I’m taking general chemistry 2 this half of the summer. We are currently going over coordination complexes, ligands and transition metal ions. In particular, this question is in regards to the ligand field stabilization energy.
Why does Co2+ have 7 valence electrons in the 3d orbital? Doesn’t the superscript 2+ denote that it is missing 2 electrons, and therefore isoelectric with Mn?
Well, you can find some kind of explanation why the electronic configuration of Co2+ is [Ar]3d7
here and there, but at the end of the day, you just need to * memorize it*.
When d-block elements form ions, the 4s
electrons are lost first. Source
The n+l
rule tells you the order in which atomic orbitals are filled, and according to the rule the 4s orbital is occupied before the 3d orbital because it has lower energy. Thus, the electron configuration of Mn is [Ar]3d54s2 while that of Co is [Ar]3d74s2. But, the n+l rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration. For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation Co2+ you will get the wrong electron configuration [\ce{Ar}] 3d54s2
.
As it is sometimes explained, the statement that 4s
orbital is lower in energy than 3d orbital is true only when the orbitals are unoccupied. But while you fill 3d orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the 4s orbital. Thus, when electrons are lost from Co atom, they are lost from the 4s orbital first because it is actually higher in energy when both 3d and 4s
are filled with electrons.
P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.