1D.23

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Simon Dionson 4I
Posts: 107
Joined: Sat Sep 14, 2019 12:17 am

1D.23

Postby Simon Dionson 4I » Sun Oct 27, 2019 1:03 pm

Would someone explain to me the answers of b and c? When m(l) in part b is given, is that talking about a specific orbital making the answer 2? For part c, I understand that 2p and 2s can exist, so are the orbitals of those two added together?

Charisse Vu 1H
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

Re: 1D.23

Postby Charisse Vu 1H » Sun Oct 27, 2019 3:11 pm

For b, there is actually only 1 orbital because n=4 refers to the 4th shell, l=2 refers to the d subshell, and m(l)=-2 specifies a single orbital. Therefore, part b only has 1 orbital.

For part c, there are 4 orbitals because n=2 refers to the 2nd shell. In the second shell, the s and p subshells exist. S has one orbital and p has three orbitals. If l=0, then m=0. If l=1, then m=+/-1. Therefore, part c has 4 orbitals.


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