3.2.3 Halogens Flashcards
Describe the use of acidified silver nitrate solution to identify and distinguish between halide ions
Add nitric acid to a solution to remove any ions that will react with react with silver nitrate. Add silver nitrate to a solution to produce a precipitate
Yellow = silver iodide
Cream = silver bromide
White = silver chloride
Describe the trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid
Increasing. Halide ions give away electrons (become halogen molecules). The larger the ion, the easier they lose an electrons because the outer shell is further from nucleus so attraction is less. Solid sodium halides react with concentrated sulphuric acid, forming sodium halide (solid)
Describe the trend in solubility of the silver halides in ammonia
Most soluble- chloride (dissolves in dilute ammonia)
Bromide dissolves in concentrated ammonia
Least soluble- iodide (doesn’t dissolve in ammonia
Fluorine ions give no precipitate- silver fluoride is soluble
Explain why ammonia solution is added
Distinguish between precipitates
Explain why the silver nitrate solution is acidified
To remove any other ions than may react with silver nitrate such as CO3^-
Describe the reaction of chlorine with water to produce chloride ions and chlorate (I) ions
Cl2 (g) + H2O (l) <=> HClO (aq) + HCl (aq)
Forms chloride ions and chlorate(I) ions (hydrochloric and chloric acid)
It’s a disproportionate redox reaction + reversible
Takes place to purify water for drinking and in swimming bath. Chloric acid is an oxidising agent that calls bacteria by oxidation
Describe the reaction of chlorine with water to form chloride ions and oxygen
2Cl2 (g) + 2H2O (l) —> 4HCl (aq) + O2 (g)
Occurs in sunlight
Chlorine lost rapidly from pool in sunlight so pools need frequent addition of chlorine
Explain the use of chlorine in water treatment
Reacts with water to produce chloric acid which purifies water for drinking and pools. It is an oxidising agent that kills bacteria by oxidisation. Also known as bleach
Describe the reaction of chlorine with cold, dilute, aqueous NaOH and uses of the solution formed
Cl2 (g) + 2NaOH (aq) —> NaClO (aq) + NaCl (aq) + H2O (l)
Forms sodium chlorate (I) which is an oxidising agent and the active ingredient of household bleach
Disproportionate reaction
Describe the trend in oxidising ability of the halogens down the group, including displacement reactions of halide ions in aqueous solution
Decrease. Oxidising agents need to attract electrons from other atoms. Down the group, the positive nucleus is further from the outer electrons due to the atomic radius increasing + outer shell electrons being further from nucleus. Halogens will react with metal halide in solution so the halide in the compound will be displaced by a more reactive halogen
Explain the trend in the boiling point of the group 7 elements in terms of their structure and bonding
Increase
Larger molecule
Van Der Waals forces stronger
Sodium chloride + concentrated sulphuric acid
NaCl (s) + H2SO4 (l) —>NaHSO4 (s) + HCl (g)
Steamy fumes produced
No oxidation- no state change
Acid-base reaction
Sodium bromide + concentrated sulphuric acid
NaBr (s) + H2SO4 (l) —> NaHSO4 (s) + HBr (g)
Steamy fumes of HBr
Acid-base reaction
2HBr + H2SO4 (l) —> Br2 (g) + SO2 (g) + 2H2O (l)
Brown fumes Br2
Colourless gas SO2
Redox reaction
Sodium iodide + concentrated sulphuric acid
NaI (s) + H2SO4 (l) → NaHSO4 (s) + HI (g)
Steamy fumes HI
Acid-base reaction
H2SO4 + 2I^- → SO4^2- + 2HI
2HI (g) + H2SO4 (l) → I2 (g) + SO2 (g)+ 2H2O (l)
Colourless gas SO2
I2- purple vapour
6HI (g)+ H2SO4 (l) → 3I2 (g) + S (s) + 4H2O (l)
Sulfur - yellow solid
8HI (g) + H2SO4 → 4I2 (g) + H2S (s) + 4H2O (l)
H2S - strong bad egg smell
Write an ionic equation for the reaction between chlorine and cold dilute sodium hydroxide solution
Cl2 +2HO– —> OCl– + Cl– +H2O
