化学 · CED 第二单元：分子和离子化合物的结构与性质 · 阅读约 14 分钟 · 更新于 2026-05-11

共振与形式电荷 — AP 化学

AP 化学 · CED 第二单元：分子和离子化合物的结构与性质 · 14 min read

1. 计算形式电荷 ★★☆☆☆ ⏱ 4 min

形式电荷将原子在路易斯结构中“拥有”的价电子数，与中性孤立原子的价电子数进行比较。计算公式为：

text{FC} = V - left(N + frac{B}{2}right)

Where $V$ = 中性自由原子的价电子数，$N$ = 该原子上的非键（孤对）电子数， and $B$ = 该原子共享的成键电子总数。一个关键验证：所有形式电荷的总和必须始终等于该物种的净电荷。

📐 Worked Example

计算硫氰酸根离子 $SCN^-$ (连接方式：S-C-N) 中，当结构为S-C单键、C-N三键时，每个原子的形式电荷。验证你的结果是否符合该离子的净电荷。

确定每个中性原子的价电子数：
硫: $V=6$, 碳: $V=4$, 氮: $V=5$
统计每个原子的非键电子和成键电子数：
硫有 3 对孤对 ($N=6$) 和 2 个成键电子 ($B=2$); 碳没有孤对 ($N=0$)，有 8 个成键电子 ($B=8$); 氮有 1 对孤对 ($N=2$) 和 6 个成键电子 ($B=6$)
计算每个原子的形式电荷：
$begin{aligned} text{FC}_S &= 6 - left(6 + frac{2}{2}right) = -1 ext{FC}_C &= 4 - left(0 + frac{8}{2}right) = 0 \text{FC}_N &= 5 - left(2 + frac{6}{2}right) = 0 end{aligned}$
Sum the formal charges: $-1 + 0 + 0 = -1$, which matches the net charge of $SCN^-$.

Exam tip: 计算完单个形式电荷后，请务必立即计算所有形式电荷的总和。总和不匹配说明你数错了电子，请在评估共振贡献结构稳定性前修正该错误。

2. 共振贡献结构与共振杂化体 ★★☆☆☆ ⏱ 3 min

杂化体中π电子在多个键上离域，降低了分子的总能量，因此比任何单个贡献结构都更稳定。标准表示法是在贡献结构之间使用单个双箭头 $\leftrightarrow$，绝对不能使用平衡箭头（平衡箭头表示物种间相互转化，而共振不存在这种情况）。

📐 Worked Example

Draw all valid resonance contributors for the nitrite ion $NO_2^-$, with nitrogen as the central atom.

Calculate total valence electrons: $5$ (N) + $2 \times 6$ (O) + $1$ (ion charge) = 18 total electrons.
Draw the first contributor: one O forms a double bond with N, giving formal charges: left O = 0, N = +1, right O = -1. Sum of FC matches the ion charge of -1, all atoms have full octets.
Generate the second contributor by moving the pi electron pair from the left double bond to the right O, turning the right single bond into a double bond. This gives formal charges: right O = 0, N = +1, left O = -1, which also satisfies all rules.
Separate the two contributors with a double-headed arrow. The actual resonance hybrid has delocalized pi electrons spread evenly across both N-O bonds, so both bonds are identical.

Exam tip: 切勿在共振贡献结构之间使用平衡箭头。AP考试阅卷人会因为这个常见错误扣分，因为这说明你误解了共振的含义。

3. Identifying Major Resonance Contributors ★★★☆☆ ⏱ 4 min

Not all resonance contributors are equally stable. The most stable (major) contributor contributes more to the resonance hybrid, while less stable (minor) contributors contribute less. There are three hierarchical rules for ranking stability:

First, eliminate any contributors where period 2 nonmetals have incomplete octets (full octets are always prioritized over favorable formal charge).
Contributors with smaller absolute values of formal charge are more stable than those with large charges.
For contributors with similar formal charge magnitudes, the contributor that places negative formal charge on the most electronegative atom (and positive formal charge on the least electronegative atom) is more stable.

📐 Worked Example

Identify the major resonance contributor for $SCN^-$ (connectivity S-C-N) from three valid contributors: 1) Single S-C, triple C-N (FC: S = -1, C = 0, N = 0); 2) Double S-C, double C-N (FC: S = 0, C = 0, N = -1); 3) Triple S-C, single C-N (FC: S = +1, C = 0, N = -2)

Check octets: All three contributors have full octets for all atoms, so none are eliminated.
Compare absolute formal charge magnitudes: Contributor 3 has a total absolute charge of 3, which is much larger than the total of 1 for contributors 1 and 2. Eliminate contributor 3 as minor.
Compare contributors 1 and 2: Both have a total absolute charge of 1. Nitrogen (electronegativity 3.04) is more electronegative than sulfur (2.58).
Negative formal charge is more stable on the more electronegative atom, so Contributor 2 is the major resonance contributor.

Exam tip: Do not prioritize 'all formal charges equal zero' over the electronegativity rule. A small negative charge on a very electronegative atom is more stable than a negative charge on a less electronegative atom, even if the latter gives more zero formal charges.

4. AP-Style Worked Practice Problems ★★★★☆ ⏱ 3 min

📐 Worked Example

Free Response: The azide ion $N_3^-$ is linear with three connected nitrogen atoms: N-N-N. (a) Draw all valid resonance contributors. (b) Identify major/minor contributors. (c) Compare hybrid bond lengths to typical N-N single and triple bonds.

Part (a): Total valence electrons = 16. Three valid contributors, all with full octets and sum FC = -1: $[:N^-=N^{+}=N^-:]^-$, $[:N^--N^{+}≡N:]^-$, $[:N≡N^{+}-N^-:]^-$
Part (b): Contributors 2 and 3 are major, contributor 1 is minor. Contributor 1 has total absolute FC magnitude of 3, while 2 and 3 have magnitude 2. Smaller absolute FC gives more stable contributors.
Part (c): Both N-N bonds in the hybrid are identical, with an average bond order of 2. Their length is between the longer N-N single bond (bond order 1) and shorter N≡N triple bond (bond order 3).

Common Pitfalls

Why: 学生将共振离域与两种物种相互转化的可逆化学反应混淆

Why: 学生同时学习两种电荷分配方法，混淆了它们的记账规则

Why: Textbooks that display multiple contributors separately often lead to this misinterpretation

Why: Students memorize 'smaller formal charges are more stable' and apply it even when an atom has an incomplete octet

Why: Students assume negative charge always prefers larger atoms, even when applying formal charge rules

Quick Reference Cheatsheet

← 返回章节主页

某道题卡住了？
拍照或粘贴题目 — 小欧（我们的 AI 学习助手）会一步步讲解并配示意图。
免费试用小欧 →

共振与形式电荷 — AP 化学

1. 计算形式电荷 ★★☆☆☆ ⏱ 4 min

2. 共振贡献结构与共振杂化体 ★★☆☆☆ ⏱ 3 min

3. Identifying Major Resonance Contributors ★★★☆☆ ⏱ 4 min

4. AP-Style Worked Practice Problems ★★★★☆ ⏱ 3 min

Common Pitfalls

Quick Reference Cheatsheet

更多学习指南