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Earlier in class, why was Cu2+'s electron configuration [Ar](3d9) instead of [Ar](4s1)(3d8)? Why did Lavelle take an electron from one s orbital and one of the d orbital instead of taking 2 electrons from the d orbital?
Its easier if you just think about it in the order you write out the configuration. 3d comes before 4s when you write out the configuration. So, you would remove the 4s1, and then one from the 3d.
It would be [Ar]3d8 if Cu was just any other element; however, we have to remember that Cu is one of the exceptions. The electron configuration of neutral Cu is [Ar]3d10 4s1, and so the 2 electrons that are removed for Cu2+ is the 1 electron from the 4s orbital, then 1 electron from the 3d orbital leaving it as [Ar]3d9.
my TA said to just always write the electron configuration as if it had no charge, and depending on if the charge is positive or negative, add or subtract electrons from the end. So in this case, 3d10 is to the left of 4s1. We take off one electron in the 4s because it is the rightmost one and then work our way back taking off another from 3d.
Always take from the highest energy state. Once you get to the D orbital, you write the electron configuration 3d4s and so forth (4d5s, 5d6s). You write it like this because at this point the s orbital is higher energy than the d orbital. Always take from the highest, so start from the outside and go in
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