Topic 4B: Elements of group 7 (halogens)

• 9. understand reasons for the trends in physical state at room temperature for Group 7 elements

  Fluorine is often ignored as it reacts differently to chlorine, bromine and iodine. Astatine is ignored as it only exists as radioactive isotopes- half life about 8hrs. Halogens exist as diatomic molecules. F2: very pale yellow gas, highly reactive. Cl2: greenish, reactive gas, poisonous in high conc. Br2: red liquid, that gives off dense brown/orange fumes. I2: shiny grey solid sublimes to purple gas. Astatine is a solid. Br2(l)->Br2(g) carry out in fume cupboard. I2(s)->I2(g) sublimes
• 9. understand reasons for the trends in melting and boiling temperatures, physical state at room temperature, and electronegativity for Group 7 elements

  Trends in mp/bp: increases down the group, as the molecules become larger they have more electrons and so have larger London forces (therefore are more polarisable) between the molecules. As the intermolecular forces get larger more energy has to be put in to break the forces, this increases the mp/bp. Covalent bonds are not broken.
• 9. understand reasons for the trends in melting and boiling temperatures, physical state at room temperature, and electronegativity for Group 7 elements

  Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself.

  Trends in electronegativity: As one goes down the group, the atomic radii increases due to the increasing number of shells. The nucleus is therefore less able to attract the bonding pair of electrons. F2 is the most electronegative element. Group 7 has the highest electronegativity in the Periodic table.
• 10. understand reasons for the trend in reactivity of Group 7 elements down the group

  The reactivity of the halogens decrease down the group as atoms get bigger with more shielding so they less easily attract and accept protons. They therefore form -1 ions less easily down the group. The reactions of the halogens with Fe and H show that reactivity decreases down the group. Halogens act as oxidising agents, gaining electrons to form negative ions or become slightly negative in a polar molecule.
• 11. understand the trend in reactivity of Group 7 elements in terms of the redox reactions of Cl2, Br2 and I2 with halide ions in aqueous solution, followed by the addition of an organic solvent


  A halogen that is a strong oxidising agent will displace a halogen that has a lower oxidising power from one of its compounds. The oxidising strength decreases down the group. Oxidising agents are electron acceptors. Halide ions are colourless. Halogen molecules are coloured.
• 11. understand the trend in reactivity of Group 7 elements in terms of the redox reactions of Cl2, Br2 and I2 with halide ions in aqueous solution

  Cl will displace both Br and I ions. Br will displace I ions. The colour of the solution in the test tube shows which free halogen is present in the solution. Cl=very pale green solution(often colourless). Br=yellow/orange solution. I=brown solution(sometimes black solid present). Difficult to interpret colour changes as similar colours+colours vary with conc. i.e. Br2(l)=red/brown, Br2(aq)=orange/yellow.
• 11. understand the trend in reactivity of Group 7 elements in terms of the redox reactions of Cl2, Br2 and I2 with halide ions in aqueous solution, followed by the addition of an organic solvent


  Observations if an organic solvent (cyclohexane) is added:

  the colour of the organic solvent layer in the test tube shows which free halogen is present in solution. Cl=colourless, Br=orange, I=purple. Halogens are more soluble in cyclohexane than in water so the halogen dissolves in the upper organic layer, where its colour can be seen.
• 11. understand the trend in reactivity of Group 7 elements in terms of the redox reactions of Cl2, Br2 and I2 with halide ions 


  Cl2+2KBr>2KCl+Br2 Half equations: Cl2+2e- ->2Cl- 2Br->Br2+2e- Cl has gained electrons, so it is reduced to Cl- ions. Br ions have lost electrons, so they have been oxidised to Br. In displacement reactions between halogens and halides, the halogen acts as an oxidising agent. This means that the halogen: oxidises the halide ion to the halogen, gain electrons, is reduced to form the halide ion. Oxidising ability decreases down group 7.
• 11. understand the trend in reactivity of Group 7 elements in terms of the redox reactions of Cl2, Br2 and I2 with halide ions in aqueous solution, followed by the addition of an organic solvent


