛酮常用 Tollens / Fehling、acidified \mathrm{K_2Cr_2O_7}、2,4-DNPH 橙色沉淀、甲基酮的 iodoform 反应等（教材 Ch.25–26）。
🎯 Exam focuss22 表题：同一分子式异构体的 ^{1}\mathrm{H} 与 ^{13}\mathrm{C} 峰数可完全不同。
🛠️ TechniqueMS 碎片题：断键位置与稳定碳正离子/自由基相关。
CIE 9701/42/M/J/22 Q7(b)(i)–(ii)
The three isomeric ketones with molecular formula C5H10O are pentan-2-one, pentan-3-one, 3-methylbutanone. (i) Complete Table 7.1 to show the number of peaks observed in the proton (1H) NMR spectrum and in the carbon-13 NMR spectrum for each compound listed. (ii) State all the ketones with molecular formula C5H10O that have: a doublet in their proton (1H) NMR spectrum; a singlet in their proton (1H) NMR spectrum.
(i) 按对称性数 ^1\mathrm{H} 与 ^{13}\mathrm{C} 环境数填表。(ii) 甲基邻接 CH 产生 doublet；等价甲基产生 singlet 等（按 MS）。
[M] 9701_s22_qp_42 Q7(b)。
CIE 9701/42/O/N/21 Q8(c)(i)–(ii)
(c) The mass spectrum of glutamic acid, H2NCH(CH2CH2CO2H)CO2H, is obtained. (i) State the m/e value of the molecular ion peak in this spectrum. (ii) The spectrum has peaks with m/e values of 88 and 131. Draw the structures of the ions responsible for these peaks.
(i) M_r(\mathrm{C_5H_9NO_4})=147\rightarrowm/z=147（分子离子）。(ii) 合理断键：如侧链酰鎓/亚稳碎片等（按 MS 所给结构）。
[M] 9701_w21_qp_42 Q8(c)（P4 质谱经典）。