IB Chemistry HL — Structure 2: Models of Bonding and Structure — Solutions

Q1MEDIUMPaper 12.1 Naming and Formulas

Correct formula for iron(III) sulfate?硫酸铁（iron(III) sulfate）的正确化学式？

(A) $\mathrm{FeSO_4}$
(B) $\mathrm{Fe_2(SO_4)_3}$
(C) $\mathrm{Fe_3(SO_4)_2}$
(D) $\mathrm{FeSO_3}$

Answer:答案： (B)

Iron(III) means $\mathrm{Fe^{3+}}$; sulfate is $\mathrm{SO_4^{2-}}$. To balance charge, the lowest common multiple of 3 and 2 is 6, so 2 $\mathrm{Fe^{3+}}$ (charge $+6$) balance 3 $\mathrm{SO_4^{2-}}$ (charge $-6$), giving $\mathrm{Fe_2(SO_4)_3}$. Trap (C) swaps the subscripts (a very common charge-balancing slip); (A) and (D) assume a $+2$ iron ion, which is iron(II), not iron(III).iron(III) 表示 $\mathrm{Fe^{3+}}$；硫酸根为 $\mathrm{SO_4^{2-}}$。为使电荷平衡，3 与 2 的最小公倍数是 6，因此 2 个 $\mathrm{Fe^{3+}}$（电荷 $+6$）与 3 个 $\mathrm{SO_4^{2-}}$（电荷 $-6$）平衡，得到 $\mathrm{Fe_2(SO_4)_3}$。陷阱 (C) 把下标写反了（这是电荷配平中很常见的失误）；(A)、(D) 则默认铁离子是 $+2$ 价，那是 iron(II) 而非 iron(III)。

Insight洞察 For a compound $\mathrm{M}_x(\mathrm{X})_y$ formed from $\mathrm{M}^{a+}$ and $\mathrm{X}^{b-}$, the "criss-cross" rule $x = b$, $y = a$ only gives the lowest-ratio formula after cancelling any common factor. Always check charge balance directly ($x \cdot a = y \cdot b$) rather than trusting the crossed subscripts blindly.对于由 $\mathrm{M}^{a+}$ 与 $\mathrm{X}^{b-}$ 形成的化合物 $\mathrm{M}_x(\mathrm{X})_y$，"交叉配平"法则 $x = b$、$y = a$ 只有在约去公因数后才给出最简化学式。请始终直接检验电荷平衡（$x \cdot a = y \cdot b$），而不要盲目照搬交叉后的下标。

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Q2MEDIUMPaper 12.1 Lattice Enthalpy

Which charge/radius combination gives the highest lattice enthalpy?哪种电荷/半径组合使晶格焓最高？

(A) Small charge, small radius电荷小、半径小
(B) Large charge, large radius电荷大、半径大
(C) Large charge, small radius电荷大、半径小
(D) Small charge, large radius电荷小、半径大

Answer:答案： (C)

Lattice enthalpy scales with the electrostatic attraction between ions, which increases with the product of the ionic charges and decreases as the inter-ionic distance (sum of ionic radii) increases. Large charge maximizes the numerator; small radius minimizes the separation. (C) maximizes attraction on both counts, so it gives the largest lattice enthalpy.晶格焓随离子间静电吸引力增大而增大，该吸引力随离子电荷的乘积增大而增大，随离子间距（离子半径之和）增大而减小。电荷大使分子（吸引力的"分子项"）最大；半径小使间距最小。(C) 在两方面都使吸引力最大，因此晶格焓最高。

Insight洞察 This is the same charge/radius logic used everywhere in Structure 2: it governs ionic lattice enthalpy (2.1), metallic bond strength (2.3), and (combined with electronegativity) the position of a compound on the bonding triangle (2.4). Learn it once, reuse it three times.这套电荷/半径逻辑贯穿整个 Structure 2：它同样支配离子晶格焓（2.1）、金属键强度（2.3），并与电负性一起决定化合物在成键三角形（2.4）中的位置。学一次，可以用三次。

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Q3MEDIUMPaper 12.2 VSEPR

Electron-domain geometry and molecular shape of $\mathrm{SO_3^{2-}}$?$\mathrm{SO_3^{2-}}$ 的电子域几何与分子几何？

(A) Trigonal planar; trigonal planar平面三角形；平面三角形
(B) Tetrahedral; trigonal pyramidal正四面体；三角锥形
(C) Tetrahedral; tetrahedral正四面体；正四面体
(D) Trigonal bipyramidal; seesaw三角双锥；跷跷板形

Answer:答案： (B)

S has 6 valence electrons. A Lewis structure with three S-O single bonds uses 3 of them in bonding, leaving one lone pair on S. Around S: 3 bonding pairs + 1 lone pair = 4 electron domains → tetrahedral electron-domain geometry. With one domain occupied by a lone pair, the molecular shape (the arrangement of atoms only) is trigonal pyramidal, the same shape as $\mathrm{NH_3}$. (Formal charges: S is $+1$, each O is $-1$; the three $-1$'s and one $+1$ sum to the ion's overall $-2$ charge, confirming the structure is balanced.)S 有 6 个价电子。若画出含三条 S-O 单键的路易斯结构，其中 3 个电子用于成键，S 上剩 1 对孤对电子。S 周围：3 个成键对 + 1 个孤对 = 4 个电子域 → 电子域几何为正四面体。其中一个域被孤对占据，因此分子几何（仅原子的排列方式）为三角锥形，与 $\mathrm{NH_3}$ 相同。（形式电荷：S 为 $+1$，每个 O 为 $-1$；三个 $-1$ 与一个 $+1$ 相加得到该离子的总电荷 $-2$，验证了结构的平衡。）

Insight洞察 The mark-losing trap in VSEPR questions is confusing electron-domain geometry (counts lone pairs) with molecular geometry (ignores lone pairs, describes only where the atoms are). Always state both, in that order: the electron-domain geometry is your reasoning step, and the molecular geometry is your answer to "what shape is the molecule?"VSEPR 题目最容易丢分的陷阱，是把电子域几何（计入孤对）与分子几何（忽略孤对，只描述原子的位置）搞混。务必按此顺序两者都写出：电子域几何是你的推理步骤，分子几何才是"分子是什么形状"这一问题的答案。

