There are ____ electrons in ${{t}_{2g}}$ orbitals in ${{\left[ Co{{F}_{6}} \right]}^{3-}}$ .

Hint: To find the appropriate and required solution of the given question, we need to dig deep into the concept of ${{e}_{g}}$ and ${{t}_{2g}}$ orbitals.
In the Crystal Field Theory $\left( CFT \right)$, when the splitting of the $d-orbital$ occurs, it gets divided into two parts: ${{t}_{2g}}$ that groups three of the metal atoms $d-orbital$ into a certain class while two of the orbitals belong to the ${{e}_{g}}$ class.

Complete Solution :
In the coordination complex ion, ${{\left[ Co{{F}_{6}} \right]}^{3-}}$ the central metal atom $Co$ is having an oxidation state of $+3$ and its electronic configuration required here is $3{{d}^{6}}$ .
- As we know, fluoride is a weak ligand, the electrons would prefer to go to ${{e}_{g}}$ level than get paired up. Hence, the outer electronic configuration will be $t_{2g}^{4}e_{g}^{2}$.
Therefore, conclusively it can be said that ${{\left[ Co{{F}_{6}} \right]}^{3-}}$ has $4$ electrons in the ${{t}_{2g}}$ orbitals.

Note: One is advised to remember that the $t$ in ${{t}_{2g}}$ orbitals means triply degenerate while $e$ in ${{e}_{g}}$ means doubly degenerate where degenerate means having the same energy. Also, the $g$ is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital.