Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: rleung on May 18, 2006, 06:37:57 PM
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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
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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" :)
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So, does this mean that all group 9 elements will have their d-subshell electrons removed before their s subshell ones?
Ryan
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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
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I hate to be the guy that always disagrees with people (;D) 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.
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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
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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.
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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 ;D
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Thanks! and congrats, you're a Full Member!
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alright dude that is so freakin' awesome, i thought you had to have like 100 to get that, whatever can't complain ;D
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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?
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oh my mistake i forgot that 4s came before the 3d sublevel :-[
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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.
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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! :)
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you may also would like to take a look at this...
http://www.chemicalforums.com/index.php?topic=8276.msg37587#msg37587
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3d5 4s2 is more stable than the 3d7 4s0 configuration because it describes a half-filled 3d sub-shell. The 3d5 configuration minimise electronic repulsion within the atom.
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If this were true, would that mean that all Group 9 elements exhibit this behavior? (eg. when a 2+ ion is fomed from any group 9 element, electrons come off the d-subshell first in order to create a more stable configuration, namely, a half-filled d subshell). This goes against everything I have been taught since I have always been taught that s-subshell electrons are always removed before d-subshell ones.
Ryan
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The configuration of Co (II) in solution is definately d7, you can tell from the magnetic moment. Maybe the configuration for the free gaseous ion is 4s23d5, I'm not sure, but it's an idea.
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How can you tell from a magnetic moment? 3d7 and 3d5 are both paramagnetic, so wouldn't they give you the same result?
Ryan
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3d5 4s2 is more stable than the 3d7 configuration because it describes a half-filled 3d sub-shell. The 3d5 configuration minimise electronic repulsion within the atom.
Well it depends on the ligands, if there were SFLs then the repulsion would cause the 3d5 electrons to spread accross the 3dxy 3dxz and 3dyz orbitals (lower energy) with empty 3dz^2 and 3dx^2-y^2 orbitals (higher energy), but a full 4s orbital. (Only one unpaired electron in this case).
If it were just 3d7, 5 electrons could spread accross the lowere energy orbitals, and 2 could spread accross the higher energy 3d orbitals, which would have 3 unpaired electrons!
Anyway the 3d5 4s2 configuration is only theoretical when talking about these (non-gaseous!) cations, whether this is lower energy (taking ligands into account!) or not, no one will know!
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The 4s subshell is not lower in energy then the 3d!!!
There is a difference between the energy of a subshell and the configuration enery
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The 4s subshell is not lower in energy then the 3d!!!
I need new chemistry teachers then! :D
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How can you tell from a magnetic moment? 3d7 and 3d5 are both paramagnetic, so wouldn't they give you the same result?
How can you tell from a magnetic moment? 3d7 and 3d5 are both paramagnetic, so wouldn't they give you the same result?
They are both paramagnetic, but
Eff. Mag. Moment = 2[S(S+1)]1/2 (spin only approx)
But, d5 low spin and d7 low spin give the same result (S=1/2 for both), however the Eff. Mag. Moment of d5 low spin will be temperature dependant, but d7 low spin will not be (first order orbital effects).
d5 high spin will have a slighty higher eff mag moment than high spin d7 (but this could be masked by orbital effects). The Eff. Mag. Moment of d7 high spin will be temperature dependant, but d5 high spin will not be (first order orbital effects).
Also high spin d5 gives only very very weak d-d spectra, since all d-d transitions in d5 high spin are spin forbidden, but the d-d spectra for d7 (high or low spin) is quite a bit stronger.
Guess who's been revising spectra and magnetism today...
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The 4s subshell is not lower in energy then the 3d!!!
For charged transition metals, I agree with you.
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The reason why electrons fill the 4s subshell before the 3d is because the 3d electrons are closer to the nucleus and will repel each other in the 3d more strongly then if they were in the higher energy and larger orbit of the 4s subshell. So yes, the 3d is lower in energy, but the energy of configuration would be lower if the electrons occupied the 4s.
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The reason why electrons fill the 4s subshell before the 3d is because the 3d electrons are closer to the nucleus and will repel each other in the 3d more strongly then if they were in the higher energy and larger orbit of the 4s subshell. So yes, the 3d is lower in energy, but the energy of configuration would be lower if the electrons occupied the 4s.
Ahh! Thanks Mitch, that explains a lot! ;)
So would the ligands in charged transition metal complexes repel the 3d electrons more than 4s electrons?
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How can Co(II) be 3d7 if it has lost 2 electrons?
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How can Co(II) be 3d7 if it has lost 2 electrons?
Because cobalt loses its two 4s electrons. So: [Ar]3d74s2 goes to [Ar]3d7. I'm not 100% sure though, what do you think Co(II)'s configuration is?
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How can Co(II) be 3d7 if it has lost 2 electrons?
Because cobalt loses its two 4s electrons. So: [Ar]3d74s2 goes to [Ar]3d7. I'm not 100% sure though, what do you think Co(II)'s configuration is?
makes sense to me.