electronic configuration - Find the valences of Z=26

I've tried this:

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?

Answer

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:

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

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