Why Do Transition Elements Exhibit Multiple Oxidation States?
The Role of the 4s and 3d Orbitals
- To understand why transition elements exhibit multiple oxidation states, we need to delve into their electron configurations.
- Transition elements belong to the d-block of the periodic table, meaning their defining electrons occupy the 3d sublevel.
- However, the 4s orbital, which is filled before the 3d orbital, also plays a crucial role in this behavior.
Electron Configuration of Transition Elements
The general electron configuration of transition elements is:
$$ [\text{Noble gas}] \, 4s^2 \, 3d^{(1-10)} $$
Example
- Scandium (Sc): $[Ar] \, 4s^2 , 3d^1$
- Iron (Fe): $[Ar] \, 4s^2 \, 3d^6$
- Copper (Cu): $[Ar] \, 4s^1 \, 3d^{10}$
Note
- A key point to note here is that the 4s and 3d orbitals are very close in energy.
- This proximity allows electrons to be removed from either orbital with relatively little energy difference, enabling the formation of ions with varying oxidation states.
Tip
When writing electron configurations for transition elements, remember that electrons are removed from the 4s orbital before the 3d orbital during ionization, even though the 4s orbital is filled first.
Why Multiple Oxidation States?
Successive Ionization Energies
Definition
Successive ionization energy
Successive ionization energy refers to the energy required to remove electrons one at a time from an atom.
- The ability of transition elements to exhibit multiple oxidation states stems from the small differences between their successive ionization energies.
- For transition elements, the energy required to remove the first, second, and even third electrons is relatively similar.
- This is due to two main reasons:
- The 4s and 3d electrons are close in energy.
- The nuclear charge (positive charge of the nucleus) is not fully shielded by the 3d electrons, making it easier to remove additional electrons.
Example
- Iron (Fe) can lose two electrons from the 4s orbital to form $ \text{Fe}^{2+} $ $[Ar] \, 3d^6$.
- It can also lose a third electron from the 3d orbital to form $ \text{Fe}^{3+} $ $[Ar] \, 3d^5$.
Example
- Consider manganese $Mn$, which has the electron configuration $[Ar] \, 4s^2 \, 3d^5$.
- Manganese can form ions with oxidation states ranging from +2 to +7.
- For $ \text{Mn}^{2+} $, the 4s electrons are removed, leaving $[Ar] \, 3d^5$.
- For $ \text{Mn}^{7+} $, all 4s and 3d electrons are removed, leaving only the argon core.
Stability of Oxidation States
- Not all oxidation states are equally stable.
- The stability of a particular oxidation state depends on two key factors:
- Electron Configuration:
- Certain configurations, such as half-filled ($d^5$) or fully filled ($d^{10}$) d sublevels, are particularly stable.
- Ligand Effects:
- When transition elements form compounds, the surrounding atoms or molecules (ligands) can influence the stability of different oxidation states.
- This is especially important in coordination chemistry, where ligands can stabilize higher or lower oxidation states depending on their electronic properties.
- Electron Configuration:
Example
$ \text{Fe}^{3+} $ $[Ar] \, 3d^5$ is more stable than $ \text{Fe}^{2+} $ $[Ar] \, 3d^6$ because a half-filled d sublevel is energetically favorable.
Common Mistake
- Students often assume that higher oxidation states are always less stable.
- However, the stability of an oxidation state depends on the specific element and its environment, including the influence of ligands or the type of chemical reaction.
Self review
Can you explain why $ \text{Mn}^{2+} $ is more stable than $ \text{Mn}^{3+} $, but $ \text{Fe}^{3+} $ is more stable than $ \text{Fe}^{2+} $?
