3.1.3 The halogens

A halogen atom is one electron short of a full outer shell, so two halogen atoms share a pair of electrons to form X₂. That is why chlorine, bromine and iodine exist as diatomic molecules under ordinary conditions.

Down the group the molecules have more electrons and larger electron clouds. Their temporary dipoles can therefore become larger, so the London forces between molecules become stronger and more energy is needed to separate them. Boiling points rise from Cl₂ to Br₂ to I₂.

If you look at a halogen atom, its outer configuration is ns²np⁵: it needs one more electron to complete the p sub-shell. In many reactions it gains one electron and forms an X⁻ ion.

For chlorine, [Ne]3s²3p⁵ becomes Cl⁻ with an [Ar]-like electron structure. You can also read that electron gain as reduction.

A more reactive halogen oxidises the halide ion of a less reactive halogen. So chlorine will oxidise Br⁻ and I⁻, bromine will oxidise I⁻, and iodine will not displace chloride or bromide.

Use the colour of the halogen produced for the observation. For example, adding chlorine water to bromide ions forms bromine, giving an orange/brown colour. If it were my answer, I’d write the ionic equation as well as the observation if the question gives you room.

To explain the reactivity trend, ask how strongly a halogen atom can attract an incoming electron. Down the group, that electron would enter a shell further from the nucleus and with more shielding from inner electrons.

The attraction between the nucleus and the incoming electron is therefore weaker, despite the larger nuclear charge. So you should expect oxidising power and reactivity to decrease down Group 17.

In disproportionation, the same element is both oxidised and reduced in one reaction. With chlorine, start from oxidation state 0 in Cl₂ and track the chlorine atoms into two different products.

In water, chlorine forms HCl and HClO: one chlorine goes to −1 and the other to +1. With cold dilute NaOH, you get NaCl and NaClO, giving the same −1 and +1 split.

For an evaluation, give the chemistry on both sides. Chlorination kills harmful microorganisms, so it reduces water-borne disease and gives lasting protection as the water moves through the supply system.

The drawbacks include chlorine toxicity and the possible formation of chlorinated organic compounds. You then have to make a judgement: in normal water treatment, the health benefit from controlling pathogens is usually judged to outweigh the managed chemical risks.

When you test a halide, add aqueous silver nitrate to the solution. Chloride gives a white AgCl precipitate, bromide gives cream AgBr and iodide gives yellow AgI.

Then use aqueous ammonia to tell them apart: AgCl dissolves in dilute ammonia, AgBr needs concentrated ammonia, and AgI does not dissolve. I keep the ionic equation in the simple Ag⁺ + X⁻ form because those are the particles actually making the precipitate.

• Cl⁻ → white AgCl; dissolves in dilute NH₃
• Br⁻ → cream AgBr; dissolves in concentrated NH₃
• I⁻ → yellow AgI; insoluble in NH₃