Halogens AS

F Cl Br I (At) Generally: Oxidising agents Germicides Note: Atoms are halogens Ions are halides Ions have 8 electrons by borrowing one, so single negative charge (F¯ etc.) Not very soluble in water, more soluble in organic solvent, usually cyclohexane

F Cl Br I (At) Electronic configurations  2s 2 2p 5  3s 2 3p 5  4s 2 (3d 10 )4p 5  5s 2 (4d 10 )5p 5

F Cl Br I (At) Atomic radius increases down group because of:  More shielding / repulsion  More shells  (Even though) Bigger nucleus  Ionic radius matches atomic radius

F Cl Br I (At) Boiling point (X 2 molecules) Increases down G7 Because of increased Van der Waal’s forces F & Cl gases Br liquid I & At solid

F Cl Br I (At) Reactivities (oxidising ability)  ½X 2 + e ˉ  X ˉ  Reactivity of atoms decreases down group because of:  Stronger X-X bond  Electron affinity not changing  Hydration / Lattice energy decreasing – making, say, NaI releases less energy than making, say, NaF

F Cl Br I (At) Displacement Higher-reactivity halogens will displace lower-reactivity halogens F > Cl > Br > I > (At) E.g. Cl 2 + 2NaBr  Br 2 + 2NaCl Or Cl 2 + 2Br -  2Cl - + Br 2 Displaced bromine – yellow/brown colour Displaced iodine – brown and/or black precipitate Fluorine too dangerous

F Cl Br I (At) Electronegativity and polarisation Electronegativity decreases down group  More shells so more shielding  Radius increases so less nuclear attraction for shared electrons  Bigger nucleus should attract electrons but it has less effect than other two Therefore hydrogen halides are decreasingly polar

F Cl Br I (At) Hydrogen halides (HCl etc.) All colourless gases, very soluble in water, dissociate well,  strong acids E.g. HCl (aq)  H + (aq) + Cl - (aq) Bond enthalpy decreases down group, easier to separate H + and X - So  strength of acid increases  HI > HBr > HCl These acids are oxidising agents – the hydrogen ion (H + ) can take an electron Note: conc. HF not a strong acid – bond dissociation enthalpy too high, diluted is stronger

F Cl Br I (At) Tests for ions: Test solutions with acidified silver nitrate to make a silver halide: (acidified to avoid formation of carbonates) F – no change - AgF is soluble Cl – white AgCl precipitate Br – cream AgBr precipitate I – yellow AgI precipitate

F Cl Br I (At) General reaction: AgNO 3 (aq) + Xˉ (aq)  AgX (s) + NO 3 ˉ (aq) orAg + (aq) + Xˉ (aq)  AgX (s) Check with dilute and conc. ammonia solution (NH 4 OH): Fˉ no reaction anyway AgCl precipitate soluble in dilute or concentrated ammonia AgBr precipitate soluble in concentrated ammonia only AgI precipitate insoluble, even in concentrated ammonia

F Cl Br I (At) A redox reaction: Cl 2 + H 2 O  HCl + HClO Note oxidation states: One chlorine goes from 0 to -1 in HCl Other goes 0 to +1 in HClO This is a disproportionation – one element is simultaneously oxidised and reduced.

F Cl Br I (At) Reduction by halide ions Iodide ion I - is very reducing –  “happy” to lose electron  didn’t really want it  bromide less so, etc. Decreasing reducing power up the group Electrons lost particularly to strong oxidising agents (H 2 SO 4, F 2 etc.)

F Cl Br I (At) Homework Find out the reactions of H 2 SO 4 with F - to I - and memorise

F Cl Br I (At) Halide ion reactions with concentrated H 2 SO 4 Sulphur has an oxidation number of +6 in SO 4 2 ˉ Fˉ and Clˉ cannot reduce sulphur – only white fumes of HF or HCl seen when NaF or NaCl added to conc. sulphuric acid Fˉ + H 2 SO 4  HF + HSO 4 ˉ Clˉ + H 2 SO 4  HCl + HSO 4 ˉ MEMORISE!!

F Cl Br I (At) (Brˉ) NaBr + H 2 SO 4 Brˉ can reduce sulphur from +6 to +4, producing HBr, SO 2, Br 2 Observe white fumes of HBr, invisible SO 2 (bubbles) and brown fumes of Br 2 (or possibly liquid Br 2 ) Brˉ + H 2 SO 4  HBr + HSO 4 ˉ Then 2HBr + H 2 SO 4  Br 2 + SO 2 + 2H 2 O

F Cl Br I (At) (Iˉ) NaI + H 2 SO 4 Iˉ can reduce sulphur from +6 to +4, then 0, then -2, producing HI, SO 2, S, H 2 S, I 2 White fumes of HI, invisible SO 2 and H 2 S (bubbles, distinctive smell), yellow sulphur, and purple fumes of I 2 Iˉ + H 2 SO 4  HI + HSO 4 ˉ Then 2HI + H 2 SO 4  I 2 + SO 2 + 2H 2 O then

General reaction: NaX (s) + H 2 SO 4 (aq)  NaHSO 4 (aq) + HX (g) fumes in moist air Or Xˉ + H 2 SO 4  HSO 4 ˉ + HX In these, the halide is acting as a base (proton acceptor), accepting H + E.g. NaF (s) + H 2 SO 4 (aq)  NaHSO 4 (aq) + HF (g)

F Cl Br I (At) Uses of chlorine and chlorate(I) compounds Under most conditions: H 2 O + Cl 2 Ý HCl + HClO This is a disproportionation, one chlorine atom is reduced, the other is oxidised. This makes “chlorine water” which can decompose photolytically (in sunlight): 2Cl 2 + 2H 2 O  O 2 + 4HCl

F Cl Br I (At) Chlorination Water treatment Drinking water – 0.7mg dm -3 Swimming pools more concentrated Kills bacteria, especially E.coli from bottoms.

F Cl Br I (At) Reaction with NaOH Cl 2 + 2NaOH Ý NaCl + NaClO + H 2 O This mixture of sodium chloride and sodium chlorate(I) is used as a bleach.