1) What is the atomic # for selenium?

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.

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.

1) Correct! The atomic number for selenium is indeed 34.

2) The electron configuration for selenium following the (n+l) rule is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

3) The electron configuration, grouping electrons by their "n" values, is:
[Ar] 3d10 4s2 4p4

4) The number of valence electrons for selenium is 6. Valence electrons are the electrons in the outermost shell (highest value of "n"). In selenium, the outermost shell is the 4th shell, and there are 6 electrons in the 4th shell (4s2 4p4).

The number of core electrons is the total number of electrons in all the completely filled inner shells. In the case of selenium, the 1st, 2nd, and 3rd shells are completely filled, so the number of core electrons is 28 (10 from the 3rd shell and 18 from the 2nd shell).

5) The number of unpaired electrons in selenium is 2. To determine this, we look at the electron configuration (3d10 4s2 4p4). In the 4th shell (4s and 4p orbitals), there are four electrons in total. Two of them are paired in the same orbital (4p4), while the other two are unpaired in different orbitals (one in 4s and one in 4p). Therefore, there are 2 unpaired electrons.

6) The number of filled shells in selenium is 3. In the electron configuration [Ar] 3d10 4s2 4p4, the 1st, 2nd, and 3rd shells are completely filled.

7) The number of filled orbitals in selenium is 4. In the electron configuration [Ar] 3d10 4s2 4p4, the 3d, 4s, and two of the 4p orbitals (px and pz) are filled, resulting in 4 filled orbitals.

Overall, your answers are correct!

1) Yes, the atomic number for selenium is 34.

2) The (n+l) rule is a way to determine the order in which electrons fill the orbitals. It states that the orbital with the lowest (n+l) value is filled first. If there are two orbitals with the same (n+l) value, the one with the lower n value is filled first.

To write the full electron configuration for selenium following the (n+l) rule, we start by determining the electron configuration for the preceding noble gas, which is argon (Ar). The electron configuration for argon is 1s^2 2s^2 2p^6 3s^2 3p^6.

Next, we determine the number of remaining electrons for selenium, which is 34 - 18 = 16.

Following the (n+l) rule, the orbitals are filled in order of increasing (n+l) values:

3d^10 4s^2 4p^4

So yes, the electron configuration for selenium, following the (n+l) rule, is [Ar] 3d^10 4s^2 4p^4.

3) The electron configuration can also be written by grouping electrons by their "n" values. In this case, the electron configuration for selenium would be:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4

So yes, your answer is correct.

4) The number of valence electrons for selenium is the number of electrons in its outermost shell, which is the "n" value of the highest (n) orbital. In the case of selenium, the highest "n" orbital is 4s^2 4p^4, so it has 6 valence electrons.

Core electrons are the electrons in the inner shells or orbitals. In the case of selenium, the core electrons include the electrons in the preceding noble gas configuration (Ar) - 18 in total. So the number of core electrons for selenium is 18.

5) To determine the number of unpaired electrons in an element's electron configuration, you can visualize the filling of the orbitals by using Hund's rule. Hund's rule states that orbitals of equal energy are filled with one electron each before pairing up.

By looking at the electron configuration of selenium (3d^10 4s^2 4p^4), we can see that the last two electrons are in the 4p orbital. Since there are no other electrons in the same 4p orbital, these two electrons are unpaired. Therefore, selenium has 2 unpaired electrons.

6) The number of filled shells can be determined by counting the number of "n" values present in the electron configuration. In the electron configuration for selenium ([Ar] 3d^10 4s^2 4p^4), there are three different "n" values: 1, 2, and 3. So, selenium has 3 filled shells.

7) The number of filled orbitals can be determined by counting the total number of orbitals filled in the electron configuration. In the electron configuration for selenium ([Ar] 3d^10 4s^2 4p^4), there are a total of four filled orbitals: 2 in the 3d orbital, 1 in the 4s orbital, and 1 in the 4p orbital. So, selenium has 4 filled orbitals.

Yes, your answers for the number of filled shells and filled orbitals are correct.

Thank you for your help!!!