  Redox reactions- the halogen decreases in oxidation number from 0 to -1 while the halide increases from -1 to 0. Cl2(aq)+2Br-(aq)->2Cl-(aq)+Br2(aq) Cl2(aq)+2I-(aq)->2Cl-(aq)+I2(aq) Br2(aq)+2I-(aq)->2Br-(aq)+I2(aq) Cl2+2NaBr->Br2+"NaCl. Reactions are most vigorous between elements at the bottom of Group 1 and 2 and at the top of Group 7. Br2(l)+2Na(s)->2NaBr(s) = 2Na->2Na+ +2e- Br2+ 2e- ->2Br-.
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens: i oxidation reactions with Group 1 and 2 metals ii the disproportionation reaction of chlorine

  In all reactions where halogens are reacting with metals, the metals are being oxidised. Ionic salts are produced- usually white solids. 3Cl2(g)+2Fe(s)->2FeCl3(s) = Cl2(g)+2Fe2+(s)->2Cl-(aq)+2Fe3+(aq). 2I-(g)+2Fe3+(s)->I2(aq)+2Fe2+(aq). Cl and Br can oxidise Fe2+ to Fe3+. I is not strong enough oxidising agent to do this reaction, the reaction is reversed for I.
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens:ii) the disproportionation reaction of chlorine

  Disproportionation is the name for a reaction where an element simultaneously oxidises and reduces.
• understand, in terms of changes in oxidation number, the following reactions of the halogens: ii the disproportionation reaction of chlorine

  Cl with H2O= Cl2(g)(0)+H2O(l)->HClO(aq)(+1)+HCl(-1)(aq). Cl is both simultaneously reducing and oxidising changing its oxidation number from 0 to -1 and 0 to +1. 2 acids are produced. All colourless liquids so no visible change. Br reacts in a similar way but to a lesser extent. I is insoluble in water and hardly reacts at all. HClO is a weak acid as it only dissociates slightly.
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens: ii the disproportionation reaction of chlorine


  If some universal indicator is added to the solution it will first turn red due to the acidity of both reaction products. It will then turn colourless as the HClO bleaches the colour. The pale greenish colour of these solutions is due to Cl2. Cl is used in water treatment to kill bacteria. It has been used to treat drinking water and the water in swimming pools. Cl kills pathogens responsible for cholera.
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens: iii the disproportionation reaction of chlorine with cold, dilute aqueous sodium hydroxide to form bleach

  Cl2, Br2, I2 in aq solutions will react with cold NaOH. The colour of the halogen solution will fade to colourless. Cl2(aq)(0)+2NaOH(aq)->NaCl(aq)(-1)+NaClO(aq)(+1)+H2O(l) (Disproportionation). The mixture of NaCl and NaClO is used as bleach(sodium chlorate I); to kill bacteria. Cl2  +  2e-  🡺  2Cl-. Cl2  +  4OH-  🡺  2ClO-  +  2H2O  + 2e-  Overall

  Cl2  +  2OH-  🡺  Cl-  +  ClO-  +  H2O
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens: iv the disproportionation reaction of chlorine with hot alkali


  With hot alkali disproportionation also occurs but the halogen that is oxidised goes to a higher oxidation state. 3Cl2(aq)(0)+6NaOH(aq)->5NaCl(aq)(-1)+NaClO3(aq)(+5)+3H2O(l) 3I2(aq)+6NaOH(aq)->NaI(aq)+NaIO3(aq)+3H2O(l)(sodium chlorate V) 3I2(aq)+6OH-(aq)->5I-(aq)+IO3-(aq)+3H2O. Same reaction with Br. NaClO3 used in bleaching and as a weed killer.
• 12. understand, in terms of changes in oxidation number, the following reactions of the halogens: i oxidation reactions with Group 1 and 2 metals ii the disproportionation reaction of chlorine

  NaClO: sodium chlorate(I)

  NaClO3: sodium chlorate(V)

  K2SO4:potassium sulfate(VI)

  K2SO3:potassium sulfate(IV)
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Sodium halides react with concentrated H2SO4. During this reaction two things can happen to the acid. It can be reduced or act as an acid. The reactions demonstrate the relative strengths of the halide ions as reducing agents. The halides show increasing power as reducing agents as one goes down the group. This can be clearly demonstrated in the various reactions of the solid halides with conc. H2SO4.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Reducing agent power increases down the group: CI is most reluctant to give up its electron and turn back into a molecule. Oxidising agent power decreases down the group: I- least tendency to form (-) ions + most ready to give up electron to turn back into a molecule.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Explanation of differing reducing power of halides: A reducing agent donates electrons. The reducing power of the halides increases down Group 7. They have a greater tendency to donate electrons. This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  H2SO4 -> H(+) +SO4(2-) reversible reaction.