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Q4MEDIUMPaper 12.2 Molecular Polarity

Why is $\mathrm{BF_3}$ non-polar overall despite polar bonds?为何 $\mathrm{BF_3}$ 虽有极性键但整体非极性？

(A) Polar overall; all bonds are polar.整体极性；所有键都是极性键。
(B) Non-polar; symmetric cancellation.非极性；对称抵消。
(C) Non-polar; incomplete octet on B.非极性；B 未达八隅体。
(D) Polar; F is most electronegative.极性；F 电负性最大。

Answer:答案： (B)

$\mathrm{BF_3}$ has 3 electron domains around B (3 bonding pairs, 0 lone pairs) → trigonal planar, 120° bond angles. The three identical, symmetrically arranged B-F dipoles point 120° apart and sum vectorially to zero, so the molecule has no net dipole moment even though every individual bond is polar. (C) is a real fact about boron, but it is not the reason for the non-polarity: it is the symmetry, not the octet, that cancels the dipoles.$\mathrm{BF_3}$ 中 B 周围有 3 个电子域（3 成键对，0 孤对）→ 平面三角形，键角 120°。三条完全相同、对称排列的 B-F 偶极相互间隔 120°，矢量相加为零，因此尽管每一条键都是极性的，分子整体却没有净偶极矩。(C) 虽是关于硼的真实事实，却不是非极性的原因：使偶极相消的是对称性，而非八隅体是否完整。

Insight洞察 "Bond polarity" and "molecular polarity" are graded separately on IB mark schemes. State both explicitly: (1) are the individual bonds polar (electronegativity difference)? (2) do the bond dipoles cancel by symmetry? A molecule needs a "no" on step 2 to be non-polar, regardless of the answer to step 1.在 IB 评分标准中，"键的极性"与"分子的极性"是分开给分的。请明确回答两点：(1) 各条键是否有极性（看电负性差）？(2) 键偶极是否因对称而抵消？只有第 (2) 点的答案是"是（抵消）"，分子才是非极性的，无论第 (1) 点的答案如何。

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Q5MEDIUMPaper 12.2 Covalent Network

Why does graphite conduct but diamond does not?为何石墨导电而金刚石不导电？

(A) Ionic bonds between graphite layers.石墨层间是离子键。
(B) Graphite: sp² C, one delocalized electron per atom; diamond: sp³ C, all electrons localized.石墨：sp² 碳，每原子一个离域电子；金刚石：sp³ 碳，电子全部定域。
(C) Diamond is less dense.金刚石密度更低。
(D) Graphite atoms are partially positive.石墨原子带部分正电荷。

Answer:答案： (B)

In graphite, each C forms 3 $\sigma$ bonds to neighbouring C atoms within a hexagonal layer (sp²), leaving one unhybridized p-electron per atom. These p-electrons delocalize into a $\pi$ system spread across the whole layer, free to move under an applied potential difference, hence conduction within a layer (not between layers, which are held only by weak London forces). In diamond, each C forms 4 $\sigma$ bonds (sp³) with no leftover electrons to delocalize, so there are no mobile charge carriers.石墨中每个 C 在六元环层内与相邻 C 形成 3 条 σ 键（sp² 杂化），每个原子留下 1 个未杂化的 p 电子。这些 p 电子离域形成遍布整层的 π 电子体系，在外加电势差下可自由移动，因此可沿层内导电（层间仅靠微弱的伦敦力结合，不导电）。金刚石中每个 C 形成 4 条 σ 键（sp³ 杂化），没有多余电子可供离域，因此没有可移动的电荷载体。

Insight洞察 "Both are covalent network structures of carbon" is a trap that tempts students into treating diamond and graphite as having identical properties. Structure questions almost always want you to trace the property back to the local bonding (hybridization + presence/absence of delocalized electrons), not to the broad category ("covalent network") alone."两者都是碳的共价网状结构"这句话是一个陷阱，容易诱使学生认为金刚石与石墨性质相同。结构类题目几乎总是要求你把性质追溯到局部成键方式（杂化方式 + 是否存在离域电子），而不能只停留在"共价网状结构"这一笼统类别上。

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Q6HARDPaper 12.2 Intermolecular Forces

Boiling points of $\mathrm{CH_4}$, $\mathrm{PH_3}$, $\mathrm{NH_3}$ in increasing order?$\mathrm{CH_4}$、$\mathrm{PH_3}$、$\mathrm{NH_3}$ 沸点由低到高的排序？

(A) $\mathrm{CH_4} < \mathrm{PH_3} < \mathrm{NH_3}$
(B) $\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{PH_3}$
(C) $\mathrm{NH_3} < \mathrm{PH_3} < \mathrm{CH_4}$
(D) $\mathrm{PH_3} < \mathrm{CH_4} < \mathrm{NH_3}$

Answer:答案： (A)

$\mathrm{CH_4}$ ($-161.5^\circ\mathrm{C}$) has only weak London forces and no permanent dipole: lowest boiling point. $\mathrm{PH_3}$ ($-87.7^\circ\mathrm{C}$) is a slightly polar molecule with somewhat stronger London forces than $\mathrm{CH_4}$ (more electrons), but P-H does not hydrogen-bond (P is not electronegative enough). $\mathrm{NH_3}$ ($-33.3^\circ\mathrm{C}$) hydrogen-bonds extensively (H bonded to N, interacting with lone pairs on neighbouring N atoms). Despite having fewer electrons than $\mathrm{PH_3}$ and therefore weaker London forces, the hydrogen bonding dominates and gives $\mathrm{NH_3}$ the highest boiling point of the three.$\mathrm{CH_4}$（$-161.5^\circ\mathrm{C}$）只有微弱的伦敦力、没有永久偶极：沸点最低。$\mathrm{PH_3}$（$-87.7^\circ\mathrm{C}$）是弱极性分子，电子数比 $\mathrm{CH_4}$ 多、伦敦力略强，但 P-H 不能形成氢键（P 的电负性不够）。$\mathrm{NH_3}$（$-33.3^\circ\mathrm{C}$）能广泛形成氢键（H 与 N 成键，并与相邻 N 原子的孤对相互作用）。尽管电子数比 $\mathrm{PH_3}$ 少、伦敦力更弱，但氢键占主导，使 $\mathrm{NH_3}$ 在三者中沸点最高。

Insight洞察 "Bigger molecule = higher boiling point" is only true when comparing substances with the same strongest IMF. The moment hydrogen bonding is possible for one substance and not another, it can reverse an otherwise-expected London-force trend: this NH₃/PH₃ pair is the classic exam example, alongside H₂O vs H₂S and HF vs HCl."分子越大，沸点越高"这一规律只在比较具有相同类型最强分子间作用力的物质时成立。一旦某物质能形成氢键而另一物质不能，就可能逆转本应由伦敦力决定的趋势：NH₃/PH₃ 是经典考试例子，与 H₂O 对 H₂S、HF 对 HCl 属于同一类。

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Q7HARDPaper 1HL2.3 Transition Metal Bonding

Why does Fe have a much higher melting point than K?为何 Fe 的熔点远高于 K？

(A) Fe: smaller radius + delocalized d-electrons.Fe：半径更小 + 离域 d 电子。
(B) Fe is covalent; K is metallic.Fe 是共价键；K 是金属键。
(C) K has higher nuclear charge.K 的核电荷更高。
(D) Fe conducts heat via ions; K via electrons.Fe 靠离子导热；K 靠电子导热。

Answer:答案： (A)

K is a Group 1 metal: it forms $\mathrm{K^+}$, contributing only 1 delocalized electron per atom, and has a comparatively large ionic radius. Fe is a transition (d-block) element: in addition to its 4s electron(s), its 3d electrons are also delocalized, contributing several delocalized electrons per atom, and its ionic radius is considerably smaller than K's. Both factors (higher effective charge/more delocalized electrons, and smaller radius) strengthen the electrostatic attraction between the cation lattice and the electron sea, raising the melting point sharply. (C) is false: K actually has a lower nuclear charge (Z = 19) than Fe (Z = 26); nuclear charge alone is also not the deciding factor for metallic bond strength.K 是第 1 族金属：形成 $\mathrm{K^+}$，每个原子只贡献 1 个离域电子，且离子半径相对较大。Fe 是过渡（d 区）元素：除 4s 电子外，其 3d 电子也参与离域，每个原子贡献多个离域电子，且离子半径明显小于 K。这两个因素（更高的有效贡献电荷/更多离域电子，以及更小的半径）都增强了阳离子晶格与电子海之间的静电吸引，使熔点大幅升高。(C) 错误：K 的核电荷（Z = 19）实际上比 Fe（Z = 26）更低；而且核电荷本身也不是决定金属键强度的关键因素。

Insight洞察 The HL-only content in 2.3.3 is really just "the metallic model, but count d-electrons too." Any question comparing a transition metal to an s-block metal is testing whether you remember that d-electrons delocalize in addition to s-electrons. Do not just say "more electrons": name them as delocalized d-electrons for the mark.2.3.3 中的 HL 专属内容，本质上就是"金属键模型，只不过要把 d 电子也算进去"。任何比较过渡金属与 s 区金属的题目，考的都是你是否记得 d 电子会在 s 电子之外额外离域。不要只笼统说"电子更多"，要明确点出"离域的 d 电子"才能拿到该点分数。

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Q8HARDPaper 1HL2.2 Hybridization & σ/π Bonds

Hybridization and $\sigma$/$\pi$ count for ethene, $\mathrm{CH_2{=}CH_2}$?乙烯 $\mathrm{CH_2{=}CH_2}$ 的杂化方式与 σ/π 键数目？

(A) sp²; 5$\sigma$, 1$\pi$sp²；5σ、1π
(B) sp³; 6$\sigma$, 0$\pi$sp³；6σ、0π
(C) sp; 4$\sigma$, 2$\pi$sp；4σ、2π
(D) sp²; 4$\sigma$, 2$\pi$sp²；4σ、2π

Answer:答案： (A)

Each C in ethene has 3 electron domains: 2 C-H bonds and 1 C=C double bond (a double bond counts as a single domain for hybridization purposes) → sp² hybridization, trigonal planar around each C. Counting all bonds in the molecule: 4 C-H bonds (each 1$\sigma$) + 1 C=C double bond (1$\sigma$ + 1$\pi$). Total: $4 + 1 = 5$ $\sigma$ bonds and $1$ $\pi$ bond.乙烯中每个 C 有 3 个电子域：2 条 C-H 键与 1 条 C=C 双键（就杂化而言，双键只算作一个电子域）→ sp² 杂化，每个 C 周围呈平面三角形。统计整个分子中所有的键：4 条 C-H 键（各 1σ）+ 1 条 C=C 双键（1σ + 1π）。合计：$4 + 1 = 5$ 个 σ 键，$1$ 个 π 键。

Insight洞察 The fast route: hybridization comes from the number of electron domains (2 → sp, 3 → sp², 4 → sp³) exactly as in VSEPR, treating any multiple bond as one domain. Then $\sigma$/$\pi$ counting is separate: every single bond is 1$\sigma$; every double bond is 1$\sigma$ + 1$\pi$; every triple bond is 1$\sigma$ + 2$\pi$. Losing track of which counting rule applies where is the single most common HL slip on this topic.快速解法：杂化方式由电子域数目决定（2 → sp，3 → sp²，4 → sp³），与 VSEPR 完全一致，任何重键都只算一个电子域。而 σ/π 计数是另一套规则：每条单键 = 1σ；每条双键 = 1σ + 1π；每条三键 = 1σ + 2π。分不清这两套规则各自适用的场合，是这一考点上 HL 学生最常见的失分点。

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Q9MEDIUMPaper 12.2 Coordinate Bond

What is the fourth N-H bond in $\mathrm{NH_4^+}$?$\mathrm{NH_4^+}$ 中第四条 N-H 键是什么？

