📚 A-Level Chemistry: Classification and Identification of Organic Functional Groups | A-Level 化学：有机官能团分类与识别
In organic chemistry, a functional group is a specific group of atoms within a molecule that determines its characteristic chemical reactions and physical properties. Understanding how to classify and identify these groups is essential for mastering A-Level Chemistry, particularly in CIE examinations, where structure–property–reactivity relationships are heavily tested.
Alkanes contain only C–C and C–H single bonds. They are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. Under normal conditions they are relatively unreactive, but they undergo free-radical substitution with halogens in ultraviolet light.
Alkenes contain a carbon–carbon double bond (C=C) and have the general formula CₙH₂ₙ. The double bond consists of one σ bond and one π bond, making the molecule unsaturated. Alkenes readily undergo addition reactions, such as hydrogenation, hydration and reaction with halogens.
Alkynes contain a carbon–carbon triple bond (C≡C) and have the general formula CₙH₂ₙ₋₂. A terminal alkyne, such as ethyne HC≡CH, has a slightly acidic hydrogen, allowing it to form metal acetylides.
Arenes are cyclic aromatic hydrocarbons containing a benzene ring. The delocalised π-electron system gives them extra stability, so they prefer electrophilic substitution over addition.
Alcohols contain the hydroxyl group –OH bonded to a saturated carbon atom. They are classified as primary, secondary or tertiary depending on how many carbon groups are attached to the carbon bearing the –OH group. Primary alcohols can be oxidised to aldehydes then carboxylic acids; secondary alcohols oxidise to ketones; tertiary alcohols resist oxidation.
Ethers have the general structure R–O–R′, where R and R′ are alkyl or aryl groups. They are relatively unreactive and do not form hydrogen bonds between their own molecules, which makes their boiling points lower than those of isomeric alcohols.
醚的通用结构为 R–O–R′，其中 R 和 R′ 是烷基或芳基。醚相对稳定，分子间不能形成氢键，因此其沸点低于同分异构的醇。
Phenols contain an –OH group directly attached to a benzene ring. The ring withdraws electron density from the O–H bond, so phenols are weakly acidic. They react with strong bases such as NaOH and give a characteristic purple colour with iron(III) chloride.
Aldehydes contain the –CHO group. They are easily oxidised to carboxylic acids and can be reduced to primary alcohols. A positive Tollens’ test – formation of a silver mirror – is a classic test for aldehydes.
Ketones contain a carbonyl group C=O bonded to two carbon groups. They are not readily oxidised, so they do not react with Tollens’ reagent or Fehling’s solution under normal conditions. Ketones can be reduced to secondary alcohols.
Carboxylic acids contain the –COOH group. They are weak acids, releasing H⁺ ions in water. They react with metals, bases and carbonates to form salts, and with alcohols in the presence of an acid catalyst to form esters.
Esters have the general structure RCOOR′. They are formed by condensation between a carboxylic acid and an alcohol. Esters are volatile compounds with sweet smells, and they undergo hydrolysis in acidic or alkaline conditions to give back the acid and alcohol.
Acyl chlorides contain the –COCl group. They are highly reactive derivatives of carboxylic acids. With water they hydrolyse rapidly; with alcohols they form esters; with ammonia or amines they form amides. These reactions are typically vigorous and exothermic.
Acid anhydrides have the structure (RCO)₂O. They react similarly to acyl chlorides but less violently. Ethanoic anhydride is commonly used to introduce an acetyl group, for example in the synthesis of aspirin.
Amides contain the –CONH₂ group (or substituted derivatives). They are neutral compounds. Amides can be hydrolysed by hot aqueous acid or alkali to yield a carboxylic acid and an amine or ammonia.
Amines contain the –NH₂ group (primary), –NHR (secondary) or –NR₂ (tertiary). They act as bases because the nitrogen lone pair can accept a proton. Amines react with acids to form ammonium salts, and primary amines can be prepared by nucleophilic substitution of halogenoalkanes with ammonia.
Nitro compounds contain the –NO₂ group. The nitro group is strongly electron-withdrawing, which stabilises negative charge in intermediates. Nitro compounds are important in the synthesis of dyes and pharmaceuticals, especially as precursors to aromatic amines.
Nitriles contain the –C≡N group. They can be formed by reacting halogenoalkanes with cyanide ions in ethanol under reflux. Nitriles can be reduced to primary amines or hydrolysed to carboxylic acids, making them versatile synthetic intermediates.
6. Halogenoalkanes and Sulfur Compounds | 卤代烃与含硫官能团
Halogenoalkanes contain a halogen atom (F, Cl, Br, I) bonded to a saturated carbon. The C–X bond is polarised, and the carbon is susceptible to nucleophilic attack. These compounds undergo nucleophilic substitution with hydroxide, ammonia and cyanide ions, and can also undergo elimination to form alkenes.
Sulfur-containing functional groups, such as thiols (–SH), are less common in A-Level questions but may appear in extension contexts. Thiols are weak acids, have strong odours, and can be oxidised to disulfides (–S–S–), an important linkage in protein structure.
7. Chemical Tests for Functional Groups | 官能团的化学鉴别
Chemical tests rely on distinctive reactions to identify functional groups. The table below summarises the most common tests required for CIE A-Level Chemistry.
化学鉴别依靠特征反应来识别官能团。下表总结了 CIE A-Level 化学中最常见的鉴别方法。
Test / Reagent
Observation
Functional Group
Bromine water
Orange to colourless
Alkene (C=C)
Tollens’ reagent (ammoniacal AgNO₃)
Silver mirror
Aldehyde
Fehling’s or Benedict’s solution
Red precipitate
Aldehyde
2,4-DNPH (Brady’s reagent)
Orange/yellow precipitate
Aldehyde or ketone
NaHCO₃
Effervescence (CO₂)
Carboxylic acid
Sodium metal
Colourless gas (H₂)
Alcohol or carboxylic acid
FeCl₃(aq)
Purple colour
Phenol
Acidified K₂Cr₂O₇
Orange to green
Primary or secondary alcohol; aldehyde
Iodoform test (I₂ + NaOH)
Yellow precipitate (CHI₃)
CH₃CO– or CH₃CH(OH)– group
8. Spectroscopic Identification | 波谱识别
Infrared (IR) spectroscopy identifies functional groups by their characteristic bond vibrations. For example, an O–H bond produces a broad absorption around 3300 cm⁻¹, while a C=O bond gives a strong, sharp absorption around 1700 cm⁻¹. A C≡N group shows a sharp absorption near 2200 cm⁻¹.
Mass spectrometry gives the molecular ion peak at m/z equal to the relative molecular mass, alongside fragment peaks. For example, a bromine-containing compound shows a characteristic M⁺ and (M+2)⁺ pair in roughly 1:1 ratio due to ⁷⁹Br and ⁸¹Br isotopes.
Nuclear magnetic resonance (NMR) spectroscopy provides information about the carbon–hydrogen environment. The number of signals indicates the number of distinct proton environments; integration shows the ratio of protons; and splitting patterns reveal neighbouring protons. Tetramethylsilane (TMS) is used as the internal standard at δ = 0.
One common mistake is confusing the oxidation behaviour of aldehydes and ketones. Remember that aldehydes are oxidised easily, while ketones are not. This is why Tollens’ and Fehling’s tests distinguish them.
Another pitfall is forgetting the priority order when naming compounds. If more than one functional group is present, the principal group determines the suffix. The usual priority order is: carboxylic acid > ester > aldehyde > ketone > alcohol > amine > alkene > alkyne > halogenoalkane.
Students also frequently misidentify phenol as an alcohol. Phenol is distinctly more acidic than aliphatic alcohols and shows a specific purple colour with FeCl₃, whereas ethanol does not. Always check whether the –OH group is directly attached to an aromatic ring.
Finally, pay close attention to reaction conditions. For example, oxidation of a primary alcohol gives an aldehyde only if distilled off immediately; strong heating under reflux gives the carboxylic acid. Missing these conditions in an exam answer costs marks.
Classifying and identifying organic functional groups is a central skill in A-Level Chemistry. Each functional group has a recognisable structure, predictable reactivity and characteristic analytical signals. By mastering the functional group families, their chemical tests and spectroscopic fingerprints, you can confidently tackle both synthesis and identification problems in the CIE examinations.
📚 A-Level Chemistry: Common Organic Reaction Types | 常见有机反应类型归纳
Organic chemistry is often described as the study of carbon compounds, but at its core it is the study of molecular transformations. Understanding the common reaction types — substitution, addition, elimination, oxidation, reduction, hydrolysis, polymerisation, and condensation — is essential for predicting products, proposing mechanisms, and solving synthesis problems in A-Level examinations. This guide offers a systematic review of these core reaction families under the CIE syllabus, with emphasis on conditions, reagents, and structural changes.
A substitution reaction is one in which an atom or a functional group is replaced by another atom or functional group. In aliphatic chemistry, the most significant substitution reactions are nucleophilic, occurring at saturated carbon atoms. A classic example is the reaction of a halogenoalkane with aqueous sodium hydroxide to form an alcohol: C₂H₅Br + NaOH → C₂H₅OH + NaBr. The hydroxide ion acts as a nucleophile, attacking the electron-deficient carbon and displacing the bromide ion.
In aromatic chemistry, electrophilic substitution is the dominant mechanism. Benzene, with its delocalised π-electron cloud, reacts with electrophiles such as Br₂ (in the presence of a halogen carrier), concentrated HNO₃/H₂SO₄, or alkyl halides under Friedel-Crafts conditions. The mechanism proceeds through a carbocation intermediate known as the sigma complex, followed by loss of H⁺ to restore aromaticity.
Addition reactions are characteristic of unsaturated compounds — alkenes, alkynes, carbonyls, and nitriles. In an addition reaction, two reactants combine to form a single product, with the π-bond being broken and two new σ-bonds formed. For alkenes, the typical reactions include hydrogenation (H₂/Ni), halogenation (Br₂), hydrohalogenation (HX), and hydration (steam/H₃PO₄).
The addition of hydrogen bromide to an unsymmetrical alkene follows Markovnikov’s rule: the hydrogen attaches to the carbon with more hydrogen atoms, producing the more substituted carbocation intermediate. For example, propene reacts with HBr to give 2-bromopropane as the major product. In the presence of peroxides, however, the reaction proceeds via a free-radical mechanism, giving 1-bromopropane instead.
Elimination reactions involve the removal of adjacent atoms or groups from a molecule, resulting in the formation of a π-bond (usually a C=C or C≡C). An important example in the CIE syllabus is the dehydrohalogenation of halogenoalkanes. When 2-bromopropane is heated under reflux with ethanolic potassium hydroxide, the hydroxide behaves as a base rather than a nucleophile, abstracting a proton from the β-carbon while the bromine leaves as Br⁻, yielding propene.
Elimination competes with nucleophilic substitution — aqueous conditions favour substitution while ethanolic conditions favour elimination. This is a classic exam point that students must identify from the reagents and conditions provided. The elimination product usually follows Zaitsev’s rule, forming the more substituted alkene as the major product.
Oxidation in organic chemistry is defined as an increase in the oxygen content or a decrease in the hydrogen content of a molecule. Primary alcohols are oxidised first to aldehydes and then to carboxylic acids. Secondary alcohols are oxidised to ketones, while tertiary alcohols do not undergo oxidation under mild conditions. The common oxidising agents are acidified K₂Cr₂O₇ and KMnO₄; the colour change from orange to green indicates a positive result.
For the CIE examination, the specific conditions are crucial. Gentle heating with distillation allows the aldehyde to be collected before further oxidation, whereas reflux conditions drive the oxidation all the way to the carboxylic acid. Aldehydes can be distinguished from ketones by their ready oxidation — for instance, Tollens’ reagent produces a silver mirror with aldehydes but not with ketones.
Reduction is the opposite of oxidation — it involves an increase in hydrogen content or a decrease in oxygen content. In A-Level chemistry, the most important reduction reactions are the conversion of aldehydes and ketones to alcohols using sodium tetrahydroborate(III), commonly called NaBH₄. Aldehydes are reduced to primary alcohols, while ketones are reduced to secondary alcohols. NaBH₄ is a milder reagent than lithium aluminium hydride (LiAlH₄) and can be safely used in aqueous solution.
Nitriles can be reduced to primary amines using LiAlH₄, and carboxylic acids can be reduced to primary alcohols. However, these reactions are less frequently tested in CIE than the NaBH₄ reduction of carbonyl compounds. It is also essential to write the balanced ionic equation: [BH₄]⁻ + 4RCHO + 4H₂O → 4RCH₂OH + B(OH)₄⁻.
Hydrolysis is the cleavage of a chemical bond by the action of water, often in the presence of an acid or a base. The hydrolysis of esters is an important CIE topic. Acid-catalysed hydrolysis of an ester produces a carboxylic acid and an alcohol, and the reaction is reversible. Alkaline hydrolysis, on the other hand, is irreversible because the carboxylate salt formed does not react further with the alcohol; this reaction is known as saponification.
Acid hydrolysis of amides produces a carboxylic acid and an ammonium salt, while alkaline hydrolysis produces a carboxylate salt and ammonia or an amine. The conditions determine the products. For example, the hydrolysis of N-ethylethanamide with dilute HCl yields ethanoic acid and ethylammonium chloride, whereas hydrolysis with NaOH yields sodium ethanoate and ethylamine.
Polymerisation combines many small monomer molecules into long-chain macromolecules. There are two major categories relevant to A-Level chemistry: addition polymerisation and condensation polymerisation. Addition polymerisation occurs when alkene monomers open their double bonds and join together without the loss of any small molecule. Poly(ethene) comes from ethene, and poly(propene) comes from propene. The repeating unit is identified by placing brackets around the monomer residue.
Condensation polymerisation involves monomers with two functional groups, such as dicarboxylic acids and diols or diamines, and results in the elimination of a small molecule (commonly water). A key example is the formation of Terylene from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. Students must be able to draw the repeating unit, identify the ester or amide link, and calculate the molar mass of a polymer given the degree of polymerisation.
n CH₂=CH₂ → [-CH₂-CH₂-]ₙ | n HOOC-R-COOH + n HO-R’-OH → [-OC-R-COO-R’-O-]ₙ + 2n H₂O
8. Condensation Reactions | 缩合反应
Condensation reactions involve the joining of two molecules with the simultaneous loss of a small molecule such as water or hydrogen chloride. In addition to polymerisation, important condensation reactions in the CIE syllabus include the formation of esters from carboxylic acids and alcohols, and amides from carboxylic acids and amines. These reactions are typically catalysed by concentrated H₂SO₄ and are reversible.
The formation of an ester from ethanol and ethanoic acid under acid catalysis is a classic example: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The reaction is slow at room temperature; a drop of concentrated sulfuric acid acts both as a catalyst and as a dehydrating agent to shift the equilibrium to the right. Students should also be able to name the ester formed from any given alcohol and carboxylic acid — the alcohol part forms the alkyl group and the acid part forms the alkanoate group.
9. Acid-Base Reactions of Organic Compounds | 有机化合物的酸碱反应
Although less emphasised in some textbooks, acid-base behaviour is central to several organic reactions. Carboxylic acids are weak acids: they donate a proton to form a carboxylate ion, as seen in their reaction with sodium carbonate to produce CO₂. Phenols are weaker acids than carboxylic acids but stronger than alcohols, and they dissolve in strong alkalis to form phenoxide ions. Amines are basic because the lone pair on nitrogen can accept a proton; this explains the formation of ammonium salts when amines react with acids.
The acid-base reaction between an amine and an acid is also the first step in forming an amide or a quaternary ammonium salt. Acidic and basic properties also dictate solubility, separation, and purification techniques, such as using sodium bicarbonate to separate a carboxylic acid from a phenol, as the former reacts with NaHCO₃ while the latter does not.
The table below summarises the reagents, conditions, and products of the main reaction types covered in this article. Revising from a summary table helps students quickly identify patterns across different functional groups and improves recall in the examination hall.
Organic chemistry is fundamentally the chemistry of carbon. The remarkable diversity of organic compounds arises from the unique ability of carbon atoms to form strong covalent bonds with other carbon atoms and with a variety of elements, including hydrogen, oxygen, nitrogen, and the halogens. Understanding the nature of these bonds—how they form, their geometry, and their electronic structure—is essential for predicting the properties and reactions of organic molecules. This article provides a systematic analysis of bonding in organic molecules, tailored for CIE A-Level Chemistry students.
1. The Octet Rule and Valence Electrons | 八隅体规则与价电子
The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their outermost shell, achieving a stable noble-gas electronic configuration. For carbon, which has four valence electrons (2s²2p²), this means forming four shared pairs of electrons in four covalent bonds. Hydrogen is the exception, requiring only two electrons (a duet) to fill its 1s orbital.
In covalent bonding, atoms share electron pairs rather than transferring them completely. A single covalent bond consists of one shared pair of electrons, a double bond consists of two shared pairs, and a triple bond consists of three shared pairs. Carbon readily forms single bonds in alkanes, double bonds in alkenes, and triple bonds in alkynes.
2. Orbital Hybridisation: The sp³, sp², and sp Models | 轨道杂化：sp³、sp² 与 sp 模型
To explain the observed geometries and bond angles of organic molecules, Linus Pauling introduced the concept of orbital hybridisation. When a carbon atom forms four equivalent single bonds, its 2s orbital and three 2p orbitals mix to form four equivalent sp³ hybrid orbitals. These orbitals are arranged tetrahedrally around the carbon atom at an angle of 109.5°.
For carbon atoms involved in a double bond, such as in ethene (C₂H₄), the 2s orbital mixes with only two of the three 2p orbitals, producing three sp² hybrid orbitals lying in a plane at 120° to each other. The remaining unhybridised p orbital is perpendicular to this plane and participates in π bond formation.
In a triple bond, as in ethyne (C₂H₂), the carbon atom uses sp hybridisation. The two sp hybrid orbitals are linear at 180°, and two unhybridised p orbitals are available to form two perpendicular π bonds.
在三键中，如乙炔（C₂H₂），碳原子采用 sp 杂化。两个 sp 杂化轨道呈直线排列，夹角为 180°，剩余的两个未杂化 p 轨道可用于形成两个相互垂直的 π 键。
3. Sigma (σ) and Pi (π) Bonds | σ 键与 π 键
A sigma (σ) bond is formed by head-on overlap of two atomic orbitals along the internuclear axis. This type of overlap is strong and allows free rotation about the bond axis. All single bonds are σ bonds. In ethene, the C–H and C–C single-bond components are all σ bonds; the C=C double bond consists of one σ bond and one π bond.
A pi (π) bond arises from the sideways (lateral) overlap of two parallel unhybridised p orbitals. The electron density in a π bond is concentrated above and below the plane of the nuclei. Because this sideways overlap is less efficient than head-on overlap, π bonds are weaker than σ bonds. The presence of a π bond restricts rotation about the double bond, leading to geometrical (cis-trans) isomerism in alkenes.
π 键是由两个平行的未杂化 p 轨道进行“肩并肩”侧向重叠形成的。π 键的电子云密度集中在原子核平面之上和之下。由于侧向重叠的效率低于头对头重叠，π 键弱于 σ 键。π 键的存在限制了绕双键的自由旋转，从而在烯烃中导致了几何（顺反）异构现象。
4. Bonding in Alkanes: The Pure sp³ Framework | 烷烃中的成键：纯 sp³ 骨架
In alkanes such as methane (CH₄), ethane (C₂H₆), and propane (C₃H₈), every carbon atom is sp³ hybridised and forms four σ bonds. The C–H σ bonds result from the overlap of a carbon sp³ orbital with the 1s orbital of hydrogen. The C–C σ bonds result from the overlap of two carbon sp³ orbitals. All bond angles in methane are 109.5°, giving a perfect tetrahedral shape.
Because only σ bonds are present and the electron distribution around each C–C bond is cylindrically symmetrical, free rotation occurs around C–C single bonds. This allows alkane molecules to adopt various conformations, although the staggered conformation of ethane is more stable due to reduced torsional strain.
5. Bonding in Alkenes: The sp² Framework and the π Bond | 烯烃中的成键：sp² 骨架与 π 键
Ethene (C₂H₄) is the simplest alkene. Each carbon atom is sp² hybridised. Three sp² hybrid orbitals form two C–H σ bonds and one C–C σ bond. The fourth orbital, an unhybridised 2p orbital on each carbon, overlaps sideways to form a π bond. This π bond, together with the σ bond, constitutes the C=C double bond. The H–C–H bond angle is 117° and the H–C–C angle is 121.5°, which are close to the ideal 120° of perfect trigonal planar geometry.
