You posted this just because you knew I would enjoy it
(I'm an organometallic chemist).
So the 4s energy level is higher in energy than the 3d energy level so you always remove electrons from 4s before 3d.
So now the only problem is the filling of the orbitals. So the easy answer is that 3d10 is a full d-shell and that's favorable as opposed to only have 9 electrons in the d-shell. Remember that the 3d level is low-lying (relative to 4s) and so it wants to be full and stable. Now here's why that occurs. When you fill orbitals with the metals, you run into a conundrum. The energy of 3d is almost equivalent to the energy of 4s (3d is slightly lower-lying). The first electron you fill goes into 3d because it's lower in energy than 4s. Now, the second electron encounters two possibilities: 3d or 4s. There are also two energetic terms to consider. 3d is lower in energy, but it's got a smaller radial extension and thus two electrons in 3d would be closer to each other than two electrons in 4s and thus you have a Coulombic repulsion term. That's why the next two electrons go into 4s instead of 3d. So the ground state for Sc is 3d1 4s2. Now, as you move across the transition series, you only have one option - put the electrons into the 3d orbitals.
Interesting cases arise when you encounter the possibility of achieving 3d5 or 3d10. In those cases, the d-shell is stabilized by being radially symmetric (either half- or fully-filled). Thus, one of the 4s electrons can be "promoted" or "demoted" depending on how you look at it, into the 3d shell.
Now, for your case, ground state Ni is 3d8 4s2. Ni- therefore can only place an electron in the 3d shell. But wait! Now it's 3d9 4s2! It can just promote one of the 4s electrons to the 3d shell to give a stable shell! And that's just what it does. The stabilization energy gained is more than the energy lost to Coulombic repulsion.