(A) Ionic bond.离子键。
(B) Coordinate (dative) covalent bond; identical once formed.配位（dative）共价键；形成后完全相同。
(C) Metallic bond.金属键。
(D) Coordinate bond, permanently weaker.配位键，永久较弱。

Answer:答案： (B)

In $\mathrm{NH_3}$, N has one lone pair. When $\mathrm{H^+}$ (which has no electrons of its own) bonds to N, both electrons of that lone pair are shared: this is precisely the definition of a coordinate (dative) covalent bond, where both bonding electrons originate from the same atom. Once the bond forms, there is no physical difference between it and the other three N-H bonds: all four are identical in length and strength, and the resulting $\mathrm{NH_4^+}$ ion is a perfect tetrahedron. (D) is a common misconception: the origin of the electrons differs, but the bond itself does not.在 $\mathrm{NH_3}$ 中，N 有一对孤对电子。当（本身不带电子的）$\mathrm{H^+}$ 与 N 成键时，该孤对的两个电子都被共用：这正是配位（dative）共价键的定义：两个成键电子都来自同一原子。键一旦形成，它与另外三条 N-H 键在物理上并无区别：四条键长度和强度完全相同，所得的 $\mathrm{NH_4^+}$ 离子是一个完美的正四面体。(D) 是常见的误解：电子的来源不同，但键本身并无不同。

Insight洞察 "Coordinate bond" describes only how a bond forms, never a distinct, permanently identifiable type of bond in the final structure. The exam-safe phrase is "coordinate (dative) covalent bond, indistinguishable from the other bonds once formed": always add that second clause for full marks."配位键"描述的只是键是如何形成的，绝不是最终结构中一种独立、可永久识别的键类型。考试中安全的表述是"配位（dative）共价键，一旦形成便与其他键无法区分"：务必加上后半句才能拿满分。

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Structured Response结构化解答题

SR 1MEDIUMPaper 22.1 The Ionic Model

The ionic bonding model: naming, conductivity, lattice enthalpy trends, and solution conductivity.离子键模型：命名、导电性、晶格焓趋势与溶液导电性。

(a) Formula and name.化学式与名称。

$\mathrm{Ca^{2+}}$ and $\mathrm{PO_4^{3-}}$: LCM(2, 3) = 6, so 3 $\mathrm{Ca^{2+}}$ (charge $+6$) balance 2 $\mathrm{PO_4^{3-}}$ (charge $-6$).$\mathrm{Ca^{2+}}$ 与 $\mathrm{PO_4^{3-}}$：LCM(2, 3) = 6，故 3 个 $\mathrm{Ca^{2+}}$（电荷 $+6$）与 2 个 $\mathrm{PO_4^{3-}}$（电荷 $-6$）平衡。

$$\mathrm{Ca_3(PO_4)_2} \quad \text{(calcium phosphate)}$$

(b) Solid vs molten conductivity.固态与熔融态导电性。

In the solid, ions occupy fixed positions in the lattice, held by strong electrostatic attraction, and are not free to move, so there are no mobile charge carriers and the solid does not conduct. When molten, the lattice breaks down and the ions become free to move, so they can carry charge through the liquid, and the compound conducts. (In both cases it is the ions, not electrons, that carry the current.)固态时，离子固定在晶格中的位置上，被强静电引力束缚，不能自由移动，因此没有可移动的电荷载体，固体不导电。熔融后晶格瓦解，离子变得可以自由移动，能够在液体中携带电荷，化合物因此导电。（无论哪种情况，导电的都是离子而非电子。）

Lattice enthalpy increases with the product of the ionic charges and decreases with the sum of the ionic radii. $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ each carry twice the charge of $\mathrm{Na^+}$ and $\mathrm{Cl^-}$, and both ions are also smaller than their NaCl counterparts. Both effects act in the same direction, so the electrostatic attraction (and hence the lattice enthalpy) is far greater for MgO than for NaCl, consistent with $+3795$ vs $+787~\mathrm{kJ\,mol^{-1}}$.晶格焓随离子电荷的乘积增大而增大，随离子半径之和增大而减小。$\mathrm{Mg^{2+}}$ 与 $\mathrm{O^{2-}}$ 的电荷都是 $\mathrm{Na^+}$、$\mathrm{Cl^-}$ 的两倍，且两离子半径也都比 NaCl 中对应离子更小。两个因素方向一致，因此 MgO 的静电吸引力（进而晶格焓）远大于 NaCl，与 $+3795$ 对 $+787~\mathrm{kJ\,mol^{-1}}$ 的数据相符。

(d) CaO vs MgO.CaO 与 MgO 的比较。

CaO's lattice enthalpy is less than MgO's. Both compounds involve $2+$/$2-$ ions, so charge is the same in both cases; the deciding factor is radius. $\mathrm{Ca^{2+}}$ is larger than $\mathrm{Mg^{2+}}$ (Ca is below Mg in Group 2), giving a greater inter-ionic distance and therefore a weaker electrostatic attraction. (Consistent with real values: CaO $\approx 3401~\mathrm{kJ\,mol^{-1}} <$ MgO's $3795~\mathrm{kJ\,mol^{-1}}$.)CaO 的晶格焓小于 MgO。两种化合物都是 $2+$/$2-$ 离子，电荷相同；决定因素是半径。$\mathrm{Ca^{2+}}$ 比 $\mathrm{Mg^{2+}}$ 大（Ca 在第 2 族中位于 Mg 下方），离子间距更大，静电吸引更弱。（与真实数值相符：CaO $\approx 3401~\mathrm{kJ\,mol^{-1}} <$ MgO 的 $3795~\mathrm{kJ\,mol^{-1}}$。）

(e) Identifying X and Y.判断 X 与 Y。

X is metallically bonded: delocalized electrons are already free to move in the solid, so it conducts as a solid; the non-directional metallic bond lets layers of cations slide past each other (the electron sea simply readjusts), so it is malleable rather than brittle.X 为金属键：固态中离域电子本就可以自由移动，因此固态即可导电；金属键无方向性，阳离子层可以相互滑动（电子海随之调整），因此具延展性而非脆性。