The presence of the π bond makes the double bond shorter and stronger than a single bond, but it also creates an area of high electron density above and below the plane of the molecule. This electron-rich region makes alkenes reactive towards electrophiles, which is why alkenes undergo electrophilic addition reactions—a key feature of alkene chemistry.
6. Bonding in Alkynes: The sp Framework and Two π Bonds | 炔烃中的成键：sp 骨架与两个 π 键
In ethyne (C₂H₂), each carbon atom is sp hybridised. The two sp hybrid orbitals form one C–C σ bond and one C–H σ bond, arranged linearly. Each carbon also has two unhybridised 2p orbitals that overlap sideways in two perpendicular directions, creating two π bonds. Thus, the carbon-carbon triple bond consists of one σ bond and two π bonds.
The linear geometry of alkynes (180° bond angle) places the two hydrogen atoms at opposite ends of the molecule. The triple bond is highly electron-rich and even more reactive than a double bond towards certain electrophiles. The hybridisation state also affects the acidity of the terminal hydrogen: the sp-hybridised carbon exerts greater s-character, which stabilises the negative charge on the conjugate base (acetylide ion), making terminal alkynes weakly acidic.
炔烃的直线型几何（180° 键角）使两个氢原子处于分子的两端。三键电子云密度极高，对某些亲电试剂甚至比双键更具反应活性。杂化状态还影响末端氢的酸性：sp 杂化碳具有更高的 s 轨道成分，能够稳定共轭碱（炔负离子）上的负电荷，因此末端炔烃呈现弱酸性。
7. Bonding in Benzene: The Delocalised π System | 苯中的成键：离域 π 体系
Benzene (C₆H₆) is the classic example of aromatic bonding. Each carbon atom in the hexagonal ring is sp² hybridised and forms two C–C σ bonds and one C–H σ bond. Each carbon also contributes one unhybridised 2p orbital perpendicular to the ring plane. The six p orbitals overlap sideways to form a continuous ring of electron density above and below the plane of the molecule—a delocalised π system containing six electrons.
The delocalisation of π electrons confers exceptional stability on benzene. All six C–C bonds are identical in length (139 pm), intermediate between a single bond (154 pm) and a double bond (134 pm). This equalisation and the resulting resonance stabilisation explain why benzene resists addition reactions and instead undergoes electrophilic substitution, preserving the stable aromatic ring.
When atoms of different electronegativity share electrons, the electron pair is not shared equally. The more electronegative atom attracts the bonding electrons more strongly, acquiring a partial negative charge (δ-), while the less electronegative atom acquires an equal partial positive charge (δ+). This results in a polar covalent bond, characterised by a permanent dipole moment.
C–H bonds are generally considered non-polar because the electronegativities of carbon (2.55) and hydrogen (2.20) are close. By contrast, C–O, C–N, O–H, and N–H bonds are significantly polar due to the greater electronegativity of oxygen (3.44) and nitrogen (3.04). The polarities of these bonds govern intermolecular forces, solubility, boiling points, and the reactivity of functional groups. For example, the O–H bond in alcohols and carboxylic acids is strongly polar, enabling hydrogen bonding between molecules.
Bond length is the equilibrium distance between the nuclei of two bonded atoms, and bond energy is the energy required to break one mole of a bond in the gas phase. Both properties are closely correlated with bond order. For carbon–carbon bonds:
As bond order increases, the bond becomes shorter and stronger. The additional π bonds in double and triple bonds increase bond energy substantially, although not proportionally—the π component is weaker than the σ component. This is why addition reactions across a double or triple bond are thermodynamically favourable: the energy released in forming two new strong σ bonds compensates for breaking one π bond.
10. Hybridisation Effects on Acidity and Basicity | 杂化对酸性和碱性的影响
The hybridisation state of carbon profoundly affects the acidity of hydrogen atoms attached to it. As the s-character of the hybrid orbital increases, the carbon atom holds its electrons closer to the nucleus, making the C–H bond more polar and the conjugate base more stable. The order of acidity is:
碳的杂化状态深刻影响与其相连的氢原子的酸性。随着杂化轨道 s 轨道成分的增加，碳原子将其电子拉得更靠近原子核，使 C–H 键极性增强，共轭碱更加稳定。酸性强弱顺序为：
This trend is explained by the greater electronegativity of sp-hybridised carbon (due to 50% s-character) compared with sp² (33% s) and sp³ (25% s) carbon. The acetylide ion formed from a terminal alkyne is significantly stabilised relative to alkyl or alkenyl anions, allowing terminal alkynes to be deprotonated by strong bases such as sodium amide (NaNH₂).
这一趋势可通过不同杂化态碳的电负性差异来解释：sp 杂化碳的电负性最大（具有 50% 的 s 轨道成分），sp² 次之（33% s），sp³ 最小（25% s）。相对于烷基负离子或烯基负离子，由末端炔烃形成的炔负离子被显著稳定化，因此末端炔烃可被氨基钠（NaNH₂）等强碱去质子化。
11. Intermolecular Bonding and Physical Properties | 分子间成键与物理性质
Although covalent bonds hold atoms together within a molecule, weaker intermolecular forces determine many physical properties. These forces include van der Waals forces (induced dipole interactions), permanent dipole-dipole interactions, and hydrogen bonding. Their strength depends on bond polarities and molecular structure.
In A-Level organic chemistry, this explains why ethanol (with an O–H bond) has a much higher boiling point (78 °C) than dimethyl ether (with the same molecular formula but only C–O–C bonds) at (-24 °C). Ethanol forms intermolecular hydrogen bonds, whereas dimethyl ether only exhibits weaker dipole interactions. Similarly, carboxylic acids form stable hydrogen-bonded dimers, which accounts for their relatively high boiling points.
12. Predicting Reactivity from Bonding | 从成键方式预测反应活性
The bonding model of organic molecules provides a powerful framework for predicting chemical reactivity. The distribution of σ and π bonds, the hybridisation of carbon centres, and the polarity of bonds to heteroatoms determine the sites of nucleophilic and electrophilic attack.
For example, the electron-rich π bond of an alkene attacks electrophiles; the polar C=O bond of a carbonyl group makes the carbon atom electrophilic and the oxygen atom nucleophilic; the polar O–H bond in an alcohol provides a mildly acidic proton. Recognising these bonding patterns enables students to propose sensible reaction mechanisms and to predict the major products of organic reactions—a central skill for CIE A-Level examination success.
📚 A-Level Chemistry: Chemical Reactions of Group 7 Elements | A-Level化学：第七主族元素的化学反应
The Group 7 elements, also known as the halogens, include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). These elements are highly reactive non-metals that display clear trends in their physical and chemical properties. This article focuses on their characteristic chemical reactions — with hydrogen, water, and alkali, as well as halide displacement reactions and tests — all essential for CIE A-Level Chemistry.
1. Reactivity Trend and Electron Configuration | 反应活性趋势与电子排布
The halogens all have the general outer-shell configuration ns²np⁵, meaning they require one more electron to achieve a stable noble-gas configuration. This drives their strong tendency to gain an electron and act as oxidising agents.
Going down the group, the atomic radius increases, shielding increases, and the attraction between the nucleus and an incoming electron decreases. Consequently, the ease of gaining an electron — oxidising ability — decreases down the group:
All halogens react with hydrogen to form hydrogen halides, HX, but the vigour of the reaction decreases down the group. Fluorine reacts explosively even in the dark at very low temperatures; chlorine reacts readily in sunlight; bromine reacts slowly under mild heating; iodine reacts only partially and reversibly at high temperatures.
The thermal stability of hydrogen halides also decreases down the group. HF is the most stable, while HI decomposes most readily on heating. This follows the trend in bond energy: the H–X bond length increases down the group, so the bond becomes weaker.
Chlorine, bromine, and iodine dissolve in water to form an acidic solution. Chlorine reacts with water in a disproportionation reaction, where one chlorine atom is oxidised and another is reduced:
氯、溴和碘都能溶于水生成酸性溶液。氯与水发生歧化反应，一个氯原子被氧化，另一个被还原：
Cl₂ + H₂O ⇌ HCl + HClO
In this reaction, the oxidation state of chlorine changes from 0 in Cl₂ to −1 in HCl (reduction) and +1 in HClO (oxidation). The HClO (chloric(I) acid) is an oxidising agent and bleach. The position of equilibrium lies to the left, so only a small proportion of chlorine reacts.
Bromine and iodine show the same type of behaviour but the reaction becomes progressively less extensive. Iodine reacts only very slightly with water. Fluorine, however, reacts violently with water, oxidising the oxygen in water to give oxygen gas:
4. Reaction with Alkali: Disproportionation | 与碱的反应：歧化反应
Chlorine reacts with cold, dilute sodium hydroxide to form a mixture of chloride and chlorate(I) ions:
氯气与冷、稀的氢氧化钠溶液反应生成氯化物和次氯酸盐的混合物：
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
This is a disproportionation reaction in which chlorine is simultaneously oxidised from 0 to +1 (in NaClO) and reduced from 0 to −1 (in NaCl). The product NaClO is the active ingredient in household bleach.
With hot, concentrated alkali, chlorine undergoes a more extensive disproportionation, forming chloride and chlorate(V) ions:
与热、浓碱反应时，氯的歧化更充分，生成氯化物和氯酸盐(V)：
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O
In this case, five-sixths of the chlorine is reduced to −1 and one-sixth is oxidised to +5 (in ClO₃⁻). The same pattern applies to bromine and iodine: cold dilute alkali gives X⁻ and XO⁻; hot concentrated alkali gives X⁻ and XO₃⁻.
Because oxidising ability decreases down the group, a more reactive halogen can displace a less reactive halogen from its halide salt solution. The more reactive halogen gains electrons and is reduced, while the halide ion loses electrons and is oxidised.
Bromine water + potassium iodide solution: Br₂ + 2KI → 2KBr + I₂. The solution turns brown.
溴水 + 碘化钾溶液：Br₂ + 2KI → 2KBr + I₂。溶液变为棕色。
In contrast, iodine cannot displace chlorine or bromine from their salts because iodine is a weaker oxidising agent. These displacement reactions confirm the trend in oxidising ability: Cl₂ > Br₂ > I₂.
Halide ions in aqueous solution can be identified by adding silver nitrate solution followed by ammonia. The formation of a precipitate with characteristic colour confirms the presence of a particular halide.
水溶液中的卤离子可通过加入硝酸银溶液，再加入氨水来鉴别。生成沉淀的特征颜色可确认相应卤离子的存在。
Ag⁺(aq) + X⁻(aq) → AgX(s)
Halide ion | 卤离子
Precipitate colour | 沉淀颜色
Effect of dilute NH₃ | 稀氨水作用
Effect of concentrated NH₃ | 浓氨水作用
Cl⁻
White (AgCl) | 白色
Dissolves | 溶解
Dissolves | 溶解
Br⁻
Cream (AgBr) | 奶油色
Insoluble | 不溶
Dissolves | 溶解
I⁻
Yellow (AgI) | 黄色
Insoluble | 不溶
Insoluble | 不溶
Nitric acid must be added first to remove interfering ions such as carbonate or sulfite, which would also form precipitates with silver ions. The silver halide precipitates confirm the identity and relative solubility trend: AgCl is the most soluble and AgI the least.
7. Acid–Base Character of Hydrogen Halides | 卤化氢的酸碱性
When hydrogen halides dissolve in water, they form acidic solutions. The acidity of the hydrohalic acids increases down the group: HF is a weak acid, while HCl, HBr, and HI are strong acids in water.
卤化氢溶于水后形成酸性溶液。氢卤酸的酸性随原子序数增大而增强：HF 是弱酸，而 HCl、HBr 和 HI 在水中都是强酸。
The increasing acid strength is explained by the decreasing H–X bond energy. As the bond becomes easier to break, the release of H⁺ becomes more favourable. The trend is therefore:
酸性增强的原因在于 H–X 键能逐渐降低。键越容易断裂，释放 H⁺ 的趋势越大。因此酸性顺序为：
HF < HCl < HBr < HI
In the gas phase, all hydrogen halides are covalent molecules. In aqueous solution, HCl, HBr, and HI ionise completely, whereas HF only partially ionises due to its strong H–F bond and hydrogen bonding in solution.
在气态时，所有卤化氢都是共价分子。在水溶液中，HCl、HBr 和 HI 完全电离，而 HF 仅部分电离，这是因为 H–F 键强且溶液中存在氢键作用。
8. Uses of Halogens and Their Compounds | 卤素及其化合物的用途
Halogens and their compounds have widespread industrial and everyday applications that relate directly to their chemical reactivity.
卤素及其化合物在工业生产和日常生活中有着广泛用途，这些用途与其化学活性直接相关。
Chlorine is used in water treatment to kill bacteria and to manufacture bleach (NaClO), PVC, and hydrochloric acid.
氯气用于饮用水消毒、制造漂白剂（NaClO）、聚氯乙烯（PVC）和盐酸。
Fluorine compounds, such as NaF and SnF₂, are added to toothpaste to prevent tooth decay. Teflon (PTFE) is a fluoropolymer with high thermal stability.
Bromine is used in the production of flame retardants and in silver bromide photography (AgBr is light-sensitive).
溴用于生产阻燃剂，以及溴化银感光材料（AgBr 具有光敏性）。
Iodine is an essential element for thyroid function. Iodine in the form of KI is added to table salt as an iodine supplement, and iodine solution is used as an antiseptic.
The key periodic trends in Group 7 are summarised below. These are frequently assessed in CIE A-Level examinations, both in structured questions and data-analysis tasks.
Weaker attraction for bonding electrons | 对成键电子吸引力减弱
Oxidising ability | 氧化性
Decreases | 减弱
Harder to gain an electron | 得电子更难
Reducing ability of X⁻ | X⁻的还原性
Increases | 增强
Larger ion loses electron more easily | 离子越大越易失电子
Thermal stability of HX | HX的热稳定性
Decreases | 减弱
H–X bond energy decreases | H–X 键能减弱
10. Exam-Style Problem: Applying the Trends | 考试真题演练：运用趋势规律
Consider the following reaction scheme. Chlorine gas is bubbled into an aqueous solution of potassium iodide, and the resulting mixture is shaken with an organic solvent such as hexane.
考虑以下反应设计：将氯气通入碘化钾水溶液中，所得混合液与己烷等有机溶剂振荡。
Question 1 | 问题一: Write the ionic equation for the reaction that occurs.
Question 1 | 问题一: 写出发生反应的离子方程式。
Cl₂(g) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
Question 2 | 问题二: State what would be observed in the aqueous layer and in the organic layer.
Question 2 | 问题二: 说明水层和有机层中的观察现象。
The aqueous layer would turn brown, and the organic layer would appear purple/violet, indicating the presence of iodine in the non-polar solvent.
水层变为棕色，有机层呈紫色/紫红色，说明碘溶解在非极性溶剂中。
Question 3 | 问题三: Explain, in terms of electron transfer, why this displacement occurs.
Question 3 | 问题三: 从电子转移的角度解释为何能发生该置换反应。
Chlorine is a stronger oxidising agent than iodine. Each chlorine atom gains an electron and is reduced to Cl⁻, while each iodide ion loses an electron and is oxidised to I₂.
氯的氧化性强于碘。每个氯原子获得一个电子被还原为 Cl⁻，而每个碘离子失去一个电子被氧化为 I₂。
11. Common Misconceptions and Exam Advice | 常见误区与备考建议
Many students lose marks in Group 7 questions due to a few repeated mistakes. Here are the most important points to remember.
很多学生在第七主族题目中失分是因为反复出现几个典型错误。以下是需要重点记住的要点。
Do not say ‘the halogen gains electrons because it is more reactive.’ Instead, link reactivity to oxidising ability and electron affinity trends.
不要说“因为更活泼所以卤素得电子”。应将活泼性与氧化能力及得电子趋势关联。
In disproportionation reactions, identify the oxidation states clearly. Chlorine is 0 in Cl₂, −1 in Cl⁻, +1 in ClO⁻, and +5 in ClO₃⁻.
When writing equations for the reaction with alkali, check whether the alkali is cold or hot, dilute or concentrated — this determines whether ClO⁻ or ClO₃⁻ is formed.
书写与碱反应的方程式时，需注意碱是冷是热、是稀是浓——这决定了生成 ClO⁻ 还是 ClO₃⁻。
For halide ion tests, always remember to add nitric acid first, before silver nitrate, and record the colour of the precipitate and its solubility in ammonia.
卤离子检验中，务必记住在加入硝酸银之前先加硝酸，并记录沉淀颜色及其在氨水中的溶解性。
Published by TutorHao | Chemistry Revision Series | aleveler.com
Organic chemistry is the chemistry of carbon compounds, and its apparent complexity arises from the countless ways carbon atoms can combine with one another and with other elements. Before any reaction mechanism or synthesis can be discussed, chemists must agree on a ‘language’ for drawing molecules. In the CIE A-Level specification, students are expected to interpret and interconvert several forms of structural representation — from molecular formulas to full displayed formulas, condensed formulas, skeletal formulas, and three-dimensional wedge-and-dash diagrams. Mastering these conventions is not merely a drawing exercise; it is the foundation for understanding isomerism, reactivity, and spectroscopy.
1. Molecular Formula and Empirical Formula | 分子式与最简式
The molecular formula states the exact number of atoms of each element in one molecule of a compound. For example, ethanol has the molecular formula C₂H₆O: two carbon atoms, six hydrogen atoms, and one oxygen atom per molecule. The empirical formula, by contrast, gives the simplest whole-number ratio of atoms present. For ethanol, the ratio 2 : 6 : 1 cannot be simplified, so the empirical formula is also C₂H₆O. For glucose, however, the molecular formula C₆H₁₂O₆ simplifies to an empirical formula of CH₂O.
Molecular formula = (empirical formula)ₙ where n is a positive integer
分子式 = (最简式)ₙ，其中 n 为正整数
In CIE examinations, you may be asked to deduce an empirical formula from combustion data or percentage composition by mass, and then determine the molecular formula using the relative molecular mass (Mᵣ). For instance, if a compound has empirical formula CH₂O and Mᵣ = 90, then the empirical formula mass is 12 + 2 + 16 = 30, so n = 90 ÷ 30 = 3, giving the molecular formula C₃H₆O₃. This compound could, in fact, be lactic acid — a reminder of how formula work connects to real biochemistry.
2. Full Structural Formula (Displayed Formula) | 完整结构式（图示式）
The full structural formula, also called the displayed formula, shows every atom and every bond explicitly. Each covalent bond is drawn as a line between the symbols of the two atoms it connects. This is the most informative two-dimensional representation, but it becomes cumbersome for large molecules. For ethanol, the displayed formula is:
Notice that every C−H, C−C, C−O, and O−H bond is shown individually. In an exam, when asked to ‘draw the displayed formula’, you must include all of these bonds — omitting the C−H bonds is a common and costly error. For straight-chain alkanes such as hexane, displaying every hydrogen produces a long chain of zigzag bonds: CH₃−CH₂−CH₂−CH₂−CH₂−CH₃ with all 14 hydrogens shown, which is time-consuming but unambiguous.
For alkenes, the displayed formula must also show the double bond as two parallel lines between the two sp²-hybridised carbon atoms, as in ethene: H₂C=CH₂ with each carbon carrying two hydrogen atoms drawn explicitly. Ensure that bond angles are drawn approximately correct — 120° at sp² carbons and 109.5° at sp³ carbons — though examiners rarely measure angles precisely, a realistic sketch is expected.
The condensed structural formula is a compromise between clarity and brevity. It lists atoms in the order they are bonded, grouping repeating units in parentheses where convenient. For butane, the condensed formula is CH₃CH₂CH₂CH₃, and it can also be written as CH₃(CH₂)₂CH₃. For branches, parentheses enclose the substituent: 2-methylpropane is written CH₃CH(CH₃)CH₃. This tells the reader that the middle carbon bears one hydrogen and one CH₃ side group.
Functional groups are written explicitly: ethanol as CH₃CH₂OH, propanoic acid as CH₃CH₂COOH, propanone as CH₃COCH₃, and but-2-ene as CH₃CH=CHCH₃. The carbonyl group (C=O) and carboxyl group (COOH) must be written in full so the functional group is identifiable. A common trap is writing ‘CHO’ for an alcohol — ‘CHO’ actually represents an aldehyde group (as in CH₃CHO, ethanal). The alcohol group must be written as ‘OH’ attached directly to carbon, e.g. CH₃CH₂OH.
