IB Chemistry: Analysis of Electron Sharing Reactions | IB化学：电子共享反应解析

📚 IB Chemistry: Analysis of Electron Sharing Reactions | IB化学：电子共享反应解析

In the IB Chemistry syllabus, the idea of sharing electrons sits at the heart of covalent bonding. When non-metal atoms combine, they rarely transfer electrons completely; instead, they achieve stable noble-gas configurations by sharing one or more pairs of valence electrons. This shared-pair model explains molecular structures, polarities, bond enthalpies and even the geometry of large molecules.

在 IB 化学教学大纲中，电子共享的概念居于共价键理论的核心位置。当非金属原子结合时，它们很少完全转移电子；恰恰相反，它们通过共享一对或多对价电子来达到稳定的稀有气体构型。这种共享电子对模型解释了分子结构、极性、键焓，甚至大分子的几何形状。

1. What Does ‘Sharing Electrons’ Mean? | “共享电子”究竟意味着什么？

In an ionic model, one atom donates valence electrons to another, creating oppositely charged ions. In a covalent model, however, the two nuclei compete for the same electron pair. A shared pair of electrons occupies the region between the two nuclei, and the electrostatic attraction between both nuclei and the shared negative charge holds the atoms together.

在离子模型中，一个原子将价电子捐给另一个原子，从而形成带相反电荷的离子。然而在共价模型中，两个原子核竞争同一对电子。一对共享电子占据两个原子核之间的区域，两个原子核与共享负电荷之间的静电吸引将原子维系在一起。

Because the electrons are delocalised over two centres rather than being fixed on one atom, the bond is directional. This directionality is the reason covalent molecules have distinct shapes, unlike ionic lattices which are simply close-packed arrangements.

由于电子离域在两个中心之间而不是固定于某一个原子上，共价键具有方向性。这种方向性是共价分子具有明确形状的原因；而离子晶格仅仅是密堆积排列，没有这种方向性。

2. Lewis Structures and the Octet Rule | 路易斯结构与八隅体规则

The octet rule states that main-group elements tend to share, gain or lose electrons so that they acquire eight valence electrons, mimicking the nearest noble gas. In a Lewis structure, each shared pair is drawn as a single line (e.g. H–H), while lone pairs are shown as pairs of dots around the atom.

八隅体规则指出，主族元素倾向于通过共享、获得或失去电子来拥有八个价电子，从而效仿最近的稀有气体。在路易斯结构中，每对共享电子画成一条短线（例如 H–H），孤对电子则画成原子周围成对的圆点。

To draw a reliable Lewis structure, count all valence electrons, place the central atom (usually the least electronegative), then connect atoms with single bonds. Distribute remaining electrons as lone pairs, and form double or triple bonds only if the central atom still lacks an octet.

要画出可靠的路易斯结构，需先计算所有价电子，将中心原子（通常电负性最小）置于中央，再用单键连接各原子。将其余电子分配为孤对电子；仅当中心原子仍缺少八隅体时才形成双键或三键。

Common examples include H₂O (water, with two lone pairs on oxygen), NH₃ (ammonia, with one lone pair on nitrogen) and CO₂ (carbon dioxide, with two double bonds and no lone pairs on the central carbon).

常见的例子包括 H₂O（水，氧上有两对孤对电子）、NH₃（氨，氮上有一对孤对电子）以及 CO₂（二氧化碳，中心碳原子上有两个双键，无孤对电子）。

3. Single, Double and Triple Bonds | 单键、双键与三键

When two atoms share one pair of electrons, they form a single bond; two shared pairs give a double bond, and three shared pairs give a triple bond. The bond order therefore equals the number of shared electron pairs between two atoms.

当两个原子共享一对电子时形成单键；共享两对电子形成双键；共享三对电子形成三键。因此，键级等于两个原子之间共享电子对的数目。

Consider molecular hydrogen: H· + ·H → H:H, so the bond order in H₂ is 1. Oxygen forms O₂ with a double bond (O=O) and nitrogen forms N₂ with a triple bond (N≡N), each nitrogen also retaining one lone pair.

以氢分子为例：H· + ·H → H:H，因此 H₂ 中键级为 1。氧形成含双键的 O₂（O=O），氮则形成含三键的 N₂（N≡N），且每个氮原子还保留一对孤对电子。

As bond order increases, bond length decreases and bond strength increases. The N≡N triple bond (about 945 kJ mol⁻¹) is far stronger than the N–N single bond (about 158 kJ mol⁻¹), which explains why nitrogen gas is so unreactive.

