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I've tried this: enter image description here

But its hows that the valence of Fe in normal state is 4, when in the periodic table it isn't. Why does this happen? Can you show me the answer with configuration formula in the states with energy?

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    $\begingroup$ You have the electron configuration of iron in your post. Just writing that out is only the first step. What have you done with it to attempt to figure out valences? $\endgroup$ – Ben Norris Nov 23 '14 at 11:54
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Hint: $$_{26}Fe\equiv1s^22s^22p^63s^23p^64s^23d^6\equiv [Ar]\ce{\underbrace{ ^ v }_{4s^2} \underbrace{ ^ v ,\;^ ,\;^ ,\;^ ,\;^ }_{3d^6} }$$ Think now what will be the configuration after losing electrons:

Spoiler:

! $$Fe^{2+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^ v ,\;^ ,\;^ ,\;^ ,\;^ }_{3d^6}=[Ar]\underbrace{ ^ }_{4s^2} \underbrace{ ^ ,\;^ ,\;^ ,\;^ ,\;^ }_{3d^6} }$$

! $$Fe^{3+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^ ,\;^ ,\;^ ,\;^ ,\;^ }_{3d^6} } $$

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