The skeletal formula is the most streamlined representation used in organic chemistry, and in CIE papers it frequently appears in both questions and mark schemes. In this convention, carbon atoms are represented by each vertex and each terminus of a zigzag line; hydrogen atoms attached to carbon are omitted entirely; and only heteroatoms (O, N, Cl, Br, etc.) are written explicitly, together with any hydrogens attached to them.
Consider butane, CH₃CH₂CH₂CH₃. Its skeletal formula is a simple zigzag line with four vertices (two ends plus two bends). Each vertex is understood to carry enough hydrogen atoms to give carbon its four bonds — a terminal carbon has three H’s and an internal carbon has two H’s. For cyclohexane, the skeletal formula is a regular hexagon; for benzene, a hexagon with an inscribed circle. A C=C double bond is drawn as two parallel lines in the zigzag, and it locks the geometry: groups attached to a double bond lie in one plane.
When drawing skeletal formulas in exams, three rules matter: first, do not label carbon atoms with the letter C — that defeats the purpose; second, always show heteroatoms and their attached hydrogens (e.g. −OH, −NH₂, −Cl); third, include lone pairs only when a mechanism or acid-base argument requires them. Skeletal formulas are particularly powerful for large molecules such as steroids or fatty acids, where a displayed formula would be overwhelming.
在考试中绘制骨架式时，有三条规则至关重要：第一，不要用字母 C 标注碳原子——那样就失去了意义；第二，始终画出杂原子及其所连的氢（如 −OH、−NH₂、−Cl）；第三，只有在机理或酸碱论证需要时才标出孤对电子。骨架式在处理甾体或脂肪酸这类大分子时尤其高效，因为完整结构式会过于庞大。
Organic molecules are three-dimensional, yet most representations are drawn on a flat page. The wedge-and-dash convention solves this problem elegantly. A solid wedge means the bond projects out of the plane of the paper toward the viewer; a dashed wedge (usually drawn as a set of parallel dashes) means the bond recedes behind the plane; and an ordinary thin line means the bond lies in the plane of the page.
This convention is essential at a chiral centre. Take 2-chlorobutane, CH₃CH(Cl)CH₂CH₃: the carbon-2 atom is chiral because it bears four different substituents (CH₃, Cl, H, and CH₂CH₃). To specify which enantiomer is drawn, you might place the Cl on a wedge (coming toward you), the H on a dashed wedge (going away), and the CH₃ and CH₂CH₃ in the plane. Simply swapping the Cl and H produces the opposite enantiomer — a fact that matters enormously in pharmaceutical chemistry, where one enantiomer of a drug may be therapeutic and the other toxic.
For cyclic molecules, wedges and dashes also indicate the stereochemistry at ring carbons. In cyclohexane, axial bonds drawn alternately up-and-down and equatorial bonds drawn roughly sideways on the chair conformation; but in A-Level questions, you are more likely to encounter wedge-dash notation in molecules such as 2-bromopropane or halogenoalkanes undergoing nucleophilic substitution, where the 3D outcome (inversion of configuration) may be probed.
6. Stereochemistry and E/Z Isomerism | 立体化学与 E/Z 异构
Alkenes exhibit a special type of stereoisomerism because the C=C double bond prevents free rotation. When each doubly bonded carbon carries two different groups, two arrangements are possible. The CIE specification expects you to use the E/Z system based on Cahn–Ingold–Prelog (CIP) priority rules, and for simple cases the older cis/trans labels are still accepted.
In the CIP system, each atom directly attached to the alkene carbon is assigned a priority based on atomic number: the higher the atomic number, the higher the priority. If the two priority groups lie on the same side of the double bond, the isomer is labelled Z (from German zusammen, ‘together’); if they lie on opposite sides, it is labelled E (entgegen, ‘opposite’). For but-2-ene, CH₃CH=CHCH₃, both alkene carbons carry one CH₃ (priority carbon atomic number 6) and one H (atomic number 1). In Z-but-2-ene the
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
Systematic nomenclature is the language through which chemists communicate the structure of organic molecules with precision. In the CIE A-Level syllabus, the IUPAC system is not just a list of rules — it is a logical framework that links structure to name, and name back to structure.
Common names such as “acetone” or “ethylene glycol” are convenient in everyday conversation, but they fail to reveal the molecular structure unambiguously. IUPAC nomenclature ensures that every organic compound has exactly one systematic name that can be translated directly into a structural formula.
Consider the molecular formula C₄H₁₀O: it corresponds to two different alcohols and one ether. Only systematic names — butan-1-ol, butan-2-ol and ethoxyethane — can distinguish them clearly.
2. The Foundation: Parent Chain and Prefixes | 根基：主链与前缀
The first step in naming any acyclic compound is to identify the longest continuous carbon chain. This chain gives the parent name: meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec- for 1 to 10 carbon atoms.
The suffix indicates the principal functional group, while prefixes are used for substituents. For example, a five-carbon chain with a methyl substituent on carbon 2 is named 2-methylpentane.
3. Numbering the Chain: The Lowest Locant Rule | 链编号：最低位次规则
After identifying the parent chain, we assign locants (numbers) to the carbon atoms. The chain must be numbered so that the principal functional group receives the lowest possible number.
确定主链后，需为碳原子编号。编号方向的选择必须使主要官能团获得尽可能小的位次数字。
If there are multiple substituents, the first point of difference determines the direction. Compare 1,2-dimethylpentane with 2,3-dimethylpentane — the former is correct because 1 < 2 at the first differing position.
Always give the principal group the lowest locant.
始终将主官能团放在最低位次。
Then minimise locants for substituents in alphabetical order of comparison.
然后按取代基字母顺序比较，使位次总和最小。
4. Alkanes and Branched Chains | 烷烃与支链
Alkanes are the simplest family. The suffix is -ane. For branched alkanes, each substituent is named as an alkyl group: -yl replaces -ane (methyl, ethyl, propyl, etc.).
When multiple identical substituents appear, use the multiplicative prefixes di-, tri-, tetra- and list every locant explicitly. The prefixes di-, tri- etc. are ignored when alphabetising, but iso-, sec- and tert- are not.
For unsaturated hydrocarbons, the parent chain must contain the double or triple bond, and the suffix becomes -ene or -yne. The chain is numbered to give the multiple bond the lowest locant.
对于不饱和烃，主链必须包含双键或三键，后缀变为“-烯”或“-炔”。链编号应使多重键获得最低位次。
More than one double bond is indicated as -diene, -triene, etc. Every double bond must receive a separate locant. An example is hexa-1,3-diene: the two double bonds are at positions 1 and 3.
When both a double and a triple bond appear, the suffix is ordered as -en-yne, and the double bond takes priority for lower locants — so pent-1-en-4-yne is correct, not pent-4-en-1-yne.
Halogen substituents are always treated as prefixes: fluoro-, chloro-, bromo-, iodo-. They have no priority in selecting the parent chain or the suffix; they simply modify the alkane name.
卤素取代基始终作为前缀处理：氟、氯、溴、碘。它们在选择主链或后缀时没有优先权，只是修饰烷烃名称。
Number the chain to give the lowest locants to the principal functional group first, then to multiple substituents. In 1-bromo-2-chloropropane, bromo and chloro are both substituents; the chain is numbered to give the lowest sum.
The suffix -ol identifies the presence of a hydroxyl group (–OH). The parent chain must include the carbon bearing the –OH, and this carbon receives the lowest possible locant.
后缀“-醇”标识羟基（–OH）的存在。主链必须包含连接 –OH 的碳原子，该碳获得尽可能低的位次。
If the –OH group is on a benzene ring, the compound is named as a phenol derivative. For example, C₆H₅OH is simply phenol; 2-methylphenol has a methyl group at position 2 relative to the –OH.
Polyhydric alcohols use the suffixes -diol, -triol. The simple compound HOCH₂CH₂OH is ethane-1,2-diol. Note that the final -e of ethane is retained when a vowel follows in the suffix.
Aldehydes contain the –CHO group. As the highest priority group in this pair, they take the suffix -al. Long-chain aldehydes are numbered so that the carbonyl carbon is always carbon 1.
Ketones contain C=O within the chain, given the suffix -one. The carbonyl carbon receives the lowest locant. Propanone is the systematic name for the common solvent often called acetone.
When both –CHO and –OH are present, the –CHO takes naming priority as the suffix, and –OH becomes the prefix hydroxy-. Thus HOCH₂CH₂CHO is 3-hydroxypropanal.
Carboxylic acids contain –COOH and take the suffix -oic acid. The carboxyl carbon is always carbon 1. Since it necessarily lies at the chain end, no locant is normally needed for the –COOH group itself.
Esters are named as alkyl alkanoates: The alkyl group from the alcohol comes first, followed by the alkanoate derived from the carboxylic acid. CH₃COOCH₂CH₃ is ethyl ethanoate, not ethylethanoate — do not combine them.
Amines are named by locating the longest chain bearing the –NH₂ group and adding the suffix -amine. For example, CH₃CH₂CH₂NH₂ is propan-1-amine. Secondary and tertiary amines may be named as N-substituted primary amines.
Amides contain –CONH₂ and use the suffix -amide, with the carbonyl carbon as carbon 1. Substituents on nitrogen are indicated by the prefix N-. Thus CH₃CONHCH₃ is N-methylethanamide.
11. Nitriles and Other Functional Groups | 腈与其它官能团
Nitriles contain the –C≡N group. The suffix is -nitrile, and the carbon of the –CN group is included in the parent chain number count but not as a plain carbon atom. CH₃CH₂CN is propanenitrile.
腈含有 –C≡N 基团，后缀为“-腈”，–CN 中的碳计入主链编号。CH₃CH₂CN 是丙腈。
When a nitrile coexists with a higher-priority group, –CN becomes the prefix cyano-. For example, 2-cyanopropanoic acid contains both –COOH (suffix) and –CN (prefix).
Many A-Level candidates lose marks not from a lack of doctrine but from systematic micro-errors. The most frequent ones include: selecting the wrong parent chain when two chains have equal length but different substituents; forgetting to place commas between locants and hyphens between numbers and names; and ignoring alphabetical order among prefixes.
📚 A-Level Chemistry: Sulphur and Its Oxides | A-Level化学：硫及其氧化物的性质
Sulphur is a versatile non-metal in Group 16 of the Periodic Table, with the electron configuration [Ne] 3s² 3p⁴. It exists in several allotropic forms and exhibits oxidation states of −2, +4 and +6 in its common compounds. For CIE A-Level Chemistry, the key focus is on the physical properties of sulphur, the acid–base and redox behaviour of sulphur dioxide, the industrial Contact Process for sulphur trioxide, and the environmental consequences of sulphur oxide emissions.
1. Physical Properties and Allotropes of Sulphur | 硫的物理性质与同素异形体
Sulphur is a yellow, brittle solid at room temperature. Its most stable allotrope is rhombic sulphur, which consists of crown-shaped S₈ molecules packed in an orthorhombic crystal lattice. When heated above 95.6 °C, rhombic sulphur slowly changes into monoclinic sulphur, another allotrope that is also composed of S₈ molecules but with a different crystal arrangement. Both forms are insoluble in water but dissolve in organic solvents such as carbon disulfide.
硫在室温下是一种黄色脆性固体。其最稳定的同素异形体是正交硫，由皇冠形的 S₈ 分子按正交晶格排列而成。当加热至 95.6 °C 以上时，正交硫会缓慢转变为单斜硫；单斜硫同样由 S₈ 分子构成，但晶体排列方式不同。这两种同素异形体均不溶于水，但可溶于二硫化碳等有机溶剂。
The melting point of sulphur is about 119 °C, and its boiling point is approximately 444 °C. In the solid state, S₈ rings are held together by weak van der Waals forces, which explains the low melting point despite the strong covalent bonds within each ring.
2. The Unusual Viscosity of Molten Sulphur | 熔融硫的异常黏度变化
The behaviour of sulphur when heated is unusual and is a favourite examination topic. When solid sulphur melts at 119 °C, it forms an amber, free-flowing liquid. As the temperature rises to about 160 °C, the liquid becomes darker and increasingly viscous, reaching a maximum viscosity near 190 °C. Above 190 °C, the liquid becomes thinner again until it boils at 444 °C.
硫在加热过程中表现出反常的黏度变化，这也是考试中非常喜欢考查的内容。当固态硫在 119 °C 熔化时，得到琥珀色的易流动液体。随着温度升高至约 160 °C，液体颜色加深，黏度显著增大，在约 190 °C 时黏度达到最大值。超过 190 °C 后，液体又逐渐变稀，直至 444 °C 沸腾。
This behaviour is explained in terms of ring-opening and polymerisation. The S₈ rings open when heated and join end-to-end to form long chains containing many hundreds of sulphur atoms. These long chains become entangled, raising the viscosity. At still higher temperatures, the long chains break into shorter fragments, allowing the liquid to flow more freely again.
3. Chemical Properties: Reaction with Metals and Non-metals | 硫的化学性质：与金属和非金属的反应
Sulphur reacts directly with many metals to form metal sulphides. For example, iron filings heated with sulphur produce iron(II) sulphide, and sodium reacts vigorously to form sodium sulphide. In these reactions sulphur acts as an oxidising agent, accepting electrons and adopting the −2 oxidation state.
Sulphur also burns in oxygen to form sulphur dioxide, a colourless, pungent gas. This combustion reaction is strongly exothermic and is used industrially to generate SO₂ for the manufacture of sulphuric acid.
The important oxidation states of sulphur are −2 in hydrogen sulphide and metal sulphides, +4 in SO₂ and sulphites, and +6 in SO₃ and sulphates. These oxidation states govern the redox behaviour examined in the syllabus.
4. Laboratory Preparation of Sulphur Dioxide | 二氧化硫的实验室制备
Sulphur dioxide can be prepared in the laboratory by the action of a dilute acid on a sulphite salt. A typical method involves adding dilute hydrochloric acid to solid sodium sulphite; the SO₂ gas is evolved and can be collected by upward delivery or over warm water, since it is denser than air and quite soluble in cold water.
This reaction is a simple acid–base metathesis in which the unstable sulphurous acid, H₂SO₃, decomposes immediately into SO₂ and water. The gas is collected and identified by its characteristic choking smell and by its ability to turn damp blue litmus paper red, showing its acidic nature.
5. Sulphur Dioxide as an Acidic Oxide | 二氧化硫的酸性氧化物性质
Sulphur dioxide dissolves in water to form sulphurous acid, H₂SO₃. This is a weak dibasic acid that exists only in solution. The equilibrium shown below explains why solutions of SO₂ are acidic and why the gas turns damp litmus paper red.
As an acidic oxide, SO₂ reacts with bases to form salts. With an excess of sodium hydroxide, sodium sulphite is produced. With a limited amount of alkali, the hydrogen sulphite (bisulphite) salt is formed instead.
These reactions are analogous to those of carbon dioxide with alkalis, and the same care is needed when writing balanced ionic equations. In aqueous solution, sulphite ions, SO₃²⁻, act as weak bases and accept protons from water.
The most important redox property of SO₂ is its ability to act as a reducing agent. In SO₂ the sulphur atom is in the +4 oxidation state; it can be oxidised further to the +6 state in sulphate ions, SO₄²⁻. Thus SO₂ is readily oxidised by strong oxidising agents.
When acidified potassium manganate(VII) solution is added to SO₂, the purple MnO₄⁻ ion is reduced to colourless Mn²⁺. The SO₂ is simultaneously oxidised to sulphate. This reaction decolorises the purple solution and is a simple test for SO₂.
Sulphur dioxide also reduces iron(III) ions to iron(II) ions, and bromine water to bromide ions. In each case SO₂ is oxidised to sulphate, and these reactions are useful examples in both ionic equation and oxidation-state questions.
Another application of the reducing property is the bleaching action of SO₂. It combines with coloured substances and reduces them to colourless compounds. This bleaching is usually reversible: the original colour may return when the material is exposed to air or heated, in contrast to the permanent bleaching achieved by chlorine.
7. Sulphur Dioxide as an Oxidising Agent | 二氧化硫的氧化性
Although SO₂ usually acts as a reducing agent, it can also behave as an oxidising agent when it reacts with a stronger reducing agent. The classic example is the reaction between SO₂ and hydrogen sulphide, H₂S. In this reaction, sulphur in SO₂ is reduced from +4 to 0, while sulphur in H₂S is oxidised from −2 to 0.
This reaction produces a yellow precipitate of sulphur and is sometimes called the Claus reaction, used in industry to recover sulphur from hydrogen sulphide in natural gas. In the laboratory, mixing the two gases immediately produces a yellow deposit on the walls of the container, a clear demonstration that SO₂ can act as an oxidising agent.
In terms of electron transfer, the SO₂ molecule gains four electrons overall (2 × 2) as two sulphur atoms in H₂S each lose two electrons. The stoichiometry is important: one mole of SO₂ reacts with two moles of H₂S to give three moles of sulphur atoms.
8. Sulphur Trioxide and the Contact Process | 三氧化硫与接触法
Sulphur trioxide, SO₃, is the anhydride of sulphuric acid. It is produced industrially by the oxidation of SO₂ in the Contact Process. The reaction is reversible and exothermic:
The industrial conditions are a temperature of about 450 °C, a pressure of about 2 atm, and the use of vanadium(V) oxide, V₂O₅, as a catalyst. The catalyst provides an alternative reaction pathway with lower activation energy, allowing equilibrium to be reached more quickly at a moderate temperature.
工业条件为：约 450 °C 的温度、约 2 atm 的压力，并使用五氧化二钒 V₂O₅ 作为催化剂。催化剂提供了活化能更低的反应路径，使体系能在中等温度下更快达到平衡。
A low temperature favours the forward exothermic reaction and improves the equilibrium yield, but it makes the rate too slow. A compromise temperature of 450 °C is therefore chosen. High pressure favours the side with fewer gas molecules, which is the product side, but the yield at 2 atm is already over 95%, so very high pressures are unnecessary and uneconomic.
The SO₃ formed is not absorbed directly in water because the reaction is violently exothermic and produces a corrosive mist. Instead, SO₃ is first dissolved in concentrated H₂SO₄ to form oleum, H₂S₂O₇, which is then carefully diluted with water to give concentrated sulphuric acid.
Sulphur trioxide itself is a colourless, fuming liquid or solid at room temperature. As an acidic oxide, it reacts vigorously with water to form sulphuric acid and with basic oxides to form sulphates.
When coal and other fossil fuels containing sulphur impurities are burned, sulphur dioxide is released into the atmosphere. In the air, SO₂ can dissolve directly in water droplets to form sulphurous acid, and it can also be oxidised slowly to SO₃, which then dissolves in water to form sulphuric acid.
当含硫杂质的煤和其他化石燃料燃烧时，二氧化硫
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
📚 A-Level Chemistry: Properties and Uses of Sulfuric Acid | A-Level化学：硫酸的性质与用途
Sulfuric acid, H₂SO₄, is one of the most important industrial chemicals in the world. In A-Level chemistry, you need to understand its physical properties, its behaviour as an acid, a dehydrating agent and an oxidising agent, and how it is manufactured by the Contact Process.
Pure sulfuric acid is a dense, colourless, oily liquid with a molar mass of 98.1 g mol⁻¹. It has a high boiling point of about 337 °C, which means it is much less volatile than water and can be used to displace more volatile acids from their salts.
纯硫酸是一种密度大、无色、油状的液体，摩尔质量为98.1 g mol⁻¹。它的沸点约为337 °C，比水难挥发得多，因此可以用来从盐中置换出更易挥发的酸。
The molecule has a tetrahedral arrangement around the sulfur atom. Sulfur is bonded to two hydroxyl groups and two oxygen atoms by double bonds. In aqueous solution, sulfuric acid ionises in two stages because it is a dibasic acid.
The first ionisation is complete, and the second is partial. Therefore dilute sulfuric acid contains H⁺, HSO₄⁻ and SO₄²⁻ ions, along with water molecules.
第一步电离完全，第二步电离部分进行。因此稀硫酸中含有H⁺、HSO₄⁻和SO₄²⁻离子，以及水分子。
2. Acidic Properties: Reaction with Bases | 酸性：与碱的反应
Sulfuric acid is a strong acid, so dilute H₂SO₄ reacts readily with bases to form salts and water. Because it is dibasic, it can form two series of salts: normal sulfates and hydrogen sulfates.
With sodium hydroxide, the reaction depends on the mole ratio. With two moles of NaOH, sodium sulfate is formed:
与氢氧化钠反应时，产物取决于物质的量之比。当NaOH为两摩尔时，生成硫酸钠：
2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
With only one mole of NaOH per mole of acid, sodium hydrogen sulfate is produced:
当每摩尔酸只与一摩尔NaOH反应时，生成硫酸氢钠：
NaOH(aq) + H₂SO₄(aq) → NaHSO₄(aq) + H₂O(l)
Sulfuric acid also neutralises ammonia to give ammonium sulfate, a widely used fertiliser:
硫酸还能与氨发生中和反应，生成广泛使用的肥料——硫酸铵：
2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)
3. Reaction with Metals and Carbonates | 与金属和碳酸盐的反应
Dilute sulfuric acid reacts with reactive metals such as magnesium and zinc to release hydrogen gas. For example, with magnesium:
稀硫酸能与镁、锌等活泼金属反应并放出氢气。例如与镁的反应：
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)
However, lead and calcium are problematic because their sulfates are insoluble. The formation of an insoluble layer of PbSO₄ or CaSO₄ slows or stops the reaction.
These reactions are useful in qualitative analysis. The effervescence of CO₂ when an unknown carbonate is treated with dilute acid is a classic test.
这些反应在定性分析中很有用。用稀酸处理未知碳酸盐时产生CO₂气泡，是经典的检验方法。
4. Dehydrating Agent | 脱水剂
Concentrated sulfuric acid has a strong affinity for water. As a dehydrating agent, it removes the elements of water from covalent compounds, even when water molecules are not actually present.
For example, when concentrated H₂SO₄ is added to sucrose, C₁₂H₂₂O₁₁, the sugar is dehydrated to black carbon:
例如，将浓H₂SO₄加入蔗糖（C₁₂H₂₂O₁₁）中，蔗糖会被脱水变成黑色的碳：
C₁₂H₂₂O₁₁(s) → 12C(s) + 11H₂O(g)
The reaction produces steam, and the evolving gas also contains SO₂ from the further oxidation of carbon by concentrated sulfuric acid. The black, swollen solid is a familiar demonstration of dehydration.
Concentrated sulfuric acid also dehydrates hydrated copper(II) sulfate and many organic compounds such as ethanol. When ethanol is heated with concentrated H₂SO₄ at about 170 °C, ethene is formed:
This is why concentrated H₂SO₄ is used as a catalyst and dehydrating agent in the laboratory preparation of alkenes.
这也是浓H₂SO₄在实验室制备烯烃时被用作催化剂和脱水剂的原因。
5. Oxidising Agent | 氧化剂
Hot concentrated sulfuric acid is a powerful oxidising agent. In these reactions the sulfur in H₂SO₄ is reduced, usually from the +6 oxidation state to +4 in sulfur dioxide, SO₂.
Zinc and carbon are also oxidised. With carbon, the products are carbon dioxide, sulfur dioxide and water:
锌和碳也能被氧化。碳与浓硫酸反应生成二氧化碳、二氧化硫和水：
C(s) + 2H₂SO₄(conc) → CO₂(g) + 2SO₂(g) + 2H₂O(l)
Concentrated H₂SO₄ can also oxidise hydrogen halides. It oxidises hydrogen bromide to bromine, and hydrogen iodide to iodine, while itself being reduced to SO₂.
浓H₂SO₄还能氧化卤化氢。它可以把溴化氢氧化为溴，把碘化氢氧化为碘，自身被还原为SO₂。
2HBr + H₂SO₄(conc) → Br₂ + SO₂ + 2H₂O
Because of this oxidising behaviour, concentrated H₂SO₄ cannot be used to prepare HBr or HI from their salts, although it can be used for HCl, which is not oxidised.
由于具有氧化性，不能用浓H₂SO₄从盐中制备HBr或HI，但可以用于制备不被氧化的HCl。
6. Sulphonation and Esterification | 磺化与酯化
Sulfuric acid is an important reagent in organic chemistry. It acts as a catalyst in esterification, for example in the preparation of ethyl ethanoate from ethanol and ethanoic acid.
硫酸是有机化学中的重要试剂。它在酯化反应中作催化剂，例如由乙醇和乙酸制备乙酸乙酯。
It also participates in sulfonation. When benzene is heated with concentrated sulfuric acid, or with fuming sulfuric acid, benzenesulfonic acid is formed:
它还参与磺化反应。当苯与浓硫酸或发烟硫酸加热时，生成苯磺酸：
C₆H₆ + H₂SO₄(conc) → C₆H₅SO₃H + H₂O
This electrophilic substitution reaction is industrial important in the production of detergents and dyes. The sulfonic acid group –SO₃H makes the molecule more soluble in water.
这个亲电取代反应在工业上对洗涤剂和染料的生产很重要。磺酸基（–SO₃H）能增加分子在水中的溶解度。
In the Contact Process, concentrated sulfuric acid is also used to absorb sulfur trioxide. The SO₃ cannot be absorbed directly in water because it forms a fine mist of sulfuric acid that is difficult to condense. Instead, it is absorbed in concentrated H₂SO₄ to form oleum, H₂S₂O₇:
The oleum is then carefully diluted with water to give more sulfuric acid:
随后将发烟硫酸小心地用水稀释，得到更多的硫酸：
H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l)
7. Manufacture: The Contact Process | 工业制造：接触法
Almost all sulfuric acid is manufactured by the Contact Process. The starting material is sulfur, which is burned in air to produce sulfur dioxide:
几乎所有的硫酸都通过接触法制造。原料硫在空气中燃烧生成二氧化硫：
S(s) + O₂(g) → SO₂(g)
In many plants, SO₂ is also obtained by roasting sulfide ores such as zinc blende, ZnS. The sulfur dioxide is then oxidised to sulfur trioxide in the presence of a vanadium(V) oxide catalyst:
The forward reaction is exothermic and produces fewer moles of gas. According to Le Chatelier’s principle, low temperature and high pressure favour a high yield of SO₃. In practice, a temperature of about 450 °C is used, with a pressure of 1–2 atm and a V₂O₅ catalyst.
An intermediate temperature is chosen as a compromise: lower temperatures give a better equilibrium yield but a slower rate. The catalyst helps the reaction reach equilibrium quickly. Excess oxygen is used to drive the equilibrium to the right.
Finally, the SO₃ is absorbed in concentrated H₂SO₄ to form oleum, which is then diluted with water to form sulfuric acid of the required concentration.
最后，SO₃被吸收在浓H₂SO₄中形成发烟硫酸，再用水稀释成所需浓度的硫酸。
8. Uses in Industry and the Laboratory | 工业与实验室中的用途
Sulfuric acid is often described as the “king of chemicals” because it is used in more industrial processes than almost any other compound. Its largest use is in the manufacture of phosphate fertilisers such as ammonium sulfate and superphosphate.
Refining petroleum and removing impurities from oils.
石油精炼以及去除油品中的杂质。
Laboratory reagent for preparing gases, esters and as a dehydrating agent.
实验室中用于制备气体、酯，以及作为脱水剂。
Cleaning and pickling metal surfaces before electroplating.
电镀前对金属表面进行清洗和酸洗。
Sulfuric acid is also used in the manufacture of other acids, such as nitric acid, by displacing the more volatile acid from its salt:
硫酸还可通过置换挥发性酸的方法制备其他酸，例如硝酸：
NaNO₃(s) + H₂SO₄(conc) → NaHSO₄(s) + HNO₃(g)
9. Safety and Environmental Considerations | 安全与环境考量
Concentrated sulfuric acid is highly corrosive. It causes severe burns on contact with skin and eyes. It also reacts violently with water, releasing a great deal of heat. Therefore, when diluting, the acid must be added slowly to water with constant stirring, never water to acid.
Fumes of SO₂ and SO₃ are harmful. The Contact Process is designed to minimise the release of SO₂, and modern plants use catalytic converters or scrubbing systems to remove sulfur dioxide from waste gases.
Many raw materials for sulfuric acid, such as sulfur or sulfide ores, contain impurities. When these are burned, the SO₂ produced must be cleaned before entering the converter, because impurities can poison the vanadium(V) oxide catalyst.
In the laboratory, spills should be neutralised carefully with sodium carbonate or another weak base, and protective equipment such as chemical-resistant gloves and safety goggles should always be worn.
实验室中，硫酸泄漏应用碳酸钠等弱碱小心中和，操作时始终应佩戴耐化学品手套和护目镜等防护设备。
10. Summary of Key Reactions | 重点反应总结
The table below summarises the key reactions of sulfuric acid that you need to remember for CIE A-Level chemistry.
下表总结了你在CIE A-Level化学中需要记住的硫酸关键反应。
Type of behaviour
Reaction example
Observations
Strong acid
Mg + H₂SO₄ → MgSO₄ + H₂
Effervescence, gas burns with a pop
Dehydrating agent
C₁₂H₂₂O₁₁ → 12C + 11H₂O
Black carbon forms
Oxidising agent
Cu + 2H₂SO₄(conc) → CuSO₄ + SO₂ + 2H₂O
SO₂ smell, blue solution forms
Sulphonating agent
C₆H₆ + H₂SO₄ → C₆H₅SO₃H + H₂O
Water-soluble sulfonic acid formed
Remember that dilute H₂SO₄ is primarily an acid, while concentrated H₂SO₄ shows the additional dehydrating and oxidising properties. Being able to distinguish between these roles is essential for exams.
📚 A-Level Chemistry: Halogens and Their Compounds | A-Level化学：卤素及其化合物的用途
The halogens make up Group 17 of the periodic table: fluorine, chlorine, bromine, iodine and astatine. Their chemical behaviour is dominated by the need to gain one electron, forming halide ions X⁻. This strong oxidising ability is the reason why so many halogen-containing compounds are useful in commerce, industry, medicine and daily life.
At room temperature, fluorine is a pale yellow gas, chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid. All exist as diatomic molecules: F₂, Cl₂, Br₂ and I₂.
The outer-shell configuration of the halogens is ns²np⁵, and their chemistry is dominated by the half-equation:
卤素的外层电子构型为 ns²np⁵，它们的化学性质主要来自下面这个半反应：
X₂ + 2e⁻ → 2X⁻
The oxidising power and electronegativity decrease down the group. Thus chlorine can displace bromide and iodide ions, and bromine can displace iodide ions.
氧化能力和电负性自上而下减弱。因此氯可以置换溴离子和碘离子，溴可以置换碘离子。
Halogen
State at room temperature
Principal uses
F₂
Pale yellow gas
UF
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
📚 A-Level Chemistry: Structure and Chemical Properties of Nitrogen | A-Level化学：氮气的结构与化学性质
Nitrogen is an element of immense importance in both nature and industry. As a diatomic gas forming 78% of the Earth’s atmosphere, its structural features and unusual chemical inertness are core topics in the CIE A-Level Chemistry syllabus. This article provides a comprehensive review of the electronic structure, bonding, physical properties, chemical reactivity, and industrial applications of nitrogen, with all key points aligned to examination requirements.
1. Electronic Configuration of the Nitrogen Atom | 氮原子的电子构型
Nitrogen has an atomic number of 7. Its full electronic configuration is 1s² 2s² 2p³. The valence shell contains five electrons, with the three 2p orbitals each singly occupied according to Hund’s rule. This half-filled p-subshell gives the atom a relatively stable electronic arrangement.
The electron arrangement can be represented as 2,5 across two principal shells. The first ionisation energy of nitrogen is higher than that of carbon and oxygen, due to the extra stability of the half-filled 2p subshell and a comparatively small atomic radius.
2. The Structure of the Nitrogen Molecule | 氮气分子的结构
Two nitrogen atoms combine to form a dinitrogen molecule, N₂. A triple bond is formed between the two atoms: one sigma (σ) bond from head-on overlap of 2p orbitals, and two pi (π) bonds from side-on overlap of the remaining two pairs of p orbitals. Each nitrogen atom also retains one lone pair of electrons.
The bond length is extremely short, approximately 110 pm, and the bond energy is exceptionally high at 944 kJ mol⁻¹. This is among the strongest covalent bonds known. The bond order is 3, and the molecule is linear with no net dipole moment.
Nitrogen is a colourless, odourless, tasteless gas at room temperature. It is slightly less dense than air, with a density of 1.25 g dm⁻³. It condenses to a colourless liquid at −196 °C and freezes to a solid at −210 °C. It is sparingly soluble in water, dissolving only about 0.02 g per dm³ at room temperature.
氮气在室温下是无色、无臭、无味的气体。其密度略低于空气，为1.25 g dm⁻³。它在−196 °C时凝结为无色液体，在−210 °C时凝固为固体。氮气微溶于水，室温下每升水仅溶解约0.02克。
The weak intermolecular forces between non-polar N₂ molecules are London dispersion forces only. This explains its low boiling point and general unreactivity in the liquid phase. Liquid nitrogen is widely used as a coolant in laboratories and cryogenic applications.
Although the nitrogen atom is quite reactive, the nitrogen molecule is remarkably inert at room temperature. This apparent contradiction arises from two main factors.
尽管氮原子颇为活泼，但氮气分子在室温下却异常惰性。这一表面矛盾源于两个主要因素。
Firstly, the N≡N triple bond has an extremely high bond enthalpy of 944 kJ mol⁻¹. Breaking this bond requires a vast input of energy, so most reactions have a very high activation energy. Secondly, the bonding in N₂ is so strong that the molecule has a very stable electron cloud with a large HOMO–LUMO gap, making it resistant to attack by electrophiles and nucleophiles alike.
Kinetic stability, rather than thermodynamic instability, is the key: many nitrogen reactions are thermodynamically favourable, but proceed imperceptibly slowly at ordinary temperatures without a catalyst or high energy input.
5. Reaction with Hydrogen: The Haber Process | 与氢的反应：哈伯过程
Nitrogen reacts directly with hydrogen to form ammonia, NH₃, in the industrially crucial Haber process.
氮气与氢气直接反应生成氨NH₃，这一反应正是工业上至关重要的哈伯过程。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
Typical conditions are a temperature of 450 °C, a pressure of 200 atmospheres, and an iron catalyst promoted with potassium oxide and aluminium oxide. Le Chatelier’s principle explains why high pressure favours the forward reaction, since the forward direction involves a decrease in the number of gas moles. A moderate temperature is chosen as a compromise between rate and equilibrium yield.
The continuous removal of ammonia, use of an excess of nitrogen, and recycling of unreacted gases are all practical strategies to increase the yield and efficiency of the process.
不断移去氨气、使用过量氮气以及循环利用未反应气体，都是提高产率和过程效率的实际策略。
6. Reaction with Oxygen | 与氧的反应
At very high temperatures, such as those in a lightning discharge or a car engine, nitrogen reacts with oxygen to form nitrogen monoxide (nitric oxide).
在极高温度下，如闪电放电或汽车发动机内，氮气与氧气反应生成一氧化氮。
N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ mol⁻¹
This reaction is endothermic and requires a temperature above 2000 °C. The NO produced can then be oxidised further to nitrogen dioxide, NO₂, in the atmosphere. These nitrogen oxides are major contributors to photochemical smog and acid rain.
At ordinary temperatures, nitrogen and oxygen do not react at a measurable rate. The strong triple bond must be broken first, which demands enormous energy. This kinetic barrier is the main reason why atmospheric nitrogen is so stable in air.
7. Reaction with Metals: Formation of Nitrides | 与金属反应：生成氮化物
Although nitrogen is generally inert, it reacts with highly electropositive metals, especially magnesium, when strongly heated. For example, burning magnesium in air produces some magnesium nitride alongside the oxide.
The product magnesium nitride is an ionic compound containing the nitride ion, N³⁻. It reacts readily with water to release ammonia:
产物氮化镁为离子化合物，含氮离子N³⁻。它迅速与水反应放出氨：
Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)
This test is useful in distinguishing magnesium nitride from magnesium oxide. The similar reaction with lithium forms lithium nitride, Li₃N, which also hydrolyses to ammonia.
该反应可用于区分氮化镁与氧化镁。锂与氮气也有类似反应生成氮化锂Li₃N，同样水解产生氨。
8. Oxidation States of Nitrogen in Compounds | 氮在化合物中的氧化态
Nitrogen exhibits a wide range of oxidation states, from −3 to +5, reflecting its ability to gain or lose up to five valence electrons in covalent compounds.
氮在化合物中表现出从−3到+5的宽泛氧化态范围，反映其在共价化合物中可得失最多五个价电子的能力。
Oxidation state
Example compound
−3
NH₃, NH₄⁺
−2
N₂H₄ (hydrazine)
−1
NH₂OH
0
N₂
+1
N₂O
+2
NO
+3
HNO₂, N₂O₃
+4
NO₂
+5
HNO₃, NO₃⁻
The ability to adopt many oxidation states is a central theme in nitrogen chemistry. However, the free element N₂ at oxidation state 0 is unusually stable, which disconnects its chemical behaviour from that of its labile compounds.
9. Laboratory Preparations of Nitrogen | 实验室中氮气的制备
In the laboratory, nitrogen gas can be prepared by the thermal decomposition of ammonium nitrite or by the reduction of nitrous acid. A common method involves heating a mixture of ammonium chloride and sodium nitrite:
Alternatively, nitrogen can be produced by the oxidation of ammonia with hot copper(II) oxide, or by the thermal decomposition of sodium azide, NaN₃, though the latter is hazardous due to the explosive nature of azide salts.
These methods are rarely used industrially, because nitrogen is far cheaper to obtain by fractional distillation of liquefied air. This process exploits the difference in boiling points between nitrogen (−196 °C) and oxygen (−183 °C).
10. Economic and Environmental Significance | 经济与环境意义
Nitrogen is a vital commodity: it is used in ammonia production, fertilisers, explosives, nitric acid manufacture, food packaging, and as an inert atmosphere in chemical and metallurgical processes. Electronic industries use high-purity nitrogen during chip manufacturing to prevent oxidation of components.
However, the environmental impact of nitrogen compounds is substantial. Overuse of nitrogen fertilisers leads to eutrophication of waterways, while nitrogen oxide emissions from combustion contribute to acid rain and the greenhouse effect. Understanding the chemistry of nitrogen is essential for developing sustainable solutions.
Additionally, always use correct state symbols in equations and remember that nitrogen monoxide (NO) is not the same as nitrogen dioxide (NO₂). The former is a colourless gas; the latter is brown and highly toxic.
12. Conclusion: Master the Key Concepts | 结论：掌握关键概念
The structure of nitrogen, with its short and immensely strong triple bond, explains its physical properties and chemical inertness. Although N₂ is thermodynamically reactive in many potential reactions, kinetic barriers make it stable at room temperature. The Haber process harnesses catalysed reactivity at high pressure, while high-temperature combustion forms nitrogen oxides in the environment.
A sound grasp of electronic structure, bond energy, and the factors controlling reactivity is essential for success in the CIE A-Level Chemistry examination. Revise the equations, state symbols, and industrial conditions carefully, and you will be confident in tackling questions on nitrogen chemistry.
Group 2 metals (beryllium, magnesium, calcium, strontium, barium) form a wide range of compounds with distinct physical and chemical properties. Their uses in industry, agriculture, medicine, and daily life are directly linked to concepts in the CIE A-Level syllabus, including thermal stability, solubility trends, lattice enthalpy, and acid–base behaviour. This article systematically reviews the most important Group 2 compounds, their key reactions, and their common applications.
Magnesium oxide is a white, high-melting-point ionic solid. Its melting point is exceptionally high (approximately 2852 °C) because the Mg²⁺ and O²⁻ ions carry double charges, leading to a very strong electrostatic lattice. This thermal stability makes MgO an excellent refractory material.
Used to line furnaces and kilns in steelmaking — it withstands extreme temperatures without decomposing.
Used in electrical insulators and crucibles in laboratories.
Acts as a basic oxide, reacting slowly with water to form magnesium hydroxide.
用于炼钢炉和窑炉的内衬——可在极端高温下不分解。
用于实验室的电绝缘体和坩埚。
作为碱性氧化物，与水缓慢反应生成氢氧化镁。
MgO(s) + H₂O(l) → Mg(OH)₂(s)
2. Calcium Hydroxide (Ca(OH)₂) | 氢氧化钙
Calcium hydroxide, also known as slaked lime or limewater, is a moderately soluble base. Its aqueous solution is used as a test for carbon dioxide. In agriculture, it is used to neutralise acidic soils because it is cheap, alkaline, and less corrosive than sodium hydroxide.
Calcium carbonate, the main component of limestone, chalk, and marble, is the most widely used Group 2 compound. Its thermal decomposition follows the trend of increasing stability down the group: carbonates become more thermally stable as the cation size increases.
Used directly as a building material (marble, limestone aggregates).
Converted to calcium oxide (quicklime) in lime kilns.
Used to make cement, glass, and paper; also used as a filler in plastics and paints.
Neutralises acidic soil and acidic gases in flue-gas desulphurisation: CaCO₃ + SO₂ → CaSO₃ + CO₂.
直接作为建筑材料（大理石、石灰石骨料）。
在石灰窑中转化为氧化钙（生石灰）。
用于制造水泥、玻璃和纸张；也用作塑料和油漆的填料。
中和酸性土壤及烟气脱硫中的酸性气体：CaCO₃ + SO₂ → CaSO₃ + CO₂。
4. Calcium Oxide (CaO) | 氧化钙
Calcium oxide, or quicklime, is produced by the thermal decomposition of limestone. It is a strongly basic oxide with a high melting point (about 2613 °C). Upon reaction with water, it releases significant heat in an exothermic process to form slaked lime.
CaO(s) + H₂O(l) → Ca(OH)₂(s) ΔH is strongly negative
Used in steelmaking to remove acidic impurities such as SiO₂ and P₄O₁₀ from molten iron.
Used in the production of cement, glass, and calcium carbide (CaC₂).
Used in agriculture to condition acidic soils rapidly.
Used in the removal of moisture from gases and as a cheap drying agent.
用于炼钢中去除铁水中的酸性杂质，如SiO₂和P₄O₁₀。
用于生产水泥、玻璃和碳化钙（CaC₂）。
在农业中快速调节酸性土壤。
用于气体的除湿，作为廉价干燥剂。
5. Magnesium Hydroxide (Mg(OH)₂) | 氢氧化镁
Magnesium hydroxide is a sparingly soluble base used in medicine and industry. Its low solubility and gentle alkalinity make it safe for human consumption, unlike stronger alkalis. In antacid formulations, it neutralises excess stomach acid without producing excessive gas.
Active ingredient in milk of magnesia, used to treat indigestion and constipation.
Used in wastewater treatment to precipitate heavy metals as hydroxides.
Acts as a flame-retardant filler in plastics; on heating it releases water vapour and cools the material.
镁乳（氢氧化镁混悬液）的有效成分，用于治疗消化不良和便秘。
用于废水处理，使重金属以氢氧化物形式沉淀。
作为塑料中的阻燃填料；加热时释放水蒸气并冷却材料。
6. Calcium Sulfate (CaSO₄) | 硫酸钙
Calcium sulfate exists as anhydrous CaSO₄, gypsum (CaSO₄·2H₂O), and plaster of Paris (CaSO₄·½H₂O). Plaster of Paris is produced by heating gypsum to about 150 °C. On mixing with water, it rehydrates and hardens, making it valuable in construction and medicine.
Plaster of Paris is used for plasterboard, decorative mouldings, and protective cast for broken bones.
Gypsum is added to cement to control the setting time.
Calcium sulfate is used as a coagulant in tofu production and as a dietary calcium supplement.
熟石膏用于石膏板、装饰线条以及骨折固定石膏。
石膏加入水泥中用于控制凝固时间。
硫酸钙用作豆腐生产中的凝固剂，并可作为膳食钙补充剂。
7. Barium Sulfate (BaSO₄) | 硫酸钡
Barium sulfate is famous for being extremely insoluble in water and dilute acids. Despite Ba²⁺ being toxic, the very low solubility of BaSO₄ means that negligible amounts of barium ions enter the body. This property is exploited in medical imaging of the gastrointestinal tract.
Barium meal — a suspension of BaSO₄ is swallowed; it absorbs X-rays and outlines soft tissues in the digestive tract.
Used as a white pigment in paints and in the manufacture of lithopone (a white pigment with ZnS).
Used as a weighting agent in drilling fluids for oil and gas wells.
钡餐——吞服BaSO₄混悬液；它能吸收X射线，勾勒出消化道软组织轮廓。
用作油漆中的白色颜料，以及立德粉（与ZnS组成的白色颜料）的生产原料。
用作油气井钻井液中的加重剂。
8. Calcium Chloride (CaCl₂) | 氯化钙
Calcium chloride is highly soluble in water and strongly hygroscopic, meaning it readily absorbs moisture from the air. Its dissolution in water is exothermic, and its ability to lower the freezing point of water makes it an effective de-icing agent.
Used as a drying agent for gases, liquids, and solids in laboratories and industry.
Spread on roads in winter to melt ice and snow (freezing-point depression).
Used in concrete to accelerate the early strength gain in cold weather.
Added to swimming pool water to increase calcium hardness and prevent corrosion.
用作实验室和工业中气体、液体及固体的干燥剂。
冬季撒在道路上用于融化冰雪（凝固点降低）。
用于混凝土中，在寒冷天气下促进早期强度增长。
加入游泳池水中，增加钙硬度并防止腐蚀。
9. Magnesium Carbonate (MgCO₃) | 碳酸镁
Magnesium carbonate is a white solid with only slight solubility in water. It is less thermally stable than calcium carbonate because Mg²⁺ is smaller and has a higher charge density, thereby exerting a stronger polarising effect on the CO₃²⁻ ion.
Used as a drying agent for hands in gymnastics, climbing, and weightlifting to improve grip.
Used as a filler in rubber, plastics, and paper products.
Used in the production of magnesium oxide by thermal decomposition.
Used as an antacid and laxative in medicine.
用于体操、攀岩和举重中的防滑粉，保持手部干燥并增强抓握力。
用作橡胶、塑料和纸张产品中的填料。
通过热分解生产氧化镁。
在医药中用作抗酸剂和轻泻剂。
10. Strontium and Barium Compounds | 锶和钡的化合物
Strontium and barium compounds are less common in industry but still appear in A-Level questions. Their carbonates and sulfates show the expected solubility trend: sulfates become less soluble down the group, while hydroxides become more soluble.
The table below summarises each compound, its key property, and its important use — a compact guide for exam revision.
下表总结每种化合物、关键性质及其重要用途，是考试复习的简明指南。
化合物 Compound
关键性质 Key Property
主要用途 Main Use
MgO
极高熔点
耐火材料
Ca(OH)₂
弱碱、微溶
中和酸性土壤、检验CO₂
CaCO₃
热分解制CaO
建筑材料、制水泥
CaO
强碱性氧化物
炼钢除杂、制熟石灰
Mg(OH)₂
温和抗酸
抗酸药、阻燃剂
CaSO₄
与水硬化
石膏板、骨折固定
BaSO₄
极难溶、吸收X射线
钡餐造影
CaCl₂
强吸湿性
干燥剂、融雪剂
MgCO₃
不稳定、吸汗
防滑粉、填料
12. Exam Tips: Linking Properties to Uses | 考试技巧：从性质推断用途
In final exams, you will not be expected to memorise every use, but you must be able to predict uses from chemical properties. The key connections are listed below.
在期末考试中，你不会被要求记住每种用途，但必须能够从化学性质推断用途。关键联系如下。
High melting point → refractory material: this arises from strong ionic lattice forces and high lattice enthalpy.
Insolubility → safe for medical use: BaSO₄ is non-toxic to the body only because it is insoluble.
Alkalinity → acid neutralisation: oxides and hydroxides of Group 2 are basic and neutralise acids in industry and medicine.
Ability to absorb water → drying agent: CaCl₂ and Mg(ClO₄)₂ are used because of strong hydration of the cation.
Thermal decomposition trend → lime production: MgCO₃ decomposes at a lower temperature than CaCO₃; both produce metal oxide and CO₂.
高熔点 → 耐火材料：源于强离子晶格作用和高的晶格焓。
不溶性 → 医用安全：BaSO₄对身体无毒仅因其不溶解。
碱性 → 中和酸：第二主族氧化物和氢氧化物呈碱性，在工业和医学中用于中和酸。
吸水能力 → 干燥剂：CaCl₂和Mg(ClO₄)₂因阳离子的强水合作用而被用作干燥剂。
热分解规律 → 石灰生产：MgCO₃分解温度低于CaCO₃；两者均产生金属氧化物和CO₂。
Remember also the general solubility rule for sulfates: MgSO₄ is soluble, CaSO₄ is slightly soluble, and BaSO₄ is insoluble. This trend is often tested alongside qualitative analysis questions.
📚 A-Level Chemistry: Catalysis and Catalytic Mechanisms | A-Level 化学：催化作用与催化机制
Catalysis is one of the most fundamental concepts in chemistry, underpinning both industrial processes and biological systems. In this article, we explore the definitions, mechanisms and applications of catalysts for the CIE A-Level Chemistry syllabus, with particular attention to exam-relevant detail.
A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy. Crucially, the catalyst is chemically unchanged at the end of the reaction, meaning it can be recovered and reused.
It is specific in its action — different reactions require different catalysts.
它具有良好的选择性——不同反应需要不同的催化剂。
It does not alter the position of equilibrium in a reversible reaction; it only helps the system reach equilibrium faster.
它不改变可逆反应的平衡位置，仅仅使系统更快地达到平衡。
It is not consumed in the overall reaction, although it may participate in intermediate steps.
它在总反应中不被消耗，尽管它可能参与中间步骤。
2. Activation Energy and Reaction Profiles | 活化能与反应能量图
In an uncatalysed reaction, reactant molecules must overcome a high energy barrier known as the activation energy (Eₐ). A catalyst lowers this barrier by offering a different pathway, often involving the formation of an intermediate species that would not form in the uncatalysed reaction.
On a reaction profile diagram, the catalysed pathway shows a lower maximum peak. Consequently, a greater fraction of reactant molecules possess sufficient kinetic energy to cross the lower barrier, so the frequency of successful collisions increases dramatically.
It is essential to note that the enthalpy change (ΔH) remains identical for both catalysed and uncatalysed reactions — a catalyst does not affect the overall thermodynamics of the reaction. It only changes the kinetics.
In homogeneous catalysis, the catalyst is in the same physical state — usually aqueous or gaseous — as the reactants. This allows intimate mixing and rapid interaction between the catalyst and reactant molecules.
Iodide ions (I⁻) catalyse this decomposition. The mechanism involves a two-step process with a hypoiodite ion intermediate:
碘离子（I⁻）催化该分解反应。其机理为涉及次碘酸根离子中间体的两步过程：
H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast)
Notice that I⁻ is consumed in the first step and regenerated in the second step — this is the hallmark of homogeneous catalysis. The intermediate IO⁻ appears in the mechanism but not in the overall equation.
Other examples include acid-catalysed esterification of carboxylic acids with alcohols, and the use of nitrogen monoxide (NO) as a homogeneous catalyst in the lead chamber process for sulfuric acid manufacture.
其他示例包括羧酸与醇的酸催化酯化反应，以及铅室法制硫酸中使用一氧化氮（NO）作为均相催化剂。
4. Heterogeneous Catalysis | 多相催化
In heterogeneous catalysis, the catalyst is in a different phase from the reactants — typically a solid catalyst with gaseous or liquid reactants. The reaction occurs on the catalyst surface at active sites.
Finely divided iron is used as the catalyst at approximately 450 °C and 200 atm. The iron surface adsorbs nitrogen and hydrogen molecules, weakening their intramolecular bonds. Promoters such as Al₂O₃ and K₂O are added to increase the surface area and enhance the efficiency of the iron catalyst.
Vanadium(V) oxide (V₂O₅) is the heterogeneous catalyst. In the actual mechanism, V₂O₅ is first reduced to V₂O₄ by SO₂, and then V₂O₄ is re-oxidised back to V₂O₅ by O₂.
5. Adsorption in Heterogeneous Catalysis | 多相催化中的吸附
Adsorption is the key step in heterogeneous catalysis. In chemisorption, reactant molecules form chemical bonds with surface atoms of the catalyst. This weakens the bonds within the reactant molecules, making them more reactive.
The stages of a heterogeneous catalytic reaction can be summarised as:
多相催化反应的步骤可归纳为：
Diffusion of reactant molecules to the catalyst surface.
反应物分子向催化剂表面扩散。
Adsorption of reactants onto active sites via chemisorption.
反应物通过化学吸附在活性位点上吸附。
Reaction between adsorbed species — bonds are broken and new bonds formed.
吸附物种之间发生反应——旧键断裂，新键生成。
Desorption of product molecules from the surface.
产物分子从表面脱附。
Diffusion of products away from the catalyst surface.
产物从催化剂表面扩散离开。
Adsorption is an exothermic process. Although energy is required to break the adsorption bonds during desorption, the overall lowering of activation energy leads to a substantial net increase in reaction rate.
吸附是一个放热过程。尽管脱附时需要能量来断裂吸附键，但活化能的总体降低带来了反应速率的大幅净增长。
6. Catalytic Converters | 催化转化器
Catalytic converters in vehicle exhaust systems use heterogeneous catalysts — typically platinum, palladium and rhodium supported on a ceramic honeycomb structure. These metals provide a large surface area for gaseous reactions.
Notice that carbon monoxide is oxidised to carbon dioxide while nitrogen monoxide is reduced to nitrogen — this is a redox coupling occurring on the same catalyst surface. This is an important exam point: the same catalyst can promote both oxidation and reduction simultaneously.
Enzymes are biological catalysts, predominantly proteins, that exhibit remarkable specificity and efficiency. Each enzyme has an active site — a three-dimensional pocket with a shape complementary to its specific substrate. This is described by the lock-and-key model, later refined into the induced-fit model, in which
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
📚 Periodicity in Chemical Properties of Elements | 元素化学性质的周期性规律
The periodic table is not merely an arrangement of elements — it is a powerful tool for predicting and explaining chemical behaviour. When elements are arranged in order of increasing atomic number, a recurring pattern, known as periodicity, emerges in their physical and chemical properties. This article explores the periodic trends in chemical properties that are essential for CIE A-Level Chemistry.
Periodicity refers to the repeating trends in properties of elements across a period and down a group. These trends arise from variations in atomic structure — specifically, the number of protons, electron shielding, and the distance between the nucleus and the outermost electrons.
Across a period (left to right): nuclear charge increases, shielding remains similar, leading to stronger attraction on outer electrons.
Down a group (top to bottom): new shells are added, increasing atomic radius and shielding, which outweighs the increase in nuclear charge.
沿周期（左到右）：核电荷增加，屏蔽效应大致不变，导致对外层电子的吸引力增强。
沿族（上到下）：新增电子壳层，原子半径和屏蔽效应增大，其影响超过核电荷增加的影响。
2. Atomic Radius Trends | 原子半径的变化趋势
Atomic radius is the distance between the nucleus and the outermost electron cloud. Across Period 3, from sodium (Na) to chlorine (Cl), the atomic radius decreases steadily.
Explanation: As you move across a period, electrons are added to the same outer shell while protons increase. The increasing nuclear charge pulls the electron cloud closer to the nucleus, reducing the atomic radius.
First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. The general trend across Period 3 is an increase from Na to Ar, but with notable exceptions.
Across Period 3 (Na → Ar), ionisation energy generally increases because:
Nuclear charge increases, attracting outer electrons more strongly.
Atomic radius decreases, so the outer electrons are closer to the nucleus.
Shielding by inner electrons remains roughly constant.
沿着第三周期（Na → Ar），电离能总体增加，因为：
核电荷增加，对外层电子的吸引力更强。
原子半径减小，外层电子距原子核更近。
内层电子的屏蔽效应大致不变。
Two key exceptions in Period 3:
第三周期的两个关键异常：
Mg → Al (drop): Mg has a 3s² configuration (fully filled 3s subshell), while Al has 3s²3p¹. The 3p electron in Al is at a slightly higher energy level and is better shielded than the paired 3s electrons, making it easier to remove.
P → S (drop): P has 3p³ (half-filled, stable), while S has 3p⁴. In S, the fourth 3p electron must pair up with an existing electron in the same orbital, experiencing electron–electron repulsion, which lowers the energy needed to remove it.
P → S（下降）： P的3p³为半充满稳定构型，而S为3p⁴。S中第四个3p电子必须与同一轨道中的已有电子配对，产生电子-电子排斥，降低了移除所需能量。
4. Electronegativity Trends | 电负性的变化趋势
Electronegativity is the power of an atom to attract the bonding pair of electrons in a covalent bond. In Period 3, electronegativity increases from Na to Cl.
电负性是指原子在共价键中吸引成键电子对的能力。在第三周期中，电负性从Na到Cl递增。
The increase is explained by the same factors as ionisation energy:
Increasing nuclear charge creates a stronger pull on bonding electrons.
Decreasing atomic radius allows the nucleus to exert a stronger attraction on electrons in the bond.
这种增加与电离能的原因相同：
核电荷增加使原子核对成键电子的吸引力更强。
原子半径减小使原子核对成键电子施加更强的吸引。
Note that argon (Ar) is usually excluded from electronegativity discussions because it has a complete octet and does not typically form covalent bonds.
注意：氩（Ar）通常被排除在电负性讨论之外，因为它具有完整的八电子结构，通常不形成共价键。
5. Melting and Boiling Points Across Period 3 | 第三周期熔沸点的变化规律
The melting and boiling points across Period 3 show a characteristic pattern that reflects the type of structure and bonding present in each element.
第三周期的熔沸点呈现出特征性规律，反映了各元素的结构和键合类型。
Sodium, Magnesium, Aluminium:
These are metals with giant metallic structures. Metallic bonding strength increases with the number of delocalised electrons per atom. Na (1 delocalised electron) → Mg (2) → Al (3). More positive ions and a greater charge density strengthen the metallic bond, requiring more energy to break.
Silicon has a giant covalent (macromolecular) structure with strong Si–Si covalent bonds that must be broken throughout the entire structure. This gives silicon a very high melting point.
These exist as simple molecular structures: P₄, S₈ (and S₂ at high temperatures), and Cl₂. Between molecules, only weak van der Waals forces exist, which require very little energy to overcome. Therefore, their melting points are much lower.
6. Acid–Base Character of Period 3 Oxides | 第三周期氧化物与氢氧化物的酸碱性质
The oxides of Period 3 elements show a gradual change from strongly basic through amphoteric to acidic character. This trend is one of the most important periodic patterns in A-Level chemistry.
Na₂O and MgO are ionic oxides. They react with water to form alkaline solutions, and with acids to form salts.
碱性氧化物（Na₂O、MgO）：
Na₂O和MgO是离子型氧化物。它们与水反应生成碱性溶液，与酸反应生成盐。
Na₂O(s) + H₂O(l) → 2NaOH(aq)
MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l)
Amphoteric oxide (Al₂O₃):
Al₂O₃ reacts with both strong acids and strong bases, showing both acidic and basic behaviour.
两性氧化物（Al₂O₃）：
Al₂O₃既能与强酸反应，也能与强碱反应，兼具酸性和碱性行为。
Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)
Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2NaAl(OH)₄(aq)
Acidic oxides (SiO₂, P₄O₁₀, SO₃, Cl₂O₇):
These covalent oxides are acidic. SiO₂ is insoluble in water but reacts with alkalis. P₄O₁₀, SO₃ and Cl₂O₇ react with water to form acids of increasing strength.
7. Acid–Base Character of Period 3 Hydroxides | 第三周期氢氧化物的酸碱性质
The hydroxides of Period 3 elements show a parallel trend. NaOH and Mg(OH)₂ are basic; Al(OH)₃ is amphoteric; while Si(OH)₄ (written as H₄SiO₄ or silicic acid) and the oxoacids of P, S and Cl are acidic.
NaOH: Strongly basic, fully ionises in water releasing OH⁻.
Mg(OH)₂: Weakly basic, sparingly soluble in water.
Al(OH)₃: Amphoteric — dissolves in both acid and alkali:
NaOH： 强碱性，在水中完全电离释放OH⁻。
Mg(OH)₂： 弱碱性，微溶于水。
Al(OH)₃： 两性——既能溶于酸也能溶于碱：
Al(OH)₃(s) + 3H⁺(aq) → Al³⁺(aq) + 3H₂O(l)
Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq)
Trend explanation: As the element becomes more electronegative, the O–H bond becomes more polarised in the direction of oxygen. This increases the tendency to release H⁺, enhancing acidity.
The chlorides of Period 3 elements also display a trend from ionic to covalent character.
第三周期元素的氯化物也呈现从离子型到共价型的递变趋势。
Chloride 氯化物
NaCl
MgCl₂
AlCl₃
SiCl₄
PCl₅
S₂Cl₂
Bonding 键型
Ionic 离子型
Ionic 离子型
Covalent 共价型
Covalent 共价型
Covalent 共价型
Covalent 共价型
The change from ionic to covalent bonding reflects the decreasing electropositive character of the elements as nuclear charge increases and atomic radius decreases.
