Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
The halogens exist as diatomic molecules, X2. Down the group the molecules have more electrons, so the instantaneous dipole-induced dipole forces between them get stronger and the boiling points rise: volatility decreases from chlorine (a gas) to bromine (a liquid) to iodine (a solid). The colours deepen too. Within each molecule, the X–X bond energy decreases from Cl–Cl to I–I as the atoms get larger and the bonding pair is further from the nuclei; F–F is anomalously weak because the small atoms' lone pairs repel each other.
Boiling a halogen breaks intermolecular forces, not the X–X covalent bond. The rise in boiling point and the fall in bond energy are separate trends with separate explanations: one is about forces between molecules, the other about the bond inside each molecule.
Halogens are oxidising agents: they gain electrons to form halide ions. Oxidising power decreases down the group (Cl2 > Br2 > I2) because larger atoms attract an incoming electron less strongly. So a halogen displaces any halogen below it from a solution of its halide. The halogens react with hydrogen less and less vigorously down the group, and the thermal stability of the hydrogen halides decreases from HCl to HI because the H–X bond gets longer and weaker.
Oxidising power decreases down the group, so iodine cannot oxidise chloride or bromide ions. When explaining displacement, identify the halogen as the oxidising agent and the halide ion as the reducing agent. For thermal stability, refer to the H–X bond energy, not to intermolecular forces.
Halide ions act as reducing agents, losing electrons to form halogens. Reducing power increases down the group (I− > Br− > Cl−) because the outer electrons of a larger ion are further from the nucleus and more easily lost. With concentrated sulfuric acid, chloride gives only HCl; bromide also reduces the acid to SO2 (S: +6 → +4); iodide reduces it as far as S and H2S (S: +6 → −2). To test for halides, acidify with dilute nitric acid and add AgNO3(aq): AgCl is white, AgBr cream, AgI yellow, distinguished further by their solubility in ammonia.
Chloride ions are too weak a reducing agent to reduce concentrated sulfuric acid, so no chlorine forms; the only product is HCl. The further the sulfur is reduced, the stronger the reducing agent: iodide takes sulfur all the way to −2 in H2S. In the silver nitrate test, always acidify with nitric acid, never hydrochloric.
Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. Chlorine (oxidation number 0) disproportionates in sodium hydroxide: in cold, dilute NaOH it forms NaCl (−1) and NaClO (+1), the mixture used as bleach; in hot, concentrated NaOH it forms NaCl (−1) and NaClO3 (+5). With water, Cl2 + H2O ⇌ HCl + HClO, and the chloric(I) acid kills bacteria, which is why chlorine is used to treat drinking water.
In disproportionation the same element is both oxidised and reduced, so chlorine appears in two products with different oxidation numbers. Do not mix up the two NaOH conditions: cold dilute gives chlorate(I), NaClO; hot concentrated gives chlorate(V), NaClO3.