Group 7 Halogens
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Transcript of Group 7 Halogens
11.0 The Halogens
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”.
Name some compounds that contain a halogen
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.
Physical properties of halogens
Fluorine
11
What is unusual about the bonding in fluorine molecule? Explain.
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.
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 boiling point increases down Group 7.
Boiling point
Trends in boiling pointHalogen molecules increase in size down the group. This leads to greater van der Waals forces between molecules, increasing the energy needed to separate the molecules and therefore higher melting and boiling points.
fluorineatomic radius = 42 × 10-12 m
boiling point = -118 °C
iodine atomic radius = 115 × 10-12 m
boiling point = 184 °C
van der Waals forces
Fluorine is the most electronegative element in the periodic table.
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)
Trends in electronegativityElectronegativity of the halogens decreases down the group due to an increase in atomic radius.
fluorineatomic radius = 42 × 10-12 m
electronegativity = 4.0
iodineatomic radius = 115 × 10-12 m
electronegativity = 2.5
Increased nuclear charge has no significant effect because there are more electron shells and more shielding. Iodine atoms therefore attract electron density in a covalent bond less strongly than fluorine.
11.2 Chemical reactions of the Halogens
17
Oxidation ability
18
What is Oxidation? 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.
Oxidising power trend: Cl2 > Br2 > I2When 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)
Cl
Br
I
Oxidizing ability of halogens
fluorine
incr
easi
ng o
xidi
zing
abi
lity
iodine
bromine
chlorine
In displacement reactions between halogens and halides, the halogen acts as an oxidizing agent.
This means that the halogen:
What is the order of oxidizing ability of the halogens?
is reduced to form the halide ion.
gains electrons
oxidizes the halide ion to the halogen
Oxidizing ability of halogens
Reactions of the halogensHalogens react with metals such as sodium and iron:
They also take part in displacement reactions with halide ions, such as the reaction that is used to make bromine from potassium bromide in seawater:
halogen + hydrogen hydrogen halide
They also react with non-metals such as hydrogen:
halogen + sodium sodium halide
chlorine +potassiumbromide
potassiumchloride bromine +
Reaction with iron
Reactions with hydrogen
Chlorine and hydrogen explode in bright sunlight but react slowly in the dark.
The halogens react with hydrogen gas to product hydrogen halides. For example:
Cl2(g) + H2(g) 2HCl(g)
Iodine combines partially and very slowly with hydrogen, even on heating.
Bromine and hydrogen react slowly on heating with a platinum catalyst.
Redox reactions of halogens
Halogen displacement reactions
Halogen displacement reactionsHalogen displacement reactions are redox reactions.
Cl2 + 2KBr 2KCl + Br2
To look at the transfer of electrons in this reaction, the following two half equations can be written:
Chlorine has gained electrons, so it is reduced to Cl- ions.
What has been oxidized and what has been reduced?
2Br- Br2 + 2e-Cl2 + 2e- 2Cl-
Bromide ions have lost electrons, so they have been oxidized to bromine.
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)
Stays brown solution (no reaction)
Br–(aq)
I–(aq)
Cl2(aq) Br2(aq) I2(aq)
Cl–(aq) Stays yellow solution (no reaction)
Stays brown solution (no reaction)
Br–(aq)
Yellow solution forms (Br2 forms)
Cl2 + 2 Br-→ 2 Cl- + Br2
I–(aq)
Cl2(aq) Br2(aq) I2(aq)
Cl–(aq) Stays yellow solution (no reaction)
Stays brown solution (no reaction)
Br–(aq)
Yellow solution forms (Br2 forms)
Cl2 + 2 Br-→ 2 Cl- + Br2
Stays brown solution (no reaction)
I–(aq)
Cl2(aq) Br2(aq) I2(aq)
Cl–(aq) Stays yellow solution (no reaction)
Stays brown solution (no reaction)
Br–(aq)
Yellow solution forms (Br2 forms)
Cl2 + 2 Br-→ 2 Cl- + Br2
Stays brown solution (no reaction)
I–(aq)
Brown solution forms (I2 forms)
Cl2 + 2 I- → 2 Cl- + I2
Cl2(aq) Br2(aq) I2(aq)
Cl–(aq) Stays yellow solution (no reaction)
Stays brown solution (no reaction)
Br–(aq)
Yellow solution forms (Br2 forms)
Cl2 + 2 Br-→ 2 Cl- + Br2
Stays brown solution (no reaction)
I–(aq)
Brown solution forms (I2 forms)
Cl2 + 2 I- → 2 Cl- + I2
Brown solution forms (I2 forms)
Br2 + 2 I- → 2 Br- + I2
11.3 halides
35
HalidesWhen halogens react with metals, they form compounds called halides. Many naturally-occurring halides have industrial, household and medical applications.
caesium chloride
sodiumhexafluoroaluminate
titanium(IV) chloride
lithium iodide
potassium bromide
Halide Formula Uses
CsCl
NaAlF6
TiCl4
LiI
KBr
Extraction and separation of DNA
Electrolysis of aluminium oxide
Extraction of titanium
Electrolyte in batteries
Epilepsy treatment in animals
Identifying halide ionsHalides can be identified by their reaction with acidified silver nitrate solution to form silver halide precipitates.
Silver chloride has a low solubility in water, so it forms a white precipitate: the positive result in the test for chloride ions.
KCl(aq) + AgNO3(aq) KNO3(aq)
+ AgCl(s)
potassium chloride
silver chloride+
potassium nitrate
silver nitrate+
Identifying halide ions
Uses of halides in photographySilver halides are used in photography.
Ag+ + e- Ag
Photographic film coated with a silver halide is exposed to light, causing the halide to decompose to form silver. This appears as a black precipitate on the photographic film.
light
mask
paper coated in silver halide
silver precipitate
white paper under mask
Hydrogen halidesThe hydrogen halides are colourless gases at room temperature.
Hydrogen fluoride has an unexpectedly high boiling point compared to the other hydrogen halides. This is due to hydrogen bonding between the H–F molecules.
Hydrogen halide Boiling point (°C)
HF
HCl
HBr
HI
20
-85
-67
-35
11.3 Reactions of halide ions
41
Reducing agents
42
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 electrons
What happens when a Halide is used as a reducing agent?
Give the half equation for the reaction
Sodium halides and sulfuric acidThe sodium halides react with concentrated sulfuric acid.
The reactions of sodium halides with concentrated sulfuric acid demonstrate the relative strengths of the halide ions as reducing agents.
During this reaction two things can happen to the sulfuric acid. It can
act as an acid.
be reduced
Sodium halides and sulfuric acid
Sodium halides and sulfuric acid
Experiment
46
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–
Multiple-choice quiz
Identification of metal halides
59