Halogens, halides and disproportionation

What you’ll learn

• Explain halogen physical and oxidising-power trends.
• Predict halogen displacement and disproportionation products.
• Interpret silver-halide and reducing-agent evidence.

Understand the idea

Down Group 7, larger and more polarisable molecules have stronger London forces, raising melting and boiling temperatures. At room conditions chlorine is a pale green gas, bromine a red-brown liquid and iodine a grey-black solid with violet vapour. Aqueous bromine is orange and aqueous iodine brown; iodine becomes violet in a suitable non-polar solvent.

Halogens accept electrons, so they act as oxidising agents. Their oxidising power decreases down the group because added shielding and distance weaken attraction for an incoming electron. Chlorine can oxidise bromide and iodide; bromine can oxidise iodide. The reducing ability of halide ions increases down the group. Concentrated sulfuric acid oxidises bromide to bromine and iodide to iodine more readily than chloride; these reactions can release hazardous gases and are considered here as classroom redox evidence, not a preparation method.

In water, chlorine disproportionates into chloride and chloric(I) acid, which is relevant to water treatment. In cold dilute alkali, chloride and chlorate(I) form; hot alkali favours chloride and chlorate(V). Assign oxidation numbers to identify the same starting chlorine being both reduced and oxidised.

A halide test uses silver ions after suitable acidification with nitric acid. Silver chloride is white and dissolves in dilute ammonia; silver bromide is cream, dissolving in concentrated but not dilute ammonia; silver iodide is yellow and remains insoluble in ammonia. Hydrochloric acid would introduce chloride and could create a false positive. Hydrogen halides dissolve in water to give acidic solutions and react with ammonia to form ammonium salts.

Key formulae

$$\mathrm{Cl_2(aq)+2Br^-(aq)\rightarrow2Cl^-(aq)+Br_2(aq)}$$

$$\mathrm{Cl_2(aq)+H_2O(l)\rightleftharpoons H^+(aq)+Cl^-(aq)+HClO(aq)}$$

$$\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$$

$$\mathrm{3Cl_2+6OH^-\rightarrow5Cl^-+ClO_3^-+3H_2O}$$

Worked example

Show why $\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$ is disproportionation.

1. Chlorine in $\mathrm{Cl_2}$ has oxidation number 0.
2. It becomes −1 in $\mathrm{Cl^-}$, which is reduction.
3. In $\mathrm{ClO^-}$, oxygen is −2 and the ion is −1, so chlorine is +1: oxidation.

Answer: The same initial chlorine is reduced to −1 and oxidised to +1.