Y is ionically bonded: in the solid, ions are fixed in the lattice (no conduction); once molten, ions are free to move (conducts). Under stress, layers shift and bring like-charged ions into alignment, causing strong repulsion that shatters the crystal, hence brittle.Y 为离子键：固态时离子固定在晶格中（不导电）；熔融后离子可自由移动（导电）。受力时晶格层错位，使同号电荷离子相互对齐而强烈排斥，导致晶体碎裂，因此表现为脆性。

(f) NaCl vs $\mathrm{MgCl_2}$ conductivity.NaCl 与 $\mathrm{MgCl_2}$ 的导电性比较。

For equal moles dissolved: 1 mol NaCl gives 1 mol $\mathrm{Na^+}$ + 1 mol $\mathrm{Cl^-}$ = 2 mol of ions total. 1 mol $\mathrm{MgCl_2}$ gives 1 mol $\mathrm{Mg^{2+}}$ + 2 mol $\mathrm{Cl^-}$ = 3 mol of ions total, and the cation carries double the charge of $\mathrm{Na^+}$. The $\mathrm{MgCl_2}$ solution therefore has both more ions per mole of solute and a higher-charged cation, giving it a greater concentration of charge carriers overall. Prediction: the $\mathrm{MgCl_2}$ solution conducts more strongly.等物质的量溶解时：1 mol NaCl 生成 1 mol $\mathrm{Na^+}$ + 1 mol $\mathrm{Cl^-}$ = 共 2 mol 离子。1 mol $\mathrm{MgCl_2}$ 生成 1 mol $\mathrm{Mg^{2+}}$ + 2 mol $\mathrm{Cl^-}$ = 共 3 mol 离子，且阳离子电荷是 $\mathrm{Na^+}$ 的两倍。因此 $\mathrm{MgCl_2}$ 溶液每摩尔溶质产生的离子更多，且阳离子电荷更高，整体电荷载体浓度更大。预测：$\mathrm{MgCl_2}$ 溶液导电性更强。

Insight洞察 Every part of this question reduces to the same two levers: ion charge and ion mobility/count. (c)-(d) use charge and radius to compare lattice enthalpy; (e) uses mobility (fixed vs free) to distinguish bonding types; (f) uses both charge and count together to compare solution conductivity. Structured-response questions on 2.1 are almost always this same toolkit applied to a new pair of substances.本题各小题都可归结为同样两个杠杆：离子电荷与离子迁移率/数目。(c)-(d) 用电荷与半径比较晶格焓；(e) 用迁移率（固定 vs 自由）区分成键类型；(f) 则同时用电荷与数目比较溶液导电性。2.1 的结构化解答题几乎总是把这同一套工具应用到新的一对物质上。

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SR 2HARDPaper 2HL2.2 Resonance, Formal Charge, Hybridization

HL extensions of the covalent model: nitrate resonance and formal charge, $\mathrm{SF_4}$ expanded octet, ethyne hybridization and $\sigma$/$\pi$ bonds.共价模型的 HL 拓展：硝酸根的共振与形式电荷，$\mathrm{SF_4}$ 的扩展八隅体，乙炔的杂化与 σ/π 键。

(a) Formal charges on $\mathrm{NO_3^-}$.$\mathrm{NO_3^-}$ 的形式电荷。

Lewis structure: one N=O double bond, two N-O single bonds; N has no lone pair, each single-bonded O has 3 lone pairs, the double-bonded O has 2 lone pairs.路易斯结构：一条 N=O 双键，两条 N-O 单键；N 无孤对，每个单键 O 有 3 对孤对，双键 O 有 2 对孤对。

$$FC = (\text{valence } e^-) - (\text{lone-pair } e^-) - \tfrac{1}{2}(\text{bonding } e^-)$$

$$FC(\mathrm{N}) = 5 - 0 - \tfrac{1}{2}(8) = +1 \qquad FC(\mathrm{O_{\text{double}}}) = 6 - 4 - \tfrac{1}{2}(4) = 0 \qquad FC(\mathrm{O_{\text{single}}}) = 6 - 6 - \tfrac{1}{2}(2) = -1$$

Sum: $(+1) + (0) + (-1) + (-1) = -1$, which matches the overall charge of the nitrate ion, confirming the structure is correctly drawn.总和：$(+1) + (0) + (-1) + (-1) = -1$，与硝酸根离子的总电荷一致，确认结构画法正确。

(b) Resonance and bond length.共振与键长。

Three equivalent Lewis structures can be drawn for $\mathrm{NO_3^-}$, differing only in which N-O bond is drawn as the double bond. Since none is more "correct" than the others, the true structure is a resonance hybrid: an equal-weighted average with an effective bond order of $\tfrac{4}{3}$ for each N-O bond. All three N-O bonds are therefore equal in length, shorter than a typical N-O single bond but longer than a typical N=O double bond: intermediate between the two.$\mathrm{NO_3^-}$ 可以画出三种等价的路易斯结构，区别仅在于哪一条 N-O 键被画成双键。由于没有哪一种结构比其他更"正确"，真实结构是这些结构的共振杂化体：等权重平均，每条 N-O 键的有效键级为 $\tfrac{4}{3}$。因此三条 N-O 键长度相等，比典型的 N-O 单键更短，但比典型的 N=O 双键更长：介于两者之间。

S has 6 valence electrons; forming 4 S-F single bonds uses 4 of them, leaving 2 electrons as 1 lone pair. Around S: 4 bonding pairs + 1 lone pair = 5 electron domains → trigonal bipyramidal electron-domain geometry. With the lone pair occupying an equatorial position (to minimize repulsion), the molecular geometry is seesaw.S 有 6 个价电子；形成 4 条 S-F 单键用去 4 个，剩 2 个电子成 1 对孤对。S 周围：4 成键对 + 1 孤对 = 5 个电子域 → 电子域几何为三角双锥。孤对占据一个平伏（equatorial）位置以使排斥最小，分子几何为跷跷板形（seesaw）。

