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If you test different combinations of the halogens and their salts, you can work out a reactivity series for Group 7:

• the most reactive halogen displaces all of the other halogens from solutions of their salts, and is itself displaced by none of the others
• the least reactive halogen displaces none of the others, and is itself displaced by all of the others

It doesn’t matter whether you use sodium salts or potassium salts – it works the same for both types.

The slideshow shows what happens when chlorine, bromine and iodine are added to various halogen salts:

Halogen displacement reactions are redox reactions because the halogens gain electrons and the halide ions lose electrons.

When we consider one of the displacement reactions, we can see which element is being oxidised and which is being reduced.

bromine + potassium iodide → iodine + potassium bromide

Br₂ + 2KI → I₂ + 2KBr

As an ionic equation (ignoring the ‘spectator’ potassium ions):

Br₂ + 2I^– → I₂ + 2Br^–

We can see that the bromine has gained electrons, so it has been reduced. The iodide ions have lost electrons, so they have been oxidised.

Explaining reactivity [Higher tier only]

As we descend Group 7, the reactivity decreases. For stability, the atom needs to have a full outer shell. Group 7 elements need to gain 1 electron to have a full shell.

As a result, a negative ion is formed:

Cl + e^– → Cl^–

As we descend the group:

• there are more shells between the nucleus and the outer electron
• the force of attraction between the nucleus and outer electron decreases
• it becomes harder to gain the outer electron