a. 2d
b. 3f
c. 6p
d. 7f
This is a problem on my webassign and i have only one submission left. Please help! Will vote best answer! And if you don’t mind, an explanation would be great!
5 Answers

(b) for 3f, l=3 and n=3
but maximum value of l is n1 (here = 2)
So 3f is not possible
also
(a) 2d => l=2 and n=2
same reason so not possible
for the others l < n1 .. so they are possible

A 2d. as well as b. 3f
because in the second energy level it is impossible to have the d sublevel.
As well as the 3rd level containing an f orbital.
In order of energy level the electron configuration level goes 1s 2s 2p 3s 3p 3d and so on and so forth.
d isn’t introduced until the main energy level (n) reaches 3
This is because for each main energy level you can only have 1 subsequent sublevel [I.E. If you have your main energy level (n) = 1 you can only have an s orbital, which is always going to be spherical and elements that are in the ‘sblock’ on the periodic table can represent electron configurations as elements with an energy level of 1
using Schrodinger’s equation we can determine what orbitals are contained in an electron cloud based off of the main energy level (n)
Orbitals corresponds with the Angular momentum quantum number which corresponds wit hthe principal quantum number (n) the Angular momentum number can easily be put as n1. So Orbitals contained within the electron cloud correspond with the Angular momentum number like so
0  s
1  p
2  d
3  f
So if your Angular momentum allows you to achieve one level, the electron cloud will also include every subsequent level before it [I.E if your angular momentum number was 3, your orbitals would include d, p, and s.
So if your main energy level (n) is 1, your angular momentum number (n1) will be 0, so as a result you will always only have an s orbital.
Hope this helps and isn’t too much info.
Source(s): Chemistry Major 
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