What is a electronic configuration of scandium ion?
The electronic configuration of the cobalt is $\ce{[Ar] 3d^1 4s^2}$ the electrons with the higher energy will be snatched and the electronic configuration of Argon is very stable then with scandium you will have easely $\ce{Sc^{3+}}$. 

For the Cobalt it is a bit more difficult, like you write the configuration you cannot find a correct answer. For all element first you do first the configuration using the Klechkovsky rule and after you put all orbitals by principal quantum number growing. 

So the cobalt configuration is $\ce{[Ar] 3d^7 4s^2}$ then if you snatch two electrons from the $\ce{4s}$ orbital you have a stable configuration for the $\ce{Co(II)}$ ion. You cannot snatch them from the $\ce{3d}$ orbital (even if $\ce{[Ar] 3d^5 4s^2}$ looks stable because the $\ce{3d}$ orbital is half full then the spin is maximal) because its energy is lesser than the energy of the $\ce{4s}$ orbital. 

You can find the Cobalt at different oxydation states from $\ce{+I}$ to $\ce{+IV}$ but it depends on what you have in your solution or in your gas if you have a gas. 



NB : 
Remember that the configuration of the elements are given if gas phase, then for example the more stable configuration of the copper iron is for $\ce{Cu^+}$ and not for $\ce{Cu^2+}$, $\ce{Cu^2+}$ is stable in water, so the answer may depends on the problem you have. 



Explaination for the copper :

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^{2+}}$ ion has a greater charge density than the $\ce{Cu^+}$ ion and so forms much stronger bonds releasing more energy. 

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu^+}$ ion loses an electron to become $\ce{Cu^{2+}}$ which can then release this hydration energy.

I hope it can help you !