📚 Lewis Structures: A Step-by-Step Guide | 路易斯结构式的书写方法
Lewis structures are one of the most fundamental tools in chemistry. They show how valence electrons are arranged among atoms in a molecule and help us predict bonding, shape, and reactivity. This guide provides a clear, exam-focused method for drawing Lewis structures correctly in IB Chemistry.
路易斯结构式是化学中最基础的工具之一。它展示分子中价电子在原子间的排列方式，帮助我们预测成键、分子形状和反应活性。本指南专为 IB 化学考生提供清晰、紧扣考点的书写方法。
1. What Is a Lewis Structure? | 什么是路易斯结构式
A Lewis structure, also called a Lewis dot diagram, uses dots to represent lone-pair electrons and lines to represent shared pairs of electrons. Each line corresponds to a covalent bond containing two electrons.
路易斯结构式又称电子点式，用圆点表示孤对电子，用短线表示共用电子对。每条短线代表一根包含两个电子的共价键。
For example, the Lewis structure of water shows two O–H single bonds and two lone pairs on the oxygen atom. The dots must be placed accurately so that the total number of valence electrons matches the atom's group number.
例如，水的路易斯结构式显示两条 O–H 单键和氧原子上的两对孤对电子。点的位置必须准确，使价电子总数与原子所在族号一致。
The Lewis model was introduced by Gilbert N. Lewis in 1916. It remains essential for understanding ionic and covalent bonding, formal charge, and resonance.
路易斯模型由吉尔伯特·路易斯于 1916 年提出。它至今仍是理解离子键、共价键、形式电荷和共振的重要基础。
2. Valence Electrons and the Octet Rule | 价电子与八隅体规则
For main-group elements, the number of valence electrons is equal to the group number. Hydrogen and helium are exceptions because they follow the duet rule.
对主族元素而言，价电子数等于族号。氢和氦是例外，它们遵循“二隅体规则”。
Consider the following examples of valence-electron counts:
以下是一些常见元素的价电子数示例：
| Group / 族 | Element examples / 元素示例 | Valence electrons / 价电子数 |
| 1 | H, Li, Na | 1 |
| 14 | C, Si | 4 |
| 15 | N, P | 5 |
| 16 | O, S | 6 |
| 17 | F, Cl | 7 |
| 18 | Ne, Ar | 8 |
The octet rule states that atoms tend to gain, lose, or share electrons until they have eight valence electrons, resembling a noble gas. Hydrogen is satisfied with two electrons.
八隅体规则指出，原子倾向于通过获得、失去或共用电子达到八个价电子，从而类似稀有气体。氢原子只需要两个电子即达到稳定。
3. Step-by-Step Method for Drawing Lewis Structures | 分步书写路易斯结构式
Use the following procedure for any neutral molecule or polyatomic ion. The same steps apply to ions, but the charge is included in the total electron count.
书写任何中性分子或多原子离子时，都可以按以下步骤进行。对离子而言，只需在总电子数中加入或扣除相应电荷的电子。
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Step 1: Count the total number of valence electrons.
Add the valence electrons of all atoms. For an anion, add one electron per negative charge. For a cation, subtract one electron per positive charge.
第一步：计算价电子总数。
将所有原子的价电子相加。对于阴离子，每个负电荷加一个电子；对于阳离子，每个正电荷减去一个电子。
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Step 2: Identify the central atom.
The central atom is usually the element with the lowest electronegativity, but hydrogen is never central. Place the central atom in the middle and connect other atoms with single lines.
第二步：确定中心原子。
中心原子通常是电负性最低的元素，但氢永远不能作为中心原子。将中心原子放在中间，用单线连接其他原子。
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Step 3: Distribute remaining electrons as lone pairs.
First place lone pairs on terminal atoms to complete their octets. Then place any leftover electrons on the central atom.
第三步：分配剩余电子为孤对电子。
先为端基原子补足孤对电子以满足八隅体，再将剩余电子放在中心原子上。
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Step 4: Check the central atom's octet.
If the central atom has fewer than eight electrons, form double or triple bonds by moving lone pairs from terminal atoms into bonds.
第四步：检查中心原子的八隅体。
如果中心原子电子少于八个，则将端基原子的孤对电子移入成键区域，形成双键或三键。
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Step 5: Verify the total electron count.
Count all dots and lines as electrons. The final structure must match the total from Step 1.
