- 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
- 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
Ag+(aq) + Br-(aq) ➝ AgBr(s) cream precipitate, soluble in concentrated NH3(aq)
Ag+(aq) + I-(aq) ➝ AgI(s) Yellow precipitate, insoluble in NH3(aq)
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 is a reaction in which the same element is both oxidised and
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:📝
- State and explain the trend in boiling points of the halogens fluorine to iodine.
- How could you distinguish between NaCl, NaBr and NaI by a simple test?
- Comment on the changes in oxidation number of chlorine in the following reaction:
Cl2(aq) + H2O(l) ➝ HClO(aq) + H+(aq) + Cl-(aq)
- The boiling points increase down the group as there are more electrons leading to higher van de Waals’ forces.
- 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.
- Cl in Cl2 = 0, HClO = +1 and Cl- = -1
This is the end of the topic!
Drafted by Cherry (Chemistry)