D block elements and the cation of cobalt

Question

The configuration of cobalt in its ground state is $\ce{[Ar] 4s^2 3d^7}$. when it loses 2 electrons, it supposedly leaves from s orbital making it $\ce{[Ar] 4s^0 3d^7}$. But since the s and d orbital have almost similar energy, they tend to exchange electrons from d to s and vice versa to attain a stable configuration. Wouldn't it make sense that when cobalt loses 2 electrons, 2 electrons are lost from s orbital and then 2 electrons are donated from d orbital to the s orbital, making it $\ce {Co^{2+} = [Ar] 4s^2 3d^5}$. since 3d orbital is half filled and it is more stable than partially filled orbital?

Answer 1

This is certainly a misconception and therefore an important question. There is a difference between unstable free ions that have a small mean free path before getting electrons somewhere and ions due to oxidation states in a compound. Let us assume that we have cobalt atoms in the air that do not interact with each other before they form the well known hexagonal solid. The ground state of a cobalt atom has $4s$ orbitals for two electrons and five $3d$ orbitals for seven electrons. The latter are assumed to be degenerate. The radial extent of the $3d$ orbitals is not large compared to the $4s$ orbitals, but the $s$ orbitals have the great advantage of being closer to the nucleus ($\ell=0$) and therefore always lower in energy compared to the $3d$ orbitals. The first and second ionization energies for cobalt are 760.4 kJ/mol and 1648 kJ/mol, respectively. When these free atoms are doubly ionized, the ionized electrons will come from the $3d$ orbitals because they are higher in energy and the configuration will be [$\ce{Ar}$] $3\mathrm{d}^5$ $4\mathrm{s}^2$, two electrons remain in the $4s$ orbitals.

The misunderstanding comes from the fact that without interacting (free atom), a transition of an electron from $4s$ $\rightarrow$ $3d$ is unlikely to occur. This occurs when the $\ce{Co^2+}$ ion interacts with the environment or is in a compound and have an oxidation state due to this interaction, the $4s$ state will be delocalized due to its radial extent and the orthogonality is almost broken, a partial $4s$ charge will fill the $3d$ orbitals. This process is exactly the same for hybridization, where a charge is transferred to higher $\ell$ for interacting with neighboring atoms, creating bonds. Without such interaction, the $4s$ orbitals should always be filled in free $\ce{Co}$ atoms.