Atomic Structure Help

hello,

I need some explanation on the atomic structure and the way transition metals fill up.

I know that the 4s orbital fills up before the 3d orbital and also that the 4s electrons are lost before the 3d electrons, but here is my confusion:

The electronic structure of a Co3+ ion is (Ar)3d6. I know what the structure of argon is but why has the 3d orbital filled up first and there is no 4s orbital written. If the 4s orbital was filled up first then the structure would be (Ar)4s2,3d4, is that correct? Why has the 4s orbital not filled up first?

Thanks

Reply 1

LadySmythe

No, coz its an ion, you have to think of it as having 3 more, taking the config up to [Ar]3d7,4s2
But then you are losing 3 electrons, so using your emptying rule, you take the 4s2 and one from the 3d shell, leaving you with [Ar]3d6 :smile:

I hope that helps

Reply 2

mya369

chromium and copper are the two exceptions to the rule that the 4s shell fills up before the 3d. With these two, the 3d fills up first.

I know this for a fact, I'm not sure why though.

Reply 3

LadySmythe

mya369

chromium and copper are the two exceptions to the rule that the 4s shell fills up before the 3d. With these two, the 3d fills up first.

I know this for a fact, I'm not sure why though.

No copper and chromium are expections coz they have configurations that will be more stable. Using convension, Cu and Cr should have config as [Ar] 3d4, 4s2 and 3d9,4s2 respectivly. But atoms are more stable when shells are either half full or full, so to make these atoms more stable, they have the configs Cr = [Ar] 3d5,4s1 and Cu = [Ar]3d10,4s1

Reply 4

Gangsta Boogie

Ok, I understand it now so thanks for a quick response and for your help!

Reply 5

LadySmythe

No probs

Its made me wonder why I pretty much failed (to what i wanted anyway) the exam that this was in at AS :frown:

Reply 6

mya369

Loz17

Ok so that's the reason why. My rule is still right though surely, cr and cu are exceptions to the rule because as you explained, this means they are more stable? :confused:

I just happened to have been taught this rule but not why they are this way.

Reply 7

LadySmythe

mya369

Ok so that's the reason why. My rule is still right though surely, cr and cu are exceptions to the rule because as you explained, this means they are more stable? :confused:

I just happened to have been taught this rule but not why they are this way.

Hehe yea, its usualy better to understand if you know why :smile:

Thats what I found when I was studying it. Then, low and behold, it came up in the exam :smile:

So I thought I'd tell you to help you out a bit

Reply 8

mya369

Loz17

Hehe yea, its usualy better to understand if you know why :smile:

Thats what I found when I was studying it. Then, low and behold, it came up in the exam :smile:

So I thought I'd tell you to help you out a bit

Yeah I agree. I was just a bit confused by your 'no'. But it's all good :smile:

Reply 9

LadySmythe

mya369

Yeah I agree. I was just a bit confused by your 'no'. But it's all good :smile:

Ohh i know why i said that, you said that the 3d shells fill up 1st, thts still not the case, its just thw fact that they fill either to a half shell or a full shell, not fill the whole of 3d 1st

Reply 10

EierVonSatan

What you should be asking yourselves is that if the 4s orbital is filled up before the 3d orbitals, then why are the 4s electrons lost first when forming ions? :woo:

Reply 11

LadySmythe

EierVonSatan

What you should be asking yourselves is that if the 4s orbital is filled up before the 3d orbitals, then why are the 4s electrons lost first when forming ions? :woo:

I have wondered that, enlighten us if you can please :o:

Reply 12

EierVonSatan

Loz17

I have wondered that, enlighten us if you can please :o:

The simple answer is that the 4s electrons experience nuclear shielding from the partly filled d-orbitals and so are less tightly held, and so easier to lose :smile:

Reply 13

LadySmythe

EierVonSatan

The simple answer is that the 4s electrons experience nuclear shielding from the partly filled d-orbitals and so are less tightly held, and so easier to lose :smile:

Ahhh right, I get it

Seems so simple now lol, when you look at it (I know there will be some huge complex thng behind it beyond my mere low level of understanding, but thats nice for now)

Thank you :smile:

Reply 14

cpchem

Loz17

Actually, that's pretty much it, unless you want to get into the differing penetrative and shielding abilities of the electronic wavefunctions :tongue: