- Ionic radius increases down the group
- Reactivity decreases down the group
- Electronegativity decreases down the group
Fluorine is the most electronegative atom and it can therefore only have 1 oxidation state, -1.
This would give fluorine only the common oxide, while no oxoanions are possible.
Properties of halogens
Bond lengths and bond strengths
The smaller the bond, the stronger the bond, and it would require a lot of energy to break that bond. This is true for all molecules except for fluorine, since the bond is so small that the lone pairs would actually repel each other making it weaker.
The higher the electronegativity of the element the higher the oxidising power. Therefore, Fluorine has the highest oxidation power with Iodine being the least oxidising. Fluorine can never be oxidised itself, and it only has two oxidation number, 0 and -1.
All halogens react with Hydrogen fo produce the halide, with Fluorine being an explosive reaction, Chlorine explosive in sunlight due to the formation of radicals, bromine being a simple reaction while heating is required with Iodine.
The oxidation power of the halogens can also be checked using thiosulfate ion.
S2O32- –> SO42-
Cl2 –> Cl–
S2O32- –> S4O62-
I2 –> I–
If fluorine is the most oxidising agent it will therefore mean that the Iodide ion is the strongest reducing agent. The extra electron on Iodine is far away from the nucleus and therefore it can easily be lost, making it a good reducing agent.
All halogens have Van der Waals’ bonding and therefore these will not dissolve in water (although fluorine and chlorine will actually react with water).
On the other hand, iodine can be made to dissolve by the preparation of the tri-iodide ion, which is iodine joined to an iodide ion, to give I3–. since this has a charge, it will be able to dissolve.
Reaction with Alkali
Halogens react with alkali to give different salts, including the simple salt and oxo-salts. Since fluorine cannot have a positive valence, it reacts in a different way.
Fluorine + alkali
2F2 + 2NaOH –> F2O +2NaF + H2O @ cold temperatures
At higher temperatures the oxide will react with more alkali to give:
F2O + 2NaOH –> 2NaF + H2O +O2
Showing that F2O is an acidic oxide.
The overall reaction would then be:
2F2 + 4NaOH –> 4NaF + 2H2O + O2
Halogen + alkali
For all the other halogens the reaction goes like this:
X2 + NaOH –> NaX + NaXO + H2O
But the oxo-salt is unstable to give:
3NaXO –> 2NaX + NaXO3
Which would result in:
3X2 + 3NaOH –> 5NaX + NaXO3 + H2O
3Cl2 + 3NaOH –> 5NaCl + NaClO3 + H2O
3Br2 + 3NaOH –> 5NaBr + NaBrO3 + H2O
3I2 + 3NaOH –> 5NaI + NaIO3 + H2O
Preparation of Halogens
All halogens (except for Fluorine) can be prepared using MnO2 and sulfuric acids.
2NaCl + MnO2 + 3H2SO4 –> MnSO4 + Na2SO4 + Cl2 (collected via downward delviery through H2SO4)
2NaBr + MnO2 + 3H2SO4 –> MnSO4 + Na2SO4 + Br2 (collected via distilaltion)
2Nal + MnO2 + 3H2SO4 –> MnSO4 + Na2SO4 + l2 (collected via sublimation)
Preparation of Hydrogen halides
All acids can be prepared via the reaction of their salt with sulfuric acid.
NaCl + H2SO4 –> NaHSO4 + HCl
NaBr + H2SO4 –> NaHSO4 + HBrà Br2 if H2SO4 is concentrated
Nal + H2SO4 –> NaHSO4 + Hl à I2 if H2SO4 is concentrated
In order to avoid oxidation of HBr and HI, H3PO4 can also be used.
2NaBr + H3PO4 –> Na2HPO4 + 2HBr
2Nal + H3PO4 –> Na2HPO4 +2Hl
Test for halogens