IB Chemistry · Structure 3 · 鼎睿学苑
Classification of Matter物质的分类
Two great sorting systems in chemistry: the periodic table, which classifies elements by their electron configuration, and functional groups, which classify organic compounds by their reactive centres. This unit shows how position on the table (or the functional group present) predicts physical and chemical behaviour.化学中有两大分类体系：元素周期表（periodic table）按电子排布对元素分类，官能团（functional group）按反应活性中心对有机化合物分类。本单元说明元素在周期表中的位置（或分子所含的官能团）如何预测其物理和化学行为。
Structure 3.1
The Periodic Table: Classification of Elements元素周期表：元素的分类
The periodic table arranges every known element by increasing atomic number. Each period corresponds to the number of occupied electron shells in a ground-state atom; each group (for the s- and p-blocks) corresponds to the number of valence electrons. Elements fall into four blocks according to which subshell is being filled: the s-block (groups 1 to 2), the p-block (groups 13 to 18), the d-block (the transition elements), and the f-block (the lanthanoids and actinoids).
元素周期表按原子序数递增排列所有已知元素。周期（period）对应基态原子中被占据的电子层数；族（group，对 s 区和 p 区而言）对应价电子数。元素按正在被填充的亚层分为四个区：s 区（第 1 到 2 族）、p 区（第 13 到 18 族）、d 区（过渡元素）以及 f 区（镧系与锕系元素）。
Periodicity and Effective Nuclear Charge周期性与有效核电荷
Most periodic properties are governed by the effective nuclear charge ($Z_{eff}$): the net positive charge actually felt by a valence electron once the shielding (screening) of inner-shell electrons is taken into account. Moving across a period, protons are added to the same outer shell while inner-shell shielding stays roughly constant, so $Z_{eff}$ rises steadily. Moving down a group, a whole new outer shell is added, so the valence electrons sit farther from the nucleus and are shielded by an additional inner shell.
多数周期性性质都由有效核电荷（effective nuclear charge，$Z_{eff}$）决定：这是价电子在扣除内层电子屏蔽（shielding）效应后实际感受到的净正电荷。同一周期从左到右，质子不断加到同一外层，而内层屏蔽大致不变，因此 $Z_{eff}$ 持续升高。同一族从上到下，则整整多出一个外层，价电子离核更远，且多受一层内层电子的屏蔽。
Effective Nuclear Charge
$$Z_{eff} = Z - S$$
$Z$ = proton number; $S$ = shielding by inner-shell electrons
有效核电荷
$$Z_{eff} = Z - S$$
$Z$ = 质子数；$S$ = 内层电子的屏蔽常数
Atomic Radius and Ionic Radius原子半径与离子半径
Radius Trends
Atomic radius: decreases across a period (rising $Z_{eff}$ pulls the same shell in tighter); increases down a group (an entirely new, farther shell is occupied, which outweighs the increase in $Z_{eff}$).
Ionic radius: cations are smaller than their parent atom (fewer electrons and, often, one fewer occupied shell, e.g. Na$^+$ < Na). Anions are larger than their parent atom (extra electrons increase electron-electron repulsion with no extra protons, e.g. Cl$^-$ > Cl).
Isoelectronic series: species with the same electron count but different proton counts. For a fixed number of electrons, radius shrinks as nuclear charge rises: O$^{2-}$ > F$^-$ > Ne > Na$^+$ > Mg$^{2+}$ > Al$^{3+}$ (all 10 electrons).
半径变化规律
原子半径：同周期从左到右减小（$Z_{eff}$ 升高，把同一电子层拉得更紧）；同族从上到下增大（多占据一整层、离核更远，这一效应超过 $Z_{eff}$ 增大的影响）。
离子半径：阳离子比其母原子小（电子数更少，往往还少一个占据电子层，如 Na$^+$ < Na）；阴离子比其母原子大（电子数增多而质子数不变，电子间排斥增强，如 Cl$^-$ > Cl）。
等电子体系列：电子数相同但质子数不同的粒子。电子数固定时，核电荷越高，半径越小：O$^{2-}$ > F$^-$ > Ne > Na$^+$ > Mg$^{2+}$ > Al$^{3+}$（均为 10 个电子）。
First Ionization Energy第一电离能
First ionization energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in their ground state: $X(g) \rightarrow X^+(g) + e^-$. It generally increases across a period (rising $Z_{eff}$ holds electrons more tightly) and decreases down a group (valence electrons are farther out and more shielded).
第一电离能是从一摩尔气态基态原子中移去一摩尔电子所需的能量：$X(g) \rightarrow X^+(g) + e^-$。它在同一周期内总体随原子序数增大而升高（$Z_{eff}$ 升高使电子被抓得更紧），在同一族内则随原子序数增大而降低（价电子离核更远、屏蔽更多）。
Two Dips to Memorize
Group 2 to group 13 (e.g. Be to B, Mg to Al): the single outer electron in group 13 occupies a higher-energy p-subshell, which is also shielded slightly by the filled s-subshell beneath it. It is easier to remove than an s-electron, despite the higher nuclear charge.
Group 15 to group 16 (e.g. N to O, P to S): group 15 has a stable, half-filled p-subshell ($p^3$) with one electron in each of three orbitals. Group 16 must place two electrons in the same p-orbital; the extra electron-electron repulsion in that doubly occupied orbital makes an electron easier to remove, despite the higher nuclear charge.
两处需要记住的反常下降
第 2 族到第 13 族（如 Be 到 B，Mg 到 Al）：第 13 族多出的那一个电子位于能量更高的 p 亚层，并受到其下方已填满的 s 亚层的少量屏蔽。尽管核电荷更高，这个 p 电子仍比 s 电子更容易移走。
第 15 族到第 16 族（如 N 到 O，P 到 S）：第 15 族拥有稳定的半充满 p 亚层（$p^3$，三个轨道各占一个电子）。第 16 族必须把两个电子放进同一个 p 轨道，这个双占轨道中额外的电子间排斥，使得移走一个电子更容易，尽管核电荷更高。
Electron Affinity, Electronegativity, and Metallic Character电子亲和能、电负性与金属性
Three Related but Distinct Trends
Electron affinity: the energy change when one mole of electrons is added to one mole of gaseous atoms, $X(g) + e^- \rightarrow X^-(g)$. Generally becomes more exothermic (more negative) across a period, since higher $Z_{eff}$ attracts the incoming electron more strongly.
Electronegativity: the relative ability of a bonded atom to attract the shared pair of electrons in a covalent bond (Pauling scale, no units). Increases across a period and decreases down a group; fluorine is the most electronegative element.
Metallic character: the tendency to lose electrons and form cations. Increases down a group and decreases across a period, mirroring the fall in ionization energy and electronegativity toward the lower-left of the table.
三个相关但不同的概念
电子亲和能：一摩尔气态原子获得一摩尔电子时的能量变化，$X(g) + e^- \rightarrow X^-(g)$。同一周期内总体随原子序数增大而更放热（更负），因为更高的 $Z_{eff}$ 更强烈地吸引外来电子。
电负性：成键原子吸引共价键中共用电子对的相对能力（Pauling 标度，无单位）。同周期从左到右增大，同族从上到下减小；氟的电负性最大。
金属性：失去电子形成阳离子的倾向。同族从上到下增强，同周期从左到右减弱，这与电离能、电负性在周期表左下方降低的趋势一致。
Acid-Base Nature of Period 3 Oxides第三周期氧化物的酸碱性
The oxides of period 3 elements show a smooth transition from basic to acidic character, reflecting increasing non-metallic character (and increasingly covalent bonding) from left to right.
第三周期元素的氧化物从左到右表现出由碱性到酸性的平滑过渡，反映了非金属性（以及键的共价成分）从左到右不断增强。
Period 3 Oxides - Acid-Base Character
| Oxide | Character | Reaction |
|---|---|---|
| Na$_2$O | Strongly basic | Na$_2$O + H$_2$O $\rightarrow$ 2NaOH |
| MgO | Basic | MgO + H$_2$O $\rightarrow$ Mg(OH)$_2$ |
| Al$_2$O$_3$ | Amphoteric | reacts with acid AND base (see below) |
| SiO$_2$ | Weakly acidic | SiO$_2$ + 2NaOH $\rightarrow$ Na$_2$SiO$_3$ + H$_2$O |
| SO$_3$ | Acidic | SO$_3$ + H$_2$O $\rightarrow$ H$_2$SO$_4$ |
Al$_2$O$_3$ is amphoteric: Al$_2$O$_3$ + 6HCl $\rightarrow$ 2AlCl$_3$ + 3H$_2$O, and Al$_2$O$_3$ + 2NaOH + 3H$_2$O $\rightarrow$ 2NaAl(OH)$_4$.
第三周期氧化物的酸碱性
| 氧化物 | 酸碱性 | 反应 |
|---|---|---|
| Na$_2$O | 强碱性 | Na$_2$O + H$_2$O $\rightarrow$ 2NaOH |
| MgO | 碱性 | MgO + H$_2$O $\rightarrow$ Mg(OH)$_2$ |
| Al$_2$O$_3$ | 两性 | 既能与酸反应，又能与碱反应（见下方） |
| SiO$_2$ | 弱酸性 | SiO$_2$ + 2NaOH $\rightarrow$ Na$_2$SiO$_3$ + H$_2$O |
| SO$_3$ | 酸性 | SO$_3$ + H$_2$O $\rightarrow$ H$_2$SO$_4$ |
Al$_2$O$_3$ 是两性氧化物：Al$_2$O$_3$ + 6HCl $\rightarrow$ 2AlCl$_3$ + 3H$_2$O，同时 Al$_2$O$_3$ + 2NaOH + 3H$_2$O $\rightarrow$ 2NaAl(OH)$_4$。
HL Only - Transition Elements, Spectra, and Complexes
Line spectra: atoms absorb or emit light only at specific frequencies corresponding to transitions between quantized (discrete) energy levels, producing a line spectrum of sharp separate lines, unlike the continuous spectrum of white light. Energy levels converge at higher $n$; the frequency at which the lines converge corresponds to the ionization energy of that electron.
Transition elements: a d-block element that forms at least one stable ion with an incomplete d-subshell. By this definition, scandium (only Sc$^{3+}$, $3d^0$) and zinc (only Zn$^{2+}$, $3d^{10}$) are d-block but NOT transition elements.
Variable oxidation states: the 4s and 3d subshells are close in energy, so transition elements can lose different numbers of electrons from both, giving several stable oxidation states, e.g. Fe$^{2+}$/Fe$^{3+}$, Cu$^+$/Cu$^{2+}$.
Complex ions: a transition metal ion surrounded by ligands (species with a lone pair, e.g. H$_2$O, NH$_3$, Cl$^-$, CN$^-$) bonded by coordinate (dative) bonds, e.g. [Cu(H$_2$O)$_6$]$^{2+}$.
Colour: ligands split the five d orbitals into two groups of different energy. An electron absorbs a photon of visible light and is promoted between these split orbitals; the observed colour is complementary to the colour absorbed. Ions with an empty ($d^0$) or completely full ($d^{10}$) d-subshell, such as Sc$^{3+}$ or Zn$^{2+}$, cannot make this transition and are colourless.
Catalysis: transition metals catalyse reactions by providing a surface for adsorption (heterogeneous, e.g. Fe in the Haber process) or by using their variable oxidation states to open a lower-activation-energy pathway (homogeneous, e.g. Fe$^{2+}$/Fe$^{3+}$ catalysing the reaction between I$^-$ and S$_2$O$_8^{2-}$).
Magnetism: a species with unpaired electrons is paramagnetic (weakly attracted into a magnetic field); one with all electrons paired is diamagnetic (weakly repelled). Fe$^{3+}$ ($3d^5$, unpaired) is paramagnetic; Zn$^{2+}$ ($3d^{10}$, all paired) is diamagnetic.
HL 专属 - 过渡元素、光谱与配合物
线状光谱：原子只在与量子化（离散）能级跃迁相对应的特定频率处吸收或发射光，产生由分立谱线组成的线状光谱，不同于白光那样的连续光谱。能级随 $n$ 增大而逐渐收敛；谱线收敛处对应的频率即给出该电子的电离能。
过渡元素：指能形成至少一种 d 亚层不满的稳定离子的 d 区元素。按此定义，钪（只形成 Sc$^{3+}$，$3d^0$）与锌（只形成 Zn$^{2+}$，$3d^{10}$）属于 d 区元素，但不算过渡元素。
可变氧化态：4s 与 3d 亚层能量相近，过渡元素可从两者中失去不同数目的电子，因而呈现多种稳定氧化态，例如 Fe$^{2+}$/Fe$^{3+}$、Cu$^+$/Cu$^{2+}$。
配离子：过渡金属离子被配体（带孤对电子的分子或离子，如 H$_2$O、NH$_3$、Cl$^-$、CN$^-$）以配位键（
dative bond）包围而成，例如 [Cu(H$_2$O)$_6$]$^{2+}$。
颜色：配体使五个简并 d 轨道分裂为能量不同的两组。电子吸收一个可见光光子后在分裂的轨道间跃迁，观察到的颜色是被吸收光的互补色。d 亚层全空（$d^0$）或全满（$d^{10}$）的离子（如 Sc$^{3+}$、Zn$^{2+}$）无法发生这种跃迁，因而无色。
催化作用：过渡金属可通过提供吸附表面来催化反应（多相催化，如 Haber 法中的 Fe），也可利用可变氧化态开辟活化能更低的反应路径（均相催化，如 Fe$^{2+}$/Fe$^{3+}$ 催化 I$^-$ 与 S$_2$O$_8^{2-}$ 之间的反应）。
磁性：含未成对电子的粒子具有顺磁性（paramagnetic，被磁场微弱吸引）；电子全部成对的粒子具有逆磁性（diamagnetic，被磁场微弱排斥）。Fe$^{3+}$（$3d^5$，有未成对电子）顺磁；Zn$^{2+}$（$3d^{10}$，全部成对）逆磁。
HL Only - Anomalous Configurations: Chromium and Copper
Strict aufbau filling predicts [Ar] $4s^2\,3d^4$ for chromium and [Ar] $4s^2\,3d^9$ for copper. Both are exceptions: chromium is actually [Ar] $4s^1\,3d^5$, and copper is actually [Ar] $4s^1\,3d^{10}$.
