## 1D. 25

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

VLi_2B
Posts: 99
Joined: Sat Sep 14, 2019 12:15 am

### 1D. 25

1D.25 Which of the following subshells cannot exist in an atom: (a) 2d; (b) 4d; (c) 4g; (d) 6f?

Can someone please explain this problem to me?

ThomasNguyen_Dis1H
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am

### Re: 1D. 25

1D.25 Which of the following subshells cannot exist in an atom: (a) 2d; (b) 4d; (c) 4g; (d) 6f?

Can someone please explain this problem to me?

Sub-shells refer to the quantum number l. Since the rule for l is l=0,1,...,n-1 you can look at the n and l values and determine if the following subshells exist.
s: l=0
p: l=1
d: l=2
f: l=3
g: l=4

Kimberly Koo 2I
Posts: 99
Joined: Sat Aug 17, 2019 12:17 am

### Re: 1D. 25

Option a, 2d, does not exist because d orbitals don't exist until the third energy level. 4d can exist because this indicates 5th row transition metals in the periodic table. 4g does not exist because g orbitals only exist in energy levels 5 or greater. 6f can exist because it is in the lanthanoids/anthanoids section of the periodic table.

asannajust_1J
Posts: 105
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### Re: 1D. 25

Its important to remember during this question where the electron is both in the orbital and the periodic table is extremely helpful in working that out.

Sydney Jacobs 1C
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Joined: Thu Jul 25, 2019 12:15 am
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### Re: 1D. 25

Possible values of l include 0, 1, 2..... n-1. Therefore, when n=2, l can have the values of only 0 and 1. "d" corresponds to an l value of 2, so therefore 2d cannot exist. With n=4, l can have the values of 0, 1, 2, and 3. Therefore, 4d can exist, but 4g cannot (l=4). With n=6, l can have values of 0 up to 5. 6f can therefore exist (l=3). Hope that helps!