📚 Inorganic Chemistry: Periodicity, Group 2 and Group 7 | 无机化学：周期性、第2族与第7族
Inorganic chemistry forms a substantial portion of the AQA A-Level Chemistry syllabus, focusing on periodic trends and the behaviour of key groups. This article consolidates the essential concepts, equations and exam-ready observations needed to master this topic.
无机化学在AQA A-Level化学课程中占据重要比重，重点考察周期性规律及关键族的元素性质。本文整合核心概念、关键方程式与考试必备的观察现象，帮助你全面掌握这一专题。
1. Periodicity: Atomic Radius and Ionisation Energy | 周期性：原子半径与电离能
Across Period 3 (Na → Ar), the atomic radius decreases steadily. Although the number of protons increases, the electrons are added to the same principal quantum shell, so the increasing nuclear charge pulls the outer electrons more strongly towards the nucleus.
在第三周期（Na → Ar）中，原子半径逐渐减小。虽然质子数增加，但电子填入同一主量子壳层，因此核电荷增大使外层电子被更强烈地吸引向原子核。
First ionisation energy generally increases across the period for the same reason. However, two notable drops occur: between Mg and Al, and between P and S. The Mg → Al drop arises because Al’s outermost electron is in a 3p orbital, which is slightly higher in energy and better shielded than the 3s pair. The P → S drop reflects the extra repulsion between paired 3p electrons in sulfur.
同理，第一电离能总体呈增大趋势，但存在两个显著的下降点：Mg到Al之间，以及P到S之间。Mg→Al的下降是因为Al的最外层电子位于3p轨道，能量略高且屏蔽效应大于3s电子对；P→S的下降则源于硫原子中3p电子成对时的额外排斥力。
Na < Mg < Al < Si < P < S (slight drop) < Cl < Ar | 电离能总体增大，S处略有回落
| Element | First Ionisation Energy / kJ mol⁻¹ | 说明 |
| Na | 496 | 基线参照 |
| Mg | 738 | 3s²全满稳定 |
| Al | 578 | 3p¹能量较高，下降 |
| S | 1000 | 3p⁴成对电子排斥，低于预期 |
2. Periodicity: Electronegativity and Melting Points | 周期性：电负性与熔点
Electronegativity increases across Period 3 because the increasing nuclear charge and smaller atomic radius allow atoms to attract bonding electrons more effectively. Chlorine has the highest electronegativity in the period.
电负性沿第三周期递增，因为核电荷增大、原子半径缩小，原子吸引成键电子的能力增强。氯是第三周期电负性最高的元素。
Melting points across Period 3 show a distinctive pattern. Na, Mg and Al form metallic lattices with increasing charge density (Na⁺, Mg²⁺, Al³⁺) and more delocalised electrons, so metallic bonding strengthens. Silicon adopts a giant covalent structure with strong directional covalent bonds, giving it the highest melting point. P₄, S₈ and Cl₂ are simple molecular substances with weak van der Waals forces, hence low melting points.
第三周期的熔点呈现独特规律。Na、Mg、Al形成金属晶格，离子电荷密度递增（Na⁺、Mg²⁺、Al³⁺），离域电子增多，金属键增强。硅采取巨型共价结构，具有强方向性共价键，熔点最高。P₄、S₈、Cl₂为简单分子晶体，仅存较弱范德华力，熔点低。
Metallic → Giant covalent → Simple molecular | 金属晶体 → 巨型共价 → 简单分子
3. Period 3 Oxides: Acidic and Basic Behaviour | 第三周期氧化物：酸碱行为
Period 3 oxides display a gradual transition from basic through amphoteric to acidic behaviour. Na₂O and MgO are ionic oxides that react with water to form alkaline solutions: Na₂O + H₂O → 2NaOH. These oxides readily neutralise acids.
第三周期氧化物呈现从碱性经两性到酸性的渐变。Na₂O和MgO是离子型氧化物，与水反应生成碱性溶液：Na₂O + H₂O → 2NaOH。这些氧化物能轻易中和酸。
Al₂O₃ is amphoteric, meaning it reacts with both acids and strong bases. SiO₂ is a macromolecular acidic oxide that reacts with alkalis but not water. P₄O₁₀, SO₂ and SO₃ are acidic molecular oxides; SO₃ reacts vigorously with water to form sulfuric acid: SO₃ + H₂O → H₂SO₄.
