1D.23, part c (7th Edition)

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Chloe Thorpe 1J
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Joined: Fri Sep 28, 2018 12:16 am

1D.23, part c (7th Edition)

Postby Chloe Thorpe 1J » Tue Oct 23, 2018 10:19 pm

The question is: How many orbitals can have the following quantum numbers in an atom: c) n=2

I thought it was 3 (1s, 2s, 2p), but the textbook solutions say the answer is 4. Can someone explain this to me?

Jocelyne Milke 1G
Posts: 73
Joined: Fri Sep 28, 2018 12:25 am

Re: 1D.23, part c (7th Edition)

Postby Jocelyne Milke 1G » Wed Oct 24, 2018 12:11 am

With n=2 you could have l= 0 or 1. If your l=0, then your ml can only=0 (this is one orbital). Since l can also equal 1, your ml can be -1,0,+1. This corresponds to 3 more orbitals so the total number of orbitals would be 4.

Jamie Hsu
Posts: 31
Joined: Fri Sep 28, 2018 12:19 am

Re: 1D.23, part c (7th Edition)

Postby Jamie Hsu » Wed Oct 24, 2018 5:13 pm

n=2 can either be the 2s or 2p subshell. The 2s subshell has 1 orbital while the 2p subshell has 3 orbitals. Since it asks for the general principal quantum number of n=2, we must take into account 2s and 2p so we add 1 orbital + 3 orbitals to get 4 possible orbitals.


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