General Properties

• Each halogen has seven outer shell electrons. The outer p sub-shell therefore has five electrons.
• The halogens exist as diatomic molecules (e.g. Cl2). 
• The boiling points of the halogens increase on descending the group, as the number of electrons increases, leading to an increase in van de Waals’ forces between molecules.

• Electronegativity is a measure of the attraction of an atom for the pair of electrons within a covalent bond.
• The hydrogen halides have polar molecules (Hδ+ - Xδ-). The polarity of the molecules decreases as the X descends down the group. 
• This trend in polarity results from the decreasing electronegativity of the halogens. 
• This itself happens because:
  • The atomic radius increases, resulting in less nuclear attraction (despite the increased nuclear charge).
  • There are more electron shells between the outer electron and the nucleus, meaning shielding is increased

The relative reactivity of group 7 as oxidising agents

• The halogens are the most reactive non-metals and are strong oxidising agents.
• They become less reactive and therefore less oxidising down the group.
• The decrease in reactivity can be shown by displacement reactions of aqueous
• halides.

• Cl2(aq) + 2Br-(aq) ➝ 2Cl(aq) + Br2(aq)
• Cl2(aq) + 2I-(aq) ➝ 2Cl(aq) + I2(aq)

Bromine only oxidises Iodine:

• Br2(aq) + 2I-(aq) ➝ 2Br(aq) + I2(aq)

Testing for halide ions

Addition of aqueous silver ions using AgNO3(aq) to a solution of halide ions in dilute
nitric acid produces coloured precipitates that have different solubilities in
aqueous ammonia.

Reactions of halides with concentrated Sulfuric acid

Concentrated sulfuric acid is an oxidising agent. Halide salts react with it to
produce a range of products, depending on the halogen.

This HCL formed isn't strong enough to oxidis it further:

• NaCl(s) + H2SO4(aq) ➝ NaHSO4(s) +HCl(g) 

Some of this HBr Reduces the sulfuric acid:

• NaBr(s) + H2SO4(aq) ➝ NaHSO4(s) +HBr(g) 

• 2HBr(g) + H2SO4(aq) ➝ SO2(s) + Br2(g) + 2H2O(l)
• NaI(s) + H2SO4(aq) ➝ NaHSO4(s) +HI(g) 

The HI reduces the sulfuric acid in two ways:

• 2HI(g) + H2SO4(aq) ➝ SO2(s) + I2(g) + 2H2O(l)
• 2HI(g) + H2SO4(aq) ➝ H2S(g) + 3I2(g) + 2H2O(l)

Disproportionation

Disproportionation is a reaction in which the same element is both oxidised and
reduced.
Cl in water:

• Cl2(aq) + H2O(l) ➝ HClO(aq) + HCl(aq) 
• The Cl is reduced to -1 in HCl but oxidised to +1 in HClO

Cl in sodium hydroxide:

• Cl2(aq) + 2NaOH(aq) ➝ NaCl(aq) + NaClO(aq) + H2O(l) 
• The Cl is reduced to -1 in NaCl but oxidised to +1 in NaCLO

Questions that often appear is AS/A-level Chemistry exams:📝

1. State and explain the trend in boiling points of the halogens fluorine to iodine.
2. How could you distinguish between NaCl, NaBr and NaI by a simple test?
3. Comment on the changes in oxidation number of chlorine in the following reaction:

Cl2(aq) + H2O(l) ➝ HClO(aq) + H+(aq) + Cl-(aq)

Answers:

1. The boiling points increase down the group as there are more electrons leading to higher van de Waals’ forces.
2. Add silver ions (from AgNO3) to the solution to see either a white, cream or yellow precipitate. Further clarification can be found by noting its solubility in dilute and conc. ammonia.
3. Cl in Cl2 = 0, HClO = +1 and Cl- = -1

Reference:

https://docbrown.info/page03/The_Halogens.htm

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