Al₂O₃呈两性，既能与酸反应也能与强碱反应。SiO₂是巨型共价酸性氧化物，能与碱反应但不溶于水。P₄O₁₀、SO₂和SO₃均为酸性分子氧化物；SO₃剧烈与水反应生成硫酸：SO₃ + H₂O → H₂SO₄。
Na₂O, MgO (basic) → Al₂O₃ (amphoteric) → SiO₂, P₄O₁₀, SO₂, SO₃ (acidic)
Exam tip: You must be able to write balanced equations for Na₂O and MgO reacting with dilute HCl, and for Al₂O₃ reacting with both HCl and NaOH.
考试提示：需掌握Na₂O、MgO与稀盐酸的反应方程式，以及Al₂O₃分别与HCl和NaOH反应的方程式。
4. Group 2 Elements: Trends in Reactivity | 第2族元素：反应活性趋势
Group 2 metals (Be, Mg, Ca, Sr, Ba) are reducing agents whose reactivity increases down the group. As atomic radius increases, ionisation energies decrease, making it easier to lose the two outer s-electrons and form M²⁺ ions.
第2族金属（Be、Mg、Ca、Sr、Ba）是还原剂，反应活性随原子序数增大而增强。原子半径增大导致电离能降低，更容易失去两个外层s电子形成M²⁺离子。
Reaction with water is a key AQA focus. Magnesium reacts slowly with cold water but more readily with steam: Mg + H₂O(g) → MgO + H₂. Calcium, strontium and barium react increasingly vigorously with cold water, producing the hydroxide and hydrogen: Ca + 2H₂O → Ca(OH)₂ + H₂.
与水的反应是AQA考纲重点。镁与冷水反应缓慢，但能与水蒸气较快反应：Mg + H₂O(g) → MgO + H₂。钙、锶、钡与冷水反应越来越剧烈，生成氢氧化物和氢气：Ca + 2H₂O → Ca(OH)₂ + H₂。
All Group 2 metals react with dilute acids to release hydrogen gas, e.g. Mg + 2HCl → MgCl₂ + H₂. The rate of reaction increases down the group due to lower ionisation energy.
所有第2族金属都能与稀酸反应放出氢气，如Mg + 2HCl → MgCl₂ + H₂。由于电离能降低，反应速率随原子序数增大而加快。
5. Group 2 Hydroxides and Sulfates: Solubility Trends | 第2族氢氧化物与硫酸盐：溶解度趋势
The solubility of Group 2 hydroxides increases down the group. Mg(OH)₂ is sparingly soluble, producing a weakly alkaline suspension used as an antacid, whereas Ba(OH)₂ is freely soluble and forms a strongly alkaline solution. This trend is explained by the increasing size of the cation, which reduces its charge density and weakens the attraction to OH⁻ ions, allowing the lattice to dissolve more easily.
第2族氢氧化物的溶解度随原子序数增大而升高。Mg(OH)₂微溶于水，形成弱碱性悬浮液，可作抗酸药；而Ba(OH)₂易溶于水，形成强碱性溶液。该趋势可用阳离子半径增大解释：电荷密度降低，对OH⁻的吸引力减弱，晶格更易溶解。
In contrast, the solubility of Group 2 sulfates decreases down the group. MgSO₄ is very soluble, while BaSO₄ is virtually insoluble. The larger sulfate ion becomes increasingly difficult to accommodate as the cation size grows, and the lattice enthalpy is less exothermic relative to the hydration enthalpy.
相反，第2族硫酸盐的溶解度随原子序数增大而降低。MgSO₄易溶，而BaSO₄几乎不溶。随着阳离子增大，硫酸根大离子的晶格焓（放热程度）相对于水合焓减弱，导致溶解度下降。
| Property | Trend | Reason |
| Hydroxide solubility | Increases ↓ | Cation size ↑, charge density ↓ |
| Sulfate solubility | Decreases ↓ | Lattice enthalpy dominates |
6. Group 2 Thermal Decomposition | 第2族热分解
Group 2 carbonates and nitrates decompose on heating, and the thermal stability increases down the group. This is because the larger cations have lower charge density, which polarises the anion less, making the carbonate or nitrate ion harder to break apart.
