Electron Config of CoIII?

[andafoo]

PS Question:

What is the correct electronic config of Co3+ (CoIII ion)?

A) [Ar]4s2,3d7
B) [Ar]4s1,3d5
C) [Ar]4s2,3d5
D) [Ar]3d6


The answer is apparantly d. This question is from the Kaplan free practice exam online. Their reasoning is that the 4s energy orbital is higher than the 3d orbital and thus they are removed first. "The stability of half-filled subshells may be tempting ([Ar]4s1,3d5), but this is a less favorable configuration due to higher energy of the 4s orbital".

First, I've always thought that the 4s orbital is lower in energy than the 3d orbital. http://www.docbrown.info/page07/SSquestions/elevels2.gif (Proof). It also has been discussed in another post.

If the 4s orbital is infact lower in energy than the 3d - that destroys Kaplan's arguement and proof. Would that mean the correct answer is then [Ar]4s1,3d5?

Anyone help me out??


And another question:

I took the Kaplan free practice exam and one thing I noticed was the numbering system on the exam. Instead of restarting the numbering for 1 for each section, they continued (starting the VR at #46).

Is the real MCAT like this? Or is the first question in VR and BS #1? It kind of threw me off because it wasn't easy to determine how many questions I had left.

Also, can someone point me to a post discussing timing each question and keeping track of time? I know it's ben posted before.

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[viper420]

As far as your question goes im not really sure, as i would have chose answer B as well

But the real MCAT does in fact start over in the numbering of each section, so PS will be 1-52, VR will be 1-40 and so on

 

[andafoo]

As far as your question goes im not really sure, as i would have chose answer B as well

But the real MCAT does in fact start over in the numbering of each section, so PS will be 1-52, VR will be 1-40 and so on

#1. I find some answers to practice exams to be wrong, not exclusive to kaplan

#2. That's good to hear.

 

[spyderracing32]

You remove the 4s electrons first because they are higher energy (further away from the nucleus) that the 3d electrons. When an atom undergoes ionization it does so to achieve a more stable (lower energy) state. This goes for all atoms that contain a d orbital.

 

[spyderracing32]

The books explanation is not incorrect. D is the correct answer. It was taught to me in AP chem in HS; I'm pretty sure it still holds true today.

 

[sdb09]

This question used to bug me for the longest time until I realized that:

When you add electrons, you add to lowest n+l (which is why you will add to 4s before 3d - the lovely n+l rule)

When you remove them, you start from the highest n level.

So in this problem it is easiest to write out the entire electronic configuration before you remove the three electrons:
Co = [Ar]4s2 3d7

Now remember you are removing three electrons since it is +3, so take away the 2 from the 4s and then the 1 from the d and you have answer choice D.

 

[andafoo]

Okay, what I really want to know now is why 4s is sometimes higher in energy than 3d while other times it is not.

 

[sdb09]

So the reason it is like that has to do with the fact that when you are filling orbitals, the electrons are going to fill into the lowest energy orbitals based on the n+l rule. When you are using this rule, you know that the lower energy orbitals are going to fill first.



On the other hand, when you are removing electrons, you want to remove them from the highest n level first because this is where the electrons that are furthest out will be located. This is not saying that the 4s has higher energy than the 3d, it is more that the electrons are farther away because of the higher n level.


I hope that makes sense and that I explained it right...this concept really took me a long time to understand until I finally grasped that general trend of adding electrons with the n+l rule but then removing from the highest n.

 

[AZFutureDoc]

Sorry for being vague, but I feel that I have seen some ions that would, say, lose an S electron, and then D, but not one before the other.

I am talking about an electron configuration that is something like ...4S1 3D5. Does this only happen to get max unpaired electrons? I just feel like I remember seeing violations of the n+l rule where you can have more unpaired subshells.

 

[sdb09]

Sorry for being vague, but I feel that I have seen some ions that would, say, lose an S electron, and then D, but not one before the other.

I am talking about an electron configuration that is something like ...4S1 3D5. Does this only happen to get max unpaired electrons? I just feel like I remember seeing violations of the n+l rule where you can have more unpaired subshells.

I think you are referring to elements such as Cr. Since elements want to have either full orbitals (such as d10) or a half-filled orbital (d5), an element such as Cr will go for an electronic configuration where it will take an electron from the 4s orbital and transfer it to the 3d...this way the configuration will be [Ar]4s13d5 and the element is happier because it has a half-filled orbital.

 

[AZFutureDoc]

Ya, that is what I was talking about. Thanks.