3.2.3 Group 7

Question 1

Describe the trend in boiling point going down group seven and why that happens?

Answer 1

Boiling point increases going down the group
This is because the increasing size of atoms means there is an increasing number of electrons and therefore an increasing strength of van der Waals forces of attraction which require more energy to overcome.

Question 2

Describe the trend in electronegativity going down group 7 and explain why that happens.

Answer 2

Electronegativity decreases going down the group
This is because more occupied electron shells lead to a greater atomnic radius and a weaker attraction between nucleus and outer electrons.

Question 3

Trend in reducing power going down the group and why

Answer 3

Increases because a larger ionic radius, more shielding = a weaker attraction with the electron lost

Question 4

What are the two different types of reaction which might go on when conc sulphuric acid is added to a solid ionic halide, the sulphuric acid can act as an:

Answer 4

• acid
• oxidising agent

Question 5

What are the (ionic) euqations for sulphuric acid reacting with the halides as an acid

what is the observation for all three reactions (hint its the same)

and why does that observation happen

Answer 5

for Cl- : H+ + Cl- –> HCl
for Br- : H+ + Br- –> HBr
for I- : H+ + I- –> HI

misty fumes

because the hydrogen halide is a gas it immediately escapes from the system. If the hydrogen halide is exposed to moist air, you see it as steamy fumes.

Question 6

Which two halides are not strong enough reducing agents to reduce sulphuric acid and which two are?

Answer 6

Flourine and chloride ions aren’t string enough reducing agents.
Bromine and iodine are.

Question 7

So what always happens when flouride/chloride ions react with sulphuric acid

Answer 7

you get a hydrogen halide (sulphuric acid has just acted as a base, not an oxidising agent)

Question 8

Concentrated sulphuric acid acting as an oxidising agent with bromide ions
1)What are the products (state symbols)
2)State the two half equations (and type of reaction of each) and the observation for each reaction
3)Then state the overall (combined half equations) to give the overall ionic equation

Answer 8

1) Br (g), SO2 (g) and H2O (l)
2) 2Br- –> Br2 + 2e- (oxidation of Br-)(organge fumes)
H2SO4 + 2H+ + 2e- –> SO2 + 2H2O (reduction of H2SO4)(colourless gas turns blue litmus paper red)
3) H2SO4 + 2H+ + 2Br- –> Br2 + SO2 + 2H2O

Question 9

Concentrated sulphuric acid acting as an oxidising agent with iodide ions
1)What are the products (state symbols)
2)State the two half equations (and type of reaction of each) and the observation for each reaction

Answer 9

1) I2 (s), SO2 (g), S (s), H2S (g)
2) 2I- –> I2 + 2e- (black solid) (oxidation of iodide)
2e- + 2H+ + H2SO4 –> SO2 + 2H2O (colourless gas that turns blue litums paper red)
6e- + 6H+ + H2SO4 –> S + 4H2O (yellow solid) (reduction of H2SO4)
8e- + 8H+ + H2SO4 –> H2S + 4H2O (colourless gas, rotten egg smell) (reduction of H2SO4)

Question 10

Desribe how iodide ions reduces sulphuric acid

Answer 10

The iodide ions are powerful enough reducing agents to reduce it:

• first to sulphur dioxide (sulphur oxidation state = +4)
• then to sulphur itself (oxidation state = 0)
• and all the way to hydrogen sulphide (sulphur oxidation state = -2).

Question 11

What is added before the silver nitrate when testing for halide ions and why?

Answer 11

Nitric acid, to first remove any carbonate ions that might be present