Halogens
- Created by: JasmineR
- Created on: 30-04-16 16:10
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- Halogens
- 1) Trends
- Boiling and melting point
- Increase down the group
- As the molecules become larger they have more electrons and so larger van der Waal forces.
- Therefore, more energy needed to break the forces.
- Electronegativity
- Increases down the group
- The atomic radii increases due to the increasing number of shells.
- The nucleus is therefore less able to attract the bonding pairs of electrons
- The atomic radii increases due to the increasing number of shells.
- The relative tendency of an atom in a molecule to attract electrons in a covalent bond.
- Increases down the group
- Boiling and melting point
- 2) Displacement reactions of halide ions by halogens
- A halogen that is a strong oxidising agent will displace a halogen with a lower oxidising power from one of its compounds
- The oxidising strength decreases down the group
- Oxidising agents are electron acceptors
- Chlorine will displace both bromide and iodide ions
- Bromine will displace iodide ions
- Iodine won't displace anything
- The colour of the solution in the test tube shows which free halogen is present in the solution
- Chlorine = very pale green solution (often colourless)
- Bromine = Yellow solution
- Iodine = brown solution (sometimes black solid is present)
- Be able to write these reactions as two half equations showing oxidation or reduction
- E.g.
- 2Br- (aq) --> Br2 (aq) + 2e-
- Cl2(aq) + 2e- --> 2Cl- (aq)
- 2Br- (aq) --> Br2 (aq) + 2e-
- E.g.
- 3) The reactions of halide ions with silver nitrate
- This reaction is used as a test to identify which halide ion is present
- The test solution is made with nitric acid, and then silver nitrate soltution
- The nitric acid will react with any carbonates present to prevent the formation of Ag2CO3
- Ag2CO3 would mask the desired observations
- Results of this:
- Fluorides produce no precipitate
- Chlorides produce a white precipitate
- Bromides produce a cream precipitate
- Iodides produce a pale yellow precipitate
- The nitric acid will react with any carbonates present to prevent the formation of Ag2CO3
- The silver halide precipitates can be treated with ammonia solution to help differentiate between them if colours look similar
- Silver chloride dissolves in dilute ammonia to form a complex ion
- Silver bromide dissolves in concentrated ammonia to from a complex ion
- Silver iodide doesn't react with ammonia as it is too insoluble
- 4) The reaction of halide salts with concentrated sulphuric acid
- The halides show increasing power as reducing agents going down the group
- A reducing agent donates 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
- Fluoride and Chloride
- Fluoride and chloride ions aren't strong enough reducing agents to reduce S in H2SO4
- No redox reactions occur
- Only acid-base reactions occur
- H2SO4 plays the role of an acid (proton donor)
- Only acid-base reactions occur
- No redox reactions occur
- 1) NaF(s) + H2SO4(l) --> NaHSO4 (s) + HF(g)
- Observations: White steamy fumes of HF are evolved
- 2) NaCl (s) + H2SO4 (l) --> NaHSO4 (s) + HCl (g)
- Observations: White steamy fumes of HCl are evolved
- Fluoride and chloride ions aren't strong enough reducing agents to reduce S in H2SO4
- Bromide
- Bromide ions are stronger reducing agents than Cl- and F-
- After the initial acid-base reaction they reduce the S in H2SO4 from +6 to +4 in SO2
- 1) Acid-base step: NaBr (s) + H2SO4 (l) --> NaHSO4(s) + HBr (g)
- Observations: White steamy fumes of HBr evolved.
- 2) Redox step: 2HBr + H2SO4 --> Br2 (g) + SO2(g) + 2H2O (l)
- Observations: Red fumes of Bromine evolved and a colourless, acidic gas SO2
- Reduction production: Sulphur dioxide
- Bromide ions are stronger reducing agents than Cl- and F-
- Iodide
- Iodide ions are the strongest halide reducing agents
- They can reduce S from +6 in H2SO4 to +4 in SO2, to 0 in S and -2 in H2S
- 1) NaI (s) + H2SO4(l) --> NaHSO4 (s) + HI (g)
- Observations: White steamy fumes of Hi are evolved
- 2) 2HI + H2SO4 --> I2(s) + SO2(g) + 2H2O (l)
- Observations: Black solid and purple fumes of iodine are evolved. A colourless, acidic gas of SO2
- 3) 6HI + H2SO4 --> 3I2 + S (s) + 4H2O (l)
- Observations: A yellow solid of sulphur
- 4) 8HI + H2SO4 --> 4I2 (s) + H2S(g) + 4H2O (l)
- Observations: H2S (hydrogen sulphide), a gas with a bad egg smell
- Reduction products: sulphur dioxide, sulphur and hydrogen sulphide
- Note that the H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the 3 redox steps
- Iodide ions are the strongest halide reducing agents
- The halides show increasing power as reducing agents going down the group
- 5) The disproportionation reactions of chlorine and chlorate(I)
- Disproportionation is a reaction where an element simultaneously oxidises and reduces
- Chlorine with water:
- Cl2(aq) + H2O(l) --> HClO(aq) + HCl(aq)
- Chlorine is both simultaneously reducing and oxidising
- If universal indicator is added it will first turn red due to the acidity of both products. The it will turn colourless as the HClO bleaches the colour
- Cl2(aq) + H2O(l) --> HClO(aq) + HCl(aq)
- Chlorine with water in sunlight
- 2Cl2 + 2H2O --> 4H+ + 4Cl- + O2
- The greenish colour of these solutions is due to the Cl2
- Cl2(aq) + H2O(l) --> HClO(aq) + HCl(aq)
- Chlorine is both simultaneously reducing and oxidising
- If universal indicator is added it will first turn red due to the acidity of both products. The it will turn colourless as the HClO bleaches the colour
- Cl2(aq) + H2O(l) --> HClO(aq) + HCl(aq)
- The greenish colour of these solutions is due to the Cl2
- The same reaction occurs to the equilibrium mixture of chlorine water.
- The greenish colour of chlorine water fades as the Cl2 reacts and a colourless gas (O2) is produced
- 2Cl2 + 2H2O --> 4H+ + 4Cl- + O2
- Chlorine is used in water treatment to kill bacteria
- It has been used to treat drinking water and the water in swimming pools
- The benefits to health of water treatment by chlorine outweigh its toxic effects
- Reaction of chlorine with cold dilute sodium hydroxide solution
- Cl2 (and Br2 and I2) in aqueous solutions will react with cold sodium hydroxide
- The colour of the halogen solution will fade to colourless
- Cl2 (aq) + 2NaOH (aq) --> NaCl (aq) + NaClO (aq) + H2O (l)
- The mixture of NaCl and NaClO is used as bleach and to disinfect/kill bacteria
- 6) Naming chlorate and sulphates
- The various forms of sulfur and chlorine compounds where oxygen is combined are all called sulfates and chlorates
- The relevant oxidation number is given in roman numberals
- If asked to name these compounds remember to add the oxidation number
- E.g.
- NaClO: sodium chlorate(I)
- NaClO3: sodium chlorate(V)
- K2SO4: potassium sulfate(VI)
- K2SO3: potassium sulfate(IV)
- 1) Trends
- Electronegativity
- Increases down the group
- The atomic radii increases due to the increasing number of shells.
- The nucleus is therefore less able to attract the bonding pairs of electrons
- The atomic radii increases due to the increasing number of shells.
- The relative tendency of an atom in a molecule to attract electrons in a covalent bond.
- Increases down the group
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