Homework Help: Chemistry: Electron Configuration of Manganese Cation

1. Jun 22, 2012

HunterDX77M

1. The problem statement, all variables and given/known data

Write the ground state electron configuration of Mn2+.

2. Relevant equations

N/A

3. 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!

2. Jun 23, 2012

Saitama

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.

3. Jun 23, 2012

HunterDX77M

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.

4. Jun 23, 2012

Infinitum

Hi HunterDX

The electronic configuration for Mn is correct

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?

5. Jun 23, 2012

HunterDX77M

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?

6. Jun 23, 2012

Infinitum

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.

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

7. Jun 23, 2012

HunterDX77M

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+:
$$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$$

8. Jun 23, 2012

Infinitum

Yep. You got it.

9. Jun 23, 2012

HunterDX77M

Thank you for your help. I appreciate it. Peace and have a nice day.