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In one of the review sessions, it was asked what the maximum number of electrons would be present when n=2 or when n=3. I thought about how if n=2 then l=1 or l=0, which corresponds to the p-block and s-block, respectively. This would mean that the maximum electrons would be 6 (in the p-block). Is this the right way to answer the question or am I mixing concepts?
I believe the maximum number of electrons that would be present if n=2 would be 8 electrons because when n=2, l could be 0 or 1 (s or p), therefore there is an s orbital and a p orbital in the 2nd shell and s can hold 2 electrons and p can hold 6 so 2 + 6 = 8. This would mean that the maximum number of electrons in the third shell (when n=3) would be 2 + 6 + 10 = 18 electrons ( 2 electrons for 3s, 6 for 3p, and 10 for 3d)
The way this is solved is by counting the total number of electrons in all the orbitals with a certain n. For n=3 there could 3s 3p and 3d so thats 2+6+10 so 18 possible
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