 So we're going to look at disproportionation reactions and these are a type of redox reaction where a compound which is in an intermediate oxidation state forms products in a different oxidation state which is higher and lower. So if we use this representation for a disproportionation reaction if we start with A which has some oxidation number let's say N. Now when the reaction happens the products formed with A will usually have different oxidation states in which one is greater than N and one is less than N. So for A it is going from an intermediate oxidation state to a higher oxidation state and a lower oxidation state as the reaction happens. So let's take an example of a reaction in which this happens and here you can see that TIN which is in the form of SN2 plus is going to SN4 plus and here since this is in free state the oxidation number will be 0. So in one case the oxidation number has reduced from 2 to 0 and in the other case it has increased from 2 to 4. So this reaction is an example of a disproportionation reaction and interestingly this was also the first disproportionation reaction that was ever studied in the late 1700s. Now let's take one more example which is this reaction of H3PO3 decomposing to give H3PO4 and pH3. So now let's look at the oxidation state of phosphorus in all three of these. So we know that the oxidation state for hydrogen is plus 1 and for oxygen it is usually minus 2 and we know that this is a neutral species so the sum of all the oxidation numbers should be 0. So we can write down 3 times the oxidation number of hydrogen which is 1 plus the oxidation number of phosphorus let's say it is x plus the oxidation number of oxygen which is 3 times minus 2 should be equal to 0. So if you solve for x here you get x to be plus 3. So the oxidation state of this phosphorus is plus 3. Now if you consider the case of H3PO4 again we know that hydrogen will be plus 1 and oxygen will be minus 2. So we can write down 3 times the oxidation number of hydrogen which is 1 plus let's say the oxidation number is x in the case of this phosphorus plus 4 times minus 2 should be equal to 0 because this is a neutral species. So if you again solve for x here we get x to be plus 5. Now in the last case which is of pH3 we know that hydrogen has an oxidation state of plus 1 and there are 3 hydrogens here so which means phosphorus here has an oxidation state of minus 3. So in this case phosphorus went from a plus 3 state to plus 5 in one case and it reduced to minus 3 in the other case which is why this reaction is also an example of a disproportionation reaction. Disproportionation reactions like these are mostly observed in cases where one of the species shows variable oxidation states. So let's see some more examples of this. Apart from phosphorus sulfur also shows multiple oxidation states. So like here if we look at the oxidation numbers of sulfur here it will be 0 because it is a free species. Now here let's say we want to calculate what will be the oxidation number of sulfur. Let's say it is x so we have 2x plus 3 times the oxidation number of oxygen which is negative 2 is equal to the charge here which is minus 2. So if you solve for x you get x equal to plus 2 so that means the oxidation state of sulfur here is plus 2 and here since the charge is too negative the oxidation state of sulfur is as it is minus 2. So we have sulfur going from 0 to plus 2 that is increasing and 0 to minus 2 that is decreasing and so this also is an example of a disproportionation reaction. One more example is of chlorine in this reaction. So again if we were to look at the oxidation numbers chlorine goes from 0 here to minus 1 in this case and for this chlorine let's say the oxidation number is x we have x plus 3 times minus 2 which should add up to the charge of the anion which is minus 1. So if we solve for x we get x to be plus 5. So this chlorine will have an oxidation number of plus 5. So again in this case also chlorine goes from 0 to minus 1 in one case and in the other it increases from 0 to 5 which makes this also a disproportionation reaction.