1) What is the atomic # for selenium?

Question

34

2) Write the full electron configuration for selenium following the (n+l) rule?

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

Are these correct?

4) What is the number of valence electrons for selenium?
6
what is the number of core electrons?
I do not understand what core electrons are.

5) what is the number of unpaired electrons? Show how you determined this.

i think the answer is two but im not sure why.

6) What is the number of filled shells?
3
7) What is the number of filled orbitals?
4

Are these correct?

Please ignore the first two posts with these questions. My computer was acting up and submitted the question more than once. Sorry.

Answer 1

1) What is the atomic # for selenium?
34
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.
6
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?
3
right.
7) What is the number of filled orbitals?
4
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.

Answer 2

Thank you for your help!!!