Group 7; The Halogens

Question 1

What is the trend in electronegativity?

(electronegativity - the ability of an atom to pull electrons towards itself)

Answer 1

Electronegativity decreases going down the group;

• electrons are further from the nucleus; atomic radius is greater (further from the covalent bond); electrostatic attraction between bond electrons and nucleus is weaker;
• more electron shielding also.

Shielding and atomic radius have more of an influence than nuclear charge; going down group = bigger atoms = more protons but its effects are outweighed.

Question 2

What is the trend in boiling and melting points?

Answer 2

Boiling points increase going down the group:

• greater size and relative mass of molecule (it’s larger) thus more van der Waals forces between the molecules = stronger.(more electrons going down group)

Van der Waals hold together the halogen molecules; they’re diatomic.

Question 3

What is the trend in oxidising ability?

Answer 3

The oxidising ability of the halogens decreases as you go down the group; fluorine is most reactive; greatest electronegativity.

A halogen will displace a halide if the halide is below it in the group.

Cl₂ (aq) + 2NaBr (aq) → Br₂ (aq) + 2NaCl (aq)

Cl₂ (aq) + 2Na⁺ (aq) + 2Br⁻ (aq) → Br₂ (aq) + 2Na⁺ (aq) + 2Cl⁻ (aq)
(Na⁺ are struck out; spectator ions; unchanged in reaction)

Cl₂ (aq) + 2Br⁻ (aq) → Br₂ (aq) + 2Cl⁻ (aq)

Thus the colourless starting materials react to produce the red-brown colour of bromine; it has been displaced, precipitating.

Question 4

What is the trend in reducing ability of the halide ions?

Answer 4

The reducing ability of halides increases down the group:

• greater shielding
• greater atomic radius
  = easier to lose the excess electron; given away to become halogens.

Question 5

What are the product formed by reaction of NaF with H₂SO₄?

Answer 5

Fluoride ions are not powerful enough reducing agents to reduce sulphur in the sulphuric acid; the reaction stops there.

• Not a redox reaction; oxidation states of halide (Fluorine) and sulfur remain the same.

Steamy fumes of HF are given off. (acid-base)

Question 6

What are the product formed by reaction of NaCl with H₂SO₄?

Answer 6

Chloride ions are not powerful enough reducing agents to reduce sulphur in the sulphuric acid; the reaction stops there.

Not a redox reaction; oxidation states of halide (Chloride) and sulfur remain the same.

Steamy fumes of HCl are given off. (acid-base)

Question 7

What are the product formed by reaction of NaBr (solid) with H₂SO₄?

Answer 7

Bromide ions are quite a powerful reducing agent. They can reduce the sulphur in sulphuric acid to make sulphur dioxide. The bromide ions are oxidised.

1. Same first step; steamy fumes of HBr. (acid-base reaction)
2. HBr is strong enough to reduce H₂SO₄; producing choking fumes of SO₂ (colourless) and brown bromine fumes.
   (redox)

Question 8

What are the product formed by reaction of NaI (solid) with H₂SO₄?

Answer 8

Iodide ions are a powerful reducing agent. They can reduce the sulfur in sulfuric acid to make sulfur dioxide and hydrogen sulfide.

The iodide ions are oxidised.

1. Intial steamy fumes of HI. (acid-base)
2. Black solid I₂ formed. (redox)
3. Colourless SO₂ is evolved; eggy H₂S is produced.
   (during reduction of sulfur from +6 to -2, sulfur passes through ox. state 0 thus some yellow solid sulfur may be seen too)

Question 9

Why is acidified silver nitrate solution used as a reagent to identify halide ions?
State the colour of the precipitates that form and another test to confirm the halide’s prescence.

Answer 9

All metal halides (except Fluorides) react with silver ions in aqueous solution (silver nitrate) to form a precipitate of the insoluble silver halide.

Ag⁺ + X^- → AgX

Ag^⁺(aq) + Cl^⁻ (aq) → AgCl (s)

Silver Fluoride (AgF) does not form a precipitate because it is soluble in water.

1 – Dilute HNO₃ (H⁺(aq) + NO₃⁻(aq)) is added to halide solution to remove any soluble carbonate (CO₃²⁻(aq)) or hydroxide impurities.
2 – A few drops of silver nitrate are added and the halide precipitate is formed.

Question 10

What occurs when you mix chlorine with water?
What is its use in water treatment?

Answer 10

It undergoes disproportionation (where one element is both oxidised and reduced simultaneously).

0 +1 -1
Cl₂(g) + H₂O(l) → HClO(aq) + HCl
chloric(I) acid

Chloric(I) acid is a powerful oxidising agent; kills bacteria by oxidation.

Adding chlorine (or any compound containing chlorate(I) ions) to water makes it safe to drink/swim in.

Double acid product = pH levels of pools need to be regulated.

Question 11

Explain why the benefits of water treatment by chlorine outweigh its toxic effects.

Answer 11

• Kils pathogenic micro-organisms
• Prevents reinfection in supply
• Prevents algae, eliminates bad tastes and smells.
• Chlorine gas is harmful; irritating the respiratory system.
• Liquid chlorine causes severe chemical burns.
• Chlorine reacts with organic compounds found in water to form chlorinated hyrdrocarbons (e.g. chloromethane) which are carcinogenic.

However, risks from untreated water are far greater than the risk of cancer; cholera/typhoid epidemic vs. a few cancer cases.

Question 12

State the reaction of chlorine with cold, dilute aqueous NaOH and the uses of the solutions formed.

Answer 12

0 +1 -1
Cl₂(g) + 2NaOH(aq) → NaClO(aq) + NaCl + H₂O(l)
sodium chlorate(I)

Chlorine + cold dilute NaOH makes BLEACH.
Another example of disproportionation.

Sodium chlorate(I) is a powerful oxidising agent, and bleaches things by oxidising many coloured substances.

Uses:

• Water treatment
• To bleach paper/textiles
• Cleaning toilets etc.