Table 1: first ionization energies across period 3. Table 2: $d$-electron count, observed colour, and absorbed colour for four aqueous first-row transition-metal ions.表 1：第三周期各元素的第一电离能。表 2：四种第一行过渡金属水合离子的 $d$ 电子数、观察颜色与吸收颜色。
| Na | Mg | Al | Si | P | S | Cl | Ar |
|---|
| 496 | 738 | 577 | 786 | 1012 | 1000 | 1251 | 1521 |
| Ion离子 | $[\mathrm{Sc(H_2O)_6}]^{3+}$ | $[\mathrm{Cu(H_2O)_6}]^{2+}$ | $[\mathrm{Ni(H_2O)_6}]^{2+}$ | $[\mathrm{Co(H_2O)_6}]^{2+}$ |
|---|
| $d$ electron count电子数 | $3d^0$ | $3d^9$ | $3d^8$ | $3d^7$ |
|---|
| Observed colour观察颜色 | colourless无色 | pale blue淡蓝色 | green绿色 | pink粉红色 |
|---|
| Colour absorbed吸收颜色 | none无 | orange橙色 | red红色 | green绿色 |
|---|
(a) Two decreasing pairs.两对下降的相邻元素。
Scanning Table 1 for a decrease despite increasing nuclear charge: $\mathrm{IE_1}$ falls from Mg ($738$) to Al ($577$), and again from P ($1012$) to S ($1000$). Every other neighbouring pair (Na→Mg, Al→Si, Si→P, S→Cl, Cl→Ar) increases as expected.在表 1 中查找尽管核电荷增大但电离能仍下降的位置：$\mathrm{IE_1}$ 从 Mg（$738$）降到 Al（$577$），又从 P（$1012$）降到 S（$1000$）。其余每一对相邻元素（Na→Mg、Al→Si、Si→P、S→Cl、Cl→Ar）都如预期般递增。
$$\text{Mg} \rightarrow \text{Al}: 738 \rightarrow 577 \qquad \text{P} \rightarrow \text{S}: 1012 \rightarrow 1000$$
(b) Explaining the Mg to Al decrease.解释 Mg 到 Al 的下降。
Mg is $[\mathrm{Ne}]3s^2$: its outer electron is removed from a full $3s$ subshell. Al is $[\mathrm{Ne}]3s^2 3p^1$: its outer electron occupies the higher-energy $3p$ subshell, additionally shielded by the $3s^2$ electrons beneath it. Despite Al's greater nuclear charge (13 protons vs 12), this $3p$ electron is easier to remove than a paired $3s$ electron in Mg, so $\mathrm{IE_1}$ falls from 738 to $577~\mathrm{kJ\,mol^{-1}}$.Mg 的组态为 $[\mathrm{Ne}]3s^2$：其最外层电子取自充满的 $3s$ 亚层。Al 的组态为 $[\mathrm{Ne}]3s^2 3p^1$：其最外层电子处于能量更高的 $3p$ 亚层，且额外受到其下方 $3s^2$ 电子的屏蔽。尽管 Al 的核电荷更大（13 个质子对 12 个），这个 $3p$ 电子仍比 Mg 中成对的某个 $3s$ 电子更易移走，因此 $\mathrm{IE_1}$ 从 738 降至 $577~\mathrm{kJ\,mol^{-1}}$。
(c) Explaining the P to S decrease.解释 P 到 S 的下降。
P is $[\mathrm{Ne}]3s^2 3p^3$: the $3p$ subshell is exactly half-filled, with one electron in each of the three $3p$ orbitals, all with parallel spin, an arrangement with extra stability from minimized electron-electron repulsion. S is $[\mathrm{Ne}]3s^2 3p^4$: its fourth $3p$ electron must pair up in an already-occupied orbital, and the electron-electron repulsion between this pair makes it easier to remove, despite S's greater nuclear charge, so $\mathrm{IE_1}$ falls slightly from 1012 to $1000~\mathrm{kJ\,mol^{-1}}$.P 的组态为 $[\mathrm{Ne}]3s^2 3p^3$：$3p$ 亚层恰好半充满，三个 $3p$ 轨道各占据一个自旋方向相同的电子，这种排布因电子间排斥最小而具有额外稳定性。S 的组态为 $[\mathrm{Ne}]3s^2 3p^4$：其第四个 $3p$ 电子必须与已占据的轨道中的电子配对，这对成对电子间的排斥使其更容易被移走，尽管 S 的核电荷更大，因此 $\mathrm{IE_1}$ 从 1012 略降至 $1000~\mathrm{kJ\,mol^{-1}}$。
(d) Why $[\mathrm{Sc(H_2O)_6}]^{3+}$ is colourless.为何 $[\mathrm{Sc(H_2O)_6}]^{3+}$ 无色。
