Group 7

• Describe and the trend in electronegativity going down group 7.

  
  

  decreases down the group because going down the group, the atomic radius increases due to increasing number of shells so there is more shielding and a greater distance between the nucleus and bonding electron pair so there is a weaker attraction between the nucleus and bonding pair of electrons
• Describe and explain the trend in boiling point going down group 7. increases going down the groupmolecules become larger and so have more electronsgroup 7 exists with a simple covalent molecular structure (diatomic molecules)therefore larger van der waals between the moleculesso more energy is required to break these intermolecular forces
• Describe the appearance and properties of fluorine. F2very pale yellow gashighly reactive
• Describe the appearance and properties of chlorine. Cl2greenish gasreactive gaspoisonous in high concentration
• Describe the appearance and properties of bromine. Br2red/brown liquidgives off dense brown/orange poisonous fumes
• Describe the appearance and properties of iodine. I2Shiny grey solidsublimes to purple gas
• Compare the atomic radius of a halogen atom and a halogen ion. ions have a larger atomic radius than atoms (of the same group 7 element)beacuse the added electron repels the others so the radius gets larger
• Describe and explain the trend in oxidising ability of the halogens down the group. oxidising strength decreases down the groupbecause the attraction between the nucleus and incoming electrons (electrons not on the halogen aton) decreases down the groupdue to increased shielding and increased atomic radius (despite the increase in nuclear charge)
• When would one halogen displace another halogen? A halogen that is a strong oxidising agent will displace a halogen that has a lower oxidising power from one of its compounds, in aqueous solution. A more reactive halogen will always displace a less reactive halogen from its compound in aqueous solution.
• Describe the appearance of a solution where chlorine (Cl2) is present. very pale green solution (often colourless)
• Describe the appearance of a solution where bromine (Br2) is present. yellow solution
• Describe the appearance of a solution where iodine (I2) is present. brown solution (sometimes black solid present)
• Describe the observations and what is actually happening when Chlorine (aq) is added to solutions of potassium chloride, potassium bromide and potassium iodide. Added to KCl - very pale green solution, no reactionAdded to KBr - Yellow solution, Cl has displaced BrAdded to KI - brown solution, Cl has displaced I
• Describe the observations and what is actually happening when bromine (aq) is added to solutions of potassium chloride, potassium bromide and potassium iodide. Added to KCl - Yellow solution, no reactionAdded to KBr - Yellow solution, no reactionAdded to KI - Brown solution, Br has displaced I
• Describe the observations and what is actually happening when iodine (aq) is added to solutions of potassium chloride, potassium bromide and potassium iodide. Added to KCl - Brown solution, no reactionAdded to KBr - Brown solution, no reactionAdded to KI - Brown solution, no reaction
• In a halogen displacement reaction, what does the colour of the test tube show? which free halogen is present in solutionChlorine = very pale green solution (often colourless)Bromine = yellow solutionIodine = brown solution (sometimes black solid present)
• Write an ionic equation for chlorine displacing bromine from potassium bromide. Cl2(aq) + 2Br – (aq) ------> 2Cl – (aq) + Br2(aq)
• Write an ionic equation for chlorine displacing iodine from potassium iodide. Cl2(aq) + 2I – (aq) ------> 2Cl – (aq) + I2(aq)
• Write an ionic equation for bromine displacing iodine from potassium iodide. Br2(aq) + 2I – (aq) --------> 2Br – (aq) + I2(aq)
• Write two half equations to show what happens when chlorine diplaces bromine from potassium bromide. 2Br -(aq) ------> Br2 (aq) + 2e- Cl2 (aq) + 2e- ------> 2Cl- (aq)
• Describe the test for halide ions using silver nitrate. Add nitric acid to acidify solution (reacts with carbonates to avoid false positive results).Add silver nitrate (AgNO3) dropwise.If it is difficult to distinguish between the colours of the silver nitrate preciptiates, add dilute ammonia and then repeat experiment but add concentrated ammonia
• What is added before silver nitrate in the test for halide ions and why? nitric acidreacts with any carbonates present to prevent the formation of the precipitate Ag2CO3 which would give a false positive result 2HNO3 + Na2CO3 ------> 2NaNO3 + H2O + CO2
• Write an equation to show why nitric acid is added before silver nitrate solution in the test for halide ions. 2HNO3 + Na2CO3 ---------> 2NaNO3 + H2O + CO2 removes carbonate ions to prevent the formation of Ag2CO3