随着键级增大，键长缩短、键能增大。N≡N 三键（约 945 kJ mol⁻¹）远比 N–N 单键（约 158 kJ mol⁻¹）牢固，这解释了氮气为何如此不活泼。

4. Dative (Coordinate) Bonding | 配位（共价配）键

A dative covalent bond, also called a coordinate bond, forms when both electrons in the shared pair come from the same atom. The atom that donates the pair must have a lone pair and the acceptor must have an empty orbital or a positive charge.

配位共价键（也称作配位键）是指共享电子对中的两个电子都来自同一个原子。提供电子对的原子必须含有孤对电子，而接受体必须具有空轨道或正电荷。

In the ammonium ion, NH₄⁺, ammonia’s lone pair on nitrogen is donated to a hydrogen ion, H⁺. All four N–H bonds are then indistinguishable in the ion; the dative bond is no different from an ordinary covalent bond once formed.

在铵离子 NH₄⁺ 中，氨中氮上的孤对电子被捐献给一个氢离子 H⁺。此后，离子中四个 N–H 键完全无法区分；一旦形成，配位键与普通共价键没有任何区别。

Similarly, the hydronium ion H₃O⁺ forms when water donates a lone pair to H⁺. In the gas phase, BF₃ accepts a lone pair from NH₃ to form the adduct F₃B–NH₃, a classic example used in IB exam questions.

类似地，当水将一对孤对电子捐献给 H⁺ 时形成水合氢离子 H₃O⁺。在气相中，BF₃ 接受 NH₃ 的孤对电子形成加合物 F₃B–NH₃，这是 IB 考试题中的经典例子。

5. VSEPR Theory and Molecular Shape | VSEPR 理论与分子形状

The Valence-Shell Electron-Pair Repulsion (VSEPR) model states that electron domains — both bonding pairs and lone pairs — repel one another and arrange themselves as far apart as possible. The number of electron domains around the central atom determines the electron-domain geometry, while the number of bonding domains determines the molecular shape.

价层电子对互斥（VSEPR）模型认为，电子域（包括成键电子对和孤对电子）之间相互排斥，并尽量彼此远离。中心原子周围电子域的数目决定电子域几何，而成键电子域的数目决定分子形状。

Two domains give linear geometry (e.g. BeCl₂, angle 180°). Three domains give trigonal planar (e.g. BF₃, angle 120°). Four domains give tetrahedral (e.g. CH₄, angle 109.5°). Five give trigonal bipyramidal (e.g. PCl₅, angles 90° and 120°), and six give octahedral (e.g. SF₆, angle 90°).

两个电子域产生直线形（如 BeCl₂，键角 180°）。三个电子域产生平面三角形（如 BF₃，键角 120°）。四个电子域产生正四面体（如 CH₄，键角 109.5°）。五个产生三角双锥形（如 PCl₅，键角 90° 和 120°），六个产生八面体形（如 SF₆，键角 90°）。

Electron domains 电子域	Shape 形状	Bond angle 键角	Example 示例
2	Linear 直线形	180°	CO₂
3	Trigonal planar 平面三角形	120°	BF₃
4	Tetrahedral 正四面体	109.5°	CH₄
5	Trigonal bipyramidal 三角双锥形	90° / 120°	PCl₅
6	Octahedral 八面体形	90°	SF₆

Lone pairs occupy more space than bonding pairs because they are held by only one nucleus. Hence in NH₃ the H–N–H angle is compressed to about 107°, and in H₂O to about 104.5°. This repulsion hierarchy — lone-pair vs lone-pair > lone-pair vs bond-pair > bond-pair vs bond-pair — is essential for exam answers.

孤对电子比成键电子对占据更多空间，因为它只受一个原子核吸引。因此在 NH₃ 中 H–N–H 键角被压缩到约 107°，在 H₂O 中则压缩到约 104.5°。这种排斥顺序——孤对电子对孤对电子 > 孤对电子对成键电子对 > 成键电子对之间——是考试作答中必不可少的要点。

6. Bond Polarity and Electronegativity | 键的极性与电负性

Electronegativity is the tendency of an atom to attract a shared pair of electrons. In a bond between atoms of different electronegativity, the electron pair is pulled closer to the more electronegative atom, creating an unequal sharing described as a polar covalent bond.

电负性是原子吸引共享电子对的能力。在电负性不同的原子之间形成的键中，电子对被拉向电负性更大的原子，造成不均匀共享，这种键称为极性共价键。

If the electronegativity difference is very small (less than about 0.4), the bond is usually treated as non-polar, as in C–H bonds. If the difference is large (greater than about 1.7), the bond is often considered ionic, although the boundary is not absolute and many bonds are simply ‘polar covalent’.

若电负性差值很小（小于约 0.4），该键通常视为非极性，如 C–H 键。若差值很大（大于约 1.7），该键常被视为离子键；不过边界并非绝对，许多键只是“极性共价键”而已。

A molecule may be polar or non-polar depending on its shape. CO₂ has two polar C=O bonds, but they cancel because the molecule is linear, so CO₂ has zero dipole moment. H₂O is bent, its two O–H dipoles do not cancel, so water is polar. This distinction between bond polarity and molecular polarity is a favourite IB multiple-choice trap.

分子是否有极性取决于其形状。CO₂ 含有两个极性 C=O 键，但线性结构使它们相互抵消，因此 CO₂ 的偶极矩为零。H₂O 是弯折形的，两个 O–H 偶极不能抵消，因此水是极性分子。区分键的极性与分子极性是 IB 选择题中最爱设置的陷阱之一。

7. Valence Bond Theory, Hybridisation, Sigma and Pi Bonds | 价键理论、杂化轨道、sigma 键与 pi 键

Valence bond theory explains electron sharing in terms of orbital overlap. Head-on overlap of orbitals creates a sigma (σ) bond, which allows free rotation. Sideways overlap of parallel p-orbitals creates a pi (π) bond, which prevents rotation and is weaker than a σ bond.

价键理论从轨道重叠的角度解释电子共享。轨道“头对头”重叠形成 sigma（σ）键，σ 键允许自由旋转。平行 p 轨道的“肩并肩”重叠形成 pi（π）键，π 键阻碍旋转且比 σ 键弱。

To produce equivalent bonds, atomic orbitals mix to form hybrid orbitals. Four sp³ hybrids give tetrahedral CH₄; three sp² hybrids and one unhybridised p-orbital give trigonal planar C₂H₄; two sp hybrids and two unhybridised p-orbitals give linear C₂H₂.

为了生成等价的化学键，原子轨道会混合形成杂化轨道。四个 sp³ 杂化轨道产生正四面体的 CH₄；三个 sp² 杂化轨道和一个未杂化 p 轨道产生平面三角形的 C₂H₄；两个 sp 杂化轨道和两个未杂化 p 轨道产生直线形的 C₂H₂。

A double bond consists of one σ and one π bond; a triple bond consists of one σ and two π bonds. The π electrons are more exposed and more reactive, which accounts for the addition reactions of alkenes and alkynes.

双键包含一个 σ 键和一个 π 键；三键包含一个 σ 键和两个 π 键。π 电子更加暴露、更易反应，这解释了烯烃和炔烃的加成反应。

8. Resonance: Delocalised Electron Sharing | 共振：离域电子共享

Some molecules cannot be fully described by a single Lewis structure. Ozone (O₃), the nitrate ion (NO₃⁻), and benzene (C₆H₆) all have electron pairs that are delocalised over several atoms; this is represented by two or more resonance structures separated by a double-headed arrow.

某些分子无法用单一路易斯结构完整描述。臭氧（O₃）、硝酸根离子（NO₃⁻）和苯（C₆H₆）中的电子对离域在多个原子上；这用两个或多个共振结构并以双头箭头分隔来表示。

In benzene, the six C–C bonds are all identical in length, intermediate between single and double bonds. Delocalisation lowers the energy of the molecule, making it more stable than any single resonance form would suggest. IB examiners frequently ask students to explain the equal bond lengths in benzene using the delocalisation concept.

在苯中，六个 C–C 键的长度完全相同，介于单键和双键之间。离域降低了分子能量，使其比任何单一共振结构所预示的都要稳定。IB 考官经常要求学生运用离域概念解释苯中键长相等这一现象。

Remember that a resonance hybrid is not a mixture of structures that interconvert; it is a single, more accurate picture of electron density spread across the molecule.

请注意，共振杂化体并不是相互转换的几种结构的混合物；而是对分子内电子密度分布的单一、更准确的描述。

9. Electron Sharing in Giant Covalent Structures | 巨型共价结构中的电子共享

Diamond, graphite and graphene are all allotropes of carbon in which atoms share electrons in extended networks. In diamond, each carbon forms four sp³ σ bonds, giving a tetrahedral network that is exceptionally hard and electrically insulating.

金刚石、石墨和石墨烯都是碳的同素异形体，其原子在扩展网络中共享电子。在金刚石中，每个碳原子形成四个 sp³ σ 键，形成极为坚硬且不导电的四面体网络。

In graphite, each carbon forms three sp² σ bonds within planar layers, leaving a delocalised π system above and below each layer. These mobile π electrons conduct electricity, while the weak van der Waals forces between layers allow them to slide — the origin of graphite’s lubricating property.

