Question

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

Answer 1

`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.