Group 7 - The Halogens

Electronegativity
1. Decreases down the group
2. F=4.0, Cl=3.0, Br=2.8, I=2.5
Bonding
1. diatomic so have weak V der W's IM forces or attraction
  1. these increase in strength as you move down the group
2. Halogen bonds - where the electronegative halogen becomes delta -ve and it is attracted to a delta +ve on a nearby molecule.
3. Ionic in halide salts
4. Covalent in between 2 halogen atoms to form diatomic molecules and also covalent in hydrogen halides
Electronic configuration
1. Fluorine: [He] 2s2 2p5
2. Bromine: [Ar] 4s2 3d10 4p5
3. Chlorine: [Ne] 3s2 3p5
4. Iodine: [Kr] 4d10 5s2 5p5
5. Astatine: [Xe] 4f14 5d10 6s2 6p5
Disproportionation Reactions
1. Chlorine and water
  1. Cl2 + H2O --> HCl + HClO
2. Chlorine w/ cold dilute NaOH
  1. 2NaOH + Cl2 --> NaCl + NaClO + H2O
3. Chlorine w/ hot conc. NaOH
  1. 6NaOH + 3Cl2 --> 5NaCl + NaClO3 + 3H2O
Colours
1. At RTP: Fl = pale yellow, Cl = pale green, Br = dark red, I = Grey/violet, At = black
2. Colour of organic solvent solution: Fl x (reacts w/ solvents), Cl pale green, Br orange, I purple, At darker than I
3. in water: Fl = x (reacts w/ water), Cl = pale green, Br = orange, I = insoluble but forms brown sol. if excess KI is present
4. As a vapour (if not already gasseous at RTP) Br = brown, I = violet, At = darker purple / black
Half cell reactions / electrode reactions
1. 2Br-(aq) --> Br2(g) + 2e-
2. 2Cl-(aq) --> Cl2(g) + 2e-
3. 2F-(aq) --> F2(g) + 2e-
4. I-(aq) --> I(s) + e-
5. At-(aq) --> At(s) + e-
Tests
1. Ag+ ion forms insol. ppts w/ Cl-,Br- and I- ions (silver fluoride is soluable so you don't get a ppt).Add HNO3 to aqueous sol. of halide ion followed by silver nitrate sol.
  1. Ag+(aq) + Cl-(aq) --> AgCl(s) which is a white ppt.
  2. Ag+(aq) +Br-(aq) --> AgBr(s) which is a cream ppt
  3. Ag+(aq) +I-(aq) --> AgI(s) which is a yellow ppt.
  4. Nitric acid is added to ensure any carbonate or hydroxide ions are removed as CO2 or H20
2. Add dilute ammonia sol. to te ppt formed in the AgNO3(aq) test. The Cl ppt will dissolve, the Br ppt is partially soluable, and the I ppt will not dissovle.
  1. NH3 + AgCl --> [Ag(NH3)2]+ + Cl-
3. Add conc. ammonia sol. to the ppt formed in the AgNO3 test. Cl and Br ppts will dissovle bu the I ppt will not.
  1. NH3 + AgBr --> [Ag(NH3)2]+ + Br-
4. Add conc. sulphuric acid to a solid sample of the halide. The concentrated sulphuric acid gives a hydrogen ion to the halide ion to produce a hydrogen halide. Because this is a gas, it immediately escapes from the system. If the hydrogen halide is exposed to moist air, you see it as steamy fumes.
  1. Steamy acidic fumes of HF if F- ions were present
  2. Steamy white acidic fumes of HCl if Cl- was present. Turn blue litmus red.
  3. steamy white acidic fumes (of HBr) contaminated with brown bromine vapour. This is because Br = a strong enough reducing agent to reduce the H2SO4 which in the process oxidises the Br- ions creating Br gas
    1. 2Br- --> Br2(g) + 2e-
  4. Some steamy fumes (of HI), but lots of purple iodine vapour. I = an even more powerful reducing agent than Br so they reduce the H2SO4 first to SO2, then to S, then to H2S (which has a rotten egg smell)
    1. H2SO4 + 8H+ + 8I- --> 4I2 + H2S + 4H2O
States at RTP
1. Fl + Cl = gas Br = liquid I + At = solid
Displacement reactions
1. If a more reactive halogen is placed into a solution containing a less reactive halide a displacement reaction is seen. The more reactive halogens - i.e. the strongest oxidising agents - will displace the more reactive halides - i.e. the strongest reducing agents - from sols. of their ions.
2. Fluorine = most reactive, Astatine = least reactive
Reaction with metals
1. With Na
  1. All react w/ Na to produce Na halides. In each case you get an orange flame and a white solid.
2. With Fe(II)
  1. With the exception of iodine, iron burns in the halogen vapour to give iron(III) halides. Iodine is less reactive, and only produces iron(II) iodide.
    1. Fluorine: Cold iron wool burns in cold fluorine to give iron(III) fluoride. Anhydrous iron(III) fluoride is variously described as white or pale green. This reaction is very rapid.
      1. 2Fe + 3F2 --> 2FeF3
    2. Chlorine: If you pass chlorine gas over hot iron, the iron burns to form iron(III) chloride. Iron(III) chloride forms black crystals if it is anhydrous. Any trace of water present in the apparatus, or in the chlorine, reacts with this to give reddish-brown colours.
      1. 2Fe + 3Cl2 --> 2FeCl3
    3. Bromine: If you pass bromine vapour over hot iron, a similar but slightly less vigorous reaction happens, this time producing iron(III) bromide. Anhydrous iron(III) bromide is usually produced as a reddish-brown solid.
      1. 2Fe + 3Br2 --> 2FeBr3
    4. Iodine: The reaction between hot iron and iodine vapour only produces iron(II) iodide, and is much less vigorous. The iron(II) iodide is said to be grey. If you do this reaction, the product is always contaminated with iodine, so it is difficult to be sure.
      1. Fe + 2I2 --> FeI2
3. W/ Fe(III)
  1. Chlorine and bromine are strong enough oxidising agents to oxidise iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions; the bromine to bromide ions: 2Fe2+ + Cl2 --> 2Fe3+ + 2Cl-. This is the same reaction with bromine. The very pale green solution containing the iron(II) ions will tur into a yellow or orange solution containing iron(III) ions.Iodine isn't a strong enough oxidising agent to oxidise iron(II) ions, so there is no reaction. In fact the reverse reaction happens. Iron(III) ions are strong enough oxidising agents to oxidise iodide ions to iodine: 2Fe3+ + 2I- --> 2Fe2+ + I2
BP's and MP's
1. BP's: Fl=-188, Cl=-35, Br=59, I=184, At=337
2. MP's: Fl=-220, Cl=-101, Br=-7, I=114, At=302
Polarisability
1. More and more easily polarised anions as you move down group 7

Nächster

Group 7 - The Halogens

Beschreibung

Zusammenfassung der Ressource

ähnlicher Inhalt

	Erstellt von J000D97 vor fast 11 Jahre