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Fe(26) has eight(8) valence electrons, but it's valency is 2 and 3. How can I calculate this?

Actually, iron does not have 8 per se, it only has 2, but you could say that it has 8 - here's why.

Remember that valance electrons are the electrons located in the outermost shell of an atom. If you take iron, its is 26, which means that a neutral iron atom has 26 electrons.

The of iron is

##"Fe": 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(color(red)(2)) 3d^(6)##

Iron's outermost shell is the n = 4 shell, the ##color(red)(2)## electrons that occupy it being located in the 4s subshell. These two electrons are iron's valence electrons.

Transition metals however can use electrons located in their inner shells as as well. For iron, when the two electrons located in the 4s subshell are removed, it has a +2 and this electron configuration

##"Fe"^(2+): 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6)##

Now the n = 3 becomes the outermost shell; iron can lose electrons from this shell as well, more specifically from the the 3d subshell which holds 6 electrons.

When one electron from the 3d subshell is removed, iron has a +3 and the electron configuration

##"Fe"^(3+)": 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5)##

+2 and +3 are the most common oxidation states of iron, but it can have oxidation states ranging from -2 to +6.

Read more about iron's valence electrons here:



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