从离子键到共价键的变化反映了元素正电性随核电荷增加和原子半径减小而逐渐减弱。
Reaction of chlorides with water:
氯化物与水的反应：
NaCl and MgCl₂ dissolve in water to form neutral or slightly acidic solutions. NaCl produces a neutral solution; MgCl₂ hydrolyses slightly to give a weakly acidic solution.
AlCl₃ (anhydrous) reacts vigorously with water, producing fumes of HCl and Al(OH)₃.
SiCl₄, PCl₅ and S₂Cl₂ undergo complete hydrolysis with water, producing steamy fumes of HCl and the corresponding oxide or oxoacid.
9. Periodic Trends in Groups 2 and 7 | 第二族和第七族的周期趋势
Periodicity applies vertically too. In Group 2 (alkaline earth metals) and Group 7 (halogens), clear trends exist.
周期性在纵向上同样适用。在第二族（碱土金属）和第七族（卤素）中，存在明显的趋势。
Group 2 trends (down the group):
第二族趋势（沿族向下）：
Atomic radius increases due to additional electron shells.
First ionisation energy decreases because increased shielding outweighs increased nuclear charge.
Reactivity with water increases (e.g., Be reacts slowly, Mg slowly with steam, Ca rapidly with cold water).
Solubility of sulfates decreases (BaSO₄ is effectively insoluble), whereas solubility of hydroxides increases.
原子半径因新增电子壳层而增大。
第一电离能降低，因为屏蔽效应增强超过了核电荷增加的影响。
与水反应活性增强（例如，Be缓慢反应，Mg与水蒸气缓慢反应，Ca与冷水迅速反应）。
硫酸盐溶解度递减（BaSO₄几乎不溶），而氢氧化物溶解度递增。
Group 7 trends (down the group):
第七族趋势（沿族向下）：
Atomic radius increases.
Electronegativity decreases.
Oxidising ability decreases: F₂ is the strongest oxidising agent; I₂ is the weakest.
Volatility decreases: boiling points increase from F₂ to I₂.
原子半径增大。
电负性减小。
氧化能力减弱：F₂是最强氧化剂；I₂最弱。
挥发性降低：从F₂到I₂沸点升高。
10. Exam-Focused Summary | 考试重点总结
The following points are the most frequently tested in CIE A-Level chemistry examinations:
以下要点是CIE A-Level化学考试中最常考查的内容：
Across Period 3: atomic radius decreases, first ionisation energy generally increases (with Mg→Al and P→S drops), and electronegativity increases.
Na₂O and MgO are basic; Al₂O₃ is amphoteric; SiO₂, P₄O₁₀, SO₃ and Cl₂O₇ are acidic.
Melting points: Na < Mg < Al < Si > P₄ > S₈ > Cl₂.
AlCl₃ is covalent and hydrolyses in water; SiCl₄ and PCl₅ undergo complete hydrolysis to produce HCl fumes.
Group 2: reactivity with water increases down the group.
Group 7: oxidising power decreases down the group.
第三周期从左到右：原子半径减小，第一电离能总体增加（但Mg→Al和P→S出现下降），电负性增加。
Na₂O和MgO为碱性；Al₂O₃为两性；SiO₂、P₄O₁₀、SO₃和Cl₂O₇为酸性。
熔点：Na < Mg < Al < Si > P₄ > S₈ > Cl₂。
AlCl₃为共价化合物，在水中水解；SiCl₄和PCl₅发生完全水解，产生HCl烟雾。
第二族：沿族向下与水反应活性增强。
第七族：沿族向下氧化能力减弱。
Exam tip: Always explain trends in terms of nuclear charge, shielding, and atomic radius. When comparing ionisation energies, remember to mention the stabilising effect of half-filled and fully filled subshells.
Reaction kinetics is the study of how fast chemical reactions occur and what factors control that speed. It provides a quantitative framework for understanding reaction mechanisms, rate laws, and the energy requirements of chemical change.
The rate of a reaction is the change in concentration of a reactant or product per unit time. The average rate over an interval is calculated as Δ[substance]/Δt, but chemists often need the instantaneous rate, which is the gradient of a concentration–time graph at a particular moment.
For a general reaction A → B, the rate can be expressed as:
对于一般反应 A → B，速率可以表示为：
rate = −d[A]/dt = d[B]/dt
Because reactant concentrations decrease, the negative sign makes the rate positive. Rates are always positive quantities and are measured in mol dm⁻³ s⁻¹.
由于反应物浓度减小，负号使速率为正值。速率始终是正量，单位为 mol dm⁻³ s⁻¹。
2. Measuring Reaction Rates | 测定反应速率
Experimental methods for following a reaction must be fast enough to capture the concentration change at the chosen time scale. Common techniques include measuring gas volume, monitoring colour change using a colorimeter, and removing samples for titration.
Gas volume: A gas syringe measures the volume of gaseous product evolved at regular intervals.
Colorimetry: The intensity of a coloured species is directly proportional to its concentration (Beer–Lambert law).
Titration: Quenched samples are titrated to determine the concentration of a reactant or product.
气体体积法：用气筒以固定时间间隔测量生成气体的体积。
比色法：有色物质的颜色强度与其浓度成正比（比尔-朗伯定律）。
滴定法：对淬灭样品进行滴定，以确定反应物或产物的浓度。
3. The Rate Equation | 速率方程
The rate equation relates the rate of a reaction to the concentration of species raised to powers called orders. For a reaction aA + bB → products, the rate equation is generally:
速率方程将反应速率与各物种浓度的幂（称为级数）联系起来。对于反应 aA + bB → 产物，速率方程通常为：
rate = k[A]m[B]n
Here, k is the rate constant, and m and n are the orders of reaction with respect to A and B. The values of m and n must be determined experimentally; they are not derived from the stoichiometric coefficients.
其中 k 是速率常数，m 和 n 分别是关于 A 和 B 的反应级数。m 和 n 的值必须通过实验确定，不能由化学计量系数推导。
The overall order is m + n. If the concentration of a reactant is doubled and the rate doubles, the order with respect to that reactant is 1. If the rate quadruples, the order is 2; if the rate does not change, the order is 0.
总级数为 m + n。若某反应物浓度加倍后速率也加倍，则该反应物的级数为1；若速率变为四倍，则级数为2；若速率不变，则级数为0。
4. Zero-Order Reactions | 零级反应
In a zero-order reaction, the rate is independent of the concentration of the reactant. The rate equation is rate = k, and a concentration–time graph gives a straight line with a negative slope.
The half-life of a zero-order reaction decreases as the initial concentration decreases because the amount consumed per unit time is fixed.
零级反应的半衰期随初始浓度降低而减小，因为单位时间内消耗的量为固定值。
5. First-Order Reactions | 一级反应
For a first-order reaction, rate = k[A]. The concentration–time curve is exponential, and the half-life is constant, independent of the initial concentration.
对于一级反应，rate = k[A]。浓度-时间曲线呈指数形式，半衰期恒定，与初始浓度无关。
The integrated rate law and half-life are:
积分速率定律和半衰期为：
ln[A]t = ln[A]0 − kt
t1/2 = ln 2 / k
A plot of ln[A] versus time is linear, with slope −k. This is a reliable way to test whether a reaction is first order.
ln[A] 对时间作图呈直线，斜率为 −k。这是检验一级反应的可靠方法。
6. Second-Order Reactions | 二级反应
When the overall order is 2, the reaction may be second order in one reactant (rate = k[A]²) or first order in each of two reactants (rate = k[A][B]). For a single reactant, the integrated rate law is:
For this case, a plot of 1/[A] versus time gives a straight line. The half-life for a second-order reaction increases with decreasing concentration.
在这种情况下，1/[A] 对时间作图呈直线。二级反应的半衰期随浓度降低而增大。
7. Collision Theory | 碰撞理论
For a reaction to occur, reactant particles must collide with sufficient energy and the correct orientation. Only a fraction of collisions are successful; these are called effective collisions.
The minimum energy required to break bonds is the activation energy (Eₐ). The rate of reaction depends on the fraction of collisions that have at least this energy.
断裂化学键所需的最低能量称为活化能（Eₐ）。反应速率取决于至少具有该能量碰撞的比例。
Table 1 summarises the factors that affect reaction rate:
表1总结了影响反应速率的因素：
Factor
Effect
Concentration / pressure
Increases collision frequency
Temperature
Increases collision energy and frequency
Catalyst
Provides an alternative route with lower Eₐ
Surface area
Increases number of exposed particles
因素
影响
浓度/压强
增加碰撞频率
温度
增加碰撞能量和频率
催化剂
提供活化能更低的新路径
表面积
增加暴露粒子数
8. Boltzmann Distribution | 玻尔兹曼分布
The Maxwell–Boltzmann distribution shows the spread of energies among particles at a given temperature. The area under the curve represents the total number of particles, and the shaded region to the right of Eₐ indicates the number with energy ≥ Eₐ.
When temperature increases, the curve flattens and shifts to the right, so the fraction of particles exceeding Eₐ increases significantly. This explains why a small rise in temperature causes a large increase in rate.
The Arrhenius equation links the rate constant, temperature, and activation energy:
阿伦尼乌斯方程将速率常数、温度和活化能联系起来：
k = A e−Eₐ/(RT)
A is the pre-exponential (frequency) factor, and R is the gas constant. Taking natural logarithms gives:
A 是指前因子（频率因子），R 是气体常数。取自然对数得到：
ln k = ln A − Eₐ/(RT)
Thus, plotting ln k against 1/T gives a straight line with slope −Eₐ/R. This allows experimental determination of Eₐ.
因此，以 ln k 对 1/T 作图得直线，斜率为 −Eₐ/R。这使得实验测定 Eₐ 成为可能。
10. Catalysis | 催化作用
A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy. It takes part in the reaction but is regenerated unchanged at the end.
催化剂通过提供活化能较低的新路径来加快反应。它参与反应，但在反应结束时保持不变地再生。
Homogeneous catalysis occurs when the catalyst and reactants are in the same phase, such as acid catalysis in ester hydrolysis. Heterogeneous catalysis involves a different phase, typical of solid catalysts in gas-phase reactions.
Catalysts are important in industrial processes because they reduce energy costs and increase productivity without being consumed.
催化剂在工业过程中非常重要，因为它们在自身不被消耗的情况下降低能源成本并提高生产率。
11. Rate-Determining Step | 速率控制步骤
Reactions often occur in a series of elementary steps known as a reaction mechanism. The slowest step in this sequence is the rate-determining step (RDS) because it limits the overall rate.
The rate equation can be predicted from the RDS. For example, if the RDS is A + B → X, the predicted rate law is rate = k[A][B]. Comparing the predicted rate law with the experimental one helps confirm or reject a proposed mechanism.
速率方程可以从RDS预测。例如，如果RDS是A + B → X，则预测的速率定律为 rate = k[A][B]。将预测的速率定律与实验测得的结果比较，有助于确认或排除所提出的机理。
12. Molecularity | 反应分子数
Molecularity refers to the number of reactant particles involved in an elementary step. A unimolecular step involves one molecule, a bimolecular step involves two, and a termolecular step involves three (rare due to low probability).
Reaction order is defined experimentally for the overall reaction, while molecularity applies to a single elementary step. They are equal only when the elementary step is the entire reaction.
📚 A-Level Chemistry: The Effect of Concentration on Reaction Rate | A-Level 化学：浓度对反应速率的影响
In any chemical reaction, the rate measures how quickly reactants are converted into products. For most reactions, changing the concentration of a dissolved or gaseous reactant produces a clear and measurable change in rate. In A-Level Chemistry, concentration is one of the four key factors — together with temperature, pressure and catalysts — that control reaction kinetics, and it is closely linked to collision theory.
Reaction rate is usually defined as the change in concentration of a reactant or product per unit time. The units for reaction rate are typically mol dm⁻³ s⁻¹.
The minus sign for a reactant is used because the concentration of A decreases with time. At A-Level, you are also expected to think about the initial rate: the instantaneous rate at the moment the reaction begins.
对反应物 A 而言，速率可写成：Rate = −Δ[A]/Δt；对产物 P 而言，速率可写成：Rate = Δ[P]/Δt。反应物浓度随时间下降，因此需要负号。在 A-Level 中，你还需要理解“初始速率”的概念：即反应刚开始那一瞬间的瞬时速率。
2. Collision Theory | 碰撞理论
Collision theory states that reacting particles must collide before they can react. However, not every collision leads to a reaction. A collision is successful only when the colliding particles have total kinetic energy at least equal to the activation energy Eₐ, and when they collide with the correct orientation.
The rate of reaction depends on two things: how often collisions occur, called the collision frequency, and what fraction of those collisions are successful. Concentration mainly changes the collision frequency.
📚 A-Level Chemistry | Periodic Trends in Physical Properties of Elements | A-Level 化学：元素物理性质的周期性规律
The periodic table is not merely a catalogue of elements; it is a powerful framework that organises elements according to their atomic structure. When elements are arranged by increasing atomic number, their physical properties—such as atomic radius, ionisation energy, melting point, and electrical conductivity—display recurring, predictable patterns. These repeating trends are known as periodicity, and they form a cornerstone of the CIE A-Level Chemistry syllabus.
This article will systematically explore each key physical property across Period 2 and Period 3, explaining the underlying reasons in terms of nuclear charge, shielding, and electron-electron repulsion. By mastering these trends, you will be able to predict and compare the properties of unfamiliar elements with confidence.
Across a period, the atomic radius decreases steadily. From sodium to argon in Period 3, for example, the covalent radius falls from about 186 pm for sodium to approximately 98 pm for chlorine. This decrease occurs because electrons are added to the same principal energy level, while the nuclear charge (proton number) increases. The increasing positive charge pulls the electron cloud closer to the nucleus, and the shielding effect from inner shells remains roughly constant because the additional electrons enter the same outer shell.
Consequently, the effective nuclear charge experienced by the outermost electrons increases, leading to stronger attraction and a smaller atomic radius. This trend is consistent across both Period 2 and Period 3, and it directly influences many other periodic properties.
Trend: Atomic radius decreases across a period as effective nuclear charge increases.
趋势：随着有效核电荷增大，同一周期内原子半径减小。
2. Ionic Radius | 离子半径
Ionic radius follows distinct patterns depending on whether the ion is a cation or an anion. Positive ions (cations) are always smaller than their parent atoms. When a metal atom loses its outer electrons, the remaining electron shells are fewer, so the ionic radius drops sharply. For instance, the sodium atom has a radius of 186 pm, but the Na⁺ ion has a radius of only about 98 pm. Moreover, in a cation, the number of protons now exceeds the number of electrons, so each remaining electron is pulled more strongly towards the nucleus.
Negative ions (anions) are always larger than their parent atoms. When a non-metal atom gains electrons, the electron-electron repulsion in the outer shell increases, causing the electron cloud to expand. For example, the chlorine atom has a radius of approximately 98 pm, while the Cl⁻ ion expands to about 181 pm.
For isoelectronic ions—species with the same number of electrons, such as O²⁻, F⁻, Na⁺, Mg²⁺, and Al³⁺—the ionic radius decreases as the number of protons increases. A greater nuclear charge with the same electron count pulls the electrons inwards more effectively.
First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. The general trend across a period is an increase in first ionisation energy. This is because the nuclear charge rises, atomic radius falls, and the shielding effect stays almost constant; thus the outermost electrons are held more tightly, requiring more energy to remove.
However, there are two notable exceptions within each period. In Period 2, boron has a lower first ionisation energy than beryllium. Beryllium has a filled 2s subshell, while boron’s outer electron enters the 2p subshell. The 2p electron is slightly higher in energy and better shielded by the filled 2s electrons, so it is easier to remove. Similarly, oxygen has a lower first ionisation energy than nitrogen. Nitrogen has a half-filled 2p subshell (three unpaired electrons), which is unusually stable, whereas oxygen has one electron pair in the 2p subshell. The electron-electron repulsion in oxygen’s paired 2p orbital makes removing an electron slightly easier.
The same pattern appears in Period 3: aluminium is lower than magnesium, and sulfur is lower than phosphorus. These exceptions are classic examination questions, so you should be able to explain them clearly with reference to subshell structure and electron pairing.
Note the dips at Al and S. These are the two exceptions you must be prepared to describe in an exam.
注意铝和硫处的下降。这是你必须在考试中能够描述的两个例外。
4. Second and Third Ionisation Energies | 第二和第三电离能
Ionisation energy does not stop at the first electron. The second ionisation energy is the energy needed to remove a second electron from each gaseous 1+ ion, and the third ionisation energy applies to gaseous 2+ ions. For a given element, successive ionisation energies always increase because the remaining electrons experience a greater effective nuclear charge after each removal.
More importantly, a very large jump in ionisation energy signals that an electron is being removed from a much lower, closer energy level, often a full inner shell. For example, the difference between the tenth and eleventh ionisation energies of sodium is enormous, confirming that sodium’s first ten electrons are from inner shells while the single outer electron is entirely responsible for its chemical behaviour.
In the CIE exam, you may be asked to deduce the number of outer electrons from a table of successive ionisation energies. Watch for the large jump: the number of electrons removed before that jump equals the number of valence electrons.
Electron affinity is the energy change when an electron is added to a gaseous atom. The first electron affinity of most atoms is negative, meaning energy is released. This is because the added electron is attracted by the nucleus, and for many non-metals the process is exothermic. For example, the first electron affinity of chlorine is about -349 kJ mol⁻¹.
Across a period, the first electron affinity becomes more negative from left to right, reflecting the increasing nuclear charge and smaller atomic radius. However, the second electron affinity is always positive because adding an electron to a negatively charged ion requires energy to overcome the strong electrostatic repulsion. For instance, adding a second electron to O⁻ to form O²⁻ absorbs significant energy.
This distinction between endothermic and exothermic steps is essential when constructing Born-Haber cycles and explaining the stability of ionic compounds.
区分吸热与放热步骤，对于构建 Born-Haber 循环以及解释离子化合物的稳定性至关重要。
6. Electronegativity | 电负性
Electronegativity is the ability of an atom in a covalent bond to attract the bonding electrons towards itself. Across a period, electronegativity increases from left to right. This trend mirrors the increase in effective nuclear charge and the decrease in atomic radius. In Period 3, sodium has a very low electronegativity (0.93), while chlorine has a high value (3.16), and argon is not assigned a value because it forms virtually no covalent bonds.
Electronegativity differences between two atoms determine the polarity of a covalent bond. A large difference leads to an ionic bond, whereas a small difference gives a polar covalent bond, and an identical or very close value results in a non-polar covalent bond. Understanding this spectrum is vital for predicting bond type and molecular behaviour.
The melting and boiling points of elements across a period do not follow a simple monotonic trend; instead, they reflect the type of structure and bonding in each element. In Period 3, sodium, magnesium, and aluminium are metals with metallic bonding. Their melting points increase from sodium to aluminium because the number of delocalised electrons per atom increases and the ionic charge of the metal cations rises (Na⁺ to Al³⁺). These factors strengthen the metallic bond, requiring more energy to break.
Silicon is a giant covalent (macromolecular) structure. Each silicon atom forms four strong covalent bonds, giving silicon a very high melting point of about 1414 °C. Phosphorus (white phosphorus), sulfur, chlorine, and argon exist as simple molecular solids or gases. Their melting points are low because only weak van der Waals’ forces hold the molecules together, and the energy required to overcome these intermolecular forces is small.
Within the simple molecules, sulfur has a higher melting point than phosphorus or chlorine because the S₈ ring is larger and more polarizable, giving stronger van der Waals’ forces. Argon, being a monatomic gas, has the lowest melting point of the period.
Electrical conductivity across a period is closely linked to electron availability. Metals such as sodium, magnesium, and aluminium conduct electricity in both solid and molten states because they contain delocalised electrons that are free to move. In Period 3, sodium is a fair conductor, magnesium is better, and aluminium is the best metallic conductor among the period’s elements, correlating with its three delocalised electrons per atom.
Silicon is a metalloid: it is a semiconductor, with conductivity between that of a metal and an insulator. Its conductivity increases sharply with temperature because heating excites electrons from the valence band into the conduction band. Non-metals such as phosphorus, sulfur, chlorine, and argon do not conduct electricity because they lack delocalised electrons. Sulfur and phosphorus molecular solids are insulators in all states.
A common exam question is to explain why aluminium conducts better than sodium, or why silicon is a semiconductor rather than a metal. Remember to connect conductivity to delocalised electrons, lattice structure, and energy bands.
