Reducing Power Increases Down the Group
To reduce something, the halide ion needs to lose an electron from its outer shell. How easy this is depends on the attraction between the nucleus and the outer electrons.
As you go down the group, the attraction gets weaker because:
1. The ions get bigger, so the electrons are further away from the positive nucleus
2. There are extra inner electron shells, so there's a greater shielding effect.
An example of them doing this is the good ald halogen/halide displacement reactions.
Reactions with H2SO4 & NaF/NaCl with H2SO4
All halides react with concentrated sulphuric acid to give a hydrogen halide as a product to start wth. But what happens next depends on which halide you've got.
Reaction of NaF or NaCl with H2SO4
NaF+ H2SO4 ---> NaHSO4 + HF
NaCl + H2SO4 ---> NaHSO4 + HCl
Hydrogen fluoride (HF) or hydrogen chloride gas (HCl) is formed. You'll see misty fumes as the gas comes into contact moisture in the air.
But HF and HCl aren't strong enough reducing agents to reduce the sulphuric acid, so the reaction stops there.
It's not a redox reaction - the oxidation states of the halide and sulphur stay the same (-1 and +6).
Reaction of NaBr with H2SO4
NaBr + H2SO4 ---> NaHSO4 + HBr
2HBr + H2SO4 ----> Br2 + SO2 + 2H2O
-1 +6 0 +4
The first reaction gives misty fumes of hydrogen bromide gas (HBr).
But the HBr is a stronger reducing agent than HCl and reacts with H2SO4 in a redox reaction.
The reaction produces choking of SO2 and orange fumes Br2.
Reaction of NaI and H2SO4
NaI + H2SO4 ---> NaHSO4 + HI
2HI + H2SO4 ---> I2 + SO2 + 2H2O
-1 +6 0 +4
6HI + SO2 ----> H2S + 3I2 + 2H2O
-1 +4 -2 0
Same initial reaction giving HI gas
The HI then reduces H2SO4 like above.
But HI (being well'ard as far as reducing agents go) keeps going and reduces the SO2 to H2S.
H2S is toxic and smells of bad eggs.
Test for Halides
The test for halides is dead easy. First you add dilute nitric acid to remove ions which might interfere with the test. Then you just add dilver nitrate solution (AgNO3). A precipitate is formed (of silver halide).
Ag+ + X- ----> AgX (where X is F, Cl Br or I)
The colour of the precipitate identifies the halide.
Fluoride (F-) = no precipitate, Chloride (Cl-) = white precipitate), Bromide (Br-) = cream precipitate, Iodidie (I-) = yellow precipiate
Then to be extra sure, you can test your results by adding ammonia solution. Each silver halide has a different solubility in ammonia.
Chloride (Cl-)= white precipitate, dissolves in dilute NH3, Bromide (Br-) = cream precipitate, dissolves in conc. NH3, Iodide (I-) = yellow precipitate, insoluble in conc. NH3