Why the exception occurs: a half-filled ($d^5$, one electron in each of five orbitals) or completely filled ($d^{10}$) d-subshell is an unusually stable, lower-energy arrangement, because it minimizes electron-electron repulsion between orbitals. Promoting a single 4s electron into the 3d subshell to reach that more stable arrangement costs less energy than it releases, so the atom adopts the anomalous configuration.
The exception does not change ion-formation rules: electrons are still removed from 4s before 3d when a transition metal ion forms, exactly as for every other first-row transition element. This is easy to misapply precisely because the neutral atom's configuration already looks unusual.
Classification consequence: Cr and Cu remain ordinary d-block elements and transition elements by the IB definition, since each forms stable ions with an incomplete d-subshell (e.g. Cr$^{3+}$ and Cu$^{2+}$, below). The anomaly affects only the neutral-atom configuration, not which category the element belongs to.
HL 专属 - 反常电子组态：铬与铜
严格按能量最低原理（
aufbau）推测，铬应为 [Ar] $4s^2\,3d^4$，铜应为 [Ar] $4s^2\,3d^9$。但两者都是例外：铬实际为 [Ar] $4s^1\,3d^5$，铜实际为 [Ar] $4s^1\,3d^{10}$。
反常原因：半充满（$d^5$，五个轨道各占一个电子）或全充满（$d^{10}$）的 d 亚层是异常稳定、能量更低的排布方式，因为它使轨道间的电子排斥最小化。把一个 4s 电子提升到 3d 亚层以达到这种更稳定的排布，所付出的能量代价小于由此获得的稳定化能量，因此原子采用了这种反常组态。
该反常不改变离子形成规则：过渡金属形成离子时，仍然先失去 4s 电子，再失去 3d 电子，与其他第一行过渡元素完全相同。正因为基态原子本身的组态已经看起来不合常规，这条规则在考试中很容易被误用。
对分类的影响：按 IB 的定义，Cr 与 Cu 仍然是普通的 d 区元素与过渡元素，因为二者各自都能形成 d 亚层不满的稳定离子（如下文的 Cr$^{3+}$ 与 Cu$^{2+}$）。这一反常只影响基态原子的电子组态，不影响该元素所属的分类。
Worked Example - Predicting Ion Configurations for Cr and Cu
Give the full electron configurations of Cr, Cr$^{3+}$, Cu, and Cu$^{2+}$.
Step 1 - Ground-state Cr
Aufbau would predict [Ar] $4s^2\,3d^4$, but the half-filled 3d subshell is more stable, so one 4s electron shifts into 3d: Cr is [Ar] $4s^1\,3d^5$.
Step 2 - Cr$^{3+}$
Remove the single $4s^1$ electron first, leaving [Ar] $3d^5$, then remove two more electrons from 3d, giving Cr$^{3+}$ = [Ar] $3d^3$. A common mistake is removing all three electrons from 3d directly, since Cr already "looks" d-heavy; that gives the wrong answer [Ar] $4s^1\,3d^2$. The 4s electron is removed first no matter how few electrons occupy it.
Step 3 - Ground-state Cu and Cu$^{2+}$
By the same logic, Cu is [Ar] $4s^1\,3d^{10}$, not the aufbau prediction [Ar] $4s^2\,3d^9$. Removing the single $4s^1$ electron first leaves [Ar] $3d^{10}$; the second electron for Cu$^{2+}$ must then come from 3d, since 4s is already empty, giving Cu$^{2+}$ = [Ar] $3d^9$.
Step 4 - Why this matters
Despite the anomalous neutral-atom configurations, the ion-formation rule (4s before 3d) never changes. Only the ground-state starting point looks unusual, not the mechanism of ionization, and both Cr and Cu still satisfy the IB definition of a transition element via the incomplete d-subshells of Cr$^{3+}$ and Cu$^{2+}$.
例题 - 预测 Cr 与 Cu 的离子电子组态
写出 Cr、Cr$^{3+}$、Cu、Cu$^{2+}$ 的完整电子组态。
第 1 步 - Cr 的基态组态
按能量最低原理应为 [Ar] $4s^2\,3d^4$，但半充满的 3d 亚层更稳定，因此一个 4s 电子转移到 3d：Cr 实际为 [Ar] $4s^1\,3d^5$。
第 2 步 - Cr$^{3+}$
先移去唯一的 $4s^1$ 电子，得到 [Ar] $3d^5$，再从 3d 中移去两个电子，得到 Cr$^{3+}$ = [Ar] $3d^3$。常见错误是直接从 3d 中移去全部三个电子（因为 Cr 组态"看起来"已经以 d 为主），得到错误答案 [Ar] $4s^1\,3d^2$。无论 4s 上剩几个电子，都要先移去 4s 电子。
第 3 步 - Cu 的基态组态与 Cu$^{2+}$
同理，Cu 实际为 [Ar] $4s^1\,3d^{10}$，而非按能量最低原理预测的 [Ar] $4s^2\,3d^9$。先移去唯一的 $4s^1$ 电子，得到 [Ar] $3d^{10}$；由于 4s 此时已空，Cu$^{2+}$ 的第二个电子必须来自 3d，得到 Cu$^{2+}$ = [Ar] $3d^9$。
第 4 步 - 为何这很重要
尽管基态原子的组态反常，离子形成规则（先 4s 后 3d）从未改变。看起来不寻常的只是基态的起点，而不是电离的机制；Cr 与 Cu 依然通过 Cr$^{3+}$ 与 Cu$^{2+}$ 的 d 亚层不满，满足 IB 对过渡元素的定义。
Worked Example - A Period 3 Ionization Energy Anomaly
Explain why the first ionization energy of aluminium (578 kJ mol$^{-1}$) is lower than that of magnesium (738 kJ mol$^{-1}$), even though Al has a greater nuclear charge.
Step 1 - Electron configurations
Mg: $1s^2\,2s^2\,2p^6\,3s^2$. Al: $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$.
Step 2 - Compare the outermost electron
Al's outermost electron occupies the 3p subshell, which is higher in energy than the filled 3s subshell of Mg, and is slightly shielded by the 3s$^2$ pair.
Step 3 - Conclusion
Despite Al's higher nuclear charge (+13 vs +12), its 3p electron is easier to remove than Mg's 3s electron, so Al has the lower first ionization energy. This is the group 2 to group 13 dip.
例题 - 第三周期电离能反常
解释为何铝的第一电离能（578 kJ mol$^{-1}$）低于镁（738 kJ mol$^{-1}$），尽管 Al 的核电荷更大。
第 1 步 - 电子组态
Mg：$1s^2\,2s^2\,2p^6\,3s^2$。Al：$1s^2\,2s^2\,2p^6\,3s^2\,3p^1$。
第 2 步 - 比较最外层电子
Al 最外层电子位于 3p 亚层，能量高于 Mg 已填满的 3s 亚层，并受到 3s$^2$ 电子对的少量屏蔽。
第 3 步 - 结论
尽管 Al 的核电荷更高（+13 对 +12），其 3p 电子仍比 Mg 的 3s 电子更容易移走，因此 Al 的第一电离能更低。这就是第 2 族到第 13 族的反常下降。
Which pair shows a first-ionization-energy decrease caused by electron pairing within a p-orbital (not by a change of subshell)?下列哪一对相邻元素，其第一电离能下降是由 p 轨道内电子成对（而非亚层改变）引起的？
Correct! Oxygen's $2p^4$ configuration forces two electrons into the same p-orbital. The extra electron-electron repulsion makes removal easier than from nitrogen's stable, half-filled $2p^3$, even though O has one more proton.正确！氧的 $2p^4$ 组态迫使两个电子进入同一个 p 轨道，额外的电子间排斥使其比氮稳定的半充满 $2p^3$ 更容易失去电子，即使 O 多一个质子。
Be to B and Mg to Al are subshell-change dips (s to p). Ne to Na is a group-18-to-group-1 drop across a period boundary (new shell). N to O is the p-orbital pairing dip. Answer: (C).Be 到 B 与 Mg 到 Al 是亚层改变（s 到 p）造成的下降。Ne 到 Na 是跨周期（进入新电子层）造成的下降。N 到 O 才是 p 轨道电子成对造成的下降。答案：(C)。
(HL) Which ion would you expect to be colourless in aqueous solution?（HL）下列哪种离子在水溶液中预计是无色的？
Correct! Sc$^{3+}$ has an empty ($3d^0$) d-subshell, so there are no d electrons available to be promoted between the split d orbitals. With no visible light absorbed, the ion is colourless.正确！Sc$^{3+}$ 的 d 亚层是空的（$3d^0$），没有 d 电子可以在分裂的 d 轨道间跃迁。由于不吸收可见光，该离子无色。
Colour requires a partially filled d-subshell so an electron can be promoted between split d orbitals. Sc$^{3+}$ ($3d^0$) has none available. Answer: (A).显色需要 d 亚层部分填充，才能让电子在分裂的 d 轨道间跃迁。Sc$^{3+}$（$3d^0$）没有可跃迁的 d 电子。答案：(A)。
Worked Example — Ranking an Isoelectronic Series by Radius
Rank N$^{3-}$, O$^{2-}$, F$^-$, and Na$^+$ in order of decreasing radius. All four species have 10 electrons.
Step 1 — Confirm they are isoelectronic
N$^{3-}$: 7 + 3 = 10. O$^{2-}$: 8 + 2 = 10. F$^-$: 9 + 1 = 10. Na$^+$: 11 − 1 = 10. All four have exactly 10 electrons, so the only variable affecting radius is nuclear charge.
Step 2 — Apply the isoelectronic-series rule
With electron count fixed, more protons pull the same 10 electrons in more tightly, shrinking the radius. Proton numbers: N = 7, O = 8, F = 9, Na = 11.
Step 3 — Order by increasing proton number = decreasing radius
$$\text{N}^{3-} > \text{O}^{2-} > \text{F}^{-} > \text{Na}^{+}$$
Notice this ranking needs no data booklet values at all — only the electron count (fixed) and the proton count (from the periodic table). This is a purely reasoning-based question, not a lookup one.
例题 — 按半径排列等电子系列
将 N$^{3-}$、O$^{2-}$、F$^-$、Na$^+$ 按半径从大到小排序。这四种粒子都有 10 个电子。
第 1 步 — 确认它们互为等电子体
N$^{3-}$：7 + 3 = 10。O$^{2-}$：8 + 2 = 10。F$^-$：9 + 1 = 10。Na$^+$：11 − 1 = 10。四者电子数都恰好是 10，因此影响半径的唯一变量是核电荷。
第 2 步 — 应用等电子系列规则
电子数固定时，质子越多，对同样 10 个电子的吸引越强，半径就越小。质子数：N = 7，O = 8，F = 9，Na = 11。
第 3 步 — 按质子数递增（即半径递减）排序
$$\text{N}^{3-} > \text{O}^{2-} > \text{F}^{-} > \text{Na}^{+}$$
注意这道题完全不需要查数据手册的数值——只需要电子数（固定）和质子数（来自周期表）。这是一道纯推理题，而不是查表题。
Structure 3.2
Functional Groups: Classification of Organic Compounds官能团：有机化合物的分类
Organic compounds are classified into homologous series: families of compounds sharing a general formula, differing from one another by a fixed CH$_2$ increment, showing similar chemical properties, and showing physical properties (boiling point, viscosity) that grade smoothly as chain length grows. Each member of a series contains one or more functional groups, the specific arrangements of atoms responsible for a compound's characteristic reactivity.
