[rleung]

Hi,

I was just reviewing some old notes in chemistry, and I came across two things about electron configuration that confused me.  I have in my notes the following electron configuration of Co2+

[Ar]4s23d5

I thought it would be [Ar]3d7 since the electrons from the highest energy shell should be removed first, shouldn't it?  Did I just copy this down wrong?  I "could" see why it might be that answer for this special case since removing two electrons from the d-shell would lead to a rather stable half-filled d subshell..

Thanks.

Ryan

[rctrackstar2007]

Hi,

I was just reviewing some old notes in chemistry, and I came across two things about electron configuration that confused me.  I have in my notes the following electron configuration of Co2+

[Ar]4s23d5

I thought it would be [Ar]3d7 since the electrons from the highest energy shell should be removed first, shouldn't it?  Did I just copy this down wrong?  I "could" see why it might be that answer for this special case since removing two electrons from the d-shell would lead to a rather stable half-filled d subshell..

Thanks.

Ryan

i believe you hit it with "could" 

[rleung]

So, does this mean that all group 9 elements will have their d-subshell electrons removed before their s subshell ones?

Ryan

[tennis freak]

that depends on which d-shel, 3-4-5, what ever it may be and if there are s or p shells after them. it is the further away that the electrons are more likely to be taken so it depends on the last shell but don't forget that if you have sp³ that the s will give up an electron first so as to stay stable

[Will]

I hate to be the guy that always disagrees with people () but I am sure that when cations are formed, they lose their outer most electron i.e. the electron with the highest value for n (i.e the number of the shell) and then the highest l value (s, p, d, f).

So I would say rleung is right to believe the electron configuration of the Co²⁺ ion is [Ar]3d⁷ because the two 4s electrons are lost first. I can't imagine any element losing their 3-something electrons before their 4-something electrons- same goes for 4&5 etc.

In fact in the case of Co²⁺, I think [Ar]3d⁷ would be stabler than the other one because when it is charged you have to think about the ligands bonding to the cobalt, and the large amount of repulsion between these and the theoretical 4s electrons.

3d⁵ wouldn't be any more stable in the ion because I think it would be unlikely that one electron would occupy each orbital anyway, since the difference in energy levels between the orbitals; I imagine there would be 4 or 5 electrons occupying the lowest 3 energy orbitals rather than 3, assuming it is d²sp³ hybridised, so 3d⁷ wouldn't be any worse.

[tennis freak]

sorry, maybe i misspoke what i was meaning to say was that they would lose the 3 somethings first as long as there was nothing after them in a higher shell like a 4s. sorry if i caused some confusion, your way was just a simpler version i think, correct me if i am wrong. to your last point however, if you have 2 electrons in the 4s orbital say and 3 in the 4p orbitals (one each) then it will most likely lose one of the 4s electrons rather than one of the 4p electrons so as to keep an electron in each orbital and keeping it stable

[Will]

ooooops! Sorry tennis freak, i thought you were agreeing with rctrackstar2007. I just re-read what you wrote and you are completely right (Its late here so I can't be 100% sure about anything right now!)

to your last point however, if you have 2 electrons in the 4s orbital say and 3 in the 4p orbitals (one each) then it will most likely lose one of the 4s electrons rather than one of the 4p electrons so as to keep an electron in each orbital and keeping it stable

I think I explained my last point very badly! I wasn't talking about 4p and 4s electrons.
I was trying to explain why having [Ar]3d⁵ configuration is no more stable than [Ar]3d⁷ to any significant effect in a d-block element ion, due to ligands and the different energy levels of the d-orbitals. I don't even know if I'm right on that point!
Not that that matters since the element would lose its outer electrons regardless of whether its 'inner' electron configuration is more or less stable.

[tennis freak]

ooooops! Sorry tennis freak, i thought you were agreeing with rctrackstar2007. I just re-read what you wrote and you are completely right (Its late here so I can't be 100% sure about anything right now!)

quite understandable will, i can see how you might make that mistake

[Will]

Thanks! and congrats, you're a Full Member!

[tennis freak]

alright dude that is so freakin' awesome, i thought you had to have like 100 to get that, whatever can't complain

[jdurg]

Guys, the 4s subshell is a LOWER ENERGY SHELL than the 3d shell.  It is thusly closer to the nucleus than the 3d shell is.  Once you get past the first couple of rows on the periodic table you can COMPLETELY forget about that initial number indicating which shell has a higher energy.  Once you get down to lanthanum, neodymium, platinum, gold, etc. the orders REALLY get screwy.

When you look at the periodic table, the order of the energy levels in terms of lowest energy orbital to highest energy orbital is as follows:

1s
2s
2p
3s
3p
4s
3d
4p
5s
4d
5p
6s
4f
5d
6p
7s
5f
6d
7p

So with Cobalt, the electronic configuration is more properly written as [Ar]4s2,3d7.  Those 4s2 electrons are very well protected by the 3d7 electrons.  If you want to make this a Co+2 ion, what would be easier?  Removing two inner s-shell electrons, or two outer d-shell electrons and creating a stable arrangement of each shell half-filled?

[tennis freak]

oh my mistake i forgot that 4s came before the 3d sublevel

[Yggdrasil]

Even though the 4s energy shell has a lower potential energy than the 3d shell, it still has a greater electron density away from the nucleus (recall that there are various radial nodes in the 4s orbital).  Therefore, though a "fluke" of quantum mechanics, it becomes easier to remove electrons from the 4s orbital even though it has a lower potential energy than the 3d orbitals.

[Will]

Guys, the 4s subshell is a LOWER ENERGY SHELL than the 3d shell.  It is thusly closer to the nucleus than the 3d shell is.  Once you get past the first couple of rows on the periodic table you can COMPLETELY forget about that initial number indicating which shell has a higher energy.  Once you get down to lanthanum, neodymium, platinum, gold, etc. the orders REALLY get screwy.

So with Cobalt, the electronic configuration is more properly written as [Ar]4s2,3d7.  Those 4s2 electrons are very well protected by the 3d7 electrons.  If you want to make this a Co+2 ion, what would be easier?  Removing two inner s-shell electrons, or two outer d-shell electrons and creating a stable arrangement of each shell half-filled?

Just because 4s is lower energy than 3d (thats why it gets 'filled up' first) it doesn't mean it is closer to the nucleus. In fact, the electron shells go out in order from 1 to 7 etc, its just the energy of threse orbitals that is so screwy. There is a relatively simple order in which the shells get filled, except lanthanum, like you said and actinium, and thorium etc. Therefore the 4s electrons are the valence electrons and will get lost before the 3d electrons, thats why the first row of d-block elements usually form +2 ions, not +1 (except Cu which has only 1 electron in the 4s shell, by coincidence?).

Yggdrasil has explained it much better than me, but I attempted anyway!

[xiankai]

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http://www.chemicalforums.com/index.php?topic=8276.msg37587#msg37587