Copper oxidation states
Unpaired Electrons of d-orbitals
To fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in a d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s-orbital electrons since the 3d-orbital is located before the 4s-orbital in the periodic table.
For example: Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and therefore the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3.
The formula for determining oxidation states would be (with the exception of copper and chromium):
Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons
In the image above, the blue-boxed area is the d block, or also known as transition metals.
Scandium is one of the two elements in the first transition metal period which has only one oxidation state (zinc is the other, with an oxidation state of +2). All the other elements have at least two different oxidation states. Manganese, which is in the middle of the period, has the highest number of oxidation states, and indeed the highest oxidation state in the whole period since it has five unpaired electrons (see table below).
It was mentioned previously that both copper and chromium do not follow the general formula for transition metal oxidation states. This is because copper has 9 d-electrons, which would produce 4 paired d-electrons and 1 unpaired d-electron. Since copper is just 1 electron short of having a completely full d-orbital, it steals an electron from the s-orbital, allowing it to have 10 d-electrons. Likewise, chromium has 4 d-electrons, only 1 short of having a half-filled d-orbital, so it steals an electron from the s-orbital, allowing chromium to have 5 d-electrons.
Rules About Transition Metals
- Free elements (elements that are not combined with other elements) have an oxidation state of zero, e.g., the oxidation state of Cr (chromium) is 0.
- For ions, the oxidation state is equal to the charge of the ion, e.g., the ion Fe3+ (ferric ion) has an oxidation state of +3.
- The oxidation state of a neutral compound is zero, e.g., What is the oxidation state of Fe in FeCl3?
- Answer: Cl has an oxidation state of -1. Since there are 3 Cl atoms the negative charge is -3. Since FeCl3 has no overall charge, the compound have a neutral charge, and therefore the oxidation state of Fe is +3.