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11.0 The Halogens Text book p166 to 173 1. AQA AS Specification LessonsTopics 1 How and why does the atomic radius and electronegativity change in Gp.

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Presentation on theme: "11.0 The Halogens Text book p166 to 173 1. AQA AS Specification LessonsTopics 1 How and why does the atomic radius and electronegativity change in Gp."— Presentation transcript:

1 11.0 The Halogens Text book p166 to 173 1

2 AQA AS Specification LessonsTopics 1 How and why does the atomic radius and electronegativity change in Gp 7. What effect does this have on the boiling point? 2 To understand that the ability of the halogens (from fluorine to iodine) to oxidise decreases down the group (e.g. the displacement reactions with halide ions in aqueous solution) 3 understand the trend in reducing ability of the halide ions know the different products formed by reaction of NaX and H 2 SO 4 4 understand why acidified silver nitrate solution is used as a reagent to identify and distinguish between F - Cl -, Br - and I - know the trend in solubility of the silver halides in ammonia 5 know the reactions of chlorine with water and the use of chlorine in water treatment /appreciate that the benefits to health of water treatment by chlorine outweigh its toxic effects/know the reaction of chlorine with cold, dilute, aqueous NaOH and the uses of the solutions formed 2

3 Halogens 3 What are the Halogens, what does their name mean, and where are they found in the Periodic Table? The Halogens are elements that are found in Group 7 (VII) of the PT. The name means “salt former”. Name some compounds that contain a halogen What is the valency (OXIDATION NUMBER) of the halogens?

4 Fluorine, F 2, is a pale yellow gas at room temperature. Appearance fluorine video

5 Chlorine, Cl 2, is a pale green gas at room temperature. Chlorine video

6 Bromine, Br 2, is a dark red liquid at room temperature. It is the only liquid non-metal. Bromine video

7 Bromine is volatile and readily forms a dark red vapour.

8 Iodine, I 2, forms shiny black crystals at room temperature. iodine video

9 When warmed, iodine crystals sublime (turn directly to a gas), forming a purple vapour.

10 Fluorine 10 What is unusual about the bonding in fluorine molecule? Explain. Summarize the physical trends in Group VII, draw stick graphs for the data and explain the trends. Compared to the other halogens, the F-F bond is very weak. This is because the fluorine atoms are very small and there is a lot of repulsion between the bonding electrons. Compared to the other halogens, the F-F bond is very weak. This is because the fluorine atoms are very small and there is a lot of repulsion between the bonding electrons.

11 Going down the group, there are more filled energy levels between the nucleus and the outer electrons. This results in the outer electrons being shielded more from the attraction of the nucleus. Going down the group, there are more filled energy levels between the nucleus and the outer electrons. This results in the outer electrons being shielded more from the attraction of the nucleus. The atomic radius increases down Group 7. Atomic Radius

12 The strength of the instantaneous dipole−induced dipole forces between the molecules increases as the size of the molecules increases. The boiling point increases down Group 7. Boiling point

13 Fluorine is the most electronegative element in the periodic table. Electronegativity decreases down Group 7. Electronegativity The atomic radius increases, the outer electrons are more shielded, so bonding electrons are less strongly attracted to the nucleus. (Fig 2 p167)

14 11.2 Chemical reactions of the Halogens P 168 ( Lister) P138 (Atkinson) 14

15 Oxidation ability 15 What is Oxidation? Oxidation is the loss of electrons. What is an oxidizing agent? An oxidizing agent is an electron acceptor, the agent is reduced during the course of the reaction. This forms a redox reaction. An oxidizing agent is an electron acceptor, the agent is reduced during the course of the reaction. This forms a redox reaction.

