[3.2.3] Group 7, the Halogens Flashcards
Trend in Properties and Uses of Chlorine & Chlorate (I)
Describe and explain the trend in melting and boiling points in group 7.
- Increase down the group.
- As the molecules become larger, they have more electrons and so have larger van der Waals forces between the molecules.
- As the intermolecular forces get larger, more energy has to be put into break the forces.
- This increases the melting and boiling points.
Describe and explain the trend in electronegativity in group 7.
- As one goes down the group the electronegativity of the elements decreases.
- As one goes down the group, the atomic radii increases due to the increasing number of shells.
- The nucleus is therefore less able to attract the bonding pair of electrons.
Describe the trend in the oxidising ability of the halogens.
- The oxidising strength decreases down the group.
- (Oxidising agents are electron acceptors).
Fill in the gaps in the table to show whether a displacement reaction takes place between each of the halogens and what you would observe.
- A strong oxidising agent will displace a halogen with a lower oxidising power from one of its compounds.
- Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions.
Write the oxidation and reduction half-equations for each of the displacement reactions between halide ions in group 7.
With each oxidation and reduction half-equation, form a combined half-equation.
DISPLACEMENT REACTION BETWEEN CHLORINE AND BROMINE
- Chlorine is a stronger oxidising agent than bromine ∴ displaces bromine.
- OXIDATION HALF-EQUATION
- 2Br⁻ (aq) -> Br₂ (aq) + 2e⁻
- REDUCTION HALF-EQUATION
- Cl₂ (aq) + 2e⁻ -> 2Cl⁻ (aq)
- COMBINED HALF-EQUATION
- Cl₂ (aq) + 2Br⁻ (aq) -> 2Cl⁻ (aq) + Br₂ (aq)
DISPLACEMENT REACTION BETWEEN CHLORINE AND IODINE
- Chlorine is a stronger oxidising agent than iodine ∴ displaces iodine.
- OXIDATION HALF-EQUATION
- 2I⁻ (aq) -> I₂ (aq) + 2e⁻
- REDUCTION HALF-EQUATION
- Cl₂ (aq) + 2e⁻ -> 2Cl⁻ (aq)
- COMBINED HALF-EQUATION
- Cl₂ (aq) + 2I⁻ (aq) -> 2Cl⁻ (aq) + I₂ (aq)
DISPLACEMENT REACTION BETWEEN BROMINE AND IODINE
- Bromine is a stronger oxidising agent than iodine ∴ displaces iodine.
- OXIDATION HALF-EQUATION
- 2I⁻ (aq) -> I₂ (aq) + 2e⁻
- REDUCTION HALF-EQUATION
- Br₂ (aq) + 2e⁻ -> 2Br⁻ (aq)
- COMBINED HALF-EQUATION
- Br₂ (aq) + 2I⁻ (aq) -> 2Br⁻ (aq) + I₂ (aq)
How does the use of acidified silver nitrate solution identify and distinguish between halide ions?
- Fluorides produce no precipitate.
- Chlorides produce a white precipitate.
- Ag⁺ (aq) + Cl⁻ (aq) -> AgCl (s)
- Bromides produce a cream precipitate.
- Ag⁺ (aq) + Br⁻ (aq) -> AgBr (s)
- Chlorides produce a pale yellow precipitate.
- Ag⁺ (aq) + l⁻ (aq) -> Agl (s)
Explain why silver nitrate has to be acidified when identifying and distinguishing between halide ions.
- It is acidified with nitric acid which reacts with any carbonates present to prevent the formation of the precipitate Ag₂CO₃.
- This precipitate would mask the desired observation.
What’s the reason for adding ammonia to silver halides?
Describe the trend in solubility of the silver halides in ammonia.
- The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar.
- Silver chloride dissolves in dilute ammonia to form a complex ion.
- Silver bromide dissolves in concentration ammonia to form a complex ion.
- Silver iodide does not react with ammonia - it is too insoluble.
- ∴ solubility of the silver halides in ammonia decreases down group 7.
Describe and explain the trend in the reducing ability of the halide ions.
- The halides show increasing power as reducing agents as one goes down the group.
- (A reducing agent donates electrons).
- The reducing power of the halides increases down group 7 because they have a greater tendency to donate electrons.
- This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller.
Write equations to show the reactions of solid sodium fluoride and chloride with concentrated sulfuric acid.
What would you observe in these reactions?
- F⁻ & Cl⁻ ions are not strong enough reducing agents to reduce the S in H₂SO₄ so no redox reactions occur. Only acid-base reactions occur.
-
NaF (s) + H₂SO₄ (I) -> NaHSO₄ (s)+ HF (g)
- OBSERVATIONS: White steamy fumes of HF are evolved.