有机化合物按同系列（homologous series）分类：同一系列的化合物具有相同的通式，彼此相差固定数目的 CH$_2$ 单元，化学性质相似，物理性质（沸点、粘度）随链长增加而平滑变化。系列中的每个成员含有一个或多个官能团（functional group），即决定化合物特征反应活性的特定原子排列。
Functional Groups and Homologous Series官能团与同系列
Key Homologous Series
| Series | General formula | Group / suffix | Example |
|---|---|---|---|
| Alkane | C$_n$H$_{2n+2}$ | C-C, C-H (-ane) | propane, C$_3$H$_8$ |
| Alkene | C$_n$H$_{2n}$ | C=C (-ene) | propene, C$_3$H$_6$ |
| Alkyne | C$_n$H$_{2n-2}$ | C≡C (-yne) | propyne, C$_3$H$_4$ |
| Halogenoalkane | C$_n$H$_{2n+1}$X | C-X (halogeno-) | chloroethane, C$_2$H$_5$Cl |
| Alcohol | C$_n$H$_{2n+1}$OH | -OH (-ol) | ethanol, C$_2$H$_5$OH |
| Ether | C$_n$H$_{2n+2}$O | C-O-C (alkoxy-) | methoxymethane, CH$_3$OCH$_3$ |
| Aldehyde | C$_n$H$_{2n}$O | -CHO (-al) | ethanal, CH$_3$CHO |
| Ketone | C$_n$H$_{2n}$O | C=O (-one) | propanone, CH$_3$COCH$_3$ |
| Carboxylic acid | C$_n$H$_{2n}$O$_2$ | -COOH (-oic acid) | ethanoic acid, CH$_3$COOH |
| Ester | C$_n$H$_{2n}$O$_2$ | -COO- (-oate) | ethyl ethanoate, CH$_3$COOC$_2$H$_5$ |
| Amine | C$_n$H$_{2n+3}$N | -NH$_2$ (amino-) | ethylamine, C$_2$H$_5$NH$_2$ |
| Arene | C$_n$H$_{2n-6}$ | benzene ring | methylbenzene, C$_6$H$_5$CH$_3$ |
主要同系列一览
| 系列 | 通式 | 官能团 / 后缀 | 例子 |
|---|---|---|---|
| 烷烃（alkane） | C$_n$H$_{2n+2}$ | C-C、C-H（-ane） | propane（丙烷），C$_3$H$_8$ |
| 烯烃（alkene） | C$_n$H$_{2n}$ | C=C（-ene） | propene（丙烯），C$_3$H$_6$ |
| 炔烃（alkyne） | C$_n$H$_{2n-2}$ | C≡C（-yne） | propyne（丙炔），C$_3$H$_4$ |
| 卤代烷（halogenoalkane） | C$_n$H$_{2n+1}$X | C-X（halogeno-） | chloroethane（氯乙烷），C$_2$H$_5$Cl |
| 醇（alcohol） | C$_n$H$_{2n+1}$OH | -OH（-ol） | ethanol（乙醇），C$_2$H$_5$OH |
| 醚（ether） | C$_n$H$_{2n+2}$O | C-O-C（alkoxy-） | methoxymethane（甲氧基甲烷），CH$_3$OCH$_3$ |
| 醛（aldehyde） | C$_n$H$_{2n}$O | -CHO（-al） | ethanal（乙醛），CH$_3$CHO |
| 酮（ketone） | C$_n$H$_{2n}$O | C=O（-one） | propanone（丙酮），CH$_3$COCH$_3$ |
| 羧酸（carboxylic acid） | C$_n$H$_{2n}$O$_2$ | -COOH（-oic acid） | ethanoic acid（乙酸），CH$_3$COOH |
| 酯（ester） | C$_n$H$_{2n}$O$_2$ | -COO-（-oate） | ethyl ethanoate（乙酸乙酯），CH$_3$COOC$_2$H$_5$ |
| 胺（amine） | C$_n$H$_{2n+3}$N | -NH$_2$（amino-） | ethylamine（乙胺），C$_2$H$_5$NH$_2$ |
| 芳香烃（arene） | C$_n$H$_{2n-6}$ | 苯环 | methylbenzene（甲苯），C$_6$H$_5$CH$_3$ |
IUPAC NomenclatureIUPAC 命名法
To name a compound: (1) identify the longest continuous carbon chain that contains the principal functional group; (2) number the chain to give the lowest possible locant to the principal group, then to substituents; (3) name substituents as prefixes, listed alphabetically with locants; (4) express the principal functional group as a suffix (carboxylic acids, esters, and amides instead determine the parent name directly).
化合物命名步骤：（1）找出包含主官能团的最长连续碳链；（2）为主链编号，使主官能团获得最小编号，其次是取代基；（3）取代基作前缀，按字母顺序列出并标明编号；（4）主官能团以后缀表示（羧酸、酯、酰胺则直接决定母体名称）。
Worked Example - Naming a Branched Haloalkane
Give the IUPAC name of CH$_3$CHClCH(CH$_3$)CH$_2$CH$_3$.
Step 1 - Find the longest chain containing both substituents
Counting carbons across the drawn structure gives a continuous chain of 5 carbons carrying both a chlorine and a methyl branch, so the parent chain is pentane. Choosing a shorter chain that puts the methyl-bearing carbon at an endpoint would drop a substituent from the chain entirely, a common source of naming errors.
Step 2 - Number for the lowest locant set
Numbering from the chlorine end gives locants {2, 3} for chloro and methyl. Numbering from the other end gives {3, 4}. Comparing the two sets at the first point of difference, 2 < 3, so numbering starts from the chlorine end: C1 is the CH$_3$ nearest the Cl.
Step 3 - Order the prefixes alphabetically
"Chloro" is listed before "methyl" (c before m in the alphabet) regardless of which locant is numerically smaller, giving 2-chloro-3-methyl- as the substituent block.
Step 4 - Assemble the name
The full name is 2-chloro-3-methylpentane. Halogens are always cited as prefixes, never as a suffix, so there is no separate suffix step here beyond the parent name "-pentane".
例题 - 为一个带支链的卤代烷命名
写出 CH$_3$CHClCH(CH$_3$)CH$_2$CH$_3$ 的 IUPAC 名称。
第 1 步 - 找出包含两个取代基的最长碳链
数一数结构中的碳原子，可得到一条连续的 5 碳链，同时带有一个氯取代基和一个甲基支链，因此母链为戊烷（pentane）。若选用更短的碳链、把带甲基的碳当作链端，会把一个取代基完全排除在主链之外，这是命名中常见的错误来源。
第 2 步 - 编号以获得最小的编号组合
从氯所在一端编号，氯和甲基的编号为 {2, 3}；从另一端编号则为 {3, 4}。在两组编号第一个不同之处比较，2 < 3，因此应从氯所在一端开始编号：C1 是靠近 Cl 的那个 CH$_3$。
第 3 步 - 按字母顺序排列前缀
"chloro"（氯代）排在 "methyl"（甲基）之前（字母 c 先于 m），无论哪个的编号数字更小，因此取代基部分写作 2-chloro-3-methyl-。
第 4 步 - 组合出完整名称
完整名称为 2-chloro-3-methylpentane（2-氯-3-甲基戊烷）。卤素在 IUPAC 命名中总是以前缀形式出现，绝不作为后缀，因此除母体名 "-pentane" 外不需要额外的后缀步骤。
Structural, Condensed, and Skeletal Formulas结构式、缩合式与骨架式
A full structural formula shows every atom and every bond explicitly. A condensed formula groups atoms carbon by carbon, e.g. CH$_3$CH$_2$OH. A skeletal formula shows only the carbon skeleton as a zig-zag line: each line-end or vertex represents a carbon atom (with hydrogens on carbon implied), while heteroatoms and functional groups are drawn explicitly.
完整结构式明确画出所有原子和所有键。缩合式按碳原子逐个分组书写，例如 CH$_3$CH$_2$OH。骨架式（skeletal formula）只用锯齿折线表示碳骨架：每条线段的端点或转折点代表一个碳原子（碳上的氢被省略不写），杂原子和官能团则明确画出。
Degree of Unsaturation不饱和度
The index of hydrogen deficiency (IHD), or degree of unsaturation, can be calculated directly from a molecular formula. Each ring or $\pi$ bond (a double bond counts 1, a triple bond counts 2) contributes 1 to the IHD.
