How can I derive the preferred oxidation states of a transition metal from its electronic configuration?
For example $\ce{Fe}\ (Z=26)$ Short formula: $\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^6\ 4s^2}$. If we make the electronic formula we get 4 single electrons ($s=1/2$), which means that the valence in normal state is 4, but in the periodic table it is $2/3/6$. How can I find its valence?
transition-metals valence-bond-theory
Annalian LoverreAnnalian Loverre
Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.
ed kinsellaed kinsella
$\begingroup$ Can you explain me what " two 4s electrons are lost first giving a valency of two" means? How can they be lost? Does that mean they go direclty to 3d? $\endgroup$ – Annalian Loverre Nov 25 '14 at 19:14
The simple valence bond model works well for nonmetals and the molecules they form. It fails completely and badly when trying to extend it to transition metal compounds. There is no easy way to explain why iron, which has four unpaired electrons in the ground state ($S = 2$) is typically found in the oxidation states $\mathrm{+II}$ and $\mathrm{+III}$ in compounds in aquaeous solution.
This behaviour can only be explained by complex molecular orbital theory. It involves a lot of delicate balances that typically cannot be explained a priori without extensive stability calculations. However, the general gist is that the orbitals are split into different energy levels depending on the compound's symmetry and the orbitals' irreducible representation. At first approximation, we can assume a stable configuration if there are no degenerate orbitals that are unevenly populated, but single or multiple unpaired electrons are not a problem. Indeed, iron(III) compounds, that are often very stable, even have a spin of $5/2$ in high-spin state. The secret thereto is even orbital population.
I am afraid that I can give you no real guidance for transition metals except learning, rote memorisation of typical cases and developing an intuition for the more complex ones.
JanJan
The electron configuration of iron is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2. Here, the 4s shell is the outermost shell. So, iron has 2 valence electrons. While reacting, these 2 electrons can go to any other electronegative element who requires these electrons to complete its outermost shell. So, this gives iron a valency of 2. Now, the one out of 2 orbitals is completely filled. So, due to repulsion this electron can also be given out. This gives iron a valency of either 2 or 3. However, iron is mostly found in valency 3 as it is mostly stable.
DakshPandeyDakshPandey
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