11.0 the halogens text book p166 to 173 1. aqa as specification lessonstopics 1 how and why does the...

11.0 The Halogens

Text book p166 to 173

1

AQA AS SpecificationLessons Topics

1 How and why does the atomic radius and electronegativity change in Gp 7. What effect does this have on the boiling point?

2 To understand that the ability of the halogens (from fluorineto iodine) to oxidise decreases down the group (e.g. thedisplacement reactions with halide ions in aqueous solution)

3 understand the trend in reducing ability of the halide ionsknow the different products formed by reaction of NaX andH2SO4

4 understand why acidified silver nitrate solution is used as areagent to identify and distinguish between F- Cl- , Br- andI- know the trend in solubility of the silver halides in ammonia

5 know the reactions of chlorine with water and the use ofchlorine in water treatment /appreciate that the benefits to health of water treatment by chlorine outweigh its toxic effects/know the reaction of chlorine with cold, dilute, aqueous NaOH and the uses of the solutions formed 2

Halogens

3

What are the Halogens, what does their name mean, and where are they found in the Periodic Table?

• The Halogens are elements that are found in Group 7 (VII) of the PT. The name means “salt former”.

• The Halogens are elements that are found in Group 7 (VII) of the PT. The name means “salt former”.

Name some compounds that contain a halogen

What is the valency (OXIDATION NUMBER) of the halogens?What is the valency (OXIDATION NUMBER) of the halogens?

Fluorine, F2, is a pale yellow gas at room temperature.

Appearance

fluorine video

Chlorine, Cl2, is a pale green gas at room temperature.

Chlorine video

Bromine, Br2, is a dark red liquid at room temperature. It is the only liquid non-metal.

Bromine video

Bromine is volatile and readily forms a dark red vapour.

Iodine, I2, forms shiny black crystals at room temperature.

iodine video

When warmed, iodine crystals sublime (turn directly to a gas), forming a purple vapour.

Fluorine

10

What is unusual about the bonding in fluorine molecule? Explain.

Summarize the physical trends in Group VII, draw stick graphs for the data and explain the trends.

Compared to the other halogens, the F-F bond is very weak.

This is because the fluorine atoms are very small and there is a lot of repulsion between the bonding electrons.

Compared to the other halogens, the F-F bond is very weak.

This is because the fluorine atoms are very small and there is a lot of repulsion between the bonding electrons.

Going down the group, there are more filled energy levels between the nucleus and the outer electrons.

This results in the outer electrons being shielded more from the attraction of the nucleus.

Going down the group, there are more filled energy levels between the nucleus and the outer electrons.

This results in the outer electrons being shielded more from the attraction of the nucleus.

The atomic radius increases down Group 7.The atomic radius increases down Group 7.

Atomic Radius

The strength of the instantaneous dipole−induced dipole forces between the molecules increases as the size of the molecules increases.

The strength of the instantaneous dipole−induced dipole forces between the molecules increases as the size of the molecules increases.

The boiling point increases down Group 7.The boiling point increases down Group 7.

Boiling point

Fluorine is the most electronegative element in the periodic table.

Fluorine is the most electronegative element in the periodic table.

Electronegativity decreases down Group 7.Electronegativity decreases down Group 7.

Electronegativity

The atomic radius increases, the outer electrons are more shielded, so bonding electrons are less strongly attracted to the nucleus. (Fig 2 p167)

The atomic radius increases, the outer electrons are more shielded, so bonding electrons are less strongly attracted to the nucleus. (Fig 2 p167)

11.2 Chemical reactions of the Halogens

P 168 ( Lister)P138 (Atkinson)

14

Oxidation ability

15

What is Oxidation? Oxidation is the loss of electrons. Oxidation is the loss of electrons.

What is an oxidizing agent? An oxidizing agent is an electron acceptor, the agent is reduced during the course of the reaction.

This forms a redox reaction.

An oxidizing agent is an electron acceptor, the agent is reduced during the course of the reaction.

This forms a redox reaction.

Oxidising power trend: Cl2 > Br2 > I2

When a halogen acts as an oxidising agent, it gains electrons (taken from the oxidised species).

X2 + 2 e- → 2 X-

Going down the group it becomes harder to gain an electron because:

atoms are larger & there is more shielding (due to extra electron shell)

Going down the group it becomes harder to gain an electron because:

atoms are larger & there is more shielding (due to extra electron shell)

Cl

Br

I

Task:

17

Read p 168 (Lister) and complete the table

React each Potassium Halide with each halogen water.•Note down the colour change (if there is any)•Write the ionic equation for the displacement.

Check you are correct by reading page 168

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq)

Br–(aq)

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq)

Br–(aq)

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Br–(aq)

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Purple in solvent layer (no reaction)

Br–(aq)

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Purple in solvent layer (no reaction)

Br–(aq)

Yellow/brown in aqueous layer

(Br2 forms)

Cl2 + 2 Br-→ 2 Cl- + Br2

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Purple in solvent layer (no reaction)

Br–(aq)

Yellow/brown in aqueous layer

(Br2 forms)

Cl2 + 2 Br-→ 2 Cl- + Br2

Purple in solvent layer (no reaction)

I–(aq)

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Purple in solvent layer (no reaction)

Br–(aq)

Yellow/brown in aqueous layer

(Br2 forms)

Cl2 + 2 Br-→ 2 Cl- + Br2

Purple in solvent layer (no reaction)

I–(aq)

Purple in solvent layer

(I2 forms)

Cl2 + 2 I- → 2 Cl- + I2

Cl2(aq) Br2(aq) I2(aq)

Cl–(aq) Stays yellow solution (no reaction)

Purple in solvent layer (no reaction)

Br–(aq)

Yellow/brown in aqueous layer

(Br2 forms)

Cl2 + 2 Br-→ 2 Cl- + Br2

Purple in solvent layer (no reaction)

I–(aq)

Purple in solvent layer

(I2 forms)

Cl2 + 2 I- → 2 Cl- + I2

Purple in solvent layer

(I2 forms)

Br2 + 2 I- → 2 Br- + I2

HSW – Extraction of Bromine homework

• Create a presentation on the extraction and uses of Bromine and of Iodine.

