Lesson Notes By Weeks and Term v4 - SHS 2

Lesson Video

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Performance objectives

• Describe reactions of halide salts with concentrated tetraoxosulphate (VI) acid.
• Explain differences in acid strength of hydrogen halides.
• Compare the reducing power of halide ions.
• Write balanced chemical equations for these reactions.
• Identify uses of halogens in Ghana.

Lesson summary

This lesson focuses on Group 17 of the Periodic Table, the Halogens. These elements, though often unseen, play a huge role in our daily lives here in Ghana. The salt we use for our jollof rice (sodium chloride), the fluoride in our Pepsodent or Colgate toothpaste that prevents tooth decay, the chlorine used by the Ghana Water Company Limited to make our water safe to drink, and the iodine added to salt to prevent goitre are all related to this fascinating group. By understanding their reactions and properties, we can better appreciate the chemistry that keeps us healthy and supports our industries.

Lesson notes

This lesson is divided into four main parts: the reducing power of halide ions, their reactions with a strong oxidizing acid, the acid strength of hydrogen halides, and the uses of the halogens. Part A: The Reducing Power of Halide Ions (X⁻) Definition: A reducing agent is a substance that donates electrons in a redox reaction and gets oxidized in the process. "Reducing power" refers to how easily it can donate these electrons. The Halide Ions: These are the ions formed when halogen atoms gain one electron (e.g., F⁻, Cl⁻, Br⁻, I⁻). The Trend: As we go down Group 17, the reducing power of the halide ions increases. > Trend in Reducing Power: I⁻ > Br⁻ > Cl⁻ > F⁻ > (Iodide is the strongest reducing agent; Fluoride is the weakest) Reasoning (Why this trend exists): Atomic/Ionic Radius: As you go down the group, each element has an additional electron shell. This means the ionic radius increases (I⁻ is much larger than F⁻). Electron Shielding: The increasing number of inner electron shells shields the outermost (valence) electrons from the positive pull of the nucleus. Nuclear Attraction: Due to the larger size and increased shielding, the nucleus has a weaker hold on the outermost electrons in larger ions like iodide (I⁻) compared to smaller ions like fluoride (F⁻). Conclusion: Because the outermost electron in an iodide ion is held less tightly, it is easier to lose or donate. Therefore, the iodide ion is the strongest reducing agent among the halides.

*Analogy:* Imagine you are trying to protect a football. If you hold it very close to your body (like the F⁻ ion), it's hard for someone to take it. If you hold it far out with an outstretched arm (like the I⁻ ion), it's much easier for someone to snatch it. The "snatching" is like oxidation (loss of the electron). Part B: Reaction of Solid Halide Salts with Concentrated Tetraoxosulphate(VI) Acid (H₂SO₄)