第五步：核对总电子数。
将所有的点和短线都计为电子。最终结构中的电子总数必须与第一步的总数一致。
4. Using Multiple Bonds | 多重键的处理
Many molecules require double or triple bonds to satisfy the octet rule. Carbon dioxide, CO₂, is a classic example.
许多分子需要通过双键或三键来满足八隅体规则。二氧化碳 CO₂ 就是一个典型例子。
The total valence-electron count for CO₂ is 4 + (2 × 6) = 16. After placing two C– erefore, two lone pairs from the oxygen atoms must be converted into two C=O double bonds.
CO₂ 的价电子总数为 4 + (2 × 6) = 16。在画出两条 C–O 单键并给氧原子补上孤对电子后，中心碳原子只有四个电子。因此，需要将氧原子上的两对孤对电子转化为两条 C=O 双键。
O=C=O
Each oxygen atom in this structure has two lone pairs and two shared pairs, giving an octet. The carbon atom has four shared pairs, also giving an octet.
在此结构中，每个氧原子有两对孤对电子和两对共用电子，满足八隅体；碳原子有四对共用电子，也满足八隅体。
Nitrogen gas, N₂, requires a triple bond. Nitrogen has five valence electrons, so two nitrogen atoms share three pairs:
氮气 N₂ 需要三键。每个氮原子有五个价电子，因此两个氮原子共用三对电子：
N≡N
Each nitrogen atom also retains one lone pair, giving it an octet.
每个氮原子还保留一对孤对电子，从而达到八隅体。
5. Formal Charge | 形式电荷
Formal charge helps you choose the most plausible Lewis structure when more than one arrangement is possible.
当存在多种可能的路易斯结构时，形式电荷可以帮助你选择最合理的结构。
Formal charge is calculated using the following equation:
形式电荷的计算公式如下：
FC = V − (L + ½B)
Here, V is the number of valence electrons in the free atom, L is the number of lone-pair electrons, and B is the number of bonding electrons around the atom.
其中，V 是自由原子的价电子数，L 是孤对电子数，B 是该原子周围的成键电子数。
For example, consider the carbonate ion CO₃²⁻. The sum of formal charges on all atoms must equal the overall charge, −2.
以碳酸根离子 CO₃²⁻ 为例，所有原子的形式电荷之和必须等于总电荷 −2。
When choosing between structures, the best structure usually has formal charges close to zero. If a negative formal charge is necessary, it should be placed on the more electronegative atom.
在多个结构之间选择时，最优结构通常形式电荷接近零。如果必须出现负的形式电荷，应将其放在电负性更大的原子上。
6. Resonance Structures | 共振结构
Sometimes no single Lewis structure accurately describes a molecule. Resonance occurs when two or more valid Lewis structures differ only in the position of electrons.
有时单一路易斯结构无法准确描述一个分子。当两个或多个有效路易斯结构仅在电子分布上不同时，就会发生共振。
Ozone, O₃, is a well-known example. The two resonance forms can be represented as:
臭氧 O₃ 是一个典型例子。它的两个共振式可以表示为：
O=O–O ↔ O–O=O
The double-headed arrow, ↔, indicates that the actual structure is a hybrid of the two resonance forms. The two terminal oxygen atoms are equivalent, and the bond lengths are equal in reality.
双向箭头 ↔ 表示真实结构是这两个共振式的杂化体。在真实臭氧分子中，两个端基氧原子是等价的，实际键长也相等。
Another important example is the nitrate ion, NO₃⁻. It has three resonance structures in which the double bond moves among the three N–O positions.
另一个重要例子是硝酸根离子 NO₃⁻。它有三个共振结构，双键在三个 N–O 键位置之间移动。
In IB Chemistry, you must be able to draw all reasonable resonance structures and use the double-headed arrow correctly.
在 IB 化学中，你需要能够画出所有合理的共振结构，并正确使用双向箭头。
7. Exceptions to the Octet Rule | 八隅体规则的例外
Some molecules do not obey the octet rule. You should recognise the three main types of exceptions.
有些分子并不遵守八隅体规则。你需要认识以下三种主要例外情况。
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Incomplete octet: The central atom has fewer than eight electrons. Boron trifluoride, BF₃, has only six electrons around boron.
缺电子八隅体：中心原子电子少于八个。三氟化硼 BF₃ 中硼原子周围只有六个电子。
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Expanded octet: Elements in period 3 and beyond can hold more than eight electrons because they have available d orbitals. Examples include PCl₅ and SF₆.
扩大的八隅体：第三周期及以后的元素因为有可利用的 d 轨道，可以容纳超过八个电子。例如 PCl₅ 和 SF₆。
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Odd-electron species: Some molecules have an unpaired electron. Nitric oxide, NO, has 11 valence electrons, so one atom cannot achieve an octet.
奇电子物种：某些分子含有一个未成对电子。一氧化氮 NO 有 11 个价电子，因此有一个原子无法满足八隅体。
When drawing these structures, do not force an octet if it violates the known behaviour of the molecule.
在书写这些结构时，不要为了满足八隅体而违背分子的已知行为。
8. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Students often lose marks on Lewis structures due to avoidable errors. The following checklist can help.
学生在路易斯结构式上失分，往往是因为一些可以避免的错误。以下清单可以帮助你。
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Incorrect valence-electron count: Recheck the total before drawing. Do not forget to adjust for ions.
价电子数计算错误：动笔前重新核对总数，离子务必调整电子数。
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Wrong central atom: Hydrogen and halogens are usually terminal, not central.
中心原子选择错误：氢和卤素通常是端基原子，而不是中心原子。
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Missing lone pairs: Every atom must be checked. Lone pairs affect shape and polarity.
遗漏孤对电子：每个原子都要检查。孤对电子会影响分子形状和极性。
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Forgetting to form multiple bonds: If the central atom lacks an octet, convert lone pairs into bonds.
忘记形成多重键：如果中心原子不满足八隅体，就把孤对电子转化为成键电子对。
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Ignoring formal charge: When several structures are possible, use formal charge to justify your choice.
忽略形式电荷：当存在多种结构时，要用形式电荷来证明你的选择。
9. Worked Examples | 实例练习
The following examples show how the step-by-step method is applied to common IB Chemistry molecules.
以下实例展示如何将分步方法应用于 IB 化学常见分子。
| Species / 物种 | Total valence electrons / 价电子总数 | Lewis structure result / 路易斯结构结果 |
| H₂O | 2(1) + 6 = 8 | Two O–H single bonds; two lone pairs on O / 两条 O–H 单键；氧上有两对孤对电子 |
| NH₄⁺ | 5 + 4(1) − 1 = 8 | Four N–H single bonds; no lone pairs / 四条 N–H 单键；无孤对电子 |
| CO₂ | 4 + 2(6) = 16 | Linear O=C=O with two lone pairs on each O / 直线形 O=C=O；每个氧有两对孤对电子 |
| NO₃⁻ | 5 + 3(6) + 1 = 24 | Three resonance structures with one N=O double bond and two N–O single bonds / 三个共振结构，含一个 N=O 双键和两个 N–O 单键 |
| SO₄²⁻ | 6 + 4(6) + 2 = 32 | Tetrahedral arrangement; multiple resonance forms / 四面体排列；存在多种共振形式 |
Notice that the total charge on an ion is reflected in the total electron count and in the sum of formal charges.
注意，离子的总电荷会体现在价电子总数中，也会体现在所有形式电荷之和中。
10. Final Exam Tips | 考场最终提示
In IB Chemistry, Lewis structures are often tested in Paper 1 and Paper 2. You must be able to draw them quickly and accurately.
在 IB 化学考试中，路易斯结构式常在 Paper 1 和 Paper 2 中出现。你必须能够快速、准确地书写它们。
Always show the lone pairs, because they are part of the structure. Do not combine structures into an ambiguous formula.
始终标出孤对电子，因为它们是结构的一部分。不要将多个结构混成一个模糊的表达式。
Practise with molecules and ions such as H₂O, CO₂, NH₃, CH₄, NO₃⁻, CO₃²⁻, and SO₄²⁻. Applying the same five steps every time will reduce mistakes and improve your speed.
多练习 H₂O、CO₂、NH₃、CH₄、NO₃⁻、CO₃²⁻ 和 SO₄²⁻ 等分子和离子。每次坚持使用相同的五个步骤，可以减少错误并提高速度。
Finally, remember that a Lewis structure is a model. It predicts bonding and geometry, but the real electron distribution may require resonance or expanded octets.
最后要记住，路易斯结构式是一种模型。它能预测成键和几何，但真实的电子分布可能需要共振或扩大的八隅体来解释。
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