氢缺指数（index of hydrogen deficiency，IHD），即不饱和度，可直接由分子式算出。每一个环或每一个 $\pi$ 键（双键记 1，三键记 2）都对 IHD 贡献 1。
Index of Hydrogen Deficiency
$$\text{IHD} = \frac{2C + 2 + N - H}{2}$$
halogens are counted as H; oxygen and sulfur are ignored
氢缺指数
$$\text{IHD} = \frac{2C + 2 + N - H}{2}$$
卤素按 H 计算；氧和硫不计入公式
Worked Example — Using IHD to Narrow Down a Structure
A compound has molecular formula C$_4$H$_6$O. Calculate its IHD and state what combinations of rings/$\pi$ bonds are consistent with that value.
Step 1 — Apply the formula (O is ignored)
$$\text{IHD} = \frac{2(4) + 2 + 0 - 6}{2} = \frac{8 + 2 - 6}{2} = \frac{4}{2} = 2$$
Step 2 — Translate IHD = 2 into structural possibilities
A value of 2 could mean: two $\pi$ bonds and no ring (e.g. a diene, or a C=C plus a C=O), one ring and one $\pi$ bond (e.g. a cyclic ketone), or one triple bond (which alone counts as 2).
Step 3 — Why this matters
IHD narrows down candidate structures before you even start drawing them, but it never uniquely identifies one structure by itself — it must be combined with other evidence (IR, NMR, or a description of reactivity) to settle which arrangement is actually correct.
例题 — 用 IHD 缩小结构范围
某化合物分子式为 C$_4$H$_6$O。计算其 IHD，并说明哪些"环/π 键"组合与该值相符。
第 1 步 — 代入公式（O 不计入）
$$\text{IHD} = \frac{2(4) + 2 + 0 - 6}{2} = \frac{8 + 2 - 6}{2} = \frac{4}{2} = 2$$
第 2 步 — 把 IHD = 2 转化为结构上的可能性
数值为 2 可能对应：两个 π 键、无环（例如二烯，或一个 C=C 加一个 C=O）；一个环加一个 π 键（例如环状酮）；或一个三键（三键本身就计 2）。
第 3 步 — 为何这很重要
IHD 能在动笔画结构之前就缩小候选范围，但单靠它永远无法唯一确定某个结构——必须结合其他证据（红外光谱、核磁共振，或反应活性的描述）才能确定到底是哪种排列。
Structural Isomers结构异构体
Three Types of Structural Isomerism
Chain isomers: same molecular formula, different carbon skeleton (branching), e.g. butane and 2-methylpropane (both C$_4$H$_{10}$).
Position isomers: same skeleton and same functional group, different position of that group, e.g. propan-1-ol and propan-2-ol (both C$_3$H$_8$O).
Functional group isomers: same molecular formula, entirely different functional groups, e.g. ethanol and methoxymethane (both C$_2$H$_6$O); ethanoic acid and methyl methanoate (both C$_2$H$_4$O$_2$).
结构异构的三种类型
碳链异构：分子式相同，碳骨架（支链方式）不同，例如 butane（丁烷）与 2-methylpropane（2-甲基丙烷），两者都是 C$_4$H$_{10}$。
位置异构：碳骨架与官能团种类相同，官能团所在位置不同，例如 propan-1-ol 与 propan-2-ol（两者都是 C$_3$H$_8$O）。
官能团异构：分子式相同，官能团种类完全不同，例如 ethanol（乙醇）与 methoxymethane（甲氧基甲烷），两者都是 C$_2$H$_6$O；ethanoic acid（乙酸）与 methyl methanoate（甲酸甲酯），两者都是 C$_2$H$_4$O$_2$。
HL Only - Stereoisomerism and Benzene
Stereoisomers share the same structural formula (connectivity) but differ in the spatial arrangement of atoms.
Cis-trans / E-Z isomerism: arises around a C=C double bond (or in a ring) because rotation about it is restricted, provided each doubly bonded carbon carries two different substituents. E/Z naming uses Cahn-Ingold-Prelog priority: on each carbon, rank the two substituents by atomic number. Z (zusammen) means the higher-priority groups are on the same side; E (entgegen) means they are on opposite sides.
Optical isomerism: arises at a chiral carbon, a carbon bonded to four different groups. A chiral molecule and its mirror image, called enantiomers, are non-superimposable, like left and right hands. Enantiomers share identical physical and chemical properties except that they rotate the plane of plane-polarized light in opposite directions, and they can react differently with other chiral species. A 50:50 mixture of both enantiomers (a racemic mixture) is optically inactive because the rotations cancel.
Benzene delocalization: benzene, C$_6$H$_6$, is a planar ring in which the six p-orbitals (one per carbon) overlap sideways to form a delocalized $\pi$ system above and below the ring. All six C-C bonds are equal in length, intermediate between a single and a double bond. This delocalization stabilizes the ring, so benzene typically undergoes electrophilic substitution rather than the addition reactions typical of alkenes.
HL 专属 - 立体异构与苯
立体异构体连接方式（结构式）相同，只是原子的空间排列不同。
顺反 / E-Z 异构：由于绕 C=C 双键（或环内某键）不能自由旋转，只要每个双键碳都连有两个不同的取代基，就会出现这种异构。E/Z 命名依据 Cahn-Ingold-Prelog 优先级规则：在每个碳上，按原子序数比较两个取代基。Z（zusammen，意为"共同"）表示两个优先级较高的基团在同侧；E（entgegen，意为"相对"）表示它们在异侧。
光学异构：产生于手性碳（
chiral carbon），即连接四个互不相同基团的碳原子。手性分子与其镜像互为对映体（enantiomers），二者不能重合，如同左右手。对映体的物理、化学性质完全相同，唯独使平面偏振光的旋转方向相反，并且与其他手性物质的反应可能不同。等量（50:50）的两种对映体混合而成的外消旋混合物（racemic mixture）因旋光相互抵消而不显光学活性。
苯的离域：苯（C$_6$H$_6$）是一个平面环状分子，六个 p 轨道（每个碳一个）侧面重叠，在环上、下方形成离域的 $\pi$ 电子体系。六条 C-C 键长度完全相等，介于单键与双键之间。这种离域使苯环更稳定，因此苯通常发生亲电取代反应，而不像烯烃那样容易发生加成反应。
Worked Example - Chirality and E/Z Isomerism
(a) Does 2-bromobutane, CH$_3$CHBrCH$_2$CH$_3$, have a chiral carbon? (b) Does but-2-ene, CH$_3$CH=CHCH$_3$, show E/Z isomerism?
Step 1 - Check C2 of 2-bromobutane
The four groups on C2 are Br, H, CH$_3$, and CH$_2$CH$_3$ - all four are different, so C2 is a chiral centre.
Step 2 - Check the double-bond carbons of but-2-ene
Each double-bond carbon carries CH$_3$ and H, two different groups, so E/Z isomerism is possible.
Step 3 - Apply CIP priority
On each carbon, CH$_3$ (attached atom C) outranks H (attached atom H). The Z-isomer has both CH$_3$ groups on the same side (the cis form); the E-isomer has them on opposite sides (the trans form).
例题 - 手性与 E/Z 异构
（a）2-溴丁烷 CH$_3$CHBrCH$_2$CH$_3$ 是否含有手性碳？（b）丁-2-烯 CH$_3$CH=CHCH$_3$ 是否表现出 E/Z 异构？
第 1 步 - 检查 2-溴丁烷的 C2
C2 上连接的四个基团为 Br、H、CH$_3$ 和 CH$_2$CH$_3$，四者互不相同，因此 C2 是手性中心。
第 2 步 - 检查丁-2-烯的双键碳
每个双键碳都连有 CH$_3$ 和 H，是两个不同基团，因此可能出现 E/Z 异构。
第 3 步 - 应用 CIP 优先级
在每个碳上，CH$_3$（所连原子为 C）优先级高于 H（所连原子为 H）。Z 异构体中两个 CH$_3$ 在同侧（即顺式）；E 异构体中两个 CH$_3$ 在异侧（即反式）。
Which pair of compounds are functional group isomers of each other?下列哪一对化合物互为官能团异构体？
Correct! Ethanol (C$_2$H$_5$OH, an alcohol) and methoxymethane (CH$_3$OCH$_3$, an ether) share the molecular formula C$_2$H$_6$O but contain entirely different functional groups.正确！乙醇（C$_2$H$_5$OH，醇）和甲氧基甲烷（CH$_3$OCH$_3$，醚）分子式同为 C$_2$H$_6$O，但官能团完全不同。
The other three pairs are position isomers (same functional group, different position) or chain isomers (different branching). Only ethanol/methoxymethane swap functional groups entirely. Answer: (C).其余三对都是位置异构（官能团相同、位置不同）或碳链异构（支链方式不同）。只有 ethanol/methoxymethane 官能团完全不同。答案：(C)。
(HL) Which molecule possesses a chiral carbon?（HL）下列哪个分子含有手性碳？
Correct! The central carbon in CHFClBr is bonded to four different atoms (H, F, Cl, Br), so it is a chiral centre. In CH$_3$CHClCH$_3$, two of the four groups on the central carbon are identical (both CH$_3$), so it is not chiral.正确！CHFClBr 中心碳连接四个不同的原子（H、F、Cl、Br），是手性中心。CH$_3$CHClCH$_3$ 中心碳所连的四个基团中有两个相同（均为 CH$_3$），因此不是手性碳。
A chiral carbon needs four different substituents. Only CHFClBr (H, F, Cl, Br) qualifies. Answer: (B).手性碳需要连接四个互不相同的基团。只有 CHFClBr（H、F、Cl、Br）符合条件。答案：(B)。
Preparation
Exam Strategy考试策略
Paper 1 (Multiple Choice)Paper 1（多项选择）
Periodic trend questions are very common: memorize the two ionization-energy dips (group 2 to 13, group 15 to 16) and be ready to identify which one is being tested. Acid-base oxide questions almost always test Al$_2$O$_3$'s amphoteric behaviour. For organic questions, practise recognizing all the common functional groups on sight, and be able to classify an isomer pair as chain, position, or functional group in seconds.
周期性趋势题非常常见：把两处电离能反常下降（第 2 到 13 族，第 15 到 16 族）背熟，并能迅速判断题目考的是哪一种。氧化物酸碱性题几乎总会考 Al$_2$O$_3$ 的两性行为。有机题方面，练习一眼识别常见官能团，并能在几秒内判断一对异构体属于碳链、位置还是官能团异构。
Paper 2 (Extended Response)Paper 2（论述题）
When asked to "explain" a periodic trend, always mention effective nuclear charge (or shielding) explicitly. "Ionization energy decreases because it gets easier to remove an electron" earns few marks; "ionization energy decreases because the valence electron is farther from the nucleus and more shielded, so effective nuclear charge experienced by that electron is lower" earns full marks. For organic "explain" questions on stereoisomerism, always state the structural condition first (restricted rotation about C=C, or four different groups on a carbon) before describing the consequence.
遇到解释周期性趋势的题目，务必明确提到有效核电荷（或屏蔽效应）。只写 "电离能下降是因为更容易移走电子" 得分有限；写成 "电离能下降是因为价电子离核更远、屏蔽更强，该电子所感受到的有效核电荷更低" 才能拿满分。有机化学中解释立体异构的题目，先说明结构条件（C=C 键旋转受限，或某碳连有四个不同基团），再描述其结果。
Data Booklet数据手册
Key items: the periodic table itself (block boundaries, group and period numbers), electronegativity values, and the general formulas for each homologous series. For HL: electron configurations of the first-row transition elements and common ligand names, useful for complex-ion and colour questions.
必查内容：元素周期表本身（各区分界、族与周期编号）、电负性数值，以及各同系列的通式。HL 另需第一行过渡元素的电子组态和常见配体名称，用于配离子与颜色相关题目。
Watch Out
Common Mistakes常见错误
Mistake 1 - Explaining Trends Without Effective Nuclear Charge
Saying "atomic radius decreases because there are fewer electrons" is vague and often wrong. The correct explanation invokes effective nuclear charge: across a period, protons are added to the same shell while shielding stays roughly constant, so the increasing $Z_{eff}$ pulls the valence shell in tighter.
错误 1 - 解释趋势时不提有效核电荷
只写 "原子半径减小是因为电子变少" 既模糊又常常错误。正确的解释必须提到有效核电荷：同一周期内质子不断加到同一电子层，而屏蔽大致不变，因此不断升高的 $Z_{eff}$ 把价电子层拉得更紧。
Mistake 2 - Confusing Electronegativity with Electron Affinity
Electronegativity describes an atom's pull on a shared pair of electrons within a covalent bond (a relative, unitless scale). Electron affinity is a measured energy change (kJ mol$^{-1}$) for adding an isolated electron to a gaseous atom. The two trends run in similar directions but are not interchangeable in an explanation.
错误 2 - 把电负性和电子亲和能混为一谈
电负性描述原子在共价键中吸引共用电子对的能力（相对、无单位的标度）。电子亲和能是气态原子获得一个孤立电子时实测的能量变化（kJ mol$^{-1}$）。两者变化方向相似，但在解释题中不能互相替代。
Mistake 3 - Assuming Cis-Trans and E-Z Labels Always Match
Cis/trans labels only work cleanly when it is obvious which substituents count as "the same." When the substituents differ, use Cahn-Ingold-Prelog priority to assign E/Z. A "cis" isomer is not automatically the Z isomer once the substituents get more complex.
错误 3 - 认为顺反标记与 E/Z 标记总是一致
只有在明显能判断哪些取代基算 "相同" 时，顺式/反式（cis/trans）标记才好用。一旦取代基更复杂，就必须用 Cahn-Ingold-Prelog 优先级来判定 E/Z。"顺式" 异构体不一定就是 Z 异构体。
Mistake 4 - Misclassifying Al$_2$O$_3$ as Purely Basic or Acidic
Because aluminium is a metal, students often assume Al$_2$O$_3$ is simply basic. It is amphoteric: it reacts with both acids (forming a salt and water) and bases (forming an aluminate complex and water). Always show both reactions if asked to justify the classification.