(d) Ethyne hybridization and bond count.乙炔的杂化与键数。

Each C in $\mathrm{H{-}C{\equiv}C{-}H}$ has only 2 electron domains (1 C-H bond + 1 C≡C triple bond, the triple bond counting as one domain) → sp hybridization at each carbon, giving the observed linear (180°) geometry.$\mathrm{H{-}C{\equiv}C{-}H}$ 中每个 C 只有 2 个电子域（1 条 C-H 键 + 1 条 C≡C 三键，三键算作一个电子域）→ 每个碳都是 sp 杂化，形成实测的直线形（180°）几何。

Bond count: 2 C-H bonds (1$\sigma$ each) + 1 C≡C triple bond (1$\sigma$ + 2$\pi$). Total: $2 + 1 = 3$ $\sigma$ bonds and $2$ $\pi$ bonds.键的统计：2 条 C-H 键（各 1σ）+ 1 条 C≡C 三键（1σ + 2π）。合计：$2 + 1 = 3$ 个 σ 键，$2$ 个 π 键。

(e) $\sigma$ vs $\pi$; restricted rotation.σ 与 π 的区别；旋转受限。

A $\sigma$ bond forms by head-on (end-to-end) overlap of orbitals directly along the internuclear axis, concentrating electron density symmetrically between the two nuclei. A $\pi$ bond forms by lateral (sideways) overlap of parallel p-orbitals, placing electron density in two lobes above and below (or in front of and behind) the bond axis, not along it.σ 键由轨道沿核间轴正面（首尾）重叠形成，电子密度对称地集中在两核之间。π 键由平行 p 轨道侧面重叠形成，电子密度分布在键轴上下（或前后）两侧的两个瓣中，而非沿轴分布。

Because a $\sigma$ bond's overlap is symmetric about the bond axis, one half of the molecule can rotate relative to the other without disturbing the orbital overlap: rotation about a pure single ($\sigma$-only) bond is essentially free. A double bond, however, includes a $\pi$ component whose overlap requires the two p-orbitals to stay parallel; rotating one CH$_2$ group by 90° relative to the other would misalign the p-orbitals and break the $\pi$ bond entirely. Rotation about a C=C double bond is therefore restricted (it costs the full $\pi$-bond energy), which is why alkenes show cis/trans isomerism while single-bonded chains do not.由于 σ 键的重叠关于键轴对称，分子的一半可以相对另一半旋转而不破坏轨道重叠：绕纯单键（只有 σ）的旋转基本是自由的。但双键还包含 π 成分，其重叠要求两个 p 轨道保持平行；若把一个 CH$_2$ 基团相对另一个旋转 90°，会使 p 轨道错位，从而彻底破坏 π 键。因此绕 C=C 双键的旋转受到限制（需要付出整个 π 键的键能），这也是烯烃存在顺式/反式异构而单键链不存在的原因。

Insight洞察 All four HL topics in this question (resonance, formal charge, expanded octets, hybridization/σ-π) share one root idea: a single classical Lewis structure is often an incomplete or ambiguous picture, and each tool refines it in a different way: formal charge picks the best single structure, resonance averages multiple valid structures, expanded octets extend the electron-counting rules for period 3+, and hybridization/σ-π describes the real 3-D orbitals underneath the 2-D drawing. Recognizing which tool a question wants is half the battle.本题涉及的四个 HL 考点（共振、形式电荷、扩展八隅体、杂化/σ-π）都源自同一个核心思想：单一的经典路易斯结构往往是不完整或含糊的图像，而每种工具都从不同角度加以完善：形式电荷用来挑选最佳的单一结构，共振对多个合理结构取平均，扩展八隅体把电子计数规则推广到第 3 周期及以下，杂化/σ-π 则描述二维图形背后真实的三维轨道。能识别一道题目想用哪一种工具，就已经解决了一半的问题。

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SR 3MEDIUMPaper 2HL2.3 / 2.4 Metals, Alloys, Polymers

Metallic bonding trends, transition-metal bonding, alloys, and addition/condensation polymers.金属键趋势、过渡金属成键、合金与加成/缩合聚合物。

(a) Na vs Al melting point.Na 与 Al 的熔点比较。

Na forms $\mathrm{Na^+}$, contributing only 1 delocalized electron per atom, and has a relatively large ionic radius. Al forms $\mathrm{Al^{3+}}$, contributing 3 delocalized electrons per atom, and has a smaller ionic radius. Both the higher charge and the smaller radius strengthen the electrostatic attraction between the cation lattice and the delocalized electron sea in Al, giving it a much stronger metallic bond and hence a far higher melting point (933 K vs 371 K).Na 形成 $\mathrm{Na^+}$，每原子仅贡献 1 个离域电子，离子半径相对较大。Al 形成 $\mathrm{Al^{3+}}$，每原子贡献 3 个离域电子，离子半径更小。更高的电荷与更小的半径都增强了 Al 中阳离子晶格与离域电子海之间的静电吸引，使其金属键强得多，因此熔点也高得多（933 K 对 371 K）。

(b) Transition-metal bonding.过渡金属成键。

d-block elements have both s- and d-electrons available for delocalization, so each atom contributes more delocalized electrons to the electron sea than a Group 1 metal of similar atomic size. More delocalized electrons per atom means a stronger electrostatic attraction to the lattice of cations, so transition metals typically have much higher melting points than Group 1 metals.d 区元素既有 s 电子也有 d 电子可供离域，因此原子尺寸相近的情况下，每个原子向电子海贡献的离域电子比第 1 族金属更多。每原子离域电子越多，对阳离子晶格的静电吸引就越强，因此过渡金属的熔点通常远高于第 1 族金属。

Cr and Ni atoms have different radii from Fe atoms. When they substitute into the iron lattice, they disrupt its otherwise regular, ordered arrangement. This makes it much harder for layers of atoms to slide smoothly past one another under stress (the mechanism that gives pure metals their malleability), so the alloy resists deformation more strongly and is harder than pure iron.Cr 与 Ni 原子的半径与 Fe 不同。当它们替代进入铁的晶格时，会打乱原本规则有序的排列。这使得原子层在受力时更难平滑地相互滑动（正是这种机制赋予纯金属延展性），因此合金抵抗形变的能力更强，比纯铁更硬。

(d) Addition vs condensation polymers.加成聚合物与缩合聚合物。

Addition polymers form when monomers containing a C=C double bond (alkenes) open that bond and link directly to one another; no atoms are lost, so atom economy is 100% (e.g., ethene → polyethene). Condensation polymers form when monomers carrying two reactive functional groups (e.g., a diol or diamine reacting with a dicarboxylic acid) react repeatedly, releasing a small molecule (usually $\mathrm{H_2O}$) at each new linkage.加成聚合物由含 C=C 双键的单体（烯烃）打开双键并直接首尾相连而成；不脱去任何原子，原子经济性为 100%（如乙烯 → 聚乙烯）。缩合聚合物由带两个反应性官能团的单体（如二元醇或二元胺与二元羧酸）反复反应而成，每形成一个新的连接就脱去一个小分子（通常是 $\mathrm{H_2O}$）。

(e) Classifying polyethene, nylon, PET.对聚乙烯、尼龙、PET 分类。

Polyethene: addition polymer (monomer is the alkene ethene; no small molecule released). Nylon: condensation polymer (polyamide from a diamine + a dicarboxylic acid); releases $\mathrm{H_2O}$ at each amide linkage. PET: condensation polymer (polyester from a diol, ethylene glycol, + a dicarboxylic acid, terephthalic acid); releases $\mathrm{H_2O}$ at each ester linkage.聚乙烯：加成聚合物（单体是烯烃乙烯；不释放小分子）。尼龙：缩合聚合物（由二元胺 + 二元羧酸生成的聚酰胺）；每个酰胺连接释放一个 $\mathrm{H_2O}$。PET：缩合聚合物（由二元醇，即乙二醇，+ 二元羧酸，即对苯二甲酸，生成的聚酯）；每个酯键连接释放一个 $\mathrm{H_2O}$。

(f) Nylon vs polyethene strength.尼龙与聚乙烯的强度比较。

Nylon chains contain amide ($\mathrm{-CONH-}$) groups, whose N-H and C=O can form hydrogen bonds between neighbouring chains: a comparatively strong intermolecular force. Polyethene chains are non-polar hydrocarbons held together only by (weaker) London dispersion forces. The stronger interchain hydrogen bonding in nylon requires more energy to separate chains or slide them past one another, giving nylon a higher tensile strength and a higher melting point than polyethene.尼龙链含有酰胺（$\mathrm{-CONH-}$）基团，其 N-H 与 C=O 可在相邻链之间形成氢键：一种相对较强的分子间作用力。聚乙烯链是非极性烃，链间仅靠（较弱的）伦敦色散力结合。尼龙链间更强的氢键需要更多能量才能使链分离或相互滑动，因此尼龙的抗拉强度和熔点都高于聚乙烯。

Insight洞察 Parts (a)-(b) are "bigger charge / more delocalized electrons / smaller radius = stronger metallic bond," parts (c)-(f) are "disrupted lattice = harder" and "stronger IMF between chains = tougher, higher-melting polymer." Structure 2.3/2.4 questions almost never ask you to derive new chemistry: they ask you to correctly identify which structural feature (lattice regularity vs interchain force vs delocalized-electron count) is doing the work in a given comparison.(a)-(b) 都是"电荷更大/离域电子更多/半径更小 = 金属键更强"，(c)-(f) 则是"晶格被打乱 = 更硬"与"链间分子间作用力更强 = 聚合物更坚韧、熔点更高"。Structure 2.3/2.4 的题目几乎从不要求你推导新的化学知识：它们要求你正确判断在给定的比较中，究竟是哪一种结构特征（晶格规则性、链间作用力，还是离域电子数目）在起决定作用。

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Data-Based Question (HL)数据题（HL）

P3-1HARDPaper 3 HLHL2.4 Bonding Triangle

Electronegativity differences and average electronegativities placed on the bonding triangle for NaF, MgO, HCl, $\mathrm{Cl_2}$, and Al.利用电负性差与平均电负性，将 NaF、MgO、HCl、$\mathrm{Cl_2}$ 与 Al 置于成键三角形中。

(a) $\Delta\chi$ and $\bar\chi$ for NaF and HCl.NaF 与 HCl 的 $\Delta\chi$ 与 $\bar\chi$。

$$\text{NaF: } \Delta\chi = 4.0 - 0.9 = 3.1, \quad \bar\chi = \frac{4.0+0.9}{2} = 2.45$$

$$\text{HCl: } \Delta\chi = 3.2 - 2.2 = 1.0, \quad \bar\chi = \frac{3.2+2.2}{2} = 2.7$$

(b) $\mathrm{Cl_2}$ and Al: same $\Delta\chi$, different regions.$\mathrm{Cl_2}$ 与 Al：$\Delta\chi$ 相同，区域却不同。

$\Delta\chi = 0$ only tells you the bond is non-polar between the two atoms: it does not by itself tell you whether the bonding is covalent or metallic. That depends on what kind of elements are bonding and on the average electronegativity, $\bar\chi$. $\mathrm{Cl_2}$: two identical non-metal atoms sharing an electron pair, with high $\bar\chi = 3.2$: a classic non-polar covalent (molecular) bond, placing it at the covalent corner of the bonding triangle. Al: identical metal atoms, with low $\bar\chi = 1.6$: here the atoms readily lose electrons to a shared delocalized "sea" rather than sharing localized pairs, which is metallic bonding, placing Al at the metallic corner. So $\Delta\chi = 0$ with high $\bar\chi$ signals covalent; $\Delta\chi = 0$ with low $\bar\chi$ signals metallic.$\Delta\chi = 0$ 只说明两原子间的键是非极性的：它本身并不能说明成键是共价还是金属。这取决于成键元素的种类以及平均电负性 $\bar\chi$。$\mathrm{Cl_2}$：两个相同的非金属原子共用一对电子，$\bar\chi = 3.2$ 较高，是典型的非极性共价（分子）键，位于成键三角形的共价角。Al：相同的金属原子，$\bar\chi = 1.6$ 较低，这里原子倾向于把电子交给共享的离域"电子海"，而非共用定域电子对，这就是金属键，因此 Al 位于成键三角形的金属角。所以 $\Delta\chi = 0$ 且 $\bar\chi$ 高，指向共价；$\Delta\chi = 0$ 且 $\bar\chi$ 低，指向金属。

