What are the oxidation number rules?

Jul 18, 2017

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 $\left(1\right)$).
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 $\left(1\right)$.

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

[

EXAMPLE

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

${\text{NH}}_{3}$ is a neutral molecule, ammonia. As a general chemist, one might predict that $\text{N}$ atom in a molecule attains a typical noble gas configuration, as ${\text{N}}^{3 -}$.

Indeed, $\text{N}$ is more electronegative than $\text{H}$, so it gets the more negative oxidation state, while $\text{H}$ gets whatever is left.

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

And as we wanted,

${\underbrace{{\overbrace{- 3}}^{{N}^{3 -}} + 3 \times {\overbrace{+ 1}}^{{H}^{+}}}}_{N {H}_{3}} = 0$,

for the neutral molecule. Thus, with two of them, the charge from ${\text{NH}}_{3}$ 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 $\text{Ag}$ 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 $\text{Ag}$ indeed commonly forms the $+ 1$ oxidation state, and is not known to use any more than $1$ valence electron.