Actual Oxidation State versus Average Oxidation State

In this JC1 webinar we want to compare Actual Oxidation State versus Average Oxidation State.

Let's use S4O62- as an example and calculate the oxidation state of sulfur.

actual oxidation state calculate OS of sulfur

Using oxidation state of -2 for oxygen as a reference, we can determine sulfur to have oxidation state of +2.5.

It might not seem unusual but once we consider the definition of oxidation state, we will realise that it is not possible.

Oxidation state of an atom is the charge that it carries as a monatomic ion for ionic compounds, or the hypothetical charge it would have if it were an ion for covalent molecules.

Check out this video for a more detailed discussion of the meaning of oxidation state.

So for sulfur to have an oxidation state of +2.5, it must lose 2.5 electrons to form an ion of +2.5 charge.

This is not possible as we cannot lose half an electron.

Hence what we are calculating is the average oxidation state of sulfur, which is good enough for us to handle most redox reactions and questions.

So how do we determine the actual oxidation state of sulfur then?

We would need to consider the lewis structure of S4O62- and break all the covalent bonds to determine the hypothetical charge of sulfur if it were an ion.

actual oxidation state lewis structure of S4O62minus

Oxygen is more electronegative than sulfur hence:
- when O-S single bond is broken, O will gain -1 charge and S will gain +1 charge;
- when O=S double bond is broken, O will gain -2 charge and S will gain +2 charge

There is no difference in electronegativity in S-S bond hence no charge is acquired when S-S bond is broken.

Therefore we can deduce that:
- the yellow sulfur that is directly bonded to 3 oxygen atoms will acquire a charge of +5 when it forms an ion. Hence oxidation state of that yellow sulfur is +5.
- the blue sulfur will not acquire any charge when all bonds around it are broken. Hence oxidation state of that blue sulfur is zero.

Notice if we sum up all the actual oxidation states of sulfur (5 + 0 + 0 + 5 = 10) and divide by the number of sulfur atoms (4), we have the average oxidation state of +2.5.

We need to know this technique of determining actual oxidation state of a specific atom in a compound as it will be useful in Organic Chemistry.

Topic: Redox Reactions, Physical Chemistry, A Level Chemistry, Singapore

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