Concentrated H₂SO₄ is a powerful oxidizing agent. Its reaction with a solid halide salt (like NaCl, KBr, or NaI) depends on the reducing power of the halide ion. We will use sodium halides as our examples. Reaction with Sodium Chloride (NaCl) or Sodium Fluoride (NaF) Halide Ion: Cl⁻ (or F⁻), a weak reducing agent. Reaction Type: It is a simple acid-base (displacement) reaction. The Cl⁻ ion is not powerful enough to reduce the sulphur in H₂SO₄ (oxidation state +6). It simply displaces a proton. Equation: `NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g)` (Sodium chloride + Conc. Sulphuric acid → Sodium hydrogen sulphate + Hydrogen chloride gas) Observations: Steamy or misty white fumes of hydrogen chloride gas are seen. No change in the oxidation state of sulphur. It remains +6. Reaction with Sodium Bromide (NaBr) Halide Ion: Br⁻, a stronger reducing agent than Cl⁻. Reaction Type: This is a two-step process involving both displacement and redox. Step 1 (Displacement): First, hydrogen bromide gas is formed, just like with NaCl. `NaBr(s) + H₂SO₄(l) → NaHSO₄(s) + HBr(g)` Step 2 (Redox): The bromide ion (in the HBr formed) is strong enough to reduce the concentrated H₂SO₄. The sulphur's oxidation state is reduced from +6 to +4. `2HBr(g) + H₂SO₄(l) → Br₂(g) + SO₂(g) + 2H₂O(l)` Observations: Steamy/misty fumes of hydrogen bromide (HBr). Red-brown fumes of bromine vapour (Br₂). A colourless gas with a sharp, choking smell (like a struck match head) which is sulphur(IV) oxide (SO₂). Reaction with Sodium Iodide (NaI) Halide Ion: I⁻, the strongest reducing agent in this series. Reaction Type: This is also a two-step process, but the redox reaction is much more extensive. The I⁻ ion is so powerful it can reduce the sulphur (+6) to several different oxidation states. Step 1 (Displacement): `NaI(s) + H₂SO₄(l) → NaHSO₄(s) + HI(g)` Step 2 (Redox): The hydrogen iodide formed reduces the H₂SO₄ vigorously. Multiple reduction products of sulphur are formed. Reduction to SO₂ (Sulphur +4): `2HI(g) + H₂SO₄(l) → I₂(s/g) + SO₂(g) + 2H₂O(l)` Reduction to S (Sulphur 0): `6HI(g) + H₂SO₄(l) → 3I₂(s/g) + S(s) + 4H₂O(l)` Reduction to H₂S (Sulphur -2): `8HI(g) + H₂SO₄(l) → 4I₂(s/g) + H₂S(g) + 4H₂O(l)` Observations (A mixture of things): Steamy/misty fumes of hydrogen iodide (HI). Purple vapour or a black solid due to iodine (I₂). Choking smell of sulphur(IV) oxide (SO₂). Yellow solid deposits of elemental sulphur (S). A very unpleasant smell of rotten eggs due to hydrogen sulphide (H₂S). Part C: Acid Strength of Hydrogen Halides (HX) Definition: The strength of an acid is determined by how easily it donates a proton (H⁺) when dissolved in water. A strong acid dissociates (breaks apart) almost completely in water. The Trend: As we go down the group, the acid strength of the hydrogen halides increases. > Trend in Acid Strength: HI > HBr > HCl >> HF > (Hydroiodic acid is the strongest; Hydrofluoric acid is a weak acid) Reasoning (Why this trend exists): The key factor is the bond dissociation enthalpy (or bond strength) of the H-X bond. H-F Bond: The fluorine atom is very small and highly electronegative, making the H-F bond very short and extremely strong. It requires a lot of energy to break this bond and release the H⁺ ion in solution. Therefore, HF does not dissociate easily and is a weak acid. H-I Bond: The iodine atom is much larger. The H-I bond is much longer and significantly weaker. It requires very little energy to break this bond. Therefore, HI dissociates almost completely in water to release H⁺ ions, making it a strong acid. Conclusion: The weaker the H-X bond, the more easily the molecule dissociates to produce H⁺ ions, and the stronger the acid. Part D: Uses of Halogens and Their Compounds

| Halogen | Key Uses | | :--------- | :------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | Fluorine | 1. In toothpaste and drinking water (as fluoride ions) to prevent tooth decay. 2. Manufacturing of Teflon (polytetrafluoroethene) for non-stick cooking pans. | | Chlorine | 1. Disinfection of drinking water by Ghana Water Company Ltd to kill harmful bacteria. 2. Manufacturing of bleach (e.g., Parazone) and disinfectants. 3. Production of PVC (polyvinyl chloride) pipes used in plumbing. | | Bromine | 1. Used in making flame retardants for plastics and textiles. 2. In the past, used in photography (as silver bromide, AgBr). | | Iodine | 1. Antiseptic for cleaning wounds (as "tincture of iodine"). 2. Added to table salt (iodized salt) to prevent goitre, a swelling of the thyroid gland. |

Evaluation guide

• Describe the reaction of halide salts and explain the differences in acid strength of
• the hydrogen halides as well as the uses of the halogens.
• Talk for L earning Approach: Using think- pair-share approach,
• Explain the differences in reducing power of the halides.