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 $\mathrm{Co}^{2+}$ have 7 valence electrons in the $3d$ orbital? Doesn't the superscript $^{2+}$ denote that it is missing 2 electrons, and therefore isoelectric with $\mathrm{Mn}$?


2 Answers 2


Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ 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 $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$. 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 $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.

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 $\ce{Co}$ atom, they are lost from the 4$s$ 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.


Electronic configuration of $\ce{Co}$ is as follows: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^7$.

When Cobalt loses 2 electrons to become $\ce{Co^{2+}}$ it loses the electrons which are in $4s^2$, not the ones in $3d^7$ because the electrons in $4s^2$ have a high reactivity.

So $\ce{Co^{2+}}$ still contains 7 electrons in it's valence shell. The new configuration of $\ce{Co^{2+}}$ is $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^7$.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$. I do also believe this is not quite correct. $\endgroup$ May 7, 2015 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.