Ligands split a transition metal's five $d$ orbitals into two groups of slightly different energy. Colour requires an electron to absorb a photon of visible light and be promoted between these split levels ($d$-$d$ transition), which needs a partially filled $d$ subshell: both an electron to promote and a vacant orbital to receive it. $\mathrm{Sc^{3+}}$ (Table 2) has $3d^0$: no $d$ electrons exist at all, so no $d$-$d$ transition is possible and no visible light is absorbed by this mechanism, so $[\mathrm{Sc(H_2O)_6}]^{3+}$ is colourless. $\mathrm{Cu^{2+}}$ ($3d^9$), $\mathrm{Ni^{2+}}$ ($3d^8$), and $\mathrm{Co^{2+}}$ ($3d^7$) all have partially filled $d$ subshells, so a $d$-$d$ transition is possible for each, and all three are coloured.配体使过渡金属的五个 $d$ 轨道分裂为能量略有差异的两组。显色要求电子吸收一个可见光光子后在这些分裂的能级间跃迁（$d$-$d$ 跃迁），这需要一个未完全充满的 $d$ 亚层：既要有可供跃迁的电子，也要有可供接纳的空轨道。$\mathrm{Sc^{3+}}$（见表 2）为 $3d^0$：完全没有 $d$ 电子，因此不可能发生 $d$-$d$ 跃迁，无法通过该机制吸收可见光，所以 $[\mathrm{Sc(H_2O)_6}]^{3+}$ 无色。$\mathrm{Cu^{2+}}$（$3d^9$）、$\mathrm{Ni^{2+}}$（$3d^8$）与 $\mathrm{Co^{2+}}$（$3d^7$）的 $d$ 亚层都未完全充满，因此每一种都可能发生 $d$-$d$ 跃迁，三者均显色。
(e) Cu blue vs Ni green; why different metals absorb different colours.Cu 呈蓝色而 Ni 呈绿色；为何不同金属吸收不同颜色。
From Table 2, $[\mathrm{Cu(H_2O)_6}]^{2+}$ absorbs orange light; orange and blue are complementary colours, so the transmitted/reflected colour observed is blue. $[\mathrm{Ni(H_2O)_6}]^{2+}$ absorbs red light; red and green are complementary colours, so the colour observed is green.由表 2 可知，$[\mathrm{Cu(H_2O)_6}]^{2+}$ 吸收橙光；橙色与蓝色互为互补色，因此观察到的透射/反射色为蓝色。$[\mathrm{Ni(H_2O)_6}]^{2+}$ 吸收红光；红色与绿色互为互补色，因此观察到的颜色为绿色。
More generally, the size of the energy gap created by $d$-orbital splitting depends on the specific metal ion (its nuclear charge, ionic radius, and number of $d$ electrons) and on the ligands surrounding it (different ligands split the $d$ orbitals by different amounts, per the spectrochemical series). A larger splitting gap requires a higher-energy, higher-frequency photon to be absorbed, and a smaller gap requires a lower-energy, lower-frequency photon. Since the absorbed frequency determines which colour is removed from white light, different metal ions, or the same metal ion with different ligands, absorb different colours of visible light and so display different observed (complementary) colours.更一般地说，$d$ 轨道分裂产生的能级差大小取决于具体的金属离子（其核电荷、离子半径与 $d$ 电子数）以及围绕它的配体（不同配体使 $d$ 轨道分裂的程度不同，遵循光谱化学序列）。分裂能级差越大，需要吸收的光子能量与频率就越高；能级差越小，所需吸收的光子能量与频率就越低。由于被吸收的频率决定了从白光中被移除的是哪种颜色，不同的金属离子，或搭配不同配体的同一金属离子，会吸收不同颜色的可见光，因而呈现出不同的观察（互补）颜色。
Insight洞察
This question links two apparently separate HL ideas through one shared logic: an exception to a general trend needs a subshell-level reason, not a nuclear-charge-only reason. Parts (a)-(c) apply that logic to ionization energy (a full or half-filled subshell resists losing an electron more than expected). Parts (d)-(e) apply the same "subshell occupancy matters more than the raw trend" logic to colour: whether a $d$ subshell is empty, partially filled, or fully split by a particular set of ligands determines whether, and which, visible light is absorbed.本题通过同一条逻辑线索，把两个看似独立的 HL 概念联系在一起：对一般趋势的例外，需要用亚层层面的原因来解释，而不能只归因于核电荷。(a)-(c) 将这一逻辑应用于电离能（充满或半充满的亚层比预期更抗拒失去电子）。(d)-(e) 则把同样"亚层占据情况比原始趋势更重要"的逻辑应用于颜色：$d$ 亚层是空的、部分充满的，还是被特定配体分裂开来，决定了是否会吸收可见光，以及吸收哪种可见光。