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

Jul 1, 2016

$5$

#### Explanation:

Start by writing out the electron configuration of a neutral manganese atom, $\text{Mn}$.

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, ${\text{Mn}}^{2 +}$, 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

$\text{Mn: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5} 4 {s}^{2}$

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 $4 s$ orbital, which is higher in energy than the $3 d$ orbitals when filled.

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

$\text{Mn: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5} \textcolor{red}{\cancel{\textcolor{b l a c k}{4 {s}^{2}}}}$

${\text{Mn}}^{2 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{\textcolor{b l u e}{5}}$

As you can see, the manganese(II) cation has a total of $\textcolor{b l u e}{5}$ electrons in its $3 d$ subshell, with one electron distributed in each of the five $3 d$ orbitals.