3.2.3 Group 7(17), the halogens Flashcards
What is group 7?
Group 7 non-metals exist as diatomic molecules (F2, Cl2, Br2, I2)
Group 7 elements are called halogens.
What are the physical properties of halogens?
Fluorine (F2): very pale yellow gas. It is highly reactive.
Chlorine (Cl2): greenish, reactive gas, poisonous in high concentrations.
Bromine (Br2): red liquid, that gives off dense brown/orange poisonous fumes
lodine (I2) : shiny grey solid sublimes to purple gas.
What is the trend in melting point and boiling point down group 7?
Increases down the group.
As the molecules become larger they have more electrons and so have larger van der waals forces between the molecules. As the intermolecular forces get larger more energy has to be put in to break the forces. This increases the melting and boiling points.
What is the trend in electronegativity down group 7?
Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself.
Decreases down the group.
As one goes down the group the atomic radii increases due to the increasing number of shells. The nucleus is therefore less able to attract the bonding pair of electrons.
What are oxidising agents?
What are reducing agents?
Oxidising agents - electron acceptors.
Reducing agents - electron donors.
What is the trend in oxidising ability of halogens down group 7?
The oxidising strength decreases down the group.
How do displacement reactions of halide ions by halogens occur?
A halogen that is a strong oxidising agent will displace a halogen that has a lower oxidising power from its compound.
The halide ion will be displaced by a more reactive halogen.
Describe which halide ions chlorine, bromine and iodine will be able to displace.
Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions; iodine will displace no ions.
List the reactions occurring and observations that can be made from the displacement of halide ions by chlorine.
The colour of the solution in the test tube shows which free halogen is present in solution.
Chlorine = very pale green solution (often colourless)
Bromine = yellow solution
lodine = brown solution (sometimes black solid present)
chlorine (aq) + chloride ions (aq) (e.g. potassium chloride)
= very pale green solution, no reaction
chlorine (aq) + bromide ions (aq) (e.g. potassium bromide)
= yellow solution, Cl has displaced Br
Cl2(aq) + 2Br^-(aq) → 2Cl^-(aq) + Br2(aq)
chlorine (aq) + iodide ions (aq) (e.g. potassium iodide)
= brown solution, Cl has displaced I
Cl2(aq) + 2I^-(aq) → 2Cl^-(aq) + I2(aq)
List the reactions occurring and observations that can be made from the displacement of halide ions by bromine.
The colour of the solution in the test tube shows which free halogen is present in solution.
Chlorine = very pale green solution (often colourless)
Bromine = yellow solution
lodine = brown solution (sometimes black solid present)
bromine (aq) + chloride ions (aq) (e.g. potassium chloride)
= yellow solution, no reaction
bromine (aq) + bromide ions (aq) (e.g. potassium bromide)
= yellow solution, no reaction
bromine (aq) + iodide ions (aq) (e.g. potassium iodide)
= brown solution, Br has displaced I
Br2(aq) + 2I^-(aq) → 2Br^-(aq) + I2(aq)
List the reactions occurring and observations that can be made from the displacement of halide ions by iodine.
The colour of the solution in the test tube shows which free halogen is present in solution.
Chlorine = very pale green solution (often colourless)
Bromine = yellow solution
lodine = brown solution (sometimes black solid present)
iodine (aq) + chloride ions (aq) (e.g. potassium chloride)
= brown solution, no reaction
iodine (aq) + bromide ions (aq) (e.g. potassium bromide)
= brown solution, no reaction
iodine (aq) + iodide ions (aq) (e.g. potassium iodide)
= brown solution, no reaction
Are halide ions oxidising or reducing agents?
Unlike halogens (atoms) being oxidising agents, halide ions act as reducing agents.
Halide ions lose electrons (oxidised) and become halogen molecules.
What is the trend in reducing ability of halide ions down group 7?
A reducing agent donates electrons.
The reducing power of the halides increases down group 7.
They have a greater tendency to donate electrons.
This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller.
- greater ionic radii - greater shielding - less attraction between nucleus and outer electrons - easier to donate
How do you test for the presence of different halide ions?
The test solution is made acidic with nitric acid, and then silver nitrate solution is added dropwise.
The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations.
2HNO3 + Na2CO3 → 2NaNO3 + H2O + CO2
- Fluorides produce no precipitate
- Chlorides produce a white precipitate
Ag^+(aq) + Cl^-(aq) → AgCl(s) - Bromides produce a cream precipitate
Ag^+(aq) + Br^-(aq) → AgBr(s) - lodides produce a pale yellow precipitate
Ag^+(aq) + I^-(aq) → AgI(s)
The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar:
- Silver chloride dissolves in dilute ammonia to form a complex ion
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]^+(aq) + Cl^-(aq)
- Silver bromide dissolves in concentrated ammonia to form a complex ion
AgBr(s) + 2NH3(aq) →[Ag(NH3)2]^+(aq) + Br^-(aq)
- Silver iodide does not react with ammonia - it is too insoluble.
How can you demonstrate the increasing reducing power of the halide ions down group 7?
The reaction of the solid halides with concentrated sulphuric acid.
Describe the reactions of fluoride with concentrated sulfuric acid.
F^- ions are not strong enough reducing agents to reduce the S in H2SO4.
No redox reactions occur. Only acid-base reactions occur.
Acid-base reaction:
NaF(s) + H2SO4(l) → NaHSO4(s) + HF(g)
Observations: White steamy fumes of HF are evolved.
