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Help with a chem question
Started by CpharmAH1
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is the answer C? 3 unpaired electrons?
i just used the aufbau principle
1s
2s 2p
3s 3p 3d
4s 4p
5s
1s,2s,2p,3s,3p,4s,3d
since 1s --> 4s = 20 electrons, there's only 3 left for 3d
3d would be:
| | | __ __ __
edit: i'm such an idiot...i meant B (3 unpaired electrons) not C haha.
1s
2s 2p
3s 3p 3d
4s 4p
5s
1s,2s,2p,3s,3p,4s,3d
since 1s --> 4s = 20 electrons, there's only 3 left for 3d
3d would be:
| | | __ __ __
edit: i'm such an idiot...i meant B (3 unpaired electrons) not C haha.
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Aufbau's as nerv12345 said, plus Hund's rule.
I don't ever draw it out like that for electron configuration, I find it easier to just use the shape of the periodic table. Unfortunately, you wouldn't have a periodic table for the test.
I don't ever draw it out like that for electron configuration, I find it easier to just use the shape of the periodic table. Unfortunately, you wouldn't have a periodic table for the test.
it's not asking you how many are needed to be filled it's asking you how many UNPAIRED electrons are there
it's not asking you how many are needed to be filled it's asking you how many UNPAIRED electrons are there
Ya I got you, but for some reason I though 7 since thats how many were left to fill it out. I need to read stuff more carefully.