Posted by Garcia on Sunday, November 11, 2012 at 9:30pm.
Add spaces to make your problem more readable. Don't run the numbers together.
1) 6s2 4f1?
2) 2s2
3) 1s2 2s2 2p6 3s2 3p6 3d2
4)4s2 3d3
5)5s1 4d5
Here is how you do this. I'll do the first one or two.
#1. Count the electrons. I count 57. Look up element #57 and I see Lu. Go to www.webelements.com and click on Lu. Scroll down to electron shell properties, click on that, and see that the ground state for Lu is [Xe]5d1 6s2 and that makes [Xe]6s2 4f1 an excited state.
#2 is He and that is the ground state.
#3 is Ca.
I'll be glad to check the others for you.
I have counted the electrons for all of the five electron configurations. I have determined that as you mentioned [Xe]6s2 4f1 is the excited state of Lu. I also managed to find that the configuration 1s2 2s2 2p6 3s2 3p6 3d2 is the excited configuration of Ca. But I am told that there are three excited electron configurations given. I cannot find the third.
Basically what I have found is:
1) excited Lu
2) He
3) excited Ca
4) V
5) Mo
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