Question

Predict the two most common oxidation states of the ions of iron. Explain your reasoning in terms of energy.

Answer 1

The two most commonoxidation states of iron are Fe(II), in which the iron ion shares two of its electrons, and Fe(III), in which it shares three electrons.

The electronic configuration of iron, Fe = [Ar] 3d6 4s2

So, electronic configuration of iron, Fe (II) = [Ar] 3d6 4s0
   electronic configuration of iron, Fe (III) = [Ar] 3d5 4s0

Fe(III) has the extra stability associated with half-filled subshells.

Because the ns and (n − 1)d subshells in these elements are similar in energy, even relatively small effects are enough to produce apparently anomalous electron configurations.

Iron's outermost shell is the n = 4 shell, the 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 valence electrons as well. For iron, when the two electrons located in the 4s subshell are removed, it has a +2 oxidation state and this electron configuration

Fe (II) = [Ar] 3d6 4s0

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 oxidation number and the electron configuration

Fe (III) = [Ar] 3d5 4s0

+2 and +3 are the most common oxidation states of iron.