What are "oxidation numbers"?

Oct 25, 2016

$\text{Oxidation number}$ is a $\text{conceptual}$ property of atoms, not molecules. However, we can make a stab at the oxidation numbers of each atom in each molecule or salt.

Explanation:

The sum of the oxidation numbers always equals the charge on the ion. If we deal with a neutral molecule, the oxidation numbers of the constituent atoms must sum to ZERO.

Most of the time, hydrogen, in its compounds has an oxidation state of $+ I$, i.e. it is conceived to have donated 1 electron, and we treat it as ${H}^{+}$; and oxygen in its compounds has an oxidation state of $- I I$, it is conceived to have accepted 2 electrons, and we treat it as ${O}^{2 -}$. (Peroxides and hydrides are the exceptions!).

So for $\text{ammonium biphosphate, } {\left(N {H}_{4}\right)}_{2} H P {O}_{4}$, the oxidation number of hydrogen is $+ I$, and that of oxygen $- I I$; with some simple algebra, I can assign a $- I I I$ state to nitrogen, and a $+ V$ state to phosphorus. If you add these oxidation states up: $2 \times - I I I + 8 \times I + V - 4 \times I I + I = 0$ you get zero as is required for a neutral compound or salt.

And for $X e O {F}_{4}$, the oxidation number of $X e = V I + \left(O = - I I , F = - I\right)$, for ${C}_{8} {H}_{10}$, the carbon has an average oxidation number of $- \frac{10}{8}$ (the terminal carbons are $- I I I$, and the methylene, $C {H}_{2}$, carbons are $- I I$, vinyl, $= C H$ carbons are $- I$; the average is $- \frac{10}{8}$.

And for $B a S {O}_{4}$, $B a = + I I , S = + V I , \mathmr{and} O = - I I$.

I have thrown a lot of facts and calculations at you. Mind you, you posed an open-ended question, which required a lot of background. The point to learn is that the sum of the oxidation numbers equals the charge on the ion. I think these calculations are well within the grasp of a 2nd year chemistry student. If there are further questions, post them, and someone will help you.