2.29

[Emily 1E]

How many electrons can have the following quantum numbers in an atom: (c) n = 2?
The answer says 8 electrons, but wouldn't n = 2 be a p orbital, which has 6 electrons? Where do the other 2 electrons come from?

[Andrew Evans - 1G]

You’re correct in that n=2 now has p orbitals, but that is in addition to the s orbital in the n=2 shell. So you have the 6 electrons from the 2p orbitals and the other two electrons are from the 2s orbital (since you need to count all the orbitals s and p within n=2).

-Andrew Evans
Section 1G

[Chem_Mod]

Hey Emily

Remember that there is also a s orbital that has n=2, which can house two electrons on its own. Thus, together with the 6 electrons within the 2p orbital, there will be 8 total electrons with n=2.

[Alexandra Wade 1L]

The question for c also does not specify the l value. Therefore we can't assume that it is only the p orbital. This is why both the s and p orbital's electrons should be counted for this question.

[Joanna Pham - 2D]

Could someone explain part a and b please? I’m confused why the answer for a is 6 and 2 for part b.

The question’s asking “how many electrons can have the following numbers in an atom.” For part a, we’re given that n=2 and l=1, so it’s referring to the 2p shell. I know that the p shell can hold a max of 6 electrons, and that’s where I assume they got the answer 6 from. But what about the 2s and 1s shell? Do we not need to add in the number of electrons from 2s and 1s?

For part b, we’re told that n=4, l=2, and ml=-2. How did they get the answer 2? Is it because the magnetic quantum number is -2?

[Celeste Martinez 1K]

Can someone explain part d from problem 2.29. since n=3 and l=2, I know that it is in the 3d shell. It then gives you the magnetic quantum number which is +1. The answer is 2 electrons, but I don't understand why it is 2 electrons. I thought the d sub shell could hold up to 10 electrons.

[Nicole Shak 1L]

I am also confused on part d, I know that it's in the 3d shell, and I'm not sure what you're supposed to do when given m1=+1.

[Jada Larson 1F]

For part (d), the problem specifically asks for the total possible electrons with the principle quantum number, n=3, and angular momentum, l=2, and magnetic quantum number, m_l=1, which gives the orientation of the orbital in space. Only one orbital can exist in that exact location in space, meaning there is only one orbital of interest. Each orbital can hold up to two electrons with two different spins, m_s, up or down. Therefore, only 2 electrons can have these quantum numbers.

[paulacamara1E]

I still don't understand letters b and d. For the explanation for letter d, what do you mean by "only one orbital can exist in that location in space"? If l=2 aren't there 5 orbitals (ml)? And doesn't n=3 and l=2 refer to row 3 in the d block? Thank you and sorry if the questions are confusing.

[Joanna Pham - 2D]

I still don't understand letters b and d. For the explanation for letter d, what do you mean by "only one orbital can exist in that location in space"? If l=2 aren't there 5 orbitals (ml)? And doesn't n=3 and l=2 refer to row 3 in the d block? Thank you and sorry if the questions are confusing.

I had trouble with this too, but one of the UA's clarified this for me. For b, we're told that n=4, l=2, and ml=-2, so the question's referring to the 4d subshell, and the d-subshell can only have a total of 5 orbitals. If you draw out the d-subshell, it would be something along the lines of

___ ___ ___ ___ ___
4d

Since ml= -l to +l, the ml values are 2,1,0,-1, and -2, so I think of it as the ml=2 value being assigned to the first bar/spot (orbital), 1 to the 2nd orbital, etc. But the question is specifically asking for how many electrons can have the quantum numbers n=4, l=2, and ml=-2. If you think of ml=-2 as the last orbital in the diagram above, because each orbital can only hold a max of TWO e-, the last orbital can only hold a max of 2 e-, so only two electrons can have those specific quantum numbers (n=4, l=2, and ml=-2).

The same explanation applies to part d.


Hope this helps!