Group 7 Revsion Guide Flashcards
(31 cards)
Halogens state at room temperature
Fluorine (F2): very pale yellow gas. It is highly reactive
Chlorine : (C l2) greenish, reactive gas, poisonous in high concentrations
Bromine (Br2) : red liquid, that gives off dense brown/orange poisonous fumes
Iodine (I2) : shiny grey solid sublimes to purple gas.
Trend in melting point and boiling point
Increase 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 into break the forces. This increases the melting and boiling points.
Trend in electronegativity
Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself.
down the group the electronegativity of the elements decreases.
the atomic radii increases due to the increasing number of shells.
The nucleus is therefore less able to attract the bonding pair of electrons.
The displacement reactions of halide ions due to oxidising strength
A halogen that is a strong oxidising agent will displace a halogen that has a lower oxidising power from one of its compounds.
The oxidising strength decreases down the group. Oxidising agents are electron acceptors
The displacement reactions of halide ions by halogens.
Chlorine
Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions
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 Iodine = brown solution (sometimes black solid present)
Ionic equasions of displacement reactions of hailide ions by halogens
Cl2(aq) + 2Br – (aq) —> 2Cl – (aq) + Br2(aq)
Cl2(aq) + 2I – (aq) —> 2Cl – (aq) + I2(aq)
Br2(aq) + 2I – (aq) —> 2Br – (aq) + I2(a
Half equasions of displacement reactions
2Br - (aq)—> Br2 (aq)+ 2e-
Cl2 (aq)+ 2e- —> 2Cl- (aq
The reactions of halide ions with silver nitrate
This reaction is used as a test to identify which halide ion is present. 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
2 HNO3 + Na2CO3 —> 2 NaNO3 + H2O + CO2
The reactions of halide ions with silver nitrate. Chloride
Chlorides produce a white precipitate Ag+(aq) + Cl- (aq) —> AgCl(s)
Bromide - reactions of hailides with silver nitrate
Bromides produce a cream precipitate Ag+(aq) + Br- (aq) AgBr(s)
Iodide reaction with silver nitrate
Iodides produce a pale yellow precipitate Ag+(aq) + I- (aq) AgI(s)
What can silver hailides be treated with if the colours look similar
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)
What happens to silver bromide in conc amonia
Silver bromide dissolves in concentrated ammonia to form a complex ion
AgBr(s) + 2NH3(aq) [Ag(NH3)2]+ (aq) + Br - (aq)
However silver iodide does not react with ammonia it is too insoluble
Explanation of differing reducing power of halides
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.
The reaction of halide salts with concentrated sulfuric acid.
The halides show increasing power as reducing agents as one goes down the
This can be clearly demonstrated in the various reactions of the solid halides with concentrated sulfuric acid.
Fluoride and chloride reaction of hailides salts with conc sulfuric acid eg with na
F- and Cl- ions are not strong enough reducing agents to reduce the S in H2SO4. No redox reactions occur. Only acid-base reactions occur.
NaF(s) + H2SO4(l) —> NaHSO4(s) + HF(g)
Observations: White steamy fumes of HF are evolved.
H2SO4 plays the NaCl(s) + H2SO4(l) —> NaHSO4(s) + HC l(g)
Observations: White steamy fumes of HCl are evolved.
Br- ions as reducing agents
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 step: NaBr(s) + H2SO4(l) —> NaHSO4(s) + HBr(g)
Redox step: 2 H+ + 2 Br - + H2SO4–> Br2(g) + SO2(g) + 2 H2O(l)
Overall equation for reduction using bromide ions
Overall equation: combining two steps above:
2NaBr + 3H2SO4 —> 2NaHSO4 + SO2 + Br2 + 2H2O
Observations: White steamy fumes of HBr are evolved. orange fumes of bromine are also evolved and a colourless, acidic gas
SO2
The reduction product is so2
Half equasions for reactions with bromide and sulfuric acid
Ox ½ equation 2Br - —> Br2 + 2e-
Re ½ equation H2SO4 + 2 H+ + 2 e- —> SO2 + 2 H2O
Iodide as a reducing agent
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.
Observations of iodide reducing sulfur
Observations: White steamy fumes of HI are evolved. Black solid and purple fumes of Iodine are also evolved
A colourless, acidic gas SO2.
A yellow solid of sulfur
H2S (Hydrogen sulfide) a gas with a bad egg smell,
Equations for reduction of sulfur using iodide
H2so4 — NaI(s) + H2SO4(l) —> NaHSO4(s) + HI(g)
So2 — 2 H+ + 2 I- + H2SO4 —> I2(s) + SO2(g) + 2 H2O(l)
S — 6 H+ + 6 I- + H2SO4 —> 3 I2 + S (s) + 4 H2O (l)
H2S — 8 H+ + 8 I- + H2SO4 —> 4 I2(s) + H2S(g) + 4 H2O(l)
Half equasions for the reduction of sulfur using iodide
Ox ½ equation 2I - —> I2 + 2e-
Re ½ equation H2SO4 + 2 H+ + 2 e- —> SO2 + 2 H2O
Re ½ equation H2SO4 + 6 H+ + 6 e- —> S + 4 H2O
Re ½ equation H2SO4 + 8 H+ + 8 e- H2S + 4 H2O
Products of reduction of sulfur using iodide
Reduction products = sulfur dioxide, sulfur
and hydrogen sulfide
