Group 7 Flashcards
Explain why iodide ions react differently from chloride ions (2 marks)
For M1 and M2, iodide ions are stronger reducing agents than chloride ions, because
M1 Relative size of ions
Iodide ions / they are larger / have more electron levels(shells) (than chloride ions) / larger atomic / ionic radius
OR electron to be lost/outer shell/level (of the iodide ion) is further the nucleus
OR iodide ion(s) / they have greater / more shielding
OR converse for chloride ion
M2 Strength of attraction for electron(s)
The electron(s) lost / outer shell / level electron from (an) iodide ion(s) less strongly held by the nucleus compared with that lost from a chloride ion
OR converse for a chloride ion
Write a half equation for the conversion of chlorine into chloride ions (1 mark)
Cl2 + 2e- —> 2Cl-
In terms of electrons, state the meaning of the term ‘reducing agent’ (1 mark)
Electron donor
In terms of electrons, state the meaning of the term ‘oxidising agent’ (1 mark)
Electron acceptor
State why the silver nitrate is acidified when testing for iodide ions (1 mark)
The silver nitrate is acidified to
• react with / remove (an)ions that would interfere with the test
• prevent the formation of other silver precipitates / insoluble silver compounds that would interfere with the test
• remove (other) ions that react with the silver nitrate
• react with / remove carbonate / hydroxide / sulfite (ions)
Explain the trend in electronegativity going down Group 7, and hence the trend in polarity of halogen bonds
-decreases as you go down the group
-due to more shielding
-less effective nuclear charge
-hence weaker electrostatic forces of attraction between outermost electron and nucleus
-so polarity of halogen bonds also decreases
H-F is the most polar bond
H-I is the least polar bond
Explain the trend in boiling point going down Group 7 in terms of their structure and bonding
-increases down the group
-atomic radius increases due to more shells
-larger surface area so stronger Van der Waals forces between molecules
-requires more energy to overcome
Explain the trend in oxidising ability of halogens going down Group 7, including displacement reactions of hallide ions in aqueous solutions
-halogens are good oxidising agents
-as they accept electrons from species being oxidised and are reduced
-decreases down group
-as ability to attract electrons decreases due to shielding and weaker electrostatic forces of attraction between nucleus and outermost electron
-halogen will displace any hallide beneath it in the periodic table
Cl2 will displace Br- and I- ions
Br2 will displace I- ions
I2 will not displace any hallide ions
Halogens are ___________ agents
Hallides are ___________ agents
Halogens are oxidising agents
Hallides are reducing agents
The trend in reducing ability of the hallide ions
-hallide ions are good reducing agents
-as they donate electrons to the species being reduced and are themselves oxidised
-reducing ability of hallides increases down group
-due to shielding and weaker electrostatic forces of attraction between nucleus and outermost electron
-makes it easier to lose electrons from larger ions
the reactions of solid sodium halides with concentrated sulfuric acid.
Fluoride and Chloride ions:
NaF + H2SO4 —> NaHSO4 + HF
(Hydrogen fluoride is a colourless, poisonous gas, but produces misty fumes when it contacts the moisture in the air)
NaCl + H2SO4 —> NaHSO4 + HCl
(Hydrogen chloride is a gas and produces steamy fumes)
Bromide ions:
NaBr + H2SO4 —> NaHSO4 + HBr
2H+ + 2Br- + H2SO4 —> Br2 + SO2 + 2H2O
(HBr initially produces steamy fumes, and then later a brown gas (bromine) and a odourless, colourless gas (SO2)
Iodide ions:
NaI + H2SO4 —> NaHSO4 + HI
2HI + H2SO4 —> I2 + SO2 + 2H2O
6HI + SO2 —> H2S + 3I2 +2H2O
(HI produces steamy fumes. Iodine is a black solid. H2S produces a rotten egg smell. Iodine gas produces purple fumes)
The use of acidified silver nitrate solution to identify and distinguish between halide ions.
-AgNO3 (aq) is used to test for hallide ions
-precipitates for different ions are difficult to distinguish
-hence dilute ammonia and then concentrated ammonia can be used to further test for ions
AgNO3 observations:
F- ions: no precipitate formed
Cl- ions: white precipitate
Br- ions: cream precipitate
I- ions: yellow precipitate
Dilute NH3 observations:
AgCl: precipitate dissolves
AgBr: no change
AgI: no change
Concentrated NH3 observations:
AgCl: precipitate dissolves
AgBr: precipitate dissolves
AgI: no change
The trend in solubility of the silver halides in ammonia.
-going down Group 7, solubility of silver hallides decreases
-silver chloride is the most soluble
-silver iodide is the least soluble
Why:
• silver nitrate solution is used to identify halide ions
• the silver nitrate solution is acidified
• ammonia solution is added.
-silver nitrate can be used to identify hallide ions, as silver hallide precipitates with different colours form
-acidifying the silver nitrate solution removes any impurities from interfering with the results (carbonate ions and hydroxide ions)
-AgCO3 or AgOH precipitates may form
-give confusing results
-ammonia solution is added as the colours of the precipitates may be difficult to distinguish
-so ions can be detected by their ability to dissolve in dilute/conc NH3
The reaction of chlorine with water to form chloride ions and chlorate(I) ions.
Cl2 + H2O ⇌ ClO- + Cl- + 2H+
Cl2 + H2O ⇌ HClO + HCl
-UV light must be absent
-disproportionation reaction
-chlorine is both oxidised and reduced