  Conc. acid contains very few ions. H2SO4 can act as an oxdising agent and be reduced. The extent of its reduction, and the products, depends on the species being oxidised.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  The H2SO4 is not strong enough an oxidising reagent to oxidise the Cl- and F- ions. No redox reactions occur. Only acid-base reactions occur. Therefore no half equations. NaF(s)+H2SO4(l)->NaHSO4(s)+HF(g). Observations: white steamy fumes of HF are evolved.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  NaCl(s) + H2SO4(l)(conc.)->NaHSO4(s)+HCl(g), white steamy fumes of HCl, can be used to make HCl. Cl- ions have low reducing power. These are acid-base reactions and not redox reactions. H2SO4 plays the role of an acid(proton donor).
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Br- ions are stronger reducing agents than Cl- and F- and after the initial acid-base reaction reduce the S in H2SO4 from +6 to +4 in SO2. Acid-base step: NaBr(s)+H2SO4(l)->NaHSO4(s)+HBr(g) Redox step: 2HBr+H2SO4->Br2(g)+SO2+2H2O(l), white steamy fumes of HBr. Brown fumes of Br also evolved and a colourless acidic gas SO2 with choking smell.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Ox 1/2 equation: 2Br- ->Br2+2e-. Re 1/2 equation: H2SO4+2H+ +2e- ->SO2+2H2O. Re product:SO2 Add 1/2 equations together to get redox reaction. Note: H2SO4 plays the role of an acid in the first step producing HBr and then acts as an oxidising agent in the second redox step.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  I- ions are the strongest halide reducing agents. They can rduce sulfur from +6 in H2SO4 to +4 in SO2, to 0 in S and -2 in H2S. NaI(s)+H2SO4(l)->NaHSO4(s)+HI(g). 2HI+H2SO4->I2(s)+SO2(g)+2H2O(l). 6HI+H2SO4->3I2+S(s)+4H2O(l). 8HI+H2SO4->4I2(s)+H2S(g)+4H2O(l). 3 redox reactions occur. Little/no HI forms as it continues to reduce.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides


  1.Ox 1/2 equation: 2I->I2+2e-

  Re ": H2SO4(+6)+2H+ +2e- ->SO2(+4)+2H2O (difference=2)

  2. Re ": H2SO4(+6)+6H+ +6E- ->S(0)+4H2O (6)

  Re ": H2SO4(+6)+8H+ +8e- ->H2S(-2)+4H2O(8)

  1 and 2 add together to make redox reaction 3.

  3. 6HI+H2SO4 . 3I2+S(s)+4H2O(l)
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  Observations: white steamy fumes of HI. Black solid and purple fumes of I. A colourless, acidic gas SO2. A yellow solid of S. H2S (hydrogen sulfide), a gas with a bad egg smell. Reduction products= sulfur dioxide, sulfur and hydrogen sulfide. Note: H2SO4 plays the role of an acid in the first step producing HI and then acts as an oxdising agent in the three redox steps.
• 13. understand the following reactions: i solid Group 1 halides with concentrated sulfuric acid, to illustrate the trend in reducing ability of the hydrogen halides

  I- reduces sulphur from +6 to 0 and -2.

  Br- reduces sulphur from +6 to +4.

  Cl- doesn't reduce sulphur.
• 13. understand the following reactions: ii precipitation reactions of the aqueous anions Cl–, Br– and I– with aqueous silver nitrate solution

  Halides can be identified by their reaction with acidified silver nitrate solution to form silver halide precipitates. KCl(aq)+AgNO3(aq)->KNO3(aq)+AgCl(s). Silver chloride has a low solubility in water, so it forms a white precipitate: the positive result in the test for Cl- ions.
• 13. understand the following reactions: precipitation reactions of the aqueous anions Cl–, Br– and I– with aqueous silver nitrate solution

  This reaction is used as a test to identify which halide ion is present. The test solution is made acidic with nitric acid, and then silver nitrate solution is added dropwise. The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate AgCO3. This would mask the desired observations. 2HNO3+Na2CO3-> 2NaNO3+H2O+CO2.
• 13. understand the following reactions: ii precipitation reactions of the aqueous anions Cl–, Br– and I– with aqueous silver nitrate solution


  F produce no precipitate. soluble in water

  Cl produce a white precipitate. Ag+(aq) +Cl-(aq)->AgCl(s)

  Br produces a cream precipitate. Ag+(aq)+Br-(aq->AgBr(s)

  I produces a pale yellow precipitate. Ag+(aq)+I-(aq)->AgI(s)
• 13. understand the following reactions ii precipitation reactions of the aqueous anions Cl–, Br– and I– with aqueous silver nitrate solution

  Effect of ammonia on silver halides: The silver halide precipitates can be treated with ammonia solution to help differentiate between them is the colours look similar.
• 13. understand the following reactions: ii precipitation reactions of the aqueous anions Cl–, Br– and I– by aqueous ammonia solution

  Silver chloride dissolves in dilute ammonia to form a complex ion. AgCl(s)+2NH3(aq)->[Ag(NH3)2]+(aq)+Cl-. colourless solution.

  Silver bromide dissolves in concentrated ammonia to form a complex ion. AgBr(s)+2NH3(aq)->[Ag(NH3)2]+(aq)+Br-(aq) colourless solution.

  Silver iodide does not react with ammonia- it is too insoluble.
• 13. understand the following reactions: ii precipitation reactions of the aqueous anions Cl–, Br– and I– with aqueous silver nitrate solution

  Effect of light on silver halides

  The precipitates (except AgI) darken in sunlight forming silver. This reaction is used in photography to form the dark bits on photographic film.
• 13. understand the following reactions: iii hydrogen halides with ammonia and with water (to produce acids)


  The hydrogen halides are colourless gases at rtp. HF has an unexpectedly high melting point compared to the other hydrogen halides. This is due to hydrogen bonding between the H-F molecules.
• 13. understand the following reactions: iii hydrogen halides with ammonia and with water (to produce acids)

  A substance that donates electrons in a reaction (i.e. is oxidized) is a reducing agent because it reduces the other reactant. The larger the halide ion, the easier it is for it to donate electrons and therefore the more reactive it is. This is because its outermost electrons are further from the attraction of the nucleus and more shielded from it by other electrons. The attraction for the outermost electrons is therefore weaker.
• 13. understand the following reactions: iii hydrogen halides with ammonia and with water (to produce acids)

  Producing hydrogen halides:

  Hydrogen halides are made by the reaction of solid soldium halide salts with phosphoric acid. NaCl(s)+H3PO4(l)->NaH2PO4(s)+HCl(g)

  Observations: white steamy fumes of the hydrogen halides. The steamy fumes of HCl are produced when the HCl meets the air because it dissolves in the moisture in the air.
• 13. understand the following reactions: iii hydrogen halides with ammonia and with water (to produce acids)

  Phosphoric acid is not an oxidising agent and so does not oxide HBr and HI. Phosphoric acid is more suitable for producing hydrogen halides than using concentrated sulfuric acid to make HCl, HBr and HI. This is because there are no extra redox reactions taking place and no other products formed. Hydrogen halides are colourless gases + exist as polar diatomic molecules. Downward delivery is used because hydrogen halides are more dense.
• 13. understand the following reactions: iii hydrogen halides with ammonia and with water (to produce acids)

  Solubility in water: The hydrogen halides are all soluble in water. They dissolve to form acidic solutions. (colourless) HCl(g) strong acid+H2O(l)->H3O+(aq)+Cl-(aq). Hydrofluoric acid HF+H2O->H3O+ +F-(weak acid). Hydroiodic acid HI+H2O->H3O+ +I- (strong acid). Hydrobromic acid HBr+H2O->H3O+ +Br- (strong acid)