Your notations are hard to read because you didn't space it properly. I have redone it below. Also, I think you should change the order, at least mentally, to what I show below.
A) 1s2 2s2 2p6 3s2 3p6 4s2 3d10
B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
C) 1s2 2s2 2p6 3s2 3p6 4s1
D) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
E) 1s2 2s2 2p6 3s2 3p2
A) 1s2 2s2 2p6 3s2 3p6 3d10 4s2
B) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
C) 1s2 2s2 2p6 3s2 3p6 4s1
D) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2
E) 1s2 2s2 2p6 3s2 3p2
Why did I reconfigure this? Because the OUTSIDE shell, which is the one used for most ionizations, is the outside shell and that is 4 instead of 3. I ALWAYS write the shells in numerical order (and not filling energy wise order) because students try to ionize the 3d first over the 4s. Any, here is my best guess how to anwer this question and basically it is by elimination.
Eliminate E since the last electrons are in the 3rd shell and we know the 4th shell is easier to pick off an electron.
Eliminate D because we PROBABLY will get +4 and not +2.
Eliminate C since +1 is the obvious choice and not +2.
Eliminate B since you probably will get +4(could get +6) and not +2.
This leaves A as the only choice and +2 electrons are handy as the two outside electrons.
Hello Dr.Bob. Could you please take a look at the following question? I know the correct answer for it, but I'm not sure how to go about solving it.
Which ground state electronic configuration will most readily ionize to a 2 + cation?
A) 1s22s22p63s23p64s23d10
B) 1s22s22p63s23p64s23d104p4
C) 1s22s22p63s23p64s1
D) 1s22s22p63s23p64s23d104p2
E) 1s22s22p63s23p2
The correct answer for this question is A). Could you please let me know why that configuration is the right one? I thought that since A) has a complete outer electron shell, then it would be more stable and thus less likely to ionize to a 2+ cation.
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