The Elements of G7 + Analysis

• Trends down group 7 (atomic radius)

  atomic radius increases down the group as nuclear charge increase is cancelled out by an increase in shielding. The atoms down the group have more shells of electrons however, and this thus increases the atomic radius.
• Trends down group 7 (boiling point)

  boiling point increases down the group. This is because:

  • atomic radius increases down the group
  • London dispersion forces increase in strength since halogens down the group have a higher number of electrons.
  • the larger the molecule down the group, the more surface area of contact between molecules (increases strength of dispersion forces)
  • Thus more energy is required to break the stronger dispersion forces.
• Trends down group 7 (electronegativity - the ability for a nucleus to attract the bonding pair of electrons)

  electronegativity decreases down group 7. This is because:

  • atomic radius increases down the group, which means that the valence electrons are held further from the nucleus.
  • shielding also increases down the group.
  • therefore, the nucleus is less able to attract bonding pairs of electrons.
• Physical properties of halogens

  Fluorine(F2) - yellow gas

  Chlorine (Cl2) - green gas and pale yellow solution colour.

  Bromine (Br2) - red/brown liquid and orange solution colour

  Iodine (I2) - dark grey solid and brown solution colour (purple when dissolved in a liquid hydrocarbon as it remains diatomic)

  Astatine (At2) black solid.
• Oxidising ability of halogens - their ability to oxidise other substances and undergo reduction themselves, thus acting as electron acceptors.
• oxidising ability of halogens decreases down the group as they are less able to act as electron acceptors. This is because:

  • both their atomic radius and shielding increase down the group.
  • the nuclear charge is thus less attracted to electrons being gained as its in a higher energy shell.
• a stronger oxidising agent will displace a halogen with a lower oxidising power in its compound.

  e.g. When Cl2 (aq) reacts with KBr (aq), chlorine displaces bromine from its compound, leaving an orange solution of bromine.
• displacement reactions are also redox reactions.
• You form an ionic equation by splitting any aqueous ionic compounds into their individual ions, and cancelling out the ions that appear on both sides.
• The reaction between a halogen and a group 1 metal is a redox reaction.

  e.g. 2Na (s) + X2 (g) = 2NaX (s)
• The reaction between a halogen and a group 2 metal is a redox reaction.

  e.g. Mg (s) + X2 (g) = MgX2 (s)
• When chlorine reacts with water, chloride and chlorate (I) ions are produced in a disproportionation reaction (absence of sunlight)

  e.g. Cl2 (aq) + H2O (l) = HCLO (aq) + HCL (aq)

  chlorine is reduced to HCL and oxidised to HCLO
• When chlorine reacts with water in sunlight, chloride ions and oxygen are made.

  e.g. 2Cl2 + 2H2O = O2 + 4HCl
• chlorine is used in water treatment to kill bacteria. Although it is toxic, the benefits of clean water outweigh its toxic effects.
• when chlorine reacts with cold dilute NaOH, it produces NaCl, NaClO, and water in a disproportionation reaction.

  e.g. Cl2 (aq) + 2NaOH (aq) = NaCl (aq) + NaClO (aq) + H2O (l)
• when chlorine reacts with hot concentrated NaOH, it produces NaClO3, NaCl, and water in a different disproportionation reaction.

  e.g. 3 Cl2 (aq) + 6 NaOH = NaClO3 (aq) + 5 NaCl (aq) + 3 H2O (l)
• when in water, hydrogen halides produce an acidic solution, because they are H+ donors.

  e.g. HX (aq) + H2O (l) = X- (aq) + H3O+ (aq)
• Down group 7, the acid strength of hydrogen halides increases, as the H-X bond becomes larger and weaker. This means they are more likely to dissociate producing H+.
• gaseous hydrogen halides react with gaseous ammonia in an acid-base reaction, making an ammonium halide solid.

  e.g. HX (g) + NH3 (g) = NH4X (s)
• test for halides - add a few drops of dilute nitric acid to the sample (removes carbonate impurities that would give a false positive white precipitate). Then add a few drops of silver nitrate solution, and the halide will produce a characteristic insoluble silver halide salt.

  e.g. X- (aq) + Ag+ (aq) = AgX (s)

  
  

  *you also need NaOH????
• Test for halides results.

  Cl- = white precipitate of AgCl (precipitate dissolves in NH3)

  Br- = cream precipitate of AgBr (precipitate dissolves in concentrated NH3 but not dilute NH3)

  I- = yellow precipitate of AgI (precipitate is insoluble in NH3)
• reducing ability of halides - their ability to reduce other substances, and undergo oxidation themselves, thus losing electrons.
• the reducing ability of halides increases down the group as they are more able to act as electron donors. This is because:

  • their ionic radius increases down the group.
  • thus their outer electrons are more easily removed as they're less attracted to the nuclear charge.
• Reactions of halides with concentrated sulphuric acid (H2SO4):

  F- = misty fumes (HF (g))

  Cl- = misty fumes (HCl (g))
• Reactions of halides with concentrated sulphuric acid (H2SO4):

  Br-:

  • misty fumes (HBr (g))
  • orange gas (Br2(g))
  • colourless acidic gas (SO2(g)) - masked by the Br2
• Reactions of halides with concentrated sulphuric acid (H2SO4):

  I-:

  • misty fumes (HI (g))
  • colourless acidic gas (SO2 (g))
  • dark grey solid/purple vapour (I2 (s/g))
  • yellow solid (S (s))
  • rotten egg smelling gas (H2S (g))
• F-/Cl- when reacted with concentrated H2SO4:

  Neither ions are strong enough reducing agents to reduce the S in H2SO4. Thus, only an acid base reaction can occur where misty fumes of the hydrogen halide are observed.

  e.g. NaX (s) + H2SO4 (l) = NaHSO4 (s) + HX(g)

  
  

  this isn't a redox reaction as the oxidation numbers of both S and X stay the same
• Br- with concentrated H2SO4:

  Br- ions are stronger reducing agents that Cl-/F-. An initial acid base reaction takes place:

  NaBr (s) + H2SO4 (l) = NaHSO4 (s) + HBr (g)

  a redox reaction where Br- reduced sulphur from +6 (H2SO4) to +4 (SO2) also occurs:

  2Br- = Br2 + 2e-

  2H+ + H2SO4 + 2e- = SO2 + 2H20

  
  

  = 2H+ + 2Br- + H2SO4 = Br2 + SO2 + 2H2O
• equation for overall reaction between NaBr ions and H2SO4

  
  

  2 NaBr (s) + 2 H2SO4 (l) = Na2SO4 (s) + Br2 (g) + SO2 (g)+ 2 H20 (l)
• I- with concentrated H2SO4:

  I -ions are stronger reducing agents that Cl-/F-/Br-. An initial acid base reaction takes place:

  NaI (s) + H2SO4 (l) = NaHSO4 (s) + HI (g)

  a number of redox reactions also occur.
• I- with concentrated H2SO4 (redox reaction 1) = I- reducing sulphur from +6 (H2SO4) to +4 (SO2)

  
  

  2I- = I2 + 2e-

  H2SO4 + 2H+ + 2e- = SO2 + 2H2O

  
  

  = 2H+ + 2I- + H2SO4 = I2 + SO2 + 2H2O
• I- with concentrated H2SO4 (redox reaction 2) = I- reducing sulphur from +6 (H2SO4) to 0 (S)

  
  

  2I- = I2 + 2e-

  H2SO4 + 6H+ + 6e- = S + 4H2O

  
  

  = 6 H+ + 6 I- + H2SO4 = 3 I2 + S + 4 H2O
• I- with concentrated H2SO4 (redox reaction 3) = I- reducing sulphur from +6 (H2SO4) to -2 (H2S)

  
  

  2I- = I2 + 2e-

  8H+ + H2SO4 + 8e- = H2S + 4H2O

  
  

  = 8 H+ + 8 I- + H2SO4 = 4 I2 + H2S + 4H2O
• **equation for overall reaction between NaI and concentrated H2SO4:

  
  

  8 NaI (s) + 5 H2SO4 (l) = 4 Na2SO4 (s)+ 4 I2 (s/g) + H2S (g) + 4 H2O (l)
• test for ammonium ions:

  • gently heat a sample containing ammonium ions in a test tube with NaOH, and ammonia gas will be produced:
  • NH4+ (aq) + OH- (aq) = NH3 (g) + H2O (l)
  • hold damp red litmus paper over the mouth of the test tube, and due to the formation of OH- ions, the red litmus paper should turn blue:
  • NH3 (g) + H2O (l) = NH4+ (aq) + OH- (aq)
• test for sulphate ions:

  • add dilute HCL to a solution of the sample being tested, to remove carbonate ions that would make the precipitate BaCO3 (false positive).
  • add BaCl solution, and if sulphate ions are present, a precipitate of barium sulphate forms:

  Ba 2+ (aq) + SO4 2- (aq) = BaSO4 (s)
• carbonate ions react with acids to produce carbon dioxide gas.

  CO3 2- (aq) + 2H+ (aq) = CO2 (g) + H2O (l)
• hydrogen carbonate ions react with acids to produce carbon dioxide gas.

  HCO3- (aq) + H+ (aq) = CO2 (g) + H2O (l)
• test for carbonate ions:

  • react your sample with HCL to produce carbon dioxide.
  • the carbon dioxide produced is tested by bubbling it through limewater (Ca(OH)2) solution, which goes from colourless to cloudy.

  CO2 (g) + Ca(OH)2 (aq) = CaCO3 (s) + H2O
• you can test for certain metal ions using precipitation with $NaOH$

  • $Fe^{2+}$ makes a green precipitate in the equation: $Fe^{2+}(aq)+$$2\ OH^-(aq)\ \rightarrow\ Fe\left(OH\right)_2(s)$
  • $Fe^{3+}$ makes a brown precipitate in the equation: $Fe^{3+}(aq)+$$3\ OH^-(aq)\ \rightarrow\ Fe\left(OH\right)_3\ (s)$
  • Copper (II) makes a blue precipitate: $Cu^{2+}(aq)+$$2\ OH^-(aq)\ \rightarrow\ Cu\left(OH\right)_2\ (s)$