错误 4 - 把 Al$_2$O$_3$ 误判为纯碱性或纯酸性
由于铝是金属，学生常常想当然地认为 Al$_2$O$_3$ 是碱性氧化物。实际上它是两性氧化物：既能与酸反应（生成盐和水），也能与碱反应（生成铝酸盐配离子和水）。题目要求说明分类理由时，两个反应都要写出来。
Review
Flashcards闪卡
Click a card to flip it.点击卡片翻面。
What do periods and groups correspond to?周期与族分别对应什么？
A period is the number of occupied electron shells; a group (s- and p-block) is the number of valence electrons.周期数等于被占据的电子层数；族（对 s 区、p 区而言）等于价电子数。
Define effective nuclear charge.什么是有效核电荷？
$Z_{eff} = Z - S$: the net positive charge felt by a valence electron once inner-shell shielding is subtracted from the proton number.$Z_{eff} = Z - S$：从质子数中减去内层屏蔽后，价电子实际感受到的净正电荷。
Why does first IE dip from Mg to Al?为什么第一电离能从 Mg 到 Al 出现下降？
Al's outer electron is in the higher-energy 3p subshell (shielded by 3s$^2$), easier to remove than Mg's 3s electron despite Al's higher nuclear charge.Al 的最外层电子位于能量更高的 3p 亚层（受 3s$^2$ 屏蔽），尽管核电荷更高，仍比 Mg 的 3s 电子更容易移走。
Why does first IE dip from N to O?为什么第一电离能从 N 到 O 出现下降？
O's $2p^4$ configuration forces two electrons into one p-orbital; the extra repulsion makes removal easier than from N's stable, half-filled $2p^3$.O 的 $2p^4$ 组态迫使两个电子挤进同一个 p 轨道，额外的排斥使其比 N 稳定的半充满 $2p^3$ 更容易失去电子。
How is Al$_2$O$_3$ classified, and why?Al$_2$O$_3$ 如何分类？为什么？
Amphoteric: it reacts with both acids and bases, sitting between the basic oxides of groups 1 to 2 and the acidic oxides of groups 14 to 16.两性氧化物：既能与酸反应也能与碱反应，介于第 1 到 2 族的碱性氧化物与第 14 到 16 族的酸性氧化物之间。
HL: What is the IB definition of a transition element, and who is excluded?HL：IB 对过渡元素的定义是什么？哪些元素被排除？
A d-block element forming at least one stable ion with an incomplete d-subshell. Sc ($3d^0$ as Sc$^{3+}$) and Zn ($3d^{10}$ as Zn$^{2+}$) are d-block but not transition elements.能形成至少一种 d 亚层不满的稳定离子的 d 区元素。Sc（Sc$^{3+}$ 为 $3d^0$）与 Zn（Zn$^{2+}$ 为 $3d^{10}$）属于 d 区但不是过渡元素。
What defines a homologous series?同系列的定义是什么？
A family of compounds with the same general formula, differing by a fixed CH$_2$ increment, with similar chemistry and gradually changing physical properties.一组具有相同通式的化合物，彼此相差固定的 CH$_2$ 单元，化学性质相似，物理性质随链长平滑变化。
HL: What makes a carbon chiral, and what special property do enantiomers show?HL：什么样的碳是手性碳？对映体有什么特殊性质？
A chiral carbon is bonded to four different groups. Its enantiomers are non-superimposable mirror images that rotate plane-polarized light in opposite directions.手性碳连接四个互不相同的基团。其对映体是不能重合的镜像，能使平面偏振光向相反方向旋转。
Assessment
Unit Quiz单元测验
1. Which lists Na, Mg, and Al in order of decreasing atomic radius?1. 下列哪一项按原子半径由大到小正确排列了 Na、Mg 和 Al？
Correct! Across period 3, effective nuclear charge rises steadily while shielding stays roughly constant, so atomic radius shrinks steadily from Na to Mg to Al.正确！第三周期从左到右，有效核电荷不断升高，而屏蔽大致不变，因此原子半径从 Na 到 Mg 到 Al 持续减小。
Atomic radius decreases steadily across a period as effective nuclear charge rises with roughly constant shielding. Answer: (A).同一周期内，有效核电荷随原子序数增大而升高，屏蔽大致不变，因此原子半径持续减小。答案：(A)。
2. Compounds A (CH$_3$CH$_2$CH$_2$OH) and B (CH$_3$CH(OH)CH$_3$) share the molecular formula C$_3$H$_8$O. What type of isomerism links A and B?2. 化合物 A（CH$_3$CH$_2$CH$_2$OH）与 B（CH$_3$CH(OH)CH$_3$）分子式同为 C$_3$H$_8$O。A 与 B 之间是哪种异构关系？
Correct! Both are alcohols on the same three-carbon chain, but the -OH is on carbon 1 in A and carbon 2 in B: the functional group is unchanged, only its position differs.正确！两者都是相同三碳链上的醇，只是 -OH 在 A 中位于碳 1，在 B 中位于碳 2：官能团种类不变，只是位置不同。
Same carbon skeleton and same functional group (-OH), just at a different position on the chain. This is position isomerism, not chain (different skeleton) or functional group (different group) isomerism. Answer: (C).碳骨架与官能团（-OH）均相同，只是在链上的位置不同。这是位置异构，而非碳链异构（骨架不同）或官能团异构（官能团不同）。答案：(C)。
3. (HL) In [Fe(CN)$_6$]$^{3-}$, iron is present as Fe$^{3+}$ ($3d^5$). Why is this ion both coloured and paramagnetic?3.（HL）[Fe(CN)$_6$]$^{3-}$ 中铁以 Fe$^{3+}$（$3d^5$）形式存在。为什么这个离子既有颜色又具有顺磁性？
Correct! The CN$^-$ ligands split the five d orbitals into two energy levels; with a partially filled ($3d^5$) subshell, an electron can absorb a photon of visible light and be promoted, giving colour. The same $3d^5$ configuration leaves unpaired electrons, which is what makes the ion paramagnetic.正确！CN$^-$ 配体使五个 d 轨道分裂为两个能级；由于 $3d^5$ 亚层部分填充，电子可以吸收一个可见光光子并跃迁，从而显色。同样的 $3d^5$ 组态留有未成对电子，这正是该离子具有顺磁性的原因。
A partially filled, ligand-split d-subshell is required for both colour (electron promotion between split orbitals) and paramagnetism (unpaired electrons). Fe$^{3+}$ ($3d^5$) satisfies both. Answer: (B).显色（电子在分裂轨道间跃迁）与顺磁性（未成对电子）都需要一个被配体分裂、且部分填充的 d 亚层。Fe$^{3+}$（$3d^5$）同时满足这两个条件。答案：(B)。