H2SO4 plays the role of an acid (proton donor).
Describe the reactions of chloride with concentrated sulfuric acid.
Cl^- ions are not strong enough reducing agents to reduce the S in H2SO4.
No redox reactions occur. Only acid-base reactions occur.
Acid-base reaction:
NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)
Observations: White steamy fumes of HCl are evolved.
H2SO4 plays the role of an acid (proton donor).
Describe the reactions of bromide with concentrated sulfuric acid.
Br^- ions are stronger reducing agents than Cl^- and F^- and after the initial acid-base reaction, the bromide ions reduce the sulfur in H2SO4 from +6 to +4 in SO2.
Acid-base reaction:
NaBr(s) + H2SO4(l) → NaHSO4(s) + HBr(g)
Observations: White steamy fumes of HBr are evolved.
Redox reaction:
2HBr + H2SO4 → Br2 + SO2 + 2H2O
2H^+ + 2Br^- + H2SO4 → Br2(g) + SO2(g) + 2H2O(l)
Observations: Orange/brown fumes of bromine are evolved.
Overall equation: combining two steps above:
2NaBr + 3H2SO4 → 2NaHSO4 + SO2 + Br + 2H2O
Ox ½ equation 2Br^- → Br2 + 2e^- (Br oxidation state -1 → 0)
Re ½ equation H2SO4 + 2H^+ + 2e^- → SO2 + 2H2O (S oxidation state +6 → +4)
H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step.
Describe the reactions of iodide with concentrated sulfuric acid.
I^- ions are the strongest halide reducing agents. They can reduce the sulfur from +6 in H2SO4 to +4 in SO2, to 0 in S and -2 in H2S.
Acid-base reaction:
NaI(s) + H2SO4(l) → NaHSO4(s) + HI(g)
Observations: White steamy fumes of HI are evolved.
Redox reaction 1:
2HI + H2SO4 → I2 + SO2 + 2H2O
2H^+ + 2I^- + H2SO4 → I2(s) + SO2(g) + 2H2O(l)
Observations: Black solid and purple fumes of iodine are evolved.
Redox reaction 2:
6HI + H2SO4 → 3I2 + S + 4H2O
6H^+ + 6I^- + H2SO4 → 3I2 + S(s) + 4H2O(l)
Observations: A yellow solid of sulfur.
Redox reaction 3:
8HI + H2SO4 → 4I2 + H2S + 4H2O
8H^+ + 8I^- + H2SO4 → 4I2(s) + H2S(g) + 4H2O(l)
Observations: H2S (hydrogen sulfide), a gas with a bad egg smell.
Ox ½ equation 2I^- → I2 + 2e^- (I oxidation state -1 → 0)
Re ½ equation H2SO4 + 2H^+ + 2e^- → SO2 + 2H2O (S oxidation state +6 → +4)
Re ½ equation H2SO4 + 6H^+ + 6e^- → S + 4H2O (S oxidation state +6 → 0)
Re ½ equation H2SO4 + 8H^+ + 8e^- → H2S + 4H2O (S oxidation state +6 → -2)
H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the three redox steps
What is disproportionation? What halogen undergoes disproportionation reactions?
Disproportionation - the name for a reaction where an element simultaneously oxidises and reduces.
Chlorine.
Describe the reaction of chlorine with water and its uses
Cl2(g) + H2O(l) ⇌ HClO(aq) + HCl(aq)
Disproportionation:
Cl2 (oxidation state 0) → HClO (oxidation state +1) (oxidised)
Cl2 (oxidation state 0) → HCl (oxidation state -1) (reduced)
Chlorine is both simultaneously reducing and oxidising.
If some universal indicator is added to the solution it will first turn red due to the acidity of both reaction products. It will then turn colourless as the HCIO bleaches the colour.
Reaction with water in sunlight
If the chlorine is bubbled through water in the presence of bright sunlight a different reaction occurs.
2Cl2 + 2H2O → 4H^+ + 4Cl^- + O2
The same reaction occurs to an equilibrium mixture of chlorine water when standing in sunlight.
The greenish colour of chlorine water fades as the Cl2 reacts and a colourless gas (O2) is produced.
The greenish colour of these solutions is due to the Cl2.
Chlorine is used in water treatment to kill bacteria. It has been used to treat drinking water and the water in swimming pools. The benefits to health of water treatment by chlorine outweigh its toxic effects.
HClO = chloric acid - oxidising agents that kills bacteria.
Describe the reaction of chlorine with cold dilute NaOH solution
Cl2 (and Br2 & I2) in aqueous solutions will react with cold sodium hydroxide. The colour of the halogen solution will fade to colourless.
Cl2(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l)
Disproportionation:
Cl2 (oxidation state 0) → NaClO (oxidation state +1) (oxidised)
Cl2 (oxidation state 0) → NaCl (oxidation state -1) (reduced)
Chlorine is both simultaneously reducing and oxidising.
The mixture of NaCl and NaClO is used as bleach and to disinfect/ kill bacteria.
How do you name chlorates/sulfates?
In IUPAC convention the various forms of sulfur and chlorine compounds where oxygen is combined are all called sulfates and chlorates with relevant oxidation number given in roman numerals. If asked to name these compounds remember to add the oxidation number.
NaCIO: sodium chlorate(I)
NaClO3: sodium chlorate(V)
K2SO4 potassium sulfate(VI)
K2SO3 potassium sulfate(IV)
