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Topic: How many electrons in the ground state of a Hg atom...  (Read 8187 times)

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Offline [V]

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How many electrons in the ground state of a Hg atom...
« 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?

Offline Schrödinger

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Re: How many electrons in the ground state of a Hg atom...
« Reply #1 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?
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Offline Borek

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Re: How many electrons in the ground state of a Hg atom...
« Reply #2 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.
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Offline [V]

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Re: How many electrons in the ground state of a Hg atom...
« Reply #3 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.

Offline Borek

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Re: How many electrons in the ground state of a Hg atom...
« Reply #4 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
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Offline [V]

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Re: How many electrons in the ground state of a Hg atom...
« Reply #5 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.

Offline [V]

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Re: How many electrons in the ground state of a Hg atom...
« Reply #6 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.

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