Pd2+ electron configuration

What is the electron configurations of Ru and Pd2+ in condensed form? I don’t know where I’m going wrong here?

I have this question on my homework and I believe its referring to Ru which I’m entering as [Kr] 4d^7 5s^1 but I can’t seem to come up with the right answer for it no matter how many similar variations I try.

Could you help me figure out what the two electron configrations for these elements are so I can see where I’m going wrong on this?

Ru3+ electron configuration whilst condensed, is actual, [Kr]4d^5. it is because of the fact, the three+ cost potential we would desire to continuously lose 3 electrons. we can start up by using removing them from the backside orbital —-> it turns into [Kr] 4d^6 because of the fact we’ve been given rid of 5s^2 that had 2 electrons, yet now we would desire to continuously lose a million greater. —–> provides us [Kr]4d^5, our very final answer. this occasion is common in transition metals.

Pd2+ electron configuration

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