Topic 6: Redox Reactions and the Halogens

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The halogens are the group 7 elements. I will discuss their distinctive properties, followed by an experimental proof for each.


Physical Properties



Standard State




Pale Yellow



Pale Green












From the above table, we can see the following patterns:


-          State changes from Gas à Solid as you go down the group, indicating a rise boiling point as you go down the group

-          A gradual darkening in colour as you go down the group.


These obvious patterns can help us predict the state and appearance of Astatine. Astatine is an extremely unstable substance, with such a short half-life it is impossible to experiment with. The increase in boiling points suggests it’s a solid, and the darkening of colour suggest it’s a grey/black. Scientific research has confirmed this.


Why the increasing boiling points?


Something you have come across in Topic 7: Van der Waals forces. An increase in number of electrons down the group means that stronger Van der Waals temporary dipoles can be formed. The stronger forces down the group means you’ll need more energy to separate molecules and change state.


Electronic Properties




Electron Configuration



1s2 2s2 2p5



1s2 2s2 2p6 3s2 3p5



1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 



1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5



*1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d10 6p5*



We can conclude from the above table that:


-          All halogens are diatomic in their natural state (Have two atoms per molecule)

-          They are all found in the p-block, with electrons in their outer shell occupying the s-orbitals, and 5 of the 6 p-orbitals.


They therefore, form halides in ionic bonds, and all will give the same formula with the same cation, because they all form 1- ions.


Reactions Of Halogens With Alkali


When halogens are reacted with Alkalis, a disproportionation reaction occurs. This is when the oxidation number of an element both increases and decreases in the same reaction. How much disproportionation that occurs though depends on if the alkali was hot or cold.




If cold, the following reaction occurs:




Cl2 (g) + 2OH- --> Cl- (aq) + ClO- (aq) + H2O (l)




If hot:


3Cl2 (g) + 6OH- --> 5Cl- (aq) + ClO3- (aq) + 3H2O (l)


In the hot alkali, the chlorine disproportionates to be reduced by -1 and oxidised by +5. This happens because the chlorate (I) formed disproportionates as well, giving the general equation above.


Whilst doing this experiment, the obvious observation is the decolourisation. The distinctive colours of bromine and iodine in solution (orange-brown) is immediately exchanged with colourless solutions of chloride and chlorate.


In the cold alkali, we get chlorate (I), and we get chlorate (V).


The Halides


Properties of halides investigated can help uncover properties of the halogens.


-Addition of Silver Nitrate


You may remember this procedure as an analytical test for ions. With halogens, it gives the following results:



Effect of adding silver nitrate

Effect of adding  Conc. Ammonia

Effect of sunlight on silver halide


White ppt.




Cream ppt.

Colourless, but with rigorous stirring



Yellow ppt.




The following reaction occurs to give the precipitates:


Ag+(aq) + X- (aq) --> AgX (s)


It is, in fact a precipitation reaction. The addition of ammonia dissolves the precipitate, by the following reaction:


AgX(aq) + 2NH3(aq) --> Ag(NH3)2X


The complex ion on the right is called diamminesilver(I) halide. It is soluble, leaving a colourless solution. Although dilute ammonia is sufficient for silver chloride, only concentrated ammonia can dissolve silver bromide.


-Reaction with Concentrated Sulphuric acid


When heated with concentrated sulphuric acid, the halogens show a trend as reducing agents:









HBr , SO2(g) , Br2


HI , SO2 (g), H2S (g) , I2


The halide ions are reducing agents. In the case for Fluoride and Chloride, they aren’t strong enough to reduce the sulphuric acid. Instead, the sulphuric acid donates a proton (H+) and an acid is released (HF or HCl). For bromide ions, it is different because they are stronger reducing agents. They reduce the sulphuric acid to SO2 gas, and are oxidised in the process (by the acid):


2Br- --> Br2 + 2e-


So, what about iodine? It is an even stronger reducing agent, and furthermore reduces the SO2 to S, then even more to H2S (Hydrogen Sulphide gas).


The tests for these gases  are:


HX (Hydrogen Halide): Bring a source of Concentrated Ammonia to the neck of the tube. White fumes should form (If it was not already fuming!)


SO2: Dip a piece of filter paper in Potassium dichromate (VI), then hold in gas. It should turn from orange to green.


H2S: Hold a strip of Lead ethanoate paper . It will turn from white to black.


We can see from the above experiments, that sulphuric is doing two things: Its acting as an acid (Donates H+ to make HX) and also acts as an oxidising agent (It oxidises bromide to bromine for example).


- Hydrogen Halides


When the experiment above is repeated with phosphoric acid, all we get are the hydrogen halides. We can experiment with them the following:


-          Their solubility:


Invert the test-tube of the hydrogen halide into water, and the level of water should rise. This is because the gas is dissolving in the water. You will be left with some residual gas, which is just air from outside.


-          Their reaction with ammonia:


This is the same as the test for hydrogen halides as described above. Here is the equation:


HX + NH3 --> NH4X


-          Their thermal stability:


When you dip a hot nichrome wire in the gas, nothing happens for HCl, but you get a decomposition for HBr and HI. The distinctive vapour colour appear (Brown and Purple respectively)


We can see from the above reaction of phosphoric acid, that it only acts as an acid, not as an oxidising agent, unlike sulphuric acid.



Click on the following link to download a few Topic 6 related pages from the Nuffield CD (I got this from the school site)