Question

What are the oxidation number rules?

Answer 1

Oxidation states are hypothetical charges we assign by assuming the bonds are completely ionic, i.e. complete transfer of valence electrons to the more electronegative atom.

Really, the core hierarchy is:

1. Charge is conserved, so that all the oxidation states in a neutral substance add up to `0`, or in an ion, add up to its charge.
2. Pure elements have an overall oxidation state of `0` (follows from `(1)`).
3. Oxidation states are predicted by the typical charge obtained with a noble gas electron configuration.
4. The more electronegative atom is given the negative oxidation state and vice versa. This one is especially important for non-obvious cases when we have to designate uncommon oxidation states to an atom that won't follow typical valencies.
5. In a bond between two identical atoms, they have the same oxidation state, whatever it may be. This rule is governed by `(1)`.

From these rules, the typical oxidation state hierarchy is derived to make it more simple (and yet, longer!):

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EXAMPLE

`["Ag"("NH"_3)_2]^(+)`

`"NH"_3` is a neutral molecule, ammonia. As a general chemist, one might predict that `"N"` atom in a molecule attains a typical noble gas configuration, as `"N"^(3-)`.

Indeed, `"N"` is more electronegative than `"H"`, so it gets the more negative oxidation state, while `"H"` gets whatever is left.

`stackrel(color(blue)(-3))"N"stackrel(" "color(blue)(3 xx +1))("H"_3)`

And as we wanted,

`underbrace(overbrace(-3)^(N^(3-)) + 3 xx overbrace(+1)^(H^(+)))_(NH_3) = 0`,

for the neutral molecule. Thus, with two of them, the charge from `"NH"_3` is still zero.

By conservation of charge, since the ion charge overall is `1+`...

`[stackrel(color(blue)(+1))"Ag"stackrel(color(blue)(" "2[-3 + (3xx+1)]))(("NH"_3)_2)]^(+)`

...it follows that `"Ag"` has an oxidation state here of `+1`.

`underbrace(overbrace((+1))^(Ag) + 2 xx overbrace([(-3) + (3 xx +1)])^(NH_3))_(["Ag"("NH"_3)_2]^(+)) = 1+`

And that makes sense because `"Ag"` indeed commonly forms the `+1` oxidation state, and is not known to use any more than `1` valence electron.