1) What is the atomic # for selenium?
The atomic number is 34, what else?
2) Write the full electron configuration for selenium following the (n+l) rule?
I must admit I don't know what the n+one rule is (although that could be n + ell rule). However the electron configuration is
[Ar]3d10 4s2 4p4
I am not sure what is meant by n+l
3) write the electron configuration, grouping electrons by their "n" values.
[Ar]3d10 4s2 4p4
I would write 2,8,18,6 for n = 1,2,3,and 4.
4) What is the number of valence electrons for selenium?
6. Those are the 4s2 4p4 and you are correct.
what is the number of core electrons?
I do not understand what core electrons are.
The core electrons are those inside the valence shell; i.e., that will be 34-6 = 28. Another way to look at it is to look at the n shells of 2,8,18,6 and take off the last 6 which leaves 2,8,and 18 = 28.
5) what is the number of unpaired electrons? Show how you determined this.
You have 4s2 4p4 in the valence shell (all of the others are paired; i.e., the 2, 8, and 18). The two in 4s2 are paired. That leaves just the 4p4 electrons to worry about. Since there are three p orbitals (the Px, Py, Pz) and Hund's rule says we fill all orbitals BEFORE we start pairing which means we have 1 in Px, 1 in Py, and 1 in Pz. The fourth one must pair up with the 1 already there in Px and there are no more electrons. Therefore, that leaves the last two electrons, 1 in Py and 1 in Pz, unpaired. So the answer is 2 unpaired electrons. So you are right!
i think the answer is two but im not sure why.
6) What is the number of filled shells?
7) What is the number of filled orbitals?
I'm not sure what the question means by filled orbitals. Is that completely filled or partially filled. Either way it is a filled orbital. I would count 1s2 as 1 orbital + 2s2 as 1, + 2p6 as 3, + 3s2 as 1, 3p6 as 3 + 3d10 as 5 + 4s2 as 1 + 4p2 as 1 and I'm not counting the last two unpaired since those orbitals are only partially filled. You can add those numbers but I think that is 16. If you interpret the question differently you can adjust the answer.
Thank you for your help!!!
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