16 Oxidising power trend: Cl 2 > Br 2 > I 2 When a halogen acts as an oxidising agent, it gains electrons (taken from the oxidised species). X 2 + 2 e - → 2 X - Going down the group it becomes harder to gain an electron because: atoms are larger & there is more shielding (due to extra electron shell) Going down the group it becomes harder to gain an electron because: atoms are larger & there is more shielding (due to extra electron shell) Cl Br I

17 Task: 17 Read p 168 (Lister) and complete the table React each Potassium Halide with each halogen water. Note down the colour change (if there is any) Write the ionic equation for the displacement. Check you are correct by reading page 168

18 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Br – (aq) I – (aq)

19 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Br – (aq) I – (aq)

20 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Br – (aq) I – (aq)

21 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Purple in solvent layer (no reaction) Br – (aq) I – (aq)

22 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Purple in solvent layer (no reaction) Br – (aq) Yellow/brown in aqueous layer (Br 2 forms) Cl 2 + 2 Br - → 2 Cl - + Br 2 I – (aq)

23 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Purple in solvent layer (no reaction) Br – (aq) Yellow/brown in aqueous layer (Br 2 forms) Cl 2 + 2 Br - → 2 Cl - + Br 2 Purple in solvent layer (no reaction) I – (aq)

24 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Purple in solvent layer (no reaction) Br – (aq) Yellow/brown in aqueous layer (Br 2 forms) Cl 2 + 2 Br - → 2 Cl - + Br 2 Purple in solvent layer (no reaction) I – (aq) Purple in solvent layer (I 2 forms) Cl 2 + 2 I - → 2 Cl - + I 2

25 Cl 2 (aq)Br 2 (aq)I 2 (aq) Cl – (aq) Stays yellow solution (no reaction) Purple in solvent layer (no reaction) Br – (aq) Yellow/brown in aqueous layer (Br 2 forms) Cl 2 + 2 Br - → 2 Cl - + Br 2 Purple in solvent layer (no reaction) I – (aq) Purple in solvent layer (I 2 forms) Cl 2 + 2 I - → 2 Cl - + I 2 Purple in solvent layer (I 2 forms) Br 2 + 2 I - → 2 Br - + I 2

26 HSW – Extraction of Bromine homework Create a presentation on the extraction and uses of Bromine and of Iodine. Use the material in the text book (page 169) and from the internet. Your presentation could be as a poster or as a powerpoint. Maximum of 6 slides. 26

27 11.3 Reactions of halide ions 27

28 Reducing agents 28 2 X – → X 2 + 2 e – When a halide ion reduces another substance, the halide is oxidised to a halogen. 2 X – → X 2 + 2 e – When a halide ion reduces another substance, the halide is oxidised to a halogen. What is reduction? Reduction is the gain of electrons What happens when a Halide is used as a reducing agent? Give the half equation for the reaction

29 Experiment 29 This experiment compares how well the halides reduce H 2 SO 4 to compare the reducing power of the halide ions. Some of the products are very toxic – hence the video clips! NaBr + Sulphuric acid NaCl + sulphuric acid NaI + sulphuric acid Watch the clips and complete the OBSERVATIONS column in the table

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32 Formation of hydrogen halides: NaX + H 2 SO 4 → NaHSO 4 + HX e.g.NaCl + H 2 SO 4 → NaHSO 4 + HCl Complete the final two columns of the table.

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34 Write half equations for: Cl – → Cl 2 Br – → Br 2 I – → I 2 H 2 SO 4 → SO 2 H 2 SO 4 → S H 2 SO 4 → H 2 S 2 Cl – → Cl 2 + 2 e – 2 Br – → Br 2 + 2 e – 2 I – → I 2 + 2 e – H 2 SO 4 + 2 H + + 2 e – → SO 2 + 2 H 2 O H 2 SO 4 + 6 H + + 6 e – → S + 4 H 2 O H 2 SO 4 + 8 H + + 8 e – → H 2 S + 4 H 2 O

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39 Cl – does not reduce H 2 SO 4 Br – reduces H 2 SO 4 from S(+6) to S(+4) I – reduces H 2 SO 4 from S(+6) to S(-2) Reducing power trend

40 Reducing power trend: Cl – < Br – < I – When a halide ion acts as a reducing agent, it loses electrons (given to the reduced species). 2 X – → X 2 + 2 e – Down the group it becomes easier to lose an electron because: ions are larger & there is more shielding (due to extra electron shell) Cl – Br – I–I–

41 Identification of metal halides 41 Complete practical ES4.4, or N-ch2-06


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