在石墨中，每个碳原子在平面层内形成三个 sp² σ 键，每一层上下方均留下离域的 π 电子体系。这些可移动的 π 电子能够导电，而层间微弱的范德华力使层可以滑动——这是石墨具有润滑性的根源。

Graphene, a single layer of graphite, is only one atom thick but extremely strong because of its extensive σ framework and stabilising π delocalisation. These structural consequences of electron sharing are strongly emphasised in the IB syllabus.

石墨烯是单层的石墨，仅一个原子厚，但由于其广泛的 σ 骨架和稳定的 π 离域而极其坚固。电子共享的这些结构后果是 IB 教学大纲中重点强调的内容。

10. Bond Enthalpies and Reaction Energetics | 键焓与反应能量学

Bond enthalpy (or bond dissociation energy) is the energy required to break one mole of a particular covalent bond in the gaseous state. Average bond enthalpies are tabulated because the exact value depends on molecular environment.

键焓（即键解离能）是指在气态下断裂一摩尔特定共价键所需的能量。由于精确值取决于分子环境，通常使用平均键焓表。

During any reaction, bonds in reactants are broken (endothermic, ΔH positive) and bonds in products are formed (exothermic, ΔH negative). The overall enthalpy change can be estimated using:

在任何反应中，反应物中的键断裂（吸热，ΔH 为正）而产物中的键形成（放热，ΔH 为负）。总焓变可以用下式估算：

ΔH = Σ(bonds broken) − Σ(bonds formed)

For example, in the hydrogenation of ethene, one C=C π bond (about 264 kJ mol⁻¹) and one H–H bond (436 kJ mol⁻¹) are broken, while one C–C and two C–H bonds are formed; the estimate is negative, confirming an exothermic reaction.

例如，在乙烯加氢反应中，断裂一个 C=C π 键（约 264 kJ mol⁻¹）和一个 H–H 键（436 kJ mol⁻¹），同时形成一个 C–C 键和两个 C–H 键；估算结果为负，确认该反应放热。

Exam questions often provide a table of average bond enthalpies and ask for ΔH. Be sure to include all bonds in the balanced equation, count each bond once, and state that bond-enthalpy estimates are less accurate than Hess’s-law calculations because averages are used.

考试题常给出平均键焓表并要求计算 ΔH。务必把配平方程式中的所有键都计入，每个键只计一次，并说明用平均键焓得到的是估算值，其精确度不如使用赫斯定律的计算，因为这里采用的是平均值。

11. Exam Focus: Common Misconceptions and Top Tips | 考试重点：常见误区与高分技巧

A frequent misconception is that ‘polar bond’ automatically means ‘polar molecule’. Always check the molecular geometry: symmetric shapes like linear AB₂ or tetrahedral AB₄ are usually non-polar even if individual bonds are polar.

常见的误区是认为“极性键”必然导致“极性分子”。务必检查分子几何构型：对称形状（如直线形 AB₂ 或正四面体 AB₄）通常是非极性的，即使其个别键是极性的。

Another error is counting the number of ‘bonds’ instead of ‘electron domains’ in VSEPR. A double bond counts as one electron domain, not two. When drawing Lewis structures, do not forget to show the formal charges where relevant, and always check the total valence-electron count.

另一个错误是在 VSEPR 中数“键数”而非“电子域数”。一个双键只算一个电子域，而不是两个。在画路易斯结构时，不要忘记在相关位置标出形式电荷，并始终核对总价电子数。

Learn the bond-angle values with their parent geometries before the exam; they are frequently tested in Paper 1.

考试前牢记各几何构型对应的键角数值；这是 Paper 1 中经常考查的内容。
Use the phrase ‘unequal sharing of electrons’ when defining a polar covalent bond, and ‘equal sharing’ for a pure covalent bond.

定义极性共价键时使用“电子共享不均”这一表述，而定义纯共价键时使用“电子共享均等”。
For dative bonding, draw a clear arrow from the donor atom’s lone pair toward the acceptor atom, and state that both electrons come from one atom.

对于配位键，用清晰的箭头从给体的孤对电子指向接受体，并说明两个电子均来自同一原子。
Always compare bond lengths or strengths using bond order first; higher bond order implies shorter and stronger bonds.

比较键长或键能时，首先比较键级；键级越高，键越短、越强。

A final reminder: the phrase ‘electron sharing’ is not only about simple covalent molecules. Mastery of Lewis structures, VSEPR, hybridisation, resonance and bond-enthalpy calculations will allow you to solve almost every bonding question IB Chemistry can offer.

最后提醒：“电子共享”并不仅仅指简单共价分子。熟练掌握路易斯结构、VSEPR、杂化轨道、共振以及键焓计算后，你几乎可以解决 IB 化学中所有关于成键的考题。

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IB Chemistry: Analysis of Electron Sharing Reactions | IB化学：电子共享反应解析