Group 7 (17), The Halogens

## Connect halogens,halides and diatomic molecules - Group 17 is the modern name for the main-group column historically called Group 7. The neutral F–I atoms have seven outer electrons, ns²np⁵. - The elemental substances in this range contain X₂ molecules. A halide ion is X⁻, formed by one-electron gain; it is different from a neutral atom or an X₂ molecule. - For example, Cl₂+2e⁻→2Cl⁻ requires two electrons per molecule. Fluorine reacts with water and is not an ordinary school aqueous-displacement reagent. ## Describe colour together with phase and solvent - The reviewed sealed-sample illustration shows chlorine gas, bromine liquid with vapour, and iodine crystals. The colours belong to the stated forms and conditions. - Aqueous iodine is brown, whereas iodine in a suitable nonpolar solvent is violet. A simple iodide solution is colourless; iodide and iodine are different species. - A colour observation should name the phase or solvent. Do not infer that iodine must always be purple or that every coloured solution contains a halide ion. ## Read boiling point and volatility as different descriptions - The rounded RSC values rise from F₂ 85 K through Cl₂ 239 K and Br₂ 332 K to I₂ 458 K. These compare boiling at the same reference-pressure convention. - At comparable conditions, the lower-boiling substances are more volatile. From F₂ to I₂, volatility generally decreases. - Boiling separates molecules into the gas phase. It does not normally break every X–X covalent bond into individual atoms. ## Explain boiling with intermolecular dispersion - F₂, Cl₂, Br₂ and I₂ are nonpolar diatomic molecules. Correlated instantaneous and induced dipoles create London dispersion attractions between molecules. - Larger, more polarizable electron clouds down this set produce stronger dispersion attractions, so separating the molecules requires more energy. - These attractions are between molecules. Keep them separate from the covalent X–X bonds inside each molecule, and do not describe all nonpolar molecules as having no attractions. ## Recognise the fluorine exception in bond strength - The supplied approximate gas-phase bond energies illustrate Cl–Cl>Br–Br>I–I. Increasing bond length and less effective overlap help explain that part of the pattern. - F–F is weaker than Cl–Cl despite fluorine’s small atoms. Repulsion between nearby lone-pair electrons is important in this exception. - Do not use a single monotonic bond-strength rule from F₂ to I₂. Bond energy and boiling temperature involve different processes and can show different trends. ## Explain the electronegativity decrease - Electronegativity describes attraction for an electron pair in a covalent bond. The supplied Pauling values decrease from F to I. - Down the group, larger size and greater shielding reduce attraction for the bonding pair despite increasing nuclear charge. - Electronegativity is not the same quantity as electron affinity or aqueous oxidising power. Redox energetics also include bond breaking, electron attachment and solvation. ## Identify the halogen as an oxidising agent - An oxidising agent accepts electrons and is itself reduced. X₂+2e⁻→2X⁻ shows this role for an elemental halogen. - Chlorine changes from oxidation state0 to−1 when reduced to chloride. The species donating those electrons is oxidised. - For the usual aqueous Cl₂/Br₂/I₂ comparisons, oxidising ability decreases Cl₂>Br₂>I₂. State the species and conditions rather than calling every form of chlorine an oxidant. ## Predict a halogen–halide displacement - Chlorine oxidises bromide:Cl₂+2Br⁻→2Cl⁻+Br₂. Chlorine is reduced while bromide is oxidised. - A stronger oxidising halogen can displace a weaker one from its halide ions under the stated conditions. Br₂ does not similarly oxidise Cl⁻ in this comparison. - The starting solution contains bromide ions, not merely bromine water. The distinction between Br⁻ and Br₂ is essential to the reaction. ## Use a phase observation without confusing ions and elements - A nonpolar solvent can help distinguish the colours of dissolved elemental halogens. Violet iodine in the specified organic phase is evidence for I₂, not for a colourless I⁻ ion. - The solvent identity and phase must be recorded. Do not assume the organic layer is always above water without density information. - Colour intensity also depends on concentration and mixtures. Use the reaction context and appropriate controls along with the observation. ## Explain the opposite reducing-agent trend - Halide ions can donate electrons and become elemental halogens. In that reaction, the halide is oxidised and acts as a reducing