Although Period 2 and Period 3 show parallel trends, there are subtle differences. Generally, atoms in Period 2 are smaller than those in Period 3 because they have fewer electron shells. For example, lithium is smaller than sodium, and fluorine is smaller than chlorine. Consequently, Period 2 elements often have higher first ionisation energies than their Period 3 counterparts, although this is not universally true across all elements due to orbital energy differences between 2p and 3p subshells.
Another difference is the greater tendency of Period 2 elements such as boron and carbon to form covalent rather than ionic compounds. The small size and high charge density of these atoms make them strongly polarising, so they tend to share electrons rather than transfer them completely. This explains why B₂O₃ is acidic and CO₂ is covalent, whereas SiO₂ is macromolecular and Al₂O₃ is amphoteric.
Students frequently make the same mistakes when answering periodicity questions. One common error is to state that atomic radius increases across a period because the number of protons increases, without mentioning the constant shielding. Another is to claim that ionisation energy increases uniformly, forgetting the drops at boron and oxygen in Period 2, or aluminium and sulfur in Period 3.
A further pitfall is confusing ionisation energy with electron affinity, or mixing up first and second electron affinities. Remember that ionisation energy is always endothermic (positive for gaseous atoms), whereas first electron affinity is usually exothermic (negative). Also, do not describe melting point trends without identifying the type of structure: metallic, giant covalent, or simple molecular.
Finally, check the charge of ions in ionic radius questions. A cation is smaller than its atom; an anion is larger. For isoelectronic series, more protons mean a smaller radius. Precision in these details earns high marks.
Periodicity in physical properties stems from three fundamental atomic factors: increasing nuclear charge, roughly constant shielding across a period, and decreasing atomic radius. These factors govern atomic and ionic radii, ionisation energies, electron affinities, and electronegativity. Meanwhile, melting points and electrical conductivity depend on the type of bonding and structure—metallic, giant covalent, or simple molecular—adopted by each element.
For your CIE A-Level examination, be ready to sketch and explain graphs of ionisation energy across a period, identify the anomalous dips, compare the conductivity of Na, Mg, Al, and Si, and discuss why sulfur has a higher melting point than chlorine. Practice these explanations aloud until they become second nature.
为备战 CIE A-Level 考试，请准备好描绘并解释同一周期内电离能的变化图、识别异常下降点、比较 Na、Mg、Al 和 Si 的电导率，并讨论为什么硫的熔点高于氯。反复练习这些解释，直到它们成为你的本能反应。
Remember that every periodic trend is ultimately a story about the balance between nuclear attraction and electron-electron repulsion. Hold onto that principle, and periodicity becomes a predictable and rewarding topic.
📚 Systematic Nomenclature of Organic Compounds in A-Level Chemistry | A-Level 化学：化合物的系统命名法
Systematic nomenclature is the international language of chemistry. In CIE A-Level Chemistry, students are expected to name organic compounds systematically according to the rules set by the International Union of Pure and Applied Chemistry (IUPAC). Mastering this skill is essential for identifying structures, predicting properties, and scoring full marks in exam questions that require precise naming.
1. The Foundation: Alkanes and the Longest Chain Rule | 基础：烷烃与最长碳链规则
The simplest family of organic compounds is the alkanes. To name an alkane, first identify the longest continuous chain of carbon atoms; this determines the parent name. The suffix “-ane” indicates a saturated hydrocarbon. Number the carbon atoms in the chain from the end that gives the lowest possible numbers to any substituents.
For example, consider a five-carbon chain with a methyl group on the second carbon:
例如，一个五碳链上，第二个碳带有一个甲基取代基：
CH₃-CH(CH₃)-CH₂-CH₂-CH₃ → 2-methylpentane
Key prefixes for alkyl groups include: methyl (one carbon), ethyl (two), propyl (three), and butyl (four). Substituents are listed alphabetically, not by numerical order. When multiple identical substituents appear, use the multiplying prefixes di-, tri-, tetra- (these do not affect alphabetical order).
2. Alkenes and Alkynes: Unsaturated Hydrocarbons | 烯烃与炔烃：不饱和烃
Alkenes contain a carbon-carbon double bond and use the suffix “-ene”; alkynes contain a triple bond and use the suffix “-yne”. The parent chain must include the multiple bond, and the chain is numbered to give the double or triple bond the lowest possible locant. The position of the multiple bond is indicated by the number of the first carbon involved in that bond.
Note that when both a multiple bond and a substituent are present, the multiple bond takes priority for low numbering. For example, CH₃-CH(CH₃)-CH=CH₂ is named 3-methylbut-1-ene, not 2-methylbut-3-ene.
Alcohols contain the hydroxyl (-OH) functional group and use the suffix “-ol”. The parent chain must include the carbon bearing the -OH group, and this carbon receives the lowest possible number. When an alcohol is named, the “-e” of the corresponding alkane is replaced by “-ol”.
Haloalkanes are named by treating the halogen as a substituent. Use the prefixes fluoro-, chloro-, bromo-, and iodo-. The chain is numbered to give the lowest combined locants to all substituents, treating the list alphabetically.
4. Aldehydes and Ketones: Carbonyl Compounds | 醛与酮：羰基化合物
Aldehydes contain a terminal carbonyl group (-CHO) and use the suffix “-al”. Since the aldehyde carbon is always at the end of the chain, it is automatically assigned position 1 and the locant is not needed in the name. Ketones contain a carbonyl group within the carbon chain and use the suffix “-one”, requiring a locant to indicate the position of the carbonyl carbon.
In naming aldehydes, the parent chain must include the -CHO carbon. Note that the aldehyde group takes priority over alcohol and halogen groups when choosing the parent chain and numbering.
命名醛时，母体链必须包含 -CHO 碳。注意，在选择母体链和编号时，醛基的优先级高于醇和卤素。
5. Carboxylic Acids and Esters | 羧酸与酯
Carboxylic acids contain the -COOH group and use the suffix “-oic acid”. The carboxyl carbon is assigned position 1, and no locant is required. For example, CH₃-CH₂-COOH is propanoic acid. When naming substituted acids, numbering begins at the carboxyl carbon.
Esters are derived from carboxylic acids and alcohols. The name consists of two parts: the alkyl group from the alcohol (ending in “-yl”) followed by the carboxylate group from the acid (ending in “-oate”).
It is a common mistake to reverse the order. Remember: write the alkyl group (from the alcohol part) first, then the “-oate” (from the acid part).
常见错误是颠倒顺序。记住：先写来自醇部分的烷基，再写来自酸部分的 “-oate”。
6. Amines and Amides: Nitrogen-Containing Compounds | 胺与酰胺：含氮化合物
Primary amines use the suffix “-amine” or the prefix “amino-“. In CIE A-Level, simple amines are typically named by adding “-amine” to the name of the alkyl group. For example, CH₃-CH₂-NH₂ is ethylamine or ethanamine. When the amino group is a substituent on a chain with a higher-priority functional group, use the prefix “amino-“.
Amides contain the -CONH₂ group and use the suffix “-amide”. The carbonyl carbon is position 1. For example, CH₃-CONH₂ is ethanamide. Substituted amides list the alkyl groups attached to nitrogen with the prefix “N-“.
Cyclic hydrocarbons are named by prefixing the corresponding alkane with “cyclo-“. For example, a six-membered saturated ring is cyclohexane, and a five-membered ring with one double bond is cyclopentene. Numbering of a substituted cycloalkane begins at one substituent and proceeds in the direction that gives the lowest possible numbers to the other substituents.
When the ring contains polar functional groups such as -OH or -COOH, the ring is considered the parent chain only if it is the largest ring. Otherwise, it may be treated as a substituent with the prefix “cycloalkyl-“.
8. Aromatic Compounds: The Benzene Ring | 芳香族化合物：苯环
Benzene is the parent name for aromatic compounds. Simple monosubstituted benzene compounds are named as “phenyl” derivatives. For example, C₆H₅-CH₃ is methylbenzene, and C₆H₅-OH is phenol. When the benzene ring carries a functional group of higher priority, the ring becomes a substituent called “phenyl”.
For disubstituted benzene, use locants 1,2- (ortho), 1,3- (meta), and 1,4- (para). Numbering starts at the substituent with alphabetical priority. For example, 1,2-dimethylbenzene is also called ortho-xylene.
When multiple functional groups are present in a molecule, only one can act as the principal functional group (reflected in the suffix). The others are treated as substituents (reflected in prefixes). The CIE syllabus requires knowledge of the following decreasing priority order:
Examiners report that the most frequent naming errors include: failing to identify the longest chain containing the principal functional group, incorrect numbering (not giving the lowest locants), neglecting to include locants for multiple bonds or substituents, and using wrong punctuation (commas between numbers, hyphens between numbers and letters).
Follow this systematic approach for every naming question:
作答每个命名题时，请遵循以下系统方法：
Identify the principal functional group and choose the parent chain.
识别主官能团并选择母体链。
Number the chain to give the principal group the lowest locant, then consider substituents and multiple bonds.
对碳链编号，使主官能团获得最小编号，再考虑取代基和多重键。
Write the name alphabetically, using multipliers where needed, and include all locants.
按字母顺序写出名称，必要时使用倍数前缀，并包含所有位置编号。
Double-check punctuation: commas separate numbers, hyphens separate numbers from words.
检查标点：逗号分隔数字，连字符连接数字与字母。
Practice drawing structures from names as well as naming from structures. This bidirectional fluency is the key to success in CIE A-Level organic chemistry questions.
📚 Applications of Equilibrium in the Chemical Industry | 平衡在化学工业中的应用
Chemical equilibrium is not merely a theoretical concept studied in textbooks; it is the central principle that governs the design and operation of many large-scale industrial processes. Understanding how equilibria respond to changes in conditions allows chemical engineers to maximise product yield, reduce energy costs, and make processes economically viable.
1. Le Chatelier’s Principle and Industrial Relevance | 勒夏特列原理与工业关联
Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts in the direction that counteracts the change. This principle provides the theoretical basis for selecting temperature, pressure, and concentration conditions in industrial reactors.
In industry, the goal is not simply to achieve a high equilibrium yield; it is also essential that the reaction proceeds at a fast enough rate. Therefore, industrial chemists must reconcile two competing factors: thermodynamic yield (determined by equilibrium position) and kinetic rate (determined by reaction speed).
2. Key Equilibrium Concepts for Industry | 工业中的关键平衡概念
Before examining specific processes, it is important to recall the fundamental quantities used to describe equilibrium:
在考察具体工艺之前，有必要回顾用于描述平衡的基本量：
Dynamic equilibrium: the forward and reverse reactions occur at equal rates, and the concentrations of reactants and products remain constant.
动态平衡：正逆反应以相等速率进行，反应物与产物的浓度保持恒定。
Equilibrium constant Kc: for a reaction aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, where square brackets denote molar concentrations at equilibrium.
平衡常数Kc：对反应 aA + bB ⇌ cC + dD，Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ，其中方括号表示平衡时的摩尔浓度。
Kp: the equilibrium constant expressed in terms of partial pressures for gaseous reactions.
Kp：对于气相反应，用分压表示的平衡常数。
Le Chatelier’s principle: the qualitative tool used to predict the direction of shift when conditions are altered.
勒夏特列原理：用于定性地预测条件改变时平衡移动方向的工具。
It is crucial to note that only temperature changes alter the value of Kc or Kp; changes in pressure or concentration shift the position of equilibrium without changing the equilibrium constant itself.
3. The Haber Process for Ammonia Synthesis | 哈伯法制氨
The Haber process is the most celebrated industrial application of equilibrium principles. Ammonia is produced from nitrogen and hydrogen according to the following reaction:
哈伯法是平衡原理最著名的工业应用。氨由氮气和氢气通过以下反应生成：
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
This reaction is exothermic and involves a decrease in the number of gas molecules from four moles to two moles. According to Le Chatelier’s principle, low temperature and high pressure would maximise the yield of ammonia.
In practice, the industrial conditions are approximately 450°C, 200 atm pressure, and a finely divided iron catalyst. The reasons for these compromises are discussed in detail in later sections.
At equilibrium, the exit gas mixture contains about 15% ammonia by volume. The ammonia is liquefied by cooling, and the unreacted nitrogen and hydrogen are recycled back into the reactor. This recycling process continuously removes product and shifts the equilibrium to the right, as discussed in the concentration section below.
The Contact process is the major industrial route for the manufacture of sulfuric acid. The key equilibrium step is the oxidation of sulfur dioxide to sulfur trioxide:
接触法是工业制造硫酸的主要途径。关键的平衡步骤是二氧化硫氧化为三氧化硫：
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ mol⁻¹
This reaction is also exothermic and involves a reduction in the number of gas molecules, making it thermodynamically favoured by low temperature and high pressure. However, the industrial conditions are typically around 450°C, 1–2 atm pressure, and a vanadium(V) oxide (V₂O₅) catalyst.
A notable aspect of the Contact process is that atmospheric pressure is sufficient to achieve a conversion of about 99.5% of sulfur dioxide. This is because the equilibrium constant is extremely large at 450°C, meaning the equilibrium position lies far to the right even at low pressure. The excellent yield eliminates the need for expensive high-pressure compression equipment.
5. Temperature Effects on Equilibrium and Rate | 温度对平衡与速率的影响
For exothermic reactions such as the Haber and Contact processes, Le Chatelier’s principle predicts that decreasing the temperature shifts the equilibrium to the right, increasing the yield of products. However, low temperatures drastically reduce the rate at which equilibrium is reached, making the process uneconomically slow.
In both processes, a compromise temperature of approximately 450°C is chosen. At this temperature, the equilibrium yield is still reasonably high, while the reaction rate is fast enough for industrial production. If the temperature were too low, catalysts would also become inefficient, and the reaction rate would be impractically slow.
For endothermic reactions, the opposite logic applies: raising the temperature increases the equilibrium yield. For example, the thermal cracking of hydrocarbons and the steam reforming of methane are endothermic processes carried out at high temperatures to improve equilibrium conversion.
Changes in pressure only affect equilibria involving gases, and only if the total number of gas molecules changes between reactants and products. Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules.
In the Haber process, high pressure dramatically improves the equilibrium yield, so a pressure of around 200 atm is employed. However, compressing gases to such pressures requires significant energy and robust equipment that must withstand high pressure without leaking. Beyond a certain point, the increased cost of compression outweighs the benefit of additional yield, which is why pressures above 350 atm are rarely used.
In the Contact process, because the equilibrium constant is so large, an excellent yield is already achieved at atmospheric pressure. Employing high pressure would require expensive compressors and high-pressure pipes for very little additional product, so it is not economically justified.
7. Concentration Effects and Product Removal | 浓度效应与产物移除
Le Chatelier’s principle predicts that increasing the concentration of reactants or decreasing the concentration of products shifts the equilibrium towards the products. Industrial processes exploit this by continuously removing product from the reaction mixture.
In the Haber process, ammonia is removed by cooling the gas mixture to a temperature at which ammonia condenses into a liquid, while nitrogen and hydrogen remain gaseous. This removal means that the equilibrium is constantly being disturbed in the forward direction, forcing more nitrogen and hydrogen to react, effectively driving the conversion to completion despite the moderate single-pass yield.
In esterification reactions, water is often removed using a Dean–Stark apparatus, or an excess of alcohol is used. Removing water or adding excess reactant shifts the equilibrium to increase ester yield. Similarly, in the Contact process, sulfur trioxide is readily absorbed in concentrated sulfuric acid to form oleum, removing SO₃ from the equilibrium gas mixture and driving further conversion.
Catalysts are used extensively in industrial equilibrium processes, but it is essential to understand that a catalyst does not change the position of equilibrium or the value of Kc. It accelerates both the forward and reverse reactions equally, so equilibrium is reached more quickly but at the same composition.
This property makes catalysts indispensable on an industrial scale. Without them, the temperatures required to achieve an acceptable rate would be so high that equilibrium yields would be minuscule. With a catalyst, a lower temperature can be used, achieving a better compromise between yield and rate.
Provides active sites for N₂ adsorption and dissociation
Contact process
Vanadium(V) oxide (V₂O₅)
Accelerates O₂ adsorption and SO₃ formation via redox cycle
Methanol synthesis
Copper/zinc oxide on alumina
Facilitates CO hydrogenation at moderate temperatures
Catalysts also degrade over time due to poisoning by impurities. For example, sulfur compounds are poisons for both the iron catalyst in the Haber process and the V₂O₅ catalyst in the Contact process. Therefore, raw materials must be purified before entering the reactor, which is an additional operational cost.
9. Using Equilibrium Constants to Predict Yield | 用平衡常数预测产率
Once the equilibrium constant Kp is known at a given temperature, it can be used to calculate the equilibrium composition of a reaction mixture under various pressures. This is a powerful quantitative tool for industrial reactor design.
For the Haber process, the equilibrium constant expression in terms of partial pressures is:
对于哈伯法，用分压表示的平衡常数表达式为：
Kp = p(NH₃)² / [p(N₂) × p(H₂)³]
At 400°C, the value of Kp is approximately 4 × 10⁻⁴ atm⁻², which is rather small, indicating that at moderate pressures the position of equilibrium lies towards the reactants. This is why high pressures are essential in the Haber process: raising the total pressure increases the partial pressures of all gases, and because the denominator of the Kp expression has a higher power (H₂³) than the numerator (NH₃²), the equilibrium must shift towards the product side to keep Kp constant.
Kc and Kp values are also temperature-dependent. For exothermic reactions, Kc decreases as temperature increases, meaning less product is formed at higher temperatures. For endothermic reactions, Kc increases with temperature. Industrial chemists use tabulated values to select the optimal operating temperature where Kc remains acceptable while the reaction rate is sufficiently fast.
10. Economic Optimisation and the Choice of Conditions | 经济优化与条件选择
Industrial processes are ultimately governed by economic considerations. A plant operator must minimise production cost per tonne of product while maximising purity and throughput. This involves balancing capital costs, energy costs, and raw material costs in a continuous optimisation exercise.
Key economic factors in deciding operating conditions include:
决定操作条件的关键经济因素包括：
Energy costs: compressing gases to high pressure and maintaining high temperatures require large amounts of energy, which is a major ongoing expense.
能源成本：将气体压缩至高压以及维持高温需要大量能量，这是一笔主要持续性开支。
Equipment costs: high-pressure reactors, thick-walled pipes, and specialized alloys are expensive to build and maintain; higher pressure demands more robust materials.
设备成本：高压反应器、厚壁管道和专用合金的建造与维护成本高昂；压力越高要求的材料强度也越高。
Catalyst lifetime: higher temperatures accelerate catalyst sintering and deactivation, increasing the frequency and cost of catalyst replacement.
催化剂寿命：较高温度加速催化剂烧结与失活，增加更换频率和费用。
Conversion per pass: higher conversion in a single pass reduces the size and cost of separation and recycling equipment.
单程转化率：单次通过中更高的转化率减小了分离和循环设备的规模与成本。
Safety and environmental compliance: high pressures and high temperatures carry safety risks, and strict regulations may impose additional costs.
安全与环保合规：高压和高温存在安全隐患，且严格法规可能带来额外成本。
The chosen conditions for the Haber process — 450°C, 200 atm, iron catalyst — represent a balanced compromise that minimises total costs. The process achieves a 15% per-pass conversion; without recycling, this would be uneconomically low. With efficient recycling of unreacted gases, an overall conversion of over 95% is achieved.
The same equilibrium principles are applied across a wide range of industrial processes beyond the two classical examples.
除两个经典实例外，同样的平衡原理还被广泛应用于各种工业过程。
Methanol synthesis is a prominent example. Carbon monoxide and hydrogen react to form methanol:
甲醇合成就是一个重要实例。一氧化碳和氢气反应生成甲醇：
CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = −91 kJ mol⁻¹
This exothermic process involves a reduction from three moles of gas to one mole, so high pressure (typically 50–100 atm) and moderate temperatures (around 250°C) are used to obtain a favourable yield. A copper-based catalyst accelerates the reaction at the relatively low temperature, which keeps Kc as large as possible.
Esterification reactions, such as the production of ethyl ethanoate, are reversible and use equilibrium principles in organic synthesis:
酯化反应（如乙酸乙酯的生产）是体内可逆反应，在有机合成中运用平衡原理：
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
To drive this equilibrium to the right, excess ethanol is used, and water is continuously removed by azeotropic distillation or by using a dehydrating agent. This is a direct application of Le Chatelier’s principle: removing a product shifts the equilibrium towards the ester.
Since the reaction is endothermic and produces more gas molecules than it consumes, Le Chatelier’s principle indicates that high temperature and low pressure favour hydrogen production. Industrial reformers operate at about 800–900°C and moderate pressures (around 20–30 atm) because higher pressure is needed downstream for storage and further processing; the pressure is a compromise between shifting the equilibrium and providing product at a usable pressure.
The principles of chemical equilibrium, especially Le Chatelier’s principle and equilibrium constants, are not abstract concepts confined to the laboratory. They are the practical tools used to design, optimise, and operate the industrial processes that supply fuels, fertilisers, plastics, pharmaceuticals, and countless other chemical products essential to modern life.
The successful industrial application of equilibrium always involves a compromise among yield, rate, catalyst activity, and economics. By manipulating temperature, pressure, concentration, and the recycling of unreacted materials, chemists and engineers harness the fundamental laws of equilibrium to achieve efficient, sustainable, and profitable production on a massive scale.
📚 A-Level Chemistry: Equilibrium Expressions and the Equilibrium Constant Kc | A-Level 化学：平衡表达式与平衡常数Kc
Chemical equilibrium is one of the most important ideas in A-Level Chemistry. Many reactions do not simply go to completion; instead, they reach a state where the forward and reverse reactions occur at the same rate. This article explains how to write equilibrium expressions and how to use the equilibrium constant, Kc, in calculations.
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
In a closed system, a reversible reaction can proceed in both the forward and reverse directions. Initially, the forward reaction is faster because the concentrations of reactants are high. As reactants are used up and products accumulate, the forward rate decreases and the reverse rate increases. Eventually, the two rates become equal.
This state is called dynamic equilibrium. It is dynamic because the reaction has not stopped; it continues at the molecular level, but there is no observable change in the concentrations of reactants and products.
Dynamic equilibrium only occurs in a closed system, and the conditions such as temperature and pressure must remain constant.
动态平衡只发生在封闭体系中，且温度、压强等条件必须保持不变。
2. The Equilibrium Constant Kc: Definition | 平衡常数Kc的定义
For a reaction at equilibrium, there is a mathematical relationship between the concentrations of reactants and products. This relationship is known as the equilibrium expression, and its value at a fixed temperature is called the equilibrium constant, Kc. The subscript ‘c’ indicates that concentrations in mol dm⁻³ are used.
Here, the square brackets mean concentration in mol dm⁻³. The coefficients a, b, c and d from the balanced equation become powers in the expression.
其中方括号表示以mol dm⁻³为单位的浓度。平衡方程式中的系数a、b、c和d在表达式中变为指数。
This expression is called the ‘mass action’ or equilibrium law. It is valid only at a specific temperature.
该表达式称为“质量作用定律”或平衡定律。它仅在特定温度下成立。
3. Writing Equilibrium Expressions | 书写平衡表达式
To write an equilibrium expression, you must first have a correctly balanced chemical equation. The products always appear in the numerator, and the reactants appear in the denominator. Each concentration is raised to the power of its stoichiometric coefficient.
Notice that the coefficient 2 for NH₃ becomes the power 2, and the coefficient 3 for H₂ becomes the power 3.
注意NH₃的系数2变成了指数2，H₂的系数3变成了指数3。
Another common example is the formation of sulfur trioxide:
另一个常见例子是三氧化硫的生成：
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Its expression is:
其表达式为：
Kc = [SO₃]² / ([SO₂]²[O₂])
In this example, SO₂ and SO₃ both have coefficients of 2, so both are squared.
在这个例子中，SO₂和SO₃的系数都为2，因此两者都要平方。
4. Conventions: Omitting Solids and Pure Liquids | 约定：忽略固体和纯液体
In a heterogeneous equilibrium, some substances are in different physical states. Only species whose concentration can change independently are included in the Kc expression. Solids and pure liquids are omitted because their concentrations are effectively constant.
For example, consider the thermal decomposition of calcium carbonate:
例如，考虑碳酸钙的热分解：
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
The equilibrium expression is simply:
其平衡表达式简化为：
Kc = [CO₂]
The solids CaCO₃ and CaO do not appear in the expression. This is because their concentrations are fixed by the amount of solid present and do not affect the equilibrium position.
Here, all species are liquids and are miscible. Because water is formed in the reaction and is not the solvent in large excess, it might be included. However, if water is present in excess as a solvent, it would be omitted. Always pay attention to the physical states given in the question.
The units of Kc depend on the form of the equilibrium expression. They are not always mol dm⁻³; they must be derived from the expression each time. To find the units, substitute the units of concentration into the expression and simplify.
So Kc has units of dm⁶ mol⁻². Different reactions give different units, so you must always calculate them.
因此Kc的单位为dm⁶ mol⁻²。不同反应会得到不同单位，因此必须每次计算。
6. What the Magnitude of Kc Tells Us | Kc的数值告诉我们什么
The size of Kc indicates how far a reaction proceeds at equilibrium. A large Kc value means that, at equilibrium, the concentration of products is much greater than that of reactants. The equilibrium lies to the right, or towards the products.
A small Kc value means that the reactants are favoured. The equilibrium lies to the left, and very little product is formed.
Kc值很小则意味着反应物占优势。平衡位置靠左，生成的产物很少。
If Kc > 10³, the equilibrium lies far to the right.
如果Kc > 10³，平衡位置远在右侧。
If Kc < 10⁻³, the equilibrium lies far to the left.
如果Kc < 10⁻³，平衡位置远在左侧。
If Kc is between roughly 10⁻³ and 10³, both reactants and products are present in significant amounts.
如果Kc约在10⁻³到10³之间，则反应物和生成物都以显著量存在。
It is important not to say that Kc ‘shows how fast’ a reaction is. A reaction with a large Kc could be very slow. Kc tells us about the position of equilibrium, not the rate.
7. Calculating Equilibrium Concentrations Using Kc | 用Kc计算平衡浓度
To calculate Kc, you need the equilibrium concentrations of all species in the expression. These are often found by constructing an ICE table: Initial, Change, Equilibrium.
This example shows how important it is to convert moles to concentrations when using Kc. Never use moles directly in the Kc expression.
这个例子说明了使用Kc时把物质的量转化为浓度是多么重要。绝不能在Kc表达式中直接使用物质的量。
8. The Reaction Quotient Qc and Predicting Direction | 反应商Qc与方向预测
The reaction quotient Qc has exactly the same form as Kc, but it is calculated using concentrations that are not necessarily at equilibrium. Comparing Qc with Kc tells you which direction the reaction will shift to reach equilibrium.
If Qc < Kc, there are more reactants relative to products than at equilibrium. The forward reaction is favoured, and more products will form.
如果Qc < Kc，意味着相对于平衡状态，反应物偏多。正反应占优势，会生成更多产物。
If Qc > Kc, there are more products relative to reactants than at equilibrium. The reverse reaction is favoured, and products will be converted back into reactants.
如果Qc > Kc，意味着相对于平衡状态，产物偏多。逆反应占优势，产物会转化回反应物。
If Qc = Kc, the system is already at equilibrium; no net change will occur.
如果Qc = Kc，体系已经处于平衡；不会发生净变化。
The reaction quotient is especially useful when you mix reactants and products in arbitrary amounts and need to predict whether the reaction will go forwards or backwards before reaching equilibrium.
反应商在我们将反应物和产物以任意量混合后、需要判断反应在达到平衡前会正向还是反向进行时尤其有用。
9. Effect of Changing Conditions on Kc | 条件改变对Kc的影响
Kc only depends on the temperature. It does not change when you alter concentration, pressure or add a catalyst. These changes may shift the position of equilibrium, but they do not change the value of Kc.
For an exothermic forward reaction, increasing the temperature decreases Kc. This is because heat is treated as a product, and the equilibrium shifts to the left, favouring reactants. For an endothermic forward reaction, increasing the temperature increases Kc.
Adding a catalyst lowers the activation energy for both the forward and reverse reactions equally. It speeds up the attainment of equilibrium but does not change the position of equilibrium or the value of Kc.
Changing pressure or concentration affects the equilibrium position according to Le Chatelier’s principle, but at a constant temperature Kc remains the same. When conditions change, the system adjusts until Qc once again equals Kc.
Many students lose marks on equilibrium questions because of careless errors. Here are some important mistakes to avoid.
许多学生在平衡题上失分是因为粗心错误。以下是一些需要避免的重要错误。
Using moles instead of concentrations. Always divide the number of moles by the total volume to get mol dm⁻³.
使用物质的量而不是浓度。务必用物质的量除以总体积，得到mol dm⁻³。
Including solids or pure liquids in the equilibrium expression. Remember to omit them.
在平衡表达式中包含固体或纯液体。记住要省略它们。
Using the wrong coefficients as powers. The powers in the Kc expression come directly from the balanced equation.
将错误的系数作为指数。Kc表达式中的指数直接来自配平的化学方程式。
Forgetting to include units. Always work out the units from the expression.
忘记写单位。务必根据表达式推导出单位。
Saying that Kc changes when concentration changes. Kc only changes with temperature.
错误地认为浓度改变会影响Kc。Kc只随温度改变。
In exam questions, always check the physical state symbols. If a substance is shown as (s) or (l) and is not in a mixture, it should be excluded from the Kc expression. Also check whether the question asks for the value only, or for both the value and units.
When asked to predict the effect of a change on Kc, think about temperature only. Changes in concentration or pressure alter the position of equilibrium but not Kc.
当被问到某个变化对Kc的影响时，只需要考虑温度。浓度或压强的改变会改变平衡位置，但不会改变Kc。
11. Summary | 总结
The equilibrium constant Kc is a fundamental quantity in chemical equilibria. It is defined by the equilibrium law and its value is constant at a fixed temperature.
平衡常数Kc是化学平衡中的基本量。它由平衡定律定义，且在固定温度下为常数。
To write the equilibrium expression, place the product concentrations in the numerator and the reactant concentrations in the denominator, each raised to the power of its stoichiometric coefficient. Omit solids and pure liquids. Be careful with units, as they differ from reaction to reaction.
The magnitude of Kc tells you whether products or reactants are favoured. The reaction quotient Qc can be used to predict the direction in which a reaction will move. Kc changes only with temperature, so any other changes to the system will not affect Kc, only the equilibrium position.
📚 Chemical Equilibrium Position Shift and Influencing Factors | 化学平衡位置的移动及影响因素
Chemical equilibrium is a cornerstone concept in A-Level physical chemistry. When a reversible reaction is confined in a closed system, the forward and reverse reactions proceed simultaneously. Equilibrium is reached when the rates of these opposing reactions become equal, and the macroscopic properties of the system remain constant. This article systematically explains how concentration, pressure, temperature, and catalysts influence the position of equilibrium, anchored in Le Chatelier’s Principle, and how these principles are applied in the Haber and Contact processes.
A reversible reaction reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The word “dynamic” is crucial: both reactions continue to occur at the microscopic level, but there is no net change in the concentrations of reactants or products. This state can only be achieved in a closed system.
Equal forward and reverse rates — not equal concentrations of reactants and products.
Constant macroscopic properties: colour, pressure, concentration, and density remain unchanged.
Equilibrium can be approached from either direction — starting with only reactants or only products leads to the same equilibrium mixture.
正逆反应速率相等——并非反应物与生成物浓度相等。
宏观性质恒定：颜色、压强、浓度和密度保持不变。
平衡可以从任意方向接近——仅由反应物或仅由生成物出发，最终得到相同的平衡混合物。
2. Le Chatelier’s Principle | 勒夏特列原理
Le Chatelier’s Principle states that when a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium position shifts in the direction that tends to counteract (oppose) the imposed change. This principle provides a qualitative tool for predicting how a system responds to external disturbances.
For example, if a system experiences an increase in temperature, the equilibrium shifts in the endothermic direction to absorb the added heat energy. If pressure increases, the system shifts towards the side with fewer moles of gas to relieve the pressure. These predictions are essential for exam questions requiring qualitative reasoning about equilibrium shifts.
Changing the concentration of a reactant or product shifts the position of equilibrium, but it does not change the value of the equilibrium constant Kc (at constant temperature).
改变反应物或生成物的浓度会使平衡位置发生移动，但在恒温条件下不会改变平衡常数 Kc 的值。
Consider the well-known equilibrium involving iron(III) thiocyanate:
考虑以下常见的硫氰酸铁平衡体系：
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) (deep red | 深红色)
Adding Fe³⁺ or SCN⁻ ions increases the concentration of a reactant, so the equilibrium shifts to the right, producing more FeSCN²⁺; the deep red colour intensifies.
Adding NaOH removes Fe³⁺ as Fe(OH)₃ precipitate, decreasing the Fe³⁺ concentration, so the equilibrium shifts to the left; the red colour fades.
Diluting the solution decreases all concentrations. In this case, there is one ion on each side, so dilution produces no shift. However, in general, diluting shifts the equilibrium towards the side with more dissolved species.
For a general gaseous reaction aA + bB ⇌ cC + dD, increasing the concentration of A or B shifts the equilibrium right; increasing the concentration of C or D shifts it left. Removing a product also shifts the equilibrium right, which is a common industrial strategy to maximise yield.
对于一般气体反应 aA + bB ⇌ cC + dD，增加 A 或 B 的浓度使平衡右移；增加 C 或 D 的浓度使平衡左移。移走生成物同样使平衡右移，这是工业中提高产率的常用策略。
4. Effect of Pressure | 压强的影响
Pressure changes only affect equilibria that involve at least one gaseous species. An increase in pressure shifts the equilibrium towards the side with fewer moles of gas molecules, because this reduces the total pressure. Conversely, a decrease in pressure shifts the equilibrium towards the side with more moles of gas. If the total number of gas moles is identical on both sides, a pressure change has no effect on the equilibrium position.
There are 4 moles of gas on the left and 2 moles on the right. Increasing the pressure shifts the equilibrium to the right, favouring the production of ammonia. This is why the Haber process operates at extremely high pressures, typically around 200 atmospheres.
Note that pressure effects are analysed using partial pressures. Doubling the total pressure doubles the partial pressure of every gaseous component, which is how the shift is driven.
注意，压强效应应通过分压来分析。总压强加倍会使每种气体组分的分压都加倍，这正是推动平衡移动的原因。
5. Effect of Temperature | 温度的影响
Temperature has a unique and crucial effect on equilibrium: it changes the value of the equilibrium constant itself. An increase in temperature shifts the equilibrium in the endothermic direction, while a decrease in temperature shifts it in the exothermic direction.
Increasing temperature: the system absorbs heat by favouring the reverse (endothermic) reaction; NH₃ yield decreases.
Decreasing temperature: the system releases heat by favouring the forward (exothermic) reaction; NH₃ yield increases.
升高温度：系统通过促进逆向（吸热）反应来吸收热量；NH₃ 产率降低。
降低温度：系统通过促进正向（放热）反应来释放热量；NH₃ 产率升高。
However, a lower temperature also lowers the rate of reaction. In industry, a compromise temperature (~450°C) is chosen to achieve a reasonable rate without sacrificing too much yield. This compromise between thermodynamics and kinetics is a classic exam theme.
A catalyst lowers the activation energy for both the forward and reverse reactions equally, increasing both rates by the same factor. Therefore, a catalyst does NOT change the position of equilibrium, nor does it change the value of Kc or Kp. It simply enables the system to reach equilibrium faster.
The effect of adding an inert gas depends on the conditions:
加入惰性气体的影响取决于条件：
Constant volume: adding an inert gas increases total pressure but does not change partial pressures of the reacting gases. No shift in equilibrium position.
Constant pressure: the container volume expands, decreasing the partial pressures of all reacting gases. The equilibrium shifts towards the side with more moles of gas.
恒容条件：加入惰性气体使总压强增大，但不改变反应气体的分压。平衡位置不变。
恒压条件：容器体积增大，所有反应气体的分压降低。平衡向气体物质的量较多的一侧移动。
7. Reaction Quotient Q vs Kc | 反应商 Q 与 Kc 的比较
The reaction quotient Q is calculated using the same expression as Kc, but with concentrations that may not be at equilibrium. Comparing Q to Kc predicts the direction of net movement:
If Q < Kc: reaction proceeds to the right (forward) | 若 Q < Kc：反应向右（正向）进行
If Q > Kc: reaction proceeds to the left (reverse) | 若 Q > Kc：反应向左（逆向）进行
If Q = Kc: the system is at equilibrium | 若 Q = Kc：体系处于平衡
For example, for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if extra N₂ is injected so that Q becomes smaller than Kc, the system must shift right to restore equilibrium. The power of this method is that it provides a quantitative prediction that complements Le Chatelier’s qualitative reasoning.
The principles of equilibrium shift are directly applied in several large-scale industrial processes.
平衡移动原理在多个大规模工业过程中得到直接应用。
Haber Process | 哈伯法（合成氨）
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
High pressure (≈200 atm) shifts equilibrium right because there are fewer gas moles on the product side.
Low temperature favours the exothermic forward reaction, but a moderate temperature of ~450°C is used to maintain an adequate rate.
An iron catalyst accelerates the attainment of equilibrium without altering its position.
NH₃ is continuously liquefied and removed, shifting the equilibrium further to the right and increasing the overall yield.
高压（约 200 atm）使平衡右移，因为产物侧气体物质的量更少。
低温有利于放热的正反应，但为了保持足够的速率，实际采用约 450°C 的中等温度。
铁催化剂加速平衡的到达，但不改变平衡位置。
氨被持续液化并移走，使平衡进一步右移，提高总产率。
Contact Process | 接触法（制硫酸）
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹
V₂O₅ is used as a catalyst; it does not affect the equilibrium position.
A temperature of ~450°C balances yield and rate.
Excess oxygen (from air) is used to drive the equilibrium to the right by increasing the concentration of a reactant, thereby improving the conversion of SO₂ to SO₃.
使用 V₂O₅ 作催化剂，不影响平衡位置。
约 450°C 的温度兼顾产率与速率。
使用过量氧气（来自空气），通过增加反应物浓度推动平衡右移，提高 SO₂ 转化为 SO₃ 的转化率。
9. Summary Table of Factors Affecting Equilibrium | 影响平衡的各因素总结表
Factor | 因素
Change | 变化
Direction of Shift | 移动方向
Effect on Kc | 对 Kc 的影响
Concentration | 浓度
Increase reactant | 增加反应物
Towards products | 向生成物
No change | 不变
Pressure | 压强
Increase | 增大
Towards fewer gas moles | 向气体物质的量少的一侧
No change | 不变
Temperature | 温度
Increase | 升高
Towards endothermic direction | 向吸热方向
Changes | 改变
Catalyst | 催化剂
Addition | 加入
No shift | 不移动
No change | 不变
Inert gas (constant V) | 惰性气体（恒容）
Addition | 加入
No shift | 不移动
No change | 不变
10. Worked Example | 典型例题
Question | 题目： The reaction below is at equilibrium in a sealed flask.
Question | 题目： 以下反应在密封烧瓶中达到平衡。
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹
(a) Predict the effect of increasing the pressure on the equilibrium position and explain your reasoning.
(a) 预测增大压强对平衡位置的影响，并解释原因。
ANSWER | 答案：Increasing the pressure shifts the equilibrium to the right. There are 3 moles of gas on the left (2SO₂ + 1O₂) and only 2 moles on the right (2SO₃). According to Le Chatelier’s Principle, the system shifts towards the side with fewer gas moles to reduce the pressure, which is the product side.
(b) Predict the effect of increasing the temperature on the equilibrium position and on the value of Kc.
(b) 预测升高温度对平衡位置及 Kc 值的影响。
ANSWER | 答案：Since the forward reaction is exothermic (ΔH = −197 kJ mol⁻¹), the reverse reaction is endothermic. Increasing the temperature shifts the equilibrium to the left, towards the endothermic direction. Because the equilibrium constant depends on temperature, Kc will decrease as the position shifts towards reactants.
(c) A vanadium(V) oxide catalyst is added. State the effect on the equilibrium position and on the rate of attainment of equilibrium.
(c) 加入五氧化二钒催化剂。说明其对平衡位置和达到平衡速率的影响。
ANSWER | 答案：The catalyst has no effect on the equilibrium position; the same equilibrium mixture is reached. It lowers the activation energy for both forward and reverse reactions, so equilibrium is reached more quickly.
(d) The volume of the flask is halved. Predict how Q compares with Kc immediately after the change.
(d) 将烧瓶体积减半。预测变化后瞬间 Q 与 Kc 的关系。
ANSWER | 答案：Halving the volume doubles the concentrations of all gases. For this reaction, Q increases: Q = [SO₃]² / ([SO₂]²[O₂]). Since concentrations double, [SO₃]² quadruples while [SO₂]²[O₂] octuples (2² × 2 = 8), so Q becomes smaller than Kc. The system must shift forward (right) to restore equilibrium.