What would the Electronic Configuration of Ru²⁺ be? I'm doing homework for Inorganic Chemistry, and I am a little confused.

The Question is: 


  What is the electronic configuration of $\ce{Ru^{2+}}$?


If I were to remove from the highest energy level it would be $\ce{[Kr]~5s^1~4d^5}$
leaving $6$ unpaired electrons.

If I were to take into account electron pairing it would be $\ce{[Kr]~4d^6}$ with $4$ unpaired electrons. 

What would be the correct approach?
You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$. 

All you need to know know is the order of energies for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the Aufbau principle (and possible pitfalls/exceptions) and another thread on this site.