## n=6,l=5

Emma Healy 2J
Posts: 100
Joined: Wed Sep 30, 2020 9:45 pm

### n=6,l=5

If n=6, would l=5 be a valid answer? I know that l can be 0,...n-1 and that l=0 corresponds to s, l=1 corresponds to p, l=2 corresponds to d, l=3 corresponds to f. Based on this, l=5 would not correspond to s, p, d, or f. So would l=5 be valid because it fits for n=6-1=5?

Posts: 155
Joined: Wed Sep 30, 2020 9:32 pm
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### Re: n=6,l=5

Yes, l=5 is a valid answer. There are other orbitals past the s, p, d, and f orbitals (for example there is a g orbital that exists, but there are no elements that have been found that can occupy this orbital). Any orbital that satisfies the 0, ..., n-1 requirement is a possible value for l.

Hope this helps!

Hailey Kang 2K
Posts: 98
Joined: Wed Sep 30, 2020 10:03 pm
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### Re: n=6,l=5

Hi!

l=5 would be a valid answer. Your l value can be all the number leading up to n-1. so for example, if n=5, all your possible l values would be 0,1,2,3,4.

Samudrala_Vaishnavi 3A
Posts: 93
Joined: Wed Sep 30, 2020 9:34 pm

### Re: n=6,l=5

Theoretically, that would work considering that the l number is one less than the n number. l is always one less than the n number considering it is a subshell of it.

sophie esherick 3H
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Joined: Wed Sep 30, 2020 9:35 pm
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### Re: n=6,l=5

Yes, l=5 is a valid answer because there are other orbitals past the s, p, d, and f orbitals. As long as the angular momentum quantum number follows the l=0,,, n-1 requirement, it is possible.

Chenning Yang Dis3l
Posts: 55
Joined: Fri Oct 09, 2020 12:16 am

### Re: n=6,l=5

Yes! l =<(n-1)

Hannah Chang 3K
Posts: 91
Joined: Sat Oct 17, 2020 12:15 am

### Re: n=6,l=5

l=5 is theoretically correct though we haven't found actual elements with h orbitals.