AS Chemistry: Group 17 — The Halogens — Revision Notes

Condensed for the final weeks. For the full explanation, use the Group 17: The Halogens study guide.

Trends down the group

Property	Trend	Reason
Melting/boiling point	Increases	More electrons → stronger van der Waals forces
Colour	Darkens: green (Cl₂) → red-brown (Br₂) → grey-black (I₂)
Oxidising power	DECREASES	Larger radius, more shielding → electron gained less easily
Reducing power of halide ion	INCREASES	Larger ion loses an electron more easily

The two power trends run in opposite directions — halogens as oxidising agents get weaker down the group, halide ions as reducing agents get stronger.

A second, easily-confused trend: X–X bond enthalpy also decreases down the group (Cl–Cl 242, Br–Br 193, I–I 151 kJ mol⁻¹), because the bonding atoms are larger, so the shared pair sits further from both nuclei in a longer, weaker bond. This happens to run in the same direction as the melting/boiling-point trend, but for an unrelated reason: melting/boiling point depends on van der Waals forces between separate molecules, while bond enthalpy is about the bond within one molecule. Keep the two explanations separate even though the numbers move the same way.

Reaction with hydrogen also tracks oxidising power down the group: H₂ + Cl₂ → 2HCl is explosive in sunlight; H₂ + Br₂ → 2HBr needs heating and a catalyst and is far less vigorous; H₂ + I₂ ⇌ 2HI is slow, needs heat and a catalyst, and does not even go to completion — it remains an equilibrium mixture. Thermal stability of the hydrogen halides follows the same pattern: HCl is stable to heat, HBr less so, and HI decomposes significantly on strong heating, because the H–X bond itself weakens down the group.

Displacement

A more reactive (stronger oxidising) halogen displaces a less reactive halide:

Cl2 + 2KBr  ->  2KCl + Br2      solution turns ORANGE
Cl2 + 2KI   ->  2KCl + I2       solution turns BROWN
Br2 + 2KI   ->  2KBr + I2       solution turns BROWN
I2 + KBr    ->  NO REACTION

Reactions with concentrated sulfuric acid — the classic distinguishing test

NaCl  + H2SO4  ->  HCl gas only            (no redox -- Cl- too weak)
NaBr  + H2SO4  ->  HBr, then SO2 + Br2     (red-brown fumes)
NaI   + H2SO4  ->  HI, then a mixture of SO2, S and H2S + I2
                                            (H2S rotten-egg smell,
                                            black/purple solid)

The sequence reflects increasing reducing power of the halide ion: chloride cannot reduce sulfuric acid at all, bromide reduces it to SO₂, iodide is the strongest reducer of the three and reduces it further still, to a mixture of sulfur dioxide, sulfur and hydrogen sulfide (not exclusively to H₂S).

Silver nitrate test

Acidify with dilute nitric acid first (removes carbonates), then add AgNO₃(aq):

Halide	Precipitate	In dilute NH₃	In concentrated NH₃
Cl⁻	White	Dissolves	Dissolves
Br⁻	Cream	Insoluble	Dissolves
I⁻	Yellow	Insoluble	Insoluble

The ammonia stage is what separates them definitively.

Disproportionation of chlorine

Chlorine is simultaneously oxidised and reduced.

COLD dilute NaOH:  Cl2 + 2NaOH -> NaCl + NaClO + H2O     (bleach)
                   Cl goes 0 -> -1 and 0 -> +1

HOT conc. NaOH:    3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O
                   Cl goes 0 -> -1 and 0 -> +5

With WATER:        Cl2 + H2O <=> HCl + HClO               (water treatment, equilibrium)

Disproportionation = the same element both oxidised and reduced in one reaction.

Exam traps

• Reversing oxidising power and halide reducing power trends.
• Forgetting to acidify with nitric acid before silver nitrate.
• Confusing cream with yellow — state colours precisely.
• Cold vs hot NaOH give different products.
• Writing halogens as single atoms — they are diatomic.

Self-test

1. Why does oxidising power decrease down Group 17?
2. What is observed when chlorine is added to potassium iodide solution?
3. Which products form when concentrated H₂SO₄ is added to sodium iodide?
4. How do you distinguish a bromide from an iodide using silver nitrate?
5. Define disproportionation and give the equation for chlorine with cold dilute NaOH.
6. Explain why the X–X bond enthalpy decreases down Group 17, and why this is a different explanation from the melting/boiling-point trend even though the two happen to move in the same direction.
7. Why does the reaction of hydrogen with iodine not go to completion, unlike its reaction with chlorine?

Answers: 1. The atomic radius increases and there is more shielding, so the incoming electron is attracted less strongly and is gained less readily. 2. The solution turns brown as iodine is displaced. 3. HI initially, then a mixture of SO₂, sulfur and H₂S (rotten-egg smell), plus iodine as a black/purple solid — iodide is a strong enough reducing agent to reduce sulfur down as far as −2, but the products are a mixture rather than H₂S alone. 4. Both give precipitates (cream and yellow); add concentrated ammonia — the bromide precipitate dissolves, the iodide does not. 5. The same element is simultaneously oxidised and reduced: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. 6. The bonding atoms get larger down the group, so the shared electron pair sits further from both nuclei in a longer, weaker bond; this is about the bond within one molecule, whereas melting/boiling point depends on van der Waals forces between separate molecules — the two trends happen to move together but for unrelated reasons. 7. Iodine is the weakest oxidising agent in the group, so the reaction is slow and incomplete, settling into an equilibrium mixture of H₂, I₂ and HI rather than going fully to product.