NaF: ionic ($\Delta\chi = 3.1$, large; does not conduct as solid, conducts molten, high $T_m = 993^\circ\mathrm{C}$).NaF：离子键（$\Delta\chi = 3.1$，很大；固态不导电，熔融导电，$T_m = 993^\circ\mathrm{C}$ 很高）。
MgO: ionic ($\Delta\chi = 2.1$, large; does not conduct as solid, conducts molten, very high $T_m = 2852^\circ\mathrm{C}$ due to $2+/2-$ charges).MgO：离子键（$\Delta\chi = 2.1$，仍然较大；固态不导电，熔融导电，因 $2+/2-$ 电荷而 $T_m = 2852^\circ\mathrm{C}$ 极高）。
HCl: polar covalent, simple molecular ($\Delta\chi = 1.0$, moderate; does not conduct in either state as a pure substance; very low $T_m = -114^\circ\mathrm{C}$, consistent with only weak intermolecular forces between molecules).HCl：极性共价，简单分子型（$\Delta\chi = 1.0$，中等；作为纯物质在固态或液态都不导电；$T_m = -114^\circ\mathrm{C}$ 很低，与分子间只有微弱作用力一致）。
$\mathrm{Cl_2}$: non-polar covalent, simple molecular ($\Delta\chi = 0$, high $\bar\chi$; non-conductor in both states; low $T_m = -101^\circ\mathrm{C}$).$\mathrm{Cl_2}$：非极性共价，简单分子型（$\Delta\chi = 0$，$\bar\chi$ 较高；两态均不导电；$T_m = -101^\circ\mathrm{C}$ 较低）。
Al: metallic ($\Delta\chi = 0$, low $\bar\chi$; conducts as a solid, the diagnostic property no ionic or molecular substance shows, and remains a moderate-to-high $T_m = 660^\circ\mathrm{C}$).Al：金属键（$\Delta\chi = 0$，$\bar\chi$ 较低；固态即可导电，这是离子型或分子型物质都不具备的判别性质，且 $T_m = 660^\circ\mathrm{C}$ 处于中高水平）。

(d) MgO's smaller $\Delta\chi$ but far higher $T_m$.MgO 的 $\Delta\chi$ 更小但 $T_m$ 却高得多。

Electronegativity difference is a measure of bond polarity, that is, how unevenly the bonding electrons are shared, not a direct measure of bond strength. The strength of an ionic bond (its lattice enthalpy, and hence its melting point) instead follows a Coulomb's-law-type relationship: attraction $\propto \dfrac{q_+ \, q_-}{r}$. $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ each carry twice the charge magnitude of $\mathrm{Na^+}$ and $\mathrm{F^-}$, so the charge product for MgO ($2 \times 2 = 4$) is four times that of NaF ($1 \times 1 = 1$) even though the ionic radii are broadly comparable. This much larger charge product outweighs MgO's smaller $\Delta\chi$, producing a far stronger electrostatic attraction, a much higher lattice enthalpy, and consequently a much higher melting point.电负性差衡量的是键的极性，即成键电子分配的不均匀程度，而不是直接衡量键的强度。离子键的强度（其晶格焓，进而其熔点）遵循库仑定律式的关系：吸引力 $\propto \dfrac{q_+ \, q_-}{r}$。$\mathrm{Mg^{2+}}$ 与 $\mathrm{O^{2-}}$ 的电荷大小都是 $\mathrm{Na^+}$、$\mathrm{F^-}$ 的两倍，因此 MgO 的电荷乘积（$2 \times 2 = 4$）是 NaF（$1 \times 1 = 1$）的四倍，而两者的离子半径大体相当。这一大得多的电荷乘积压倒了 MgO 较小的 $\Delta\chi$，产生了强得多的静电吸引、大得多的晶格焓，进而熔点也高得多。

(e) $\mathrm{SiCl_4}$ polarity.$\mathrm{SiCl_4}$ 的极性。

Si has 4 valence electrons, all used in 4 single bonds to Cl, leaving no lone pairs on Si → 4 bonding domains, 0 lone pairs → tetrahedral molecular geometry (109.5°), directly analogous to $\mathrm{CCl_4}$. Although each Si-Cl bond is polar ($\Delta\chi = 1.3$), the four identical bond dipoles are arranged symmetrically around the tetrahedral centre and cancel exactly by vector sum. Prediction: $\mathrm{SiCl_4}$ is non-polar overall, despite having polar bonds.Si 有 4 个价电子，全部用于与 Cl 形成 4 条单键，Si 上无孤对 → 4 个成键域，0 个孤对 → 分子几何为正四面体（109.5°），与 $\mathrm{CCl_4}$ 完全类似。尽管每条 Si-Cl 键都是极性的（$\Delta\chi = 1.3$），但四条相同的键偶极围绕正四面体中心对称排列，矢量和恰好为零。预测：$\mathrm{SiCl_4}$ 整体是非极性的，尽管含有极性键。

Insight洞察 The single biggest trap on bonding-triangle questions is treating $\Delta\chi$ as a one-number answer to "what kind of bonding is this?" It is not: you always need a second axis (average electronegativity, to separate metallic from covalent when $\Delta\chi$ is small; or independent property data, to confirm a classification). And even within "ionic," $\Delta\chi$ tells you nothing about lattice enthalpy or melting point: that requires falling back on the separate charge/radius (Coulomb's law) argument from 2.1, exactly as in part (d) here.成键三角形题目中最大的陷阱，是把 $\Delta\chi$ 当成回答"这是哪种成键"的唯一数字。事实并非如此：你总需要第二个维度（当 $\Delta\chi$ 很小时，用平均电负性来区分金属键与共价键；或用独立的性质数据来确认分类）。而且即使在"离子键"内部，$\Delta\chi$ 也无法告诉你晶格焓或熔点的信息：这需要回到 2.1 中独立的电荷/半径（库仑定律）论证，正如本题 (d) 所示。

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