Calculating valence of oxides Learning about Oxides. Basically when oxygen is combined with a metal.

$$\ce{FeO}$$

This is called "Iron Oxide (II)" according to my book.

Apparently, the II represents the valence. But how come? Look:

$\ce{Fe}$ has the configuration $1s^22s^22p^63s^23p^6\color{red}{4s^2}3d^6$. Since the outermost shell has $2$ electrons, the valence is $2$.

Now $\ce{O}$ has the configuration $1s^2\color{red}{2s^22p^4}$. Its outermost shell has $6$ electrons so oxygen has a valence of $6$ right?

Then how come $\ce{FeO}$ has valence $2$? Shouldn't the valence of $\ce{FeO}$ be like, the combination of the two valences? So $6 - 2 = 4$? (I read somewhere you had to subtract it).

Not only that. I am starting to think that my definition of valence is completely messed up. I just checked in here and apparently Oxygen has $-2$ valence? Why? And Iron has +2, +3, (+4), (+6)? What does all that even mean?
Valency is the combining power of an atom; so, when we talk of iron having a valency of 2, it can combine either with two atoms of valency of one; or with one atom of valency of two.  Oxygen has a valency of two and so the formula for iron (II) oxide is FeO.

In the history of chemistry valency pre-dates the idea of electron configuration which looks like the source of your confusion.

Hope that is clearer.

Edit: you mentioned that oxygen has a valency of minus 2; that's not quite right.  It has a valency of 2 but it can form an oxidation state of minus 2.  You need to separate out the ideas of valency and oxidation state.  Although the ideas are related, they are distinct.