3.2.3 Group 7, the halogens

• Group 7: Boiling Points

  > As you go down group 7 the boiling point increases

  > The size of the molecules increase as there are more electrons and shielding .

  > Therefore there is a larger contact area between the molecules.

  > Therefore there are more/stronger Van der Waals forces between the molecules.

  > More energy is required to overcome the Van der Waal forces between the molecules and so the B.p.t increase.
• Colour and State of Halogens:

  F2 ( g ) Yellow vapour

  Cl2 ( g ) Green Vapour

  Br2 ( l ) Orange Vapour / Red-brown liquid

  I2 ( s ) Purple Vapour / Black Solid / brown liquid

  At2 ( s ) Dark grey Vapour

  As you go down group 7 the vapour becomes darker , melting and boiling point increases and reactivity decreases
• Group 7 : Electronegativity

  As you go down group 7, the electronegativity decreases.

  > The halogen gets bigger in size (atomic radius increases).

  > The distance between the nucleus and the pair of electrons in the covalent bond increases.

  > The force of attraction between the nucleus and the shared pair of electrons decreases and gets weaker.

  > Therefore electronegativity decrease
• Oxidising Agent

  Are electrons acceptors ( itself is reduced )
• Group 7: Oxidising Power of Halogens

  As you go down group 7 the oxidising power (the ability to act as oxidising agent ) of the halogens decreases. Meaning they gain electrons less easily.

  Explaination:

  > The halogen atoms get bigger in size as ther is more shielding.

  > The distance between the nucleus and the electron to be gained gets bigger.

  > Therefore the oxidising power decreases.

  > The attraction gets weaker and the electron is gained less easily.
• Order of Strongest Oxidising Agent:

  Fluorine - Strongest

  Chlorine

  Bromine

  Iodide - Weakest
• Order of Strongest Reducing Agent:

  Iodide - Strongest

  Bromide

  Chloride

  Fluoride - Weakest
• Group 7: Reducing Power of Halide ions

  As you go down group 7 the reducing power (the ability to act as reducing agent ) of the halide ions increases. Meaning they lose electrons more easily.

  Explanation:

  > The halide ions get bigger in size as there is more shielding.

  > The distance between the nucleus and the electron to be lost gets bigger.

  > Therefore the reducing power increases.

  > The attraction gets weaker and the electron is lost more easily.
• Halide Reactions with Sulfuric Acid:

  Fluoride:

  Fluoride is a very weak reducing agent and doesn't donate electrons easily. The reaction with sodium fluoride and H2SO4 is an acid - base reaction.

  NaF + H2SO4 -> NaHSO4 + HF

  Steamy fumes of HF are produced

  Chloride:

  Chloride is a very weak reducing agent and doesn't donate electrons easily. The reaction with sodium chloride and H2SO4 is an acid - base reaction.

  NaCl + H2SO4 -> NaHSO4 + HCl

  Steamy fumes of HCl are produced blue litmus paper will turn red if HCl is present.
• Halide Reactions with Sulfuric Acid:

  Bromide:

  A bromide ion is a good reducing agent and gets easily oxidised

  Equation 1: ( acid - base reaction )

  NaBr + H2SO4 -> NaHSO4 + HBr

  Steamy fumes of HBr are produced.

  Equation 2: Redox reaction (dominant reaction)

  Half Equations

  (2Br- ) -> Br2 + 2e-

  ( 2H+ ) + H2SO4 + ( 2e- ) -> SO2 + 2H2O

  Full Equation

  ( 2H+ ) + ( 2Br-) + H2SO4 -> Br2 + SO2 + 2H2O

  Orange fumes of Br2 and colourless sulfur dioxide
• Halide Reactions with Sulfuric Acid: Pt1

  Iodide:

  A Iodide ion is a very good reducing agent and gets easily oxidised

  Equation 1: ( acid - base reaction )

  NaI + H2SO4 -> NaHSO4 + HI

  Steamy fumes of HI are produced.

  Equation 2: Redox reaction

  Sulfurs Ox state changes from +6 to +4

  Half Equations

  (2I- ) -> I2 + 2e-

  ( 2H+ ) + H2SO4 + ( 2e- ) -> SO2 + 2H2O

  Full Equation

  ( 2H+ ) + ( 2I-) + H2SO4 -> I2 + SO2 + 2H2O

  Purple fumes of I2, black solid from I2 and colourless sulfur dioxide
• Halide Reactions with Sulfuric Acid: Pt 2

  Iodide:

  Equation 3: Redox reaction

  Sulfurs Ox state changes from +6 to 0

  Half Equations

  (2I- ) -> I2 + 2e-

  ( 6H+ ) + H2SO4 + ( 6e- ) -> S + 4H2O

  Full Equation

  ( 6H+ ) + ( 6I-) + H2SO4 -> 3I2 + S + 4H2O

  Purple fumes of I2, black solid from I2 and yellow solid of sulfur

  Equation 4: Redox reaction

  Sulfurs Ox state changes from +6 to -2

  Half Equations

  (2I- ) -> I2 + 2e-

  ( 8H+ ) + H2SO4 + ( 8e- ) -> H2S + 4H2O

  Full Equation

  ( 8H+ ) + ( 8I-) + H2SO4 -> 4I2 + H2S + 4H2O

  Purple fumes of I2, black solid from I2 and bad smell from H2S a toxic gas.
• Disproportionation reaction is where the same species is both oxidised and reduced
• Chlorine and Water:

  Chlorine and Water react in the presence of sunlight

  2Cl2 + 2H2O ⇌ 4Cl- + O2 + 4H+
• Disproportionation reaction:

  When Chlorine reacts with water to produce hydrochloric acid and chloric acid

  Cl2 + H2O ⇌ HCl+ HClO
• Chloric acid is a weak acid and ionises (breaks up) in water

  HClO ⇌ ClO- + H+
• The Chlorate ion is a powerful oxidising agent

  and is a disinfectant/ sterilise water. So chlorine is added to water at water treatment works.

  Cl2 + H2O ⇌ HCl + HClO

  HClO ⇌ ClO- + H+

  The benefits of added chlorine outweigh the risks so it is used
• Test for Halide ions:

  Reagents: Add nitric acid - source of H+ ions

  then silver nitrate -> source of Ag+ ions

  Observations:

  F- Remains colourless as AgF is aqueous

  Cl- White precipitate as AgCl is solid

  Br- Cream precipitate as AgBr is solid

  I- Yellow precipitate as AgI is solid.

  we acidify the solution to remove any ions which could produce a ppt and false result.
• To help identify exact coloured precipitate as white, cream and yellow are similar you can use ammonia solution.

  Dilute Ammonia Solution:

  AgCl : the precipitate dissolves

  AgBr : no change

  AgI : no change

  Conc Ammonia solution:

  AgCl : the precipitate dissolves

  AgBr : the precipitate dissolves

  AgI : no change
• Reducing Agents

  Are electrons donators ( itself is oxidised )