第2族碳酸盐和硝酸盐受热分解，且热稳定性随原子序数增大而增强。这是因为较大的阳离子电荷密度低，对阴离子的极化作用弱，使碳酸根或硝酸根更难被破坏。
Carbonates: MCO₃ → MO + CO₂. For example, MgCO₃ decomposes readily at moderate heat, whereas BaCO₃ requires very strong heating. Nitrates: M(NO₃)₂ → MO + 2NO₂ + ½O₂. The brown gas NO₂ is a classic observation.
碳酸盐：MCO₃ → MO + CO₂。例如MgCO₃在中等温度下即分解，而BaCO₃需要强热。硝酸盐：M(NO₃)₂ → MO + 2NO₂ + ½O₂。生成棕色气体NO₂是典型实验现象。
Exam tip: Write down the colour change — white carbonate → white oxide, and for nitrates, the white solid turns yellow when hot (due to metal oxide formation) and brown fumes of NO₂ evolve.
考试提示：注意描述颜色变化——白色碳酸盐变为白色氧化物；硝酸盐加热时白色固体变黄（生成金属氧化物），并产生棕色NO₂气体。
7. Group 7 Halogens: Reactivity and Displacement | 第7族卤素：反应活性与置换
Halogens (F₂, Cl₂, Br₂, I₂) are oxidising agents. Oxidising power decreases down the group because atomic radius increases and electron shielding improves, so the ability to gain an electron weakens. Fluorine is the most powerful oxidising agent.
卤素（F₂、Cl₂、Br₂、I₂）均为氧化剂。氧化能力随原子序数增大而减弱，因为原子半径增大、电子屏蔽效应增强，得电子能力减弱。氟是最强的氧化剂。
This trend allows halogens to displace less reactive halides from solution. A more reactive halogen oxidises halide ions of a less reactive halogen: Cl₂ + 2KBr → 2KCl + Br₂. The solution changes from colourless to orange-brown, confirming the presence of bromine.
这一趋势使卤素能从溶液中置换出活性较弱的卤化物。较活泼的卤素氧化较不活泼卤素的离子：Cl₂ + 2KBr → 2KCl + Br₂。溶液从无色变为橙棕色，证明生成了溴。
Cl₂ + 2I⁻ → 2Cl⁻ + I₂ (brown solution) | Br₂ + 2I⁻ → 2Br⁻ + I₂ (brown solution)
Key colours for the exam: chlorine in water is pale green, bromine is orange/red-brown, and iodine is brown in solution or grey/black as a solid.
考试必记颜色：氯水呈淡绿色，溴呈橙红色，碘在溶液中呈棕色、固态呈灰黑色。
8. Halide Ions with Concentrated Sulfuric Acid | 卤离子与浓硫酸的反应
This reaction sequence distinguishes chloride, bromide and iodide ions. Concentrated H₂SO₄ acts first as an acid, then as an oxidising agent for Br⁻ and I⁻. Chloride ions are not oxidised because Cl₂ is a stronger oxidising agent than H₂SO₄.
该反应系列可用于区分氯离子、溴离子和碘离子。浓H₂SO₄先作为酸，再对Br⁻和I⁻充当氧化剂。Cl⁻不被氧化，因为Cl₂的氧化性强于H₂SO₄。
With NaCl: NaCl + H₂SO₄ → NaHSO₄ + HCl. The product HCl forms white acidic fumes (mist-like). With NaBr, further oxidation occurs: H₂SO₄ is reduced to SO₂, producing pungent SO₂ gas and orange bromine vapour. With NaI, reduction goes further to H₂S, giving a rotten-egg smell and purple iodine vapour.
与NaCl反应：NaCl + H₂SO₄ → NaHSO₄ + HCl，产物HCl形成白色酸雾。与NaBr反应时进一步氧化：H₂SO₄被还原为SO₂，产生刺激性SO₂气体和橙色溴蒸气。与NaI反应时还原程度更深，生成有臭鸡蛋气味的H₂S和紫色碘蒸气。
| Halide | Observation | Reduction product |
| Cl⁻ | White acidic fumes (HCl) | None |
| Br⁻ | Orange vapour (Br₂), pungent SO₂ | H₂SO₄ → SO₂ |
| I⁻ | Purple vapour (I₂), H₂S smell | H₂SO₄ → H₂S |
9. Testing for Halide Ions | 卤离子的检验
The standard test for halide ions uses silver nitrate solution acidified with dilute nitric acid. AgNO₃ reacts with halide ions to form precipitates of silver halides: Ag⁺ + X⁻ → AgX. The acid removes carbonate and sulfite impurities that could interfere.
卤离子的标准检验方法是用经稀硝酸酸化的硝酸银溶液。AgNO₃与卤离子反应生成卤化银沉淀：Ag⁺ + X⁻ → AgX。加入酸可除去可能干扰结果的碳酸根和亚硫酸根杂质。
Chloride gives a white precipitate (AgCl), bromide gives a cream precipitate (AgBr), and iodide gives a yellow precipitate (AgI). To confirm, add aqueous ammonia: AgCl dissolves in dilute NH₃, AgBr dissolves only in concentrated NH₃, and AgI is insoluble in both.
氯离子产生白色沉淀（AgCl），溴离子产生淡奶油色沉淀（AgBr），碘离子产生黄色沉淀（AgI）。进一步验证时加入氨水：AgCl可溶于稀氨水，AgBr仅溶于浓氨水，AgI在两者中均不溶。
AgCl (white) → dilute NH₃ dissolves | AgBr (cream) → concentrated NH₃ only | AgI (yellow) → insoluble
Soluble silver halides are photosensitive, decomposing in light: 2AgCl → 2Ag + Cl₂ (dark grey silver forms). This principle underlies traditional black-and-white photography.
卤化银具有感光性，光照下分解：2AgCl → 2Ag + Cl₂（生成深灰色银）。这一原理是传统黑白摄影的基础。
10. Uses of Chlorine and Water Treatment | 氯的用途与水质处理
Chlorine is added to public water supplies in small amounts to kill bacteria. It reacts with water in a disproportionation reaction: Cl₂ + H₂O ⇌ HClO + HCl. The hypochlorous acid (HClO) is responsible for the bactericidal action.
氯被少量添加到公共供水中以杀灭细菌。氯与水发生歧化反应：Cl₂ + H₂O ⇌ HClO + HCl。起杀菌作用的是次氯酸（HClO）。
Chlorine also disproportionates in cold dilute sodium hydroxide to form bleach: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. This mixture is used as household bleach and for disinfecting industrial wastewater.
氯在冷稀氢氧化钠中也发生歧化反应生成漂白剂：Cl₂ + 2NaOH → NaCl + NaClO + H₂O。该产物体系用于家用漂白剂和工业废水消毒。
You should note that chlorine’s oxidation state changes from 0 to −1 and +1 in both reactions, which is the defining feature of disproportionation. These equations are frequently assessed in exams.
注意氯的氧化态在这两个反应中都从0变为−1和+1，这正是歧化反应的特征。这两个方程式在考试中频繁出现，必须熟练书写。
11. Key Takeaways for Exam Success | 备考核心要点
First, memorise the trends: atomic radius, ionisation energy and electronegativity across Period 3; reactivity and solubility trends for Groups 2 and 7. Always explain trends using nuclear charge, shielding and atomic radius.
第一，牢记所有趋势：第三周期的原子半径、电离能、电负性；第2族和第7族的反应活性与溶解度趋势。解释趋势时务必从核电荷、屏蔽效应和原子半径三方面入手。
Second, practise writing balanced ionic and full equations for every reaction covered, including oxide-water, metal-acid, displacement and disproportionation reactions. Marks are often lost for missing state symbols.
第二，反复练习所有反应的离子方程式和完整方程式，包括氧化物与水、金属与酸、置换反应和歧化反应。漏写状态符号是常见失分原因。
Finally, link theory to observations. AQA rewards precise colour descriptions, gas smells and solubility test results. Learn the table of halide tests and the H₂SO₄ reaction products thoroughly.
最后，将理论与实验现象相聯系。AQA评分重视颜色描述、气体气味和溶解性测试结果的准确表述。彻底熟记卤离子检验表和浓硫酸反应产物。
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