-
NaCl (s) + H₂SO₄ (I) -> NaHSO₄ (s)+ HCl (g)
- OBSERVATIONS: White steamy fumes of HCl are evolved.
Write equations to show the reactions of solid sodium bromide with concentrated sulfuric acid.
What would you observe in these reactions?
- Br⁻ ions are stronger reducing agents than Cl⁻ and F⁻ and after the initial acid-base reaction, the bromide ions reduce the sulfur in H₂SO₄ from +6 to +4 in SO₂.
- ACID BASE STEP
- NaBr (s) + H₂SO₄ (I) -> NaHSO₄ (s) + HBr (g)
- REDOX STEP
- Ox 1/2 equation: 2Br⁻ (aq) -> Br₂ (aq) + 2e⁻
- Re 1/2 equation: H₂SO₄ + 2H⁺ + 2e⁻ -> SO₂ + 2H₂O
- Combined 1/2 equation: 2Br⁻ + H₂SO₄ + 2H⁺ -> Br₂ (g) + SO₂ (g) + 2H₂O (I)
(Note that H₂SO₄ plays the role of an acid in the first step producing HBr and then acts as an oxidising agent in the redox step)
OBSERVATIONS
- White steamy fumes of HBr are evolved.
- Orange fumes of bromine are also evolved and a colourless, acidic gas SO₂.
Write equations to show the reactions of solid sodium iodide with concentrated sulfuric acid.
What would you observe in these reactions?
- I⁻ ions are the strongest halide reducing agents. They can reduce the sulfur from +6 in H₂SO₄, to +4 in SO₂, to 0 in S and -2 in H₂S.
- ACID BASE STEP
- NaI (s) + H₂SO₄ (I) -> NaHSO₄ (s) + HI (g)
-
FIRST REDOX STEP
- Ox 1/2 equation: 2I⁻ (aq) -> I₂ (aq) + 2e⁻
- Re 1/2 equation: H₂SO₄ + 2H⁺ + 2e⁻ -> SO₂ + 2H₂O
- Combined 1/2 equation: 2I⁻ + H₂SO₄ + 2H⁺ -> I₂ (g) + SO₂ (g) + 2H₂O (I)
-
SECOND REDOX STEP
- Ox 1/2 equation: 2I⁻ (aq) -> I₂ (aq) + 2e⁻
- Re 1/2 equation: H₂SO₄ + 6H⁺ + 6e⁻ -> S + 4H₂O
- Combined 1/2 equation: 6I⁻ + H₂SO₄ + 6H⁺ -> 3I₂ (g) + S (g) + 4H₂O (I)
-
THIRD REDOX STEP
- Ox 1/2 equation: 2I⁻ (aq) -> I₂ (aq) + 2e⁻
- Re 1/2 equation: H₂SO₄ + 8H⁺ + 8e⁻ -> H₂S + 4H₂O
- Combined 1/2 equation: 8I⁻ + H₂SO₄ + 8H⁺ -> 4I₂ (g) + H₂S (g) + 4H₂O (I)
(Note that H₂SO₄ plays the role of an acid in the first step producing HI and then acts as an oxidising agent in the three redox steps)
OBSERVATIONS
- White steamy fumes of HI are evolved.
- Black solid and purple fumes of iodine are evolved.
- A colourless, acidic gas SO₂.
- A yellow solid of sulfur.
- H₂S (hydrogen sulfide), a gas with a bad egg smell.
Write an equation for the reaction of chlorine with water.
How does this reaction differ when it occurs in sunlight? What could you observe in this reaction?
CHLORINE WITH WATER
- In this reaction, chlorine is both reducing and oxidising which is why it’s known as a disproportionation reaction.
- Cl₂ (g) + H₂O (I) ⇌ HClO (aq) + HCl (aq)
REACTION WITH WATER IN SUNLIGHT
- If the chlorine is bubbled through water in the presence of bright sunlight a different reaction occurs:
- 2Cl₂ + 2H₂O -> 4H⁺ + 4Cl⁻ +O₂
- The greenish colour of chlorine water fades as the Cl₂ reacts and a colourless gas (O₂) is produced.
Write an equation to show the reaction of chlorine with cold dilute NaOH solution. What could you observe in this reaction?
What are the uses of the product formed as a result of this reaction?
- Cl₂ (aq) + 2NaOH (aq) -> NaCl (aq) + NaClO (aq) + H₂O (I)
- The greenish colour of chlorine fades to colourless.
- The mixture of NaCl and NaClO is used as bleach and to disinfect/kill bacteria.
Why is chlorine used in water treatment?
- Chlorine is used in water treatment to kill bacteria.
- It has been used to treat drinking water and water in swimming pools.
- The benefits to health of water treatment by chlorine outweigh its toxic effects.
