Why does cobalt have no negative charge?

Question

I would like to know why cobalt cannot have a negative charge (or at least why a negative charge for cobalt isn't typical). I am not sure where I have gone wrong in my reasoning.

The electronic configuration for cobalt is: $\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7$

Clearly, you could give cobalt a $+2$ charge in order to get a more stable ion as follows

$\ce{Co^{2+}}$: $[\ce{Ar}] 3d^7$

And from there, you could go further and remove $2$ more electrons from the $d$-orbital to become even more stable like this:

$\ce{Co^{4+}}$: $[\ce{Ar}] 3d^5$

If this logic is correct, then could I apply this same thinking but for adding electrons? Such that

$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7\to$ $\ce{Co^{3-}}$: $[\ce{Ar}] 4s^2 3d^{10}$

Since the $d$-orbital has been filled, the atom should have become more stable. Is it true?

Thus, am I able to argue that Cobalt has $3$ main ion charges: $+2, +4, -3$?

Answer 1

Cobalt can be found in various oxidation states including negative ones such as Co(-I) in NaCo(CO)4. In this instance however the excess electron density on the cobalt center is relieved by backbonding with the CO ligands and therefore such a low formal oxidation state of the Co atom is stabilized. On the other hand the free Co$^-$$^1$ ion cannot hold the extra electron first due to the low electronegativity of Co as TAR86 hinted out in his comment and second due to the fact that the extra electron is placed in a d orbital which is not very penetrating and thus shielded by all other electrons. Hence, it is unlikely to encounter compounds such as [Na]$^+$[Co]$^-$ where there are no ligands on the cobalt to stabilize the excess electron density. The opossite applies when you remove electrons from Co (and any other transition metal): the atom becomes positively charged and the nucleus attracts the remaining electrons even more effectively (Higher Zeff). Therefore, positive oxidation states are more common.