≫3.2 - Group 7 the Halogens ✔

Question 1

Describe and explain the trend in oxidising agent ability of the halogens down group 7:

Answer 1

• The halogens are good oxidising agents as they accept electrons from the species being oxidised and are reduced.
• The oxidising power decreases down the group as their ability to attract electrons decreases due to shielding and a greater atomic radius.

Question 2

Due to the relative oxidising strengths of a halogen, what do they do?

Answer 2

• The relative oxidising strengths mean a halogen will displace any halide beneath it in the periodic table.

Question 3

Describe and explain the trend in reducing agent ability of halide ions down group 7:

Answer 3

• Halide ions are good reducing agents as they donate electrons to the species being reduced and are themselves oxidised.
• This reducing power increases down the group as electrons are easier to lose from larger ions due to shielding and a larger ionic radius.

Question 4

Reaction of conc H₂SO₄ with NaF

Write the equation for the reaction:

Answer 4

• NaF + H₂SO₄ → NaHSO₄ + HF

Question 5

Reaction of conc H₂SO₄ with NaCl

Write the equation for the reaction:

Answer 5

• NaCl + H₂SO₄ → NaHSO₄ + HCl

Question 6

Reaction of conc H₂SO₄ with NaBr

Write the equations including the overall equation for the reaction:

Answer 6

• H₂SO₄ + NaBr → HBr + NaHSO₄
• H₂SO₄ + 2HBr → SO₂ + 2H₂O + Br₂
• You x2 the first equation to make the HBr balanced in the same ratio.
• Combining these then gives you:
• 3H₂SO₄ + 2NaBr → SO₂ + 2H₂O + Br₂ + 2NaHSO₄

Question 7

Reaction of conc H₂SO₄ with NaI

Write the three overall equations for the reaction:

Answer 7

• 3H₂SO₄ + 2NaI → SO₂ + 2NaHSO₄ + 2H₂O + I₂
• 7H₂SO₄ + 6NaI → S + 6NaHSO₄ + 4H₂O + 3I₂
• 9H₂SO₄ + 8NaI → H₂S + 8NaHSO₄ + 4H₂O + 4I₂
• The multiplication factor is the difference in oxidation number of the sulphur.
• H₂SO₄ is always 1 extra.

Question 8

What is a disproportionation reaction and give an example:

Answer 8

• Disproportionation reaction is where one molecule acts as both the oxidising and reducing agent.
• Cl₂ + H₂O → HCl + HOCl

Question 9

For reactions with sodium halides and concentrated sulfuric acid why does bromine produce 2 reactions compared to chlorine and fluorine and then iodine causes 3 reactions?

Answer 9

• The greater the reducing power, the longer the reaction as the halide is powerful enough to reduce more species.

Question 10

Describe and explain the trend in atomic radius down group 7:

Answer 10

• The atomic radius of group 7 elements increases down the group due to additional electron shells.

Question 11

Describe and explain the trend in reactivity down group 7:

Answer 11

• The group 7 elements need to gain an electron, as atomic radius increases this becomes harder as the positive attraction of the nucleus is weakened by additional shielding.
• Therefore it is harder to attract an electron so reactivity decreases down the group.

Question 12

Describe and explain the trend in ionisation energy down group 7:

Answer 12

• The first ionisation energy decreases down the group due to a greater atomic radius and increased amounts of shielding.

Question 13

Describe and explain the trend of boiling points down group 7:

Answer 13

• The group 7 elements are simple covalent molecules held together with van der Waals forces.
• The strength of these intermolecular forces increases as the Ar of the molecule increases, therefore the strength of the VDW forces increase down the group meaning more energy is required to overcome them.
• This results in a higher boiling point.

Question 14

Describe and explain the trend of electronegativity down group 7:

Answer 14

• Electronegativity decreases down group 7 due to an increasing atomic radius and amount of shielding thus reducing the electrostatic attraction to the nucleus.

Question 15

What are the states of matter and colours from fluorine-iodine?

Answer 15

• F₂(g) yellow
• Cl₂(g) yellow-green
• Br₂(l) red-brown
• I₂(s) grey-black solid but sublimes to form a violet vapour.

