# How many electrons in the ground state of a Hg atom

1. Apr 25, 2010

### [V]

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?

2. Apr 25, 2010

### Staff: Mentor

You were already told at other forum how to approach the problem. You were already told what your approach is missing. Repeating question here in hope that someone will do it for you won't help.

List all orbitals (2p, 3d and so on) that are:

1. filled in Hg
2. may have ml = +1

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methods

3. Apr 25, 2010

### [V]

I did, and I get 12. I don't know where to go from here, if someone can just walk me through to the solution thats all I need.

I am stumped, and got no replies at that other forum. I do believe a new forum will render new replies...
I appreciate your trying to guide me, but I simply do not follow what you are trying to tell me.

Can you please try explaining this another way, or can someone else please explain how to do it?

4. Apr 25, 2010

### [V]

Does nobody know how to do this?

I am inclined to believe that my answer key is wrong, and 16 is not the answer. It is either 10 or 12 I guess.

I don't see how you can possibly get 16...

5. Apr 25, 2010

### Staff: Mentor

Can you list ALL filled orbitals in Hg?

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methods

6. Apr 25, 2010

### [V]

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10

It looks like every orbital is in fact filled.
And they all could have ml=+1 except for the 1s orbital.

Now what?

7. Apr 25, 2010

### Staff: Mentor

2s can have ml=+1?

What is n for 2s? l? Possible values of ml?

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Last edited: Apr 25, 2010
8. Apr 25, 2010

### [V]

n for 2s = 2

where n=2
l= 0,1
ml can = +1 where l = 1

If I count all the subshells that can hold ml=+1 (basically everything with n=2 or higher) then multiply by two, I get a waaay bigger number.

9. Apr 25, 2010

### [V]

Sorry to be naggy, but I really need to understand this fast. I have a final tomorrow morning!

Thanks!

10. Apr 26, 2010

### Staff: Mentor

n=2 & l=1 is not 2s, it is 2p.

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methods

11. Apr 26, 2010

### [V]

I understand that. But I still do not see how to get 16.

Can you please just walk me through this?

12. Apr 26, 2010

### [V]

I think I see it. We are counting all the orbitals that are NOT "s" right? Each holds 2 electrons, thats 8x2=16?

yyaay?

13. Apr 26, 2010

### Staff: Mentor

Yes. Wasn't that hard?

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methods

14. Apr 26, 2010

### [V]

nah.

Thanks a lot Borek, I <3 you.