How many #3d# electrons does the manganese(II) ion, #"Mn"^(2+)#, have?

1 Answer
Stefan V.
Jul 1, 2016

#5#

Explanation:

Start by writing out the electron configuration of a neutral manganese atom, #"Mn"#.

Manganese is located in period 4, group 7 of the periodic table and has an atomic number equal to #25#. This means that a neutral manganese atom must have #25# electrons surrounding its nucleus.

Consequently, the manganese(II) cation, #"Mn"^(2+)#, which is formed when a neutral manganese atom loses #2# electrons, will have a total of #23# electrons surrounding its nucleus.

So, the electron configuration of manganese is

#"Mn: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2#

Now, it's very important to remember that the #2# electrons that are lost when the manganese(II) cation is formed are coming from the #4s# orbital, which is higher in energy than the #3d# orbitals when filled.

This means that the electron configuration of the manganese(II) cation will be

#"Mn: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 color(red)(cancel(color(black)(4s^2)))#

#"Mn"^(2+): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^color(blue)(5)#

As you can see, the manganese(II) cation has a total of #color(blue)(5)# electrons in its #3d# subshell, with one electron distributed in each of the five #3d# orbitals.

Related questions
See all questions in Electron Configuration
Impact of this question
43049 views around the world
You can reuse this answer
Creative Commons License