agent. - Reducing ability increases Cl⁻<Br⁻<I⁻ in the usual comparisons. The larger iodide ion holds its outer electrons less tightly in the simple size-and-shielding explanation. - This is the opposite ordering to elemental-halogen oxidising ability. Do not swap the trend for X⁻ with the trend for X₂. ## Acidify with nitric acid before the silver test - For an aqueous chloride, bromide or iodide sample, acidify with dilute HNO₃ before adding AgNO₃ solution. Nitric acid helps remove carbonate interference. - Do not use HCl to acidify this test:it introduces chloride and can produce a false AgCl result. Acidifying an existing solution is different from dissolving a solution. - The net precipitation equation is Ag⁺(aq)+X⁻(aq)→AgX(s) for these three halides. AgF is soluble, so the statement every halide gives this precipitate is too broad. ## Name the precipitate rather than a coloured solution - Silver chloride is white, silver bromide cream and silver iodide pale yellow or yellow. These observations refer to solid precipitates dispersed in the liquid. - Colour can be difficult to distinguish for small amounts or mixtures. An ammonia follow-up is useful when the stated sample and conditions permit it. - Use precise language:the precipitate forms or dissolves. The phrase silver-halide solution is misleading when describing the insoluble solid. ## Combine ammonia behaviour with colour - AgCl dissolves in dilute ammonia; AgBr generally requires concentrated ammonia; AgI remains under the usual test conditions. - The independent follow-up record therefore supports bromide. Use appropriate portions and sufficient reagent when comparing dilute and concentrated ammonia. - Do not say that iodide ions become insoluble in ammonia:the observation concerns AgI solid and its dissolution equilibrium. ## Explain ammonia dissolution with complex formation - Ammonia binds Ag⁺ to form [Ag(NH₃)₂]⁺. Lowering free Ag⁺ can shift AgCl dissolution toward more dissolved material. - The overall relation is AgCl(s)+2NH₃(aq)⇌[Ag(NH₃)₂]⁺(aq)+Cl⁻(aq). It conserves atoms and charge. - Silver remains in oxidation state+1; this is complex formation coupled to solubility, not oxidation of chloride or a loss of silver’s nuclear protons. ## Calculate silver-halide yield from the limiting ion - Ag⁺+Br⁻→AgBr(s) has a1:1 ion ratio. The supplied solutions contain2.00 mmol Ag⁺ and3.00 mmol Br⁻. - Silver ions limit the effectively complete precipitation to 2.00 mmol AgBr. The calculated solid mass is0.3756 g using the supplied molar mass. - Convert cm³ to dm³ before multiplying concentration by volume. A mass calculation must use the limiting amount, not the larger available ion amount. ## Separate acid–base formation of HCl from redox - Solid NaCl with concentrated H₂SO₄ can form HCl and sodium hydrogensulfate. In this step chloride is protonated. - Chlorine remains−1 and sulfur remains+6, so this acid–base step is not a redox reaction. Chloride is not a sufficiently strong reducing agent for the later sulfate reductions in the usual comparison. - HCl gas is colourless; white fumes in moist air are acid mist. Observations here describe supervised laboratory chemistry, not an instruction to identify gases by smelling. ## Explain the further bromide redox step - After HBr formation, the concentrated sulfuric acid can oxidise bromide/HBr to Br₂ while sulfur is reduced to SO₂. - In 2HBr+H₂SO₄→Br₂+SO₂+2H₂O, bromine goes−1→0 and sulfur+6→+4. Two electrons are transferred overall. - The balanced equation does not produce hydrogen gas. Bromine vapour is red-brown; SO₂ and hydrogen halides are colourless gases whose observations must be interpreted appropriately. ## Account for iodide reducing sulfate to sulfur dioxide - Iodide is a stronger reducing agent than bromide and can reduce concentrated sulfuric acid through several product pathways. One gives SO₂. - For 2HI+H₂SO₄→I₂+SO₂+2H₂O, sulfur changes+6→+4 and accepts two electrons; two iodides supply those electrons. - The initial protonation step and later redox step are distinct. The reaction can be a mixture of pathways, so one equation is not a claim that SO₂ is always the only sulfur product. ## Account for formation of elemental sulfur - In6HI+H₂SO₄→3I₂+S+4H₂O, sulfur changes from+6 to 0 and accepts six electrons. Six iodides each lose one electron. - The product elemental sulfur is a yellow solid. The balanced equation has four water molecules per sulfuric-acid molecule in this net pathway. - Do not confuse sulfur’s element symbol with sulfide or sulfate. Each has a different oxidation-state description and a different role in the equation. ## Account for the hydrogen sulfide pathway - In8HI+H₂SO₄→4I₂+H₂S+4H₂O, sulfur changes from+6 to−2 and accepts eight electrons. Eight iodides supply those electrons. - Hydrogen sulfide is a gas under ordinary laboratory conditions, not the solid state printed in one source equation. Iodine’s observed solid or vapour phase depends on temperature. - These concentrated-acid reactions can release hazardous gases and belong in the specified supervised fume-cupboard procedure. Deliberate smelling is not an appropriate test. ## Compare extent of reduction without reversing agents - The sulfur products SO₂, S and H₂S correspond to oxidation states+4, 0 and−2. All are reductions from sulfur’s initial+6 state. - The required electron amounts per sulfur atom are2, 6 and8 respectively. H₂S represents the largest reduction among these three pathways. - The halide donating electrons is the reducing agent and is oxidised to X₂. The sulfuric acid accepting electrons is the oxidising agent and is reduced. ## Explain cold-alkali disproportionation - Chlorine with cold dilute NaOH forms sodium chloride, sodium chlorate(I), also called hypochlorite, and water. The net ionic equation is Cl₂+2OH⁻→Cl⁻+ClO⁻+H₂O. - Chlorine changes from 0 to−1 in Cl⁻ and from 0 to+1 in ClO⁻. The same elemental reactant is both reduced and oxidised, so this is disproportionation. - The product names and oxidation states must match:chlorate(I)/hypochlorite is ClO⁻, not chlorate(V), ClO₃⁻. ## Write the chlorine-water equilibrium - Chlorine reacts with water according to Cl₂+H₂O⇌HCl+HClO. It is another disproportionation, with chlorine reaching−1 and+1. - The equilibrium arrow matters:the distribution of dissolved chlorine species depends on solution conditions. The equation is not a claim that every chlorine molecule is permanently converted. - HClO is chloric(I) acid, also called hypochlorous acid. Distinguish it from chloric(V) acid, HClO₃, and hydrochloric acid, HCl. ## Connect pH with HClO and ClO⁻ proportions - HClO⇌H⁺+ClO⁻ describes a weak-acid equilibrium. Increasing acidity favours a larger HClO fraction; raising pH favours more ClO⁻. - Both forms contribute to chlorine-based disinfection, but their effectiveness and reaction behaviour are not identical. The solution conditions influence the active-species balance. - This equilibrium applies chemical reasoning to water treatment. It does not by itself specify a suitable disinfectant concentration or replace operational monitoring. ## Evaluate chlorination as a controlled public-health process - Controlled chlorination reduces microbial disease risk and can maintain a disinfecting residual in distribution pipes. This benefit is central to the AQA evaluation. - Treatment also controls exposure and disinfection byproducts through monitoring and appropriate processes. A statement that any amount is harmless is not a justified evaluation. - Chlorine is less effective against some resistant organisms and does not remove arbitrary chemical contamination. Water safety depends on a managed treatment system, not one reaction equation alone. ## Describe sunlight effects without a universal pool-depth rule - Light can reduce available free chlorine. The supplied simplified overall model2HClO→2HCl+O₂ conserves atoms and represents one bookkeeping description of photochemical loss. - For 0.0200 mol HClO decomposing according to this model, 0.0100 mol O₂ forms. Actual photochemistry and loss rates depend on conditions. - Sunlight exposure, shading, water composition and stabilisation matter. A claim that every shallower pool necessarily loses chlorine at one fixed rate is too broad. ## Qualify the hot concentrated alkali extension - Under hot concentrated alkali conditions, the oxidised chlorine product can be chlorate(V), ClO₃⁻, alongside chloride. This is an extension to the core cold-dilute comparison. - The supplied balanced equation is3Cl₂+6OH⁻→5Cl⁻+ClO₃⁻+3H₂O. Chlorine is reduced to−1 and oxidised to+5. - Conditions belong in the answer. Do not apply the cold-dilute hypochlorite product to every alkali reaction, or call ClO₃⁻ chlorate(I). ## Choose an explanation that matches the process - Boiling a halogen changes intermolecular separation while preserving X₂ molecules. Halide displacement transfers electrons and changes oxidation states. - Proton transfer in HClO dissociation changes acid/base forms without changing chlorine’s+1 oxidation state. Complex formation in the silver test is another distinct process. - Read the species, states, conditions and task before choosing a trend explanation. A balanced equation also needs chemically appropriate species; mass and charge balance alone are not proof of a real reaction.