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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: [V] on April 22, 2010, 06:30:52 PM

Title: How many electrons in the ground state of a Hg atom...
Post by: [V] on April 22, 2010, 06:30:52 PM
 How many electrons in the ground state of a Hg atom can have the quantum number ml = +1?


The way I am trying to figure this out is as follows..

Electron configuration for HG is..
[Xe]6s^2 5d^10
n=6
l : ml
0: 0
1: -1,0,1
2: -2,-1,0,1,2
3 -3,-2,-1,0,1,2,3
4 -4,-3,-2,-1,0,1,2,3,4
5: -5,-4,-3,-2,-1,0,1,2,3,4,5

so ml can = +1 where l=1-5
Thats a total of 5 subshells that can have ml=+1.

If each subshell can hold two electrons, then the answer must be 10 right?
But apparently it is actually 16. Can someone please explain to me how it is 16?
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: Schrödinger on April 23, 2010, 01:31:27 AM
Electron configuration for HG is..
[Xe]6s^2 5d^10
Are you sure that's the electronic configuration for Hg?
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: Borek on April 23, 2010, 03:06:20 AM
If I understand correctly what you are doing, seems like you are trying to calculate electrons for n=6, and you assume all electrons for n=6 are present. That's not true. Then, you are ignoring electrons for n<6, but even for n=2 there are already electrons with ml =+1.
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: [V] on April 23, 2010, 04:04:27 AM
nvm it is
[Xe] 6s2 4f14 5d10

but I am still not sure how to go about solving this.
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: Borek on April 23, 2010, 04:43:03 AM
List all orbitals (2p, 3d and so on) that are:

1. filled in Hg
2. may have ml = +1
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: [V] on April 23, 2010, 06:35:39 PM
can someone please walk me through this, I need to know how to do this within the hour.

Thanks.
Title: Re: How many electrons in the ground state of a Hg atom...
Post by: [V] on April 24, 2010, 01:01:12 PM
A bit late now. But thanks anyways.

I would still really like to know how to do this though.