• Describe why dilute ammonia may be added after silver nitrate in the test for halide ions. The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar Silver chloride dissolves in dilute ammonia to form a complex ion AgCl(s) + 2NH3(aq) ---------> [Ag(NH3)2]+ (aq) + Cl- (aq) this gives a colourless solution
• Describe why concentrated ammonia may be added after silver nitrate in the test for halide ions. The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar. Silver bromide dissolves in concentrated ammonia to form a complex ion. AgBr(s) + 2NH3(aq) --------> [Ag(NH3)2]+ (aq) + Br - (aq) in a colourless solution
• Describe the results of the test for halide ions using silver nitrate. Fluorides produce no precipitateChlorides produce a white precipiate which dissolves in dilute ammoniaAg+(aq) + Cl-(aq) ---> AgCl(s) AgCl(s) + 2NH3(aq) ---> [Ag(NH3)2]+(aq) + Cl-(aq)Bromides produce a cream precipitate which dissolves in concentrated ammoniaAg+(aq) + Br-(aq) ---> AgBr(s) AgBr(s) + 2NH3(aq) ---> [Ag(NH3)2]+(aq) + Br-(aq)Iodides produce a yellow precipitate which does not react with ammoniaAg+(aq) + I-(aq) ---> AgI(s)
• Which silver halide dissolves in dilute ammonia solution? silver chloride
• Which silver halide dissolves in concentrated ammonia solution ? silver bromide
• Which silver halide does not dissolve/react in ammonia solution and why? silver iodideit is too insoluble
• Describe and explain the trend in reducing ability down the halides of group 7. reducing power of halides increases down the groupmeaning they have a greater tendency to donate electronsthis is because as the ions get bigger, it is easier for them to donate electronsas the attraction between the nucleus and the outer electron is smallerdue to more shielding and an increased atomic radius
• Describe the alternative test for halides using concentrated sulfuric acid. add concentrated sulfuric acid carefully to a solid halideH2SO4 displaces the weaker acids HCl, HBr and HI from their salts.going down the group, the halides become more powerful reducing agents so they can react further by reducing the sulfuric acid to lower oxidation states of sulfur (REDOX reactions occur for bromide and iodide ions)this results in different observations for each halide
• Describe what happens when solid NaF reacts with H2SO4 and why. Write an equation. What is observed? F- ions are not strong enough reducing agents to reduce the S in H2SO4. No redox reactions occur. Only acid-base reactions occur. F- is displaced from NaF NaF(s) + H2SO4(l) -------> NaHSO4(s) + HF(g) Observations: White steamy fumes of HF are evolved.
• Describe what happens when solid NaCl reacts with H2SO4 and why. Write an equation. What is observed? Cl- ions are not strong enough reducing agents to reduce the S in H2SO4. No redox reactions occur. Only acid-base reactions occur. Cl- is displaced from NaCl NaCl(s) + H2SO4(l) --------> NaHSO4(s) + HCl(g) Observations: White steamy fumes of HCl are evolved.
• Describe what happens when solid NaBr reacts with H2SO4 and why. What is observed? Br- ions are stronger reducing agents than Cl- and F- and after the initial acid-base reaction, the bromide ions reduce the sulfur in H2SO4 from +6 to +4 in SO2White steamy fumes of HBr are evolvedOrange fumes of bromine are also evolved and a colourless, acidic gas SO2
• Write equations to show the different steps when solid NaBr reacts with H2SO4. Acid- base step: NaBr(s) + H2SO4(l) -------> NaHSO4(s) + HBr(g) Redox step: 2H+ + 2Br - + H2SO4 -------> Br2(g) + SO2(g) + 2H2O(l) Overall equation: 2NaBr + 3H2SO4 --------> 2NaHSO4 + SO2 + Br2 + 2H2O
• Write the REDOX half equations for when solid NaBr reacts with H2SO4. Oxidation: 2Br - -----> Br2 + 2e- Reduction: H2SO4 + 2H+ + 2e- -------> SO2 + H2O
• Describe the role of H2SO4 in its reaction with solid NaBr. plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step. Acid-base step: NaBr(s) + H2SO4 (l) ---> NaHSO4 (s) + HBr (g)Redox step: 2H+ + 2Br- + H2SO4 ---> Br2 + SO2 + 2H2OOverall equation: 2NaBr + 3H2SO4 ---> 2NaHSO4 + SO2 + Br2 + 2H2O
• What is the reduction product from the reaction between solid NaBr and H2SO4. sulfur dioxide, SO2
• Describe what happens when solid NaI reacts with H2SO4 and why. What is observed? I- ions are the strongest halide reducing agents. They can reduce the sulfur from +6 in H2SO4 to +4 inSO2, to 0 in S and -2 in H2S. White steamy fumes of HI are evolved. Black solid and purple fumes of Iodine are also evolved A colourless, acidic gas SO2 A yellow solid of sulfur H2S (Hydrogen sulfide), a gas with a bad egg smell
• Write equations to show the different steps when solid NaI reacts with H2SO4. Acid-base step:NaI(s) + H2SO4(l) ------> NaHSO4(s) + HI(g)Redox steps: 2H+ + 2I- + H2SO4 ------> I2(s) + SO2(g) + 2H2O(l) 6H+ + 6I- +H2SO4 ------> 3I2 +S(s) + 4H2O(l) 8H+ + 8I- +H2SO4 -------> 4I2(s) + H2S(g) + 4H2O(l)