## Hw Help 1D.25

Bita Ghanei 1F
Posts: 60
Joined: Thu Feb 28, 2019 12:15 am

### Hw Help 1D.25

Which of the following subshells CANNOT exist in an atom:

(a) 2d
(b) 4d
(c) 4g
(d) 6f

Rebecca Remple 1C
Posts: 137
Joined: Wed Sep 18, 2019 12:16 am
Been upvoted: 1 time

### Re: Hw Help 1D.25

Hi Bita,

Both B and D can exist, but A and C cannot. A cannot exist because there is no 2d subshell. Instead, the lowest "d" subshell is 3d. The subshells preceding 3d are (from lowest to highest energy) 1s, 2s, 2p, 3s, 3p, and 4s. C cannot exist because the lowest "g" subshell is 5g. The subshells preceding it include all the previous ones that precede the 3d subshell and (from lowest to highest energy) 4p, 5s, 5p, 6s, 4f, 5d, and 6p. I've attached a diagram that you can reference. I hope this helps!

-Rebecca
subshells.jpg (27.31 KiB) Viewed 41 times

Sion Hwang 4D
Posts: 51
Joined: Wed Sep 18, 2019 12:21 am

### Re: Hw Help 1D.25

You can check by seeing if the subshell exists, as l=0 through n-1.
(a) 2d has an n value of 2. Therefore, l can either be 0 or 1, which corresponds to the s and p subshells. Notice that d is not a part of the angular momentum quantum number, so a) cannot exist.
(b) 4d has an n value of 3. l can be 0 through 3, which is the s, p, d, and f subshells. Since d is part of the possible subshells, b) can exist.
(c) 4g also has an n value of 3. Again, the highest l value is 3, which is an f subshell. g subshells come after the f subshell, so c) cannot exist.
(d) 6f has an n value of 6, which corresponds to l values 0 through 5. They are the s, p, d, f, and g subshells. Since f is a part of the possible subshells, d) can exist.
Hope this helps!

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

### Re: Hw Help 1D.25

I like to go about this by first associating the letter with its orbital angular momentum quantum number (l). So s -> l=0, p -> l=1, d -> l=2, and so on. Then compare that number to the first number, which is the principal quantum number (n). We know that l must be less than n, so if the second number is larger than the first number, that suborbital can't exist.