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Oxidation number of SO4

We are going to look at some examples from vanadium chemistry. If you don't know anything about vanadium, it doesn't matter in the slightest.

Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons:

The vanadium is now said to be in an oxidation state of +2.

Removal of another electron gives the V3+ ion:

The vanadium now has an oxidation state of +3.

Removal of another electron gives a more unusual looking ion, VO2+.

The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one).

The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element.

It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5.

Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1.

Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero.

What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur.

The sulphur has an oxidation state of -2.


Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.

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