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### Quantum Number l exception

Posted: **Sat Oct 21, 2017 11:08 pm**

by **Jason Muljadi 2C**

We presume that, to find l, it is literally (n-1), correct? Now, would the only exception be when n is at an s-orbital, like 2s, 3s, or 4s? For instance, at 2s, isn't l = 0 because there's not really an orbital angular momentum until we reach 2p?

### Re: Quantum Number l exception

Posted: **Sun Oct 22, 2017 10:26 am**

by **Eryn Wilkinson 3H**

The value of l is a maximum of (n-1), so it can be all numbers up to that value. If the subshell is an s orbital, l is always 0, because the s orbital can only hold 2 electrons and therefore does not have any smaller subdivisions. So if you look at the energy levels, n=1 can only have an s orbital (1s), while n=3 can have s, p, and d orbitals (3s,3p,3d).

### Re: Quantum Number l exception

Posted: **Sun Oct 22, 2017 9:18 pm**

by **Vincent Kim 2I**

The value of l is also not just (n-1), but can also include the negatives numbers too. For s: l can only be 0, while for p: l can be -1, 0, and 1.

### Re: Quantum Number l exception [ENDORSED]

Posted: **Sun Oct 22, 2017 9:27 pm**

by **Jonathan Tangonan 1E**

Just to clarify I believe that the l quantum number is going to be positive. The ml or the magnetic quantum number is when you will start getting negatives where for example where l=1, ml can be -1, 0, or 1.

### Re: Quantum Number l exception

Posted: **Mon Oct 23, 2017 2:47 pm**

by **Chem_Mod**

That is correct. L will never be negative.