How many electrons in the ground state of a Hg atom

[[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?

 

[Borek]

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

 

[[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 that's 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?

 

[[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...

 

[Borek]

Can you list ALL filled orbitals in Hg?

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methods

 

[[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?

 

[Borek]

2s can have ml=+1?

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

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[[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.

 

[[V]]

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

Thanks!

 

[Borek]

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

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methods

 

[[V]]

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

Can you please just walk me through this?

 

[[V]]

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

 

[Borek]

Yes. Wasn't that hard?

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methods

 

[[V]]

nah.

Thanks a lot Borek, I <3 you.