Question about subshells

[Astra]

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For Co (II), why is the orbital diagram [Ar]3d7 and not [Ar]4s2 3d5?

My understanding is the following:

The valence electrons are lost. These electrons are having the highest energy. Therefore, electrons in the highest orbital will be lost.

In this case, is 3d not the higher energy orbital compared to 4s?

 

[NextStepTutor_3]

This is always a confusing concept, but when removing electrons from a ground-state electron configuration (to form a cation), always remove from the subshell with the higher "n" value first. In other words, even though you typically fill the 4s subshell before 3d, you remove 4s electrons before 3d ones - because 4 is a higher principal quantum number than 3.

The reasoning behind this is out of the scope of the MCAT, but this website has a really interesting discussion of this exact issue: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

 

[Astra]

This is always a confusing concept, but when removing electrons from a ground-state electron configuration (to form a cation), always remove from the subshell with the higher "n" value first. In other words, even though you typically fill the 4s subshell before 3d, you remove 4s electrons before 3d ones - because 4 is a higher principal quantum number than 3.

The reasoning behind this is out of the scope of the MCAT, but this website has a really interesting discussion of this exact issue: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

Thanks a lot!

Are there other rules that you know of that are worth mentioning?

 

[NextStepTutor_3]

None that immediately come to mind! The classic "tricky electron configuration" problems typically ask about chromium or copper (or atoms that follow a similar pattern, like Ag). You might be familiar with this rule already - say you're asked to find the ground-state electron configuration of Cr. You'd think it would be [Ar] 4s2 3d4, but half-filled and fully-filled subshells are significantly more stable than "partially filled" subshells (like this d subshell with 4 electrons). The ground-state configuration of chromium is thus [Ar] 4s1 3d5. This allows both subshells to be half-filled and is more stable overall.

If you're already familiar with that one and the one you asked about, you should be in good shape. Just remember the Aufbau principle, Hund's rule, etc.!

 

[Astra]

None that immediately come to mind! The classic "tricky electron configuration" problems typically ask about chromium or copper (or atoms that follow a similar pattern, like Ag). You might be familiar with this rule already - say you're asked to find the ground-state electron configuration of Cr. You'd think it would be [Ar] 4s2 3d4, but half-filled and fully-filled subshells are significantly more stable than "partially filled" subshells (like this d subshell with 4 electrons). The ground-state configuration of chromium is thus [Ar] 4s1 3d5. This allows both subshells to be half-filled and is more stable overall.

If you're already familiar with that one and the one you asked about, you should be in good shape. Just remember the Aufbau principle, Hund's rule, etc.!

Thank you very much for taking the time to type all of this. Really appreciate it!!!

 

[aldol16]

The idea is that 3d is very close in energy to 4s but not above it. The reason it "appears" that you will 4s first is an illusion. That is, you're not really filling 4s. The first electron goes into 3d. Then what happens is a trade-off between pairing energy and the difference in energy levels. In this case, the latter is quite small and you end up putting the next two electrons into the 4s orbitals. It's a similar concept as high-spin/low-spin metals.