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Halogens exist as diatomic molecules and form halide ions by gaining electrons. Oxidising power decreases down the group because increased shielding and distance weaken attraction for an incoming electron. A more reactive halogen oxidises a less reactive halide: chlorine can oxidise bromide to bromine. Boiling points rise down the group because the larger molecules have more electrons and stronger London attractions. That is a different explanation from the electron-gain explanation for chemical reactivity.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Explain halogen reactivity and boiling-point trends
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: Which halogen is the strongest oxidising agent of these?
Chlorine. Oxidising ability decreases down this part of Group 17.
After watching 2: Which precipitate is silver iodide?
Yellow. It also remains insoluble in ammonia.
What you'll be able to do
Halogens exist as diatomic molecules and form halide ions by gaining electrons. Oxidising power decreases down the group because increased shielding and distance weaken attraction for an incoming electron. A more reactive halogen oxidises a less reactive halide: chlorine can oxidise bromide to bromine. Boiling points rise down the group because the larger molecules have more electrons and stronger London attractions. That is a different explanation from the electron-gain explanation for chemical reactivity.
Acidify a sample with dilute nitric acid before adding silver nitrate. Chloride, bromide and iodide form white, cream and yellow silver-halide precipitates respectively. Silver chloride dissolves in dilute ammonia; silver bromide needs concentrated ammonia; silver iodide does not dissolve in either. Nitric acid helps remove interfering carbonate without adding halide ions. A colour observation alone may be ambiguous in a coloured sample, so combine it with the ammonia result and use fresh portions for different tests.
Chlorine reacts with cold dilute sodium hydroxide to form chloride and chlorate(I), alongside water. Chlorine is both reduced and oxidised, so this is disproportionation. Hot concentrated alkali instead gives chlorate(V). Chlorine in water establishes an equilibrium producing hydrochloric acid and chloric(I) acid, explaining disinfectant action in an appropriate context. When concentrated sulfuric acid reacts with halide salts, bromide and iodide can also act as reducing agents; distinguish the initial acid–base step from subsequent redox chemistry.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: Balance chlorine reacting with cold dilute NaOH.
Cl2 + 2NaOH → NaCl + NaClO + H2O.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.