# Electronic and geometric configuration of Fe(OH)₂

I'm having serious problems in the electron configuration of iron dioxide $\ce{Fe(OH)2}$ because all I could come up with was $\ce{Fe}$ double bonded with each oxygen and single bonded with each hydrogen.

Everything seems good except for the iron that's bonded with 5 atoms (which I think is weird). Could someone you give me the right electron configuration (or geometry)?

I apologize if there's something to this that I don't know about, but my understanding of it is...

We have $\ce{Fe(OH)2}$, also called iron(II)hydroxide (iron-two-hydroxide). The reason it's not called dihydroxide, is that the number (II) describes the oxidation state of the iron, so you can easily calculate how many hydroxyl-groups (which are always -1) it needs to have an overall charge of 0.

In iron(II)hydroxide you don't have covalent bonds (double bonds, single bonds)... You have ionic bonds. This particular compound consists of a $\ce{1Fe}^{2+}\ce{+2OH-}$. They don't SHARE electrons, they are simply held together by their charges!

Ionic compounds group together in lattices. They don't have "geometric" configurations like sulfur dioxide for example. Instead, they are clustered together like in the image below: The negative dots would be the $\ce{OH-}$ and the positive dots would be $\ce{Fe}^{2+}$. Then, of course, you would have two negatives for each positive. Ion bonds are the strongest type of bonds there is.

• Thak you for your fast response ,indeed It's iron 2 hydroxide .The reason I've made the mitake is that I studied them in French .Your answer is more than clear ,It's a crystal so It has ionic bonds not covalent ones...thank you very much! – Xman Isdivine Feb 5 '13 at 20:33