n=6,l=5

[Emma Healy 2J]

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?

[Stuti Pradhan 2J]

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]

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]

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]

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]

Yes! l =<(n-1)

[Hannah Chang 3K]

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

[Elizabeth Kim 2K]

Hi! Yes, l=5 is a valid angular momentum quantum number for the principal quantum number n=6 because the maximum value for l is (n -1). Hope this helps!