Question 16

Give the equation for the reaction of chlorine with water and the subsequent reaction of the product with water:

Answer 16

• Cl₂ + H₂O ⇌ HCl + HClO (chloric (I) acid)
• HClO + H₂O ⇌ ClO⁻ + H₃O⁺
• Chloric (I) acid ionises to make chlorate (I) ions also called hypochlorite.

Question 17

What is chlorine used for and what are the issues around this:

Answer 17

• Chlorate (I) ions kill bacteria thus are used to sterilise water to drink or swim in.
• Used to treat water diseases such as cholera.
• Used in very small concentrations as this is a harmful gas if breathed in.
• However as water is treated with chlorine, some people object as they do not want to consume chlorine products.

Question 18

Give the equation for the reaction of water and chlorine with sunlight:

Answer 18

• 2Cl₂ + 2H₂O → 4HCl + O₂
• Sunlight will break down the chlorine and not release chlorate (I) ions.

Question 19

Give the equation for the production of bleach and uses of it:

Answer 19

• 2NaOH + Cl₂ → NaClO + NaCl + H₂O [Sodium chlorate (I)]
• Water treatment, bleach paper and textiles, cleaning toilets.

Question 20

Halide Identification Tests - Chlorine:

Give the relevant equations and observations when testing for chlorine with silver nitrate and ammonia:

Answer 20

• Ag⁺ + Cl⁻ → AgCl
• White ppt
• AgCl + 2NH₃ → [Ag(NH₃)₂]⁺Cl⁻
• Dissolves in weak ammonia.

Question 21

Halide Identification Tests - Bromine:

Give the relevant equations and observations when testing for bromine with silver nitrate and ammonia:

Answer 21

• Ag⁺ + Br⁻ → AgBr
• Cream ppt
• AgBr + 2NH₃ → [Ag(NH₃)₂]⁺Br⁻
• Dissolves in concentrated ammonia.

Question 22

Halide Identification Tests - Iodine:

Give the relevant equations and observations when testing for iodine with silver nitrate and ammonia:

Answer 22

• Ag⁺ + I⁻ → AgI
• Yellow ppt
• Does not dissolve in concentrated ammonia.
• Solubility of the silver halides decreases down the group.

Question 23

When testing for halide ions why is the silver nitrate solution acidified?

Answer 23

• The silver nitrate is acidified so that any other impurities that could form a precipitate are removed.

Question 24

Reaction of conc H₂SO₄ with NaF

Give the observations and the substances that cause them:

Answer 24

• Misty white fumes (HF)

Question 25

Reaction of conc H₂SO₄ with NaCl

Give the observations and the substances that cause them:

Answer 25

• Misty white fumes (HCl)

Question 26

Reaction of conc H₂SO₄ with NaBr

Give the observations and the substances that cause them:

Answer 26

• Misty white fumes (HBr)
• Red-brown vapour (Br₂)

Question 27

Reaction of conc H₂SO₄ with NaI

Give the observations and the substances that cause them:

Answer 27

• Misty white fumes (HI)
• Purple vapour (I₂)
• Yellow solid formed (S)
• Rotten egg smell (H₂S)
• Black solid formed (I₂)

Question 28

Give the ionic equation for the reaction between chlorine and cold dilute sodium hydroxide solution:

Answer 28

• Cl₂ + 2HO⁻ → OCl⁻ + Cl⁻ + H₂O

Question 29

When chlorine or bromine displace iodine in a displacement reaction what is observed?

Answer 29

• Iodine released so dark brown colour observed.

Question 30

How do you test for H₂S in the reaction of potassium iodide with sulfuric acid?

Answer 30

• Hydrogen sulfide + lead nitrate paper forms lead sulfide and nitric acid.
• Lead sulfide is a black insoluble compound thus the paper goes black.

Question 31

Why is HCl acid not used to acidify the silver nitrate?

Answer 31

• HCl is not used to acidify the silver nitrate reaction as the chloride ion would react with the silver ion to form a white ppt and give a false positive result.

Question 32

How do you test for SO₂ in the reaction of potassium bromide with sulfuric acid?

Answer 32

• Test with damp potassium dichromate paper, orange changes to green in presence of SO₂ gas.

Question 33

What colour is each halide ion in solution?

Answer 33

• F⁻ = colourless
• Cl⁻ = green
• Br⁻ = orange
• I⁻ = brown