• Use the material in the text book (page 169) and from the internet.

• Your presentation could be as a poster or as a powerpoint.

• Maximum of 6 slides.

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11.3 Reactions of halide ions

27

Reducing agents

28

2 X– → X2 + 2 e–

When a halide ion reduces another substance, the halide is oxidised to a halogen.

2 X– → X2 + 2 e–

When a halide ion reduces another substance, the halide is oxidised to a halogen.

What is reduction? Reduction is the gain of electronsReduction is the gain of electrons

What happens when a Halide is used as a reducing agent?

Give the half equation for the reaction

Experiment

29

This experiment compares how well the halides reduce H2SO4 to compare the reducing power of the halide ions. Some of the products are very toxic – hence the video clips!

NaBr + Sulphuric acid

NaCl + sulphuric acid

NaI + sulphuric acid

Watch the clips and complete the OBSERVATIONS column in the table

halide products observation reaction type equation

Cl–

Br–

I–

halide products observation reaction type equation

Cl– HCl steamy fumes

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

Formation of hydrogen halides:

NaX + H2SO4 → NaHSO4 + HX

e.g. NaCl + H2SO4 → NaHSO4 + HCl

Complete the final two columns of the table.

halide products observation reaction type equation

Cl– HCl steamy fumes acid-base NaCl + H2SO4 NaHSO4 + HCl

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

acid-base

NaBr + H2SO4 NaHSO4 + HBr

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

acid-base

NaI + H2SO4 NaHSO4 + HI

Write half equations for:

Cl– → Cl2

Br– → Br2

I– → I2

H2SO4 → SO2

H2SO4 → S

H2SO4 → H2S

2 Cl– → Cl2 + 2 e–

2 Br– → Br2 + 2 e–

2 I– → I2 + 2 e–

H2SO4 + 2 H+ + 2 e– → SO2 + 2 H2O

H2SO4 + 6 H+ + 6 e– → S + 4 H2O

H2SO4 + 8 H+ + 8 e– → H2S + 4 H2O

halide products observation reaction type equation

Cl– HCl steamy fumes acid-base NaCl + H2SO4 NaHSO4 + HCl

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

acid-base

NaBr + H2SO4 NaHSO4 + HBr

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

acid-base

NaI + H2SO4 NaHSO4 + HI

halide products observation reaction type equation

Cl– HCl steamy fumes acid-base NaCl + H2SO4 NaHSO4 + HCl

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

acid-base

NaBr + H2SO4 NaHSO4 + HBr

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

acid-base

NaI + H2SO4 NaHSO4 + HI

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

6 I– + H2SO4 + 6 H+ 3 I2 + S + 4 H2O

8 I– + H2SO4 + 8 H+ 4 I2 + H2S + 4 H2O

halide products observation reaction type equation

Cl– HCl steamy fumes acid-base NaCl + H2SO4 NaHSO4 + HCl

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

acid-base

reduction of Br–

reduction of H2SO4

NaBr + H2SO4 NaHSO4 + HBr

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

acid-base

NaI + H2SO4 NaHSO4 + HI

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

6 I– + H2SO4 + 6 H+ 3 I2 + S + 4 H2O

8 I– + H2SO4 + 8 H+ 4 I2 + H2S + 4 H2O

halide products observation reaction type equation

Cl– HCl steamy fumes acid-base NaCl + H2SO4 NaHSO4 + HCl

Br–

HBr

Br2

SO2

steamy fumes

brown fumes

colourless gas

acid-base

reduction of Br–

reduction of H2SO4

NaBr + H2SO4 NaHSO4 + HBr

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

2 Br– + H2SO4 + 2 H+ Br2 + SO2 + 2 H2O

I–

HI

I2

SO2

S

H2S

steamy fumes

purple fumes

colourless gas

yellow solid

gas (bad egg smell)

acid-base

reduction of I–

reduction of H2SO4

reduction of H2SO4

reduction of H2SO4

NaI + H2SO4 NaHSO4 + HI

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

2 I– + H2SO4 + 2 H+ I2 + SO2 + 2 H2O

6 I– + H2SO4 + 6 H+ 3 I2 + S + 4 H2O

8 I– + H2SO4 + 8 H+ 4 I2 + H2S + 4 H2O

Cl– does not reduce H2SO4

Br– reduces H2SO4 from S(+6) to S(+4)

I– reduces H2SO4 from S(+6) to S(-2)

Reducing power trend

Reducing power trend: Cl– < Br– < I–

When a halide ion acts as a reducing agent, it loses electrons (given to the reduced species).

2 X– → X2 + 2 e–

Down the group it becomes easier to lose an electron because:

ions are larger & there is more shielding (due to extra electron shell)

Cl–

Br–

I–

Identification of metal halides

41

Complete practical ES4.4, or N-ch2-06