What is the oxidation state of each element in febr2?

FeBr2 
1*z + 2*(-1) = 0 
z = 2 is the oxidation state of each element in FeBr2

To determine the oxidation state of a metal, like Fe (Iron) or anything else really, you generally won't know. But look at a periodic table and you'll see that most metals fall into the sort of "valley" section, and this part of the periodic table is called the "d-block" and it has 10 columns. These guys in the d-block of the periodic table have a special way of filling their orbitals, but everything on the left and right are a little simpler to understand. So we see that Br (Bromine) is one of the easier ones to find the oxidation state of, since it's on the right part of the periodic table (where there are 6 columns) and we know that they follow the octet rule, and since bromine is a halogen (That just means it's in the second to last column, which is next to the noble gasses) and things at the top right of the periodic table are highly electronegative, so they'll tend to gain another electron to have a complete octet to fill its orbitals and have a negative charge. Since it only had to gain one to have a complete octet, it's charge is now -1. We can now look at FeBr2 and see that overall the charge is 0. But it has two bromines which each have a charge of -1, so we see that we can set up a little algebra equation: (Charge of Iron) + (Charge of Bromines)=(Total Charge) So, we have: (Charge of Iron) + 2 (-1)=0 So then we can see that Iron is a +2 oxidation state, Iron (II) Bromide.     

Hope that helps 
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