Question

How would you explain oxidation states?

Answer 1

Oxidation state is the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge, the electron, retained by the most electronegative atom.

Explanation:

We recognize that `"oxidation number"`, `"oxidation state"` are formalisms. That is the term does not have any real physical significance, but is a convenient and useful fiction for balancing chemical equations. We formalize `"oxidation"` as the `"LOSS of ELECTRONS"`; and `"reduction"` as the `"GAIN of ELECTRONS"`. Neutral elements, that have not engaged in chemical bonding, are conceived to have an oxidation state of `0`.

When we burn carbon in air, the reactants both have `0` oxidation numbers:

`C(s) + O_2(g) rarrCO_2(g)`

The oxygen molecule is conceived to have accepted 4 electrons from the carbon to give `2xxO^(2-)`; i.e. it is reduced to `O^(-II)`. As with any chemical reaction, mass and charge are conserved. And thus formally we have `C^(IV+)`, i.e. tetravalent `C` in a `+IV` oxidation state.

We could split the reaction up into half equations:

`"Oxidation:"`

`CrarrC^(4+) + 4e^-`

`"Reduction:"`

`O_2 +4e^(-)rarr2O^(2-)`

And we add the individual reactions, to eliminate the electrons:

`C(s)+O_2(g)rarrCO_2(g)`

There should be many more answers here on Socratic that deal with such problems. You have to dig for them. See here and here and [here.](https://socratic.org/questions/how-do-you-find-the-oxidation-numbers-of-these-compounds)