Chemistry: Electron Configuration of Manganese Cation

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SUMMARY

The ground state electron configuration of the manganese cation (Mn2+) is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵. Manganese, with an atomic number of 25, loses two electrons from the 4s subshell during ionization, resulting in a total of 23 electrons, which corresponds to the electron configuration of vanadium. The discussion emphasizes the importance of the Aufbau principle in determining the order of electron removal, clarifying that electrons are removed from the highest energy level, which is the 4s shell in this case.

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  • Understanding of electron configuration and atomic structure
  • Familiarity with the Aufbau principle
  • Knowledge of transition metals and their ionization
  • Basic concepts of electron shielding and energy levels
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HunterDX77M
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Homework Statement



Write the ground state electron configuration of Mn2+.

Homework Equations



N/A

The Attempt at a Solution



Well I thought that since it had the 2+ superscript and manganese is a metal, it must have lost two electrons. Manganese has an atomic number of 25 and losing two electrons would put it at 23, the same as vanadium (V). Therefore, I thought that the electron configuration would be the same as vanadium:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3

But I am told this is wrong, and I can't see anything else that could be the answer if Mn2+ does indeed have 23 electrons. Any help is appreciated. Thank you!
 
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Before writing the configuration of Mn+2, first write the configuration for Mn.

The electrons go out from the shell having the highest energy level.
 
Pranav-Arora said:
Before writing the configuration of Mn+2, first write the configuration for Mn.

The electrons go out from the shell having the highest energy level.

Well this is what it would be for just plain old Mn:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5

Am I wrong or is the highest energy level here the 3d shell? I was using a chart like this to help me out. Does energy increase along the red arrows or just straight down. That is, is a higher period number what determines a higher energy shell? I was under the impression it was the red arrows, but now I'm not so sure.

elecfill.gif
 
Hi HunterDX :smile:

HunterDX77M said:
Well this is what it would be for just plain old Mn:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5

Am I wrong or is the highest energy level here the 3d shell? I was using a chart like this to help me out. Does energy increase along the red arrows or just straight down. That is, is a higher period number what determines a higher energy shell? I was under the impression it was the red arrows, but now I'm not so sure.

elecfill.gif

The electronic configuration for Mn is correct :smile:

And you are correct that 3d has higher energy state than 4s, by the Aufbau principle... But during ionization, the electron comes out of the outermost shell; which one is it for this case?
 
I want to say it's the 4s shell, but the Aufbau principle says that the lower levels are filled before the higher ones, correct? In the case of Mn, the 4s level is filled but the 3d is not, which now makes me confused.

So is the highest energy level simply determined by the highest period number?
 
HunterDX77M said:
I want to say it's the 4s shell, but the Aufbau principle says that the lower levels are filled before the higher ones, correct? In the case of Mn, the 4s level is filled but the 3d is not, which now makes me confused.

In Mn, the 4s is fully filled and the 3d is half filled. This is a special and a more stable configuration than having all the 7 electrons in d subshell. It applies in general to all elements, and they try to have more of fully filled/half filled subshells.

So is the highest energy level simply determined by the highest period number?

For the hydrogen atom, 4s and 3d have almost the same energies, 4s being more than 3d by a bit. Aufbau principle takes in consideration for shielding effect and orbital shapes too, that's why it places 4s below 3d in terms of energy level. If electrons from 3d are removed, the shielding gets reduced, and 4s has higher energy than the d orbital. Removing electrons from 4s will not change the effective shielding on the 3d orbital.

http://en.wikipedia.org/wiki/Electron_configuration#Ionization_of_the_transition_metals
 
Okay, so according to the article and what you're telling me, they should be removed from the 4s shell instead of the 3d shell. So for Mn2+:
1s^2 2s^2 2p^6 3s^2 3p^6 3d^5
 
HunterDX77M said:
Okay, so according to the article and what you're telling me, they should be removed from the 4s shell instead of the 3d shell. So for Mn2+:
1s^2 2s^2 2p^6 3s^2 3p^6 3d^5

Yep. You got it. :smile:
 
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Thank you for your help. I appreciate it. Peace and have a nice day.
 

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