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### 1D.23

Posted: Thu Oct 17, 2019 4:48 pm
How many orbitals can have the following quantum numbers in an atom:(a)n=2,l=1;(b)n=4,l=2,ml=22; (c)n=2;(d)n=3,l=2,ml=11?

### Re: 1D.23

Posted: Thu Oct 17, 2019 5:19 pm
Hello. The process of solving this problem is that first identify the n value to find the shell. The next step is to identify the l value to find the subshell, l=0 is the s-orbital, l=1 is the p orbital, l=2 is the d orbital, and l=3 is the f orbital. Then, identify if there is an ml value present; the ml value indicates the specific orientation of an orbital, so the ml value will indicate one specific orbital with the indicated orientation. If there is no ml value present, then the orbital type will indicate the number of orbitals with s having 1 orbital, p having 3 orbitals, d having 5 orbitals, and f having 7 orbitals. I hope this is helpful.

### Re: 1D.23

Posted: Thu Oct 17, 2019 5:20 pm
Hi,

I attached a chart that you can use for reference! I used it alongside the info Lavelle gave us, and it really helped me contextualize the information better.
In part a, since n=2 and l=1, there are 3 possible orbitals depending on m's value (ml= l, l-1, ... -l). Therefore, m can be either 1, 0, or -1 (three values).
In part b, m's value is already given. Therefore, there is only one possible orbital.
In part c, n=2. Since l= 0, 1, ... n-1, l has to be 0 or 1. If l=0, m=0 and the electron is in 2s. If l=1, there are three values for m (1, 0, -1) and the electron is in 2p. Therefore, there are 4 possible orbitals.
I'm unsure about the notation in part d, however. I will post to the forum and get back to you on that one. But besides that, this is how I would approach these sorts of questions! I hope this helps :)

-Rebecca