AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
The trend in reducing ability of the halide ions, including the reactions…
Full specification wording
Trends in properties: The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid.
Reducing power runs the opposite way to oxidising power: I⁻ gives up an electron most readily because its outer electrons are furthest from the nucleus and most shielded, so reducing power…
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 4 of 7
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Reducing power runs the opposite way to oxidising power: I⁻ gives up an electron most readily because its outer electrons are furthest from the nucleus and most shielded, so reducing power increases down the group. Add concentrated sulfuric acid to the solid sodium halides and that difference shows. With NaCl there is no redox at all, only an acid and base reaction: NaCl + H₂SO₄ → NaHSO₄ + HCl, seen as misty white fumes of hydrogen chloride. With NaBr the HBr forms first and then reduces the acid: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, giving brown bromine vapour and choking SO₂, with sulfur falling from +6 to +4. With NaI the reduction goes furthest, through SO₂ to yellow sulfur and finally to H₂S, which smells of bad eggs, and sulfur falls from +6 to −2.
What earns the marks
- Reducing power of the halide ions increasing down the group.
- Explaining that by the larger ionic radius and greater shielding, so the outer electron is lost more easily.
- Identifying misty white fumes of HCl as the only product with sodium chloride.
- Naming SO₂ and Br₂ as the additional products with sodium bromide.
- Naming H₂S, sulfur and iodine as products with sodium iodide, with sulfur reduced from +6 to −2.
Common mistakes and how to put them right
Saying chloride ions reduce sulfuric acid to sulfur dioxide, when they are too weak a reducing agent.
Calling the misty fumes chlorine rather than hydrogen chloride.
Describing hydrogen sulfide as odourless rather than smelling of bad eggs.
Giving the change in sulfur's oxidation state for the iodide reaction as +6 to 0 and stopping there.
Writing that the sulfuric acid is oxidised rather than reduced.
Exam tips
- Work down the halides in order and note what is new each time; the iodide products include all of the bromide ones.
- Quote oxidation states for sulfur in every product, since they carry the redox marks.
Try the lesson
The start of the Trends in properties: The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid. lesson, made from checked teaching notes for this specification point.
Have a go first
Which halide ion is the strongest reducing agent?
Listen
Reducing power runs the opposite way to oxidising power: I⁻ gives up an electron most readily because its outer electrons are furthest from the nucleus and most shielded, so reducing power increases down the group. Add concentrated sulfuric acid to the solid sodium halides and that difference shows. With NaCl there is no redox at all, only an acid and base reaction: NaCl + H₂SO₄ → NaHSO₄ + HCl, seen as misty white fumes of hydrogen chloride. With NaBr the HBr forms first and then reduces the acid: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, giving brown bromine vapour and choking SO₂, with sulfur falling from +6 to +4. With NaI the reduction goes furthest, through SO₂ to yellow sulfur and finally to H₂S, which smells of bad eggs, and sulfur falls from +6 to −2. A strong answer earns marks for these points. reducing power of the halide ions increasing down the group. explaining that by the larger ionic radius and greater shielding, so the outer electron is lost more easily. identifying misty white fumes of HCl as the only product with sodium chloride. naming SO₂ and Br₂ as the additional products with sodium bromide. Exam tip: Work down the halides in order and note what is new each time; the iodide products include all of the bromide ones.
Key words
Say what each word means before you turn the card.
reducing ability of halide ions
Try answering aloud or in your head before turning the card.
concentrated sulfuric acid
Try answering aloud or in your head before turning the card.
oxidation states of sulfur
Try answering aloud or in your head before turning the card.
observations and smells
Try answering aloud or in your head before turning the card.
See every definition
- reducing ability of halide ions
- Increases down the group because the larger ionic radius and greater shielding mean the outer electron is lost more easily.
- concentrated sulfuric acid
- Added to the solid sodium halides, it shows the difference in reducing power, being reduced further by bromide and furthest by iodide.
- oxidation states of sulfur
- Fall from +6 to +4 with sodium bromide, and from +6 to −2 with sodium iodide.
- observations and smells
- Misty white fumes of hydrogen chloride with sodium chloride, brown bromine vapour and choking sulfur dioxide with sodium bromide, and hydrogen sulfide smelling of bad eggs with sodium iodide.
There is more to this lesson
The full lesson goes on to check your understanding and help you earn the marks. Create a free account to carry on, with your progress saved.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Test your understanding of Trends in properties: The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid.
Active recall beats re-reading. Take an instant quiz now to test what you’ve learned and lock it into long-term memory.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge