# Question #0162a

Mar 9, 2015

No. That's not right.

Iron has the electronic structure:

$\left[A r\right] 3 {d}^{6} 4 {s}^{2}$

The 4s electrons are in the highest energy level so can be regarded as the valence electrons.

When these are lost in chemical reactions the $F {e}^{2 +}$ is formed which is :

$\left[A r\right] 3 {d}^{6}$

The +3 state can also be formed which is:

$\left[A r\right] 3 {d}^{5}$

This gives the $F {e}^{3 +}$ ion.

Mar 10, 2015

This is not right.

A valence electron for a transition metal is defined as an electron that resides outside a noble-gas core.

The electron configuration of iron is $\text{[Ar]} 4 {s}^{2} 3 {d}^{6}$.

The d electrons are valence electrons, so Fe has 8 valence electrons.

That does not mean that Fe will use all eight of its valence electrons to form compounds.

The $4 s$ electrons are highest in energy, so Fe can lose these two electrons to form Fe²⁺ ions.

Fe can also lose a $3 d$ electron to form Fe³⁺.

These are the most common oxidation states of Fe.

In some compounds, Fe has oxidation states of +4, and +6, but they are uncommon.

The number of valence electrons in a transition metal that will actually participate in a chemical reaction is hard to predict.

So, the concept of valence electrons is less useful for a transition metal than for a main group element.

Transition metal chemists use words like "d-electron count" and the "18-electron rule", but they are more advanced concepts.