HW 1D.23

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Ashley Tran 2I
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Joined: Thu Jul 11, 2019 12:17 am

HW 1D.23

Postby Ashley Tran 2I » Sat Oct 19, 2019 12:00 pm

How many orbitals can have the following quantum numbers in an atom: (a)n=2,l=1;(b)n=4,l=2,ml=-2; (c)n=2;(d)n=3,l=2,ml=-1?

Kishan Shah 2G
Posts: 132
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 1D.23

Postby Kishan Shah 2G » Sat Oct 19, 2019 12:03 pm

The first one can have 3. The second can have 1. The third can have 4. The fourth can have 1.

Helen Struble 2F
Posts: 97
Joined: Sat Aug 24, 2019 12:17 am

Re: HW 1D.23

Postby Helen Struble 2F » Sat Oct 19, 2019 3:10 pm

Let's walk through the possible quantum values, knowing that if n is given, l can be = 0, 1, 2...n-1 and m can be = l, l-1, l-2....-l. An orbital is any valid set of unique quantum numbers.


A) n=2, l=1. ml is not given, and can have the possible values of -1, 0, and 1. Therefore, since we can make three sets of valid quantum numbers, there are three possible orbitals.

B) n=4, l=2, ml=-2. Since all three quantum numbers are given to us, this only corresponds to one possible orbital. Keep in mind, two electrons of opposite spin, based on ms, can occupy this orbital.

C) n=2. If n=2, l can =0, 1. If l=0, ml can only = 0 (this is one s orbital). If l=1, ml can = -1, 0, 1 (three more possible p orbitals). So, in total there are four possible orbitals if n=2.

D) n=3, l=2, ml=-1. Since all three quantum numbers are given to us, this only corresponds to one possible orbital. Keep in mind, two electrons of opposite spin, based on ms, can occupy this orbital.


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