Question a3715

Jul 8, 2017

The question should say "which is more common", and that would be ${\text{Mn}}^{2 +}$ (such as in "Mn"("OH")_2(s)#, or just as the ${\text{Mn}}^{2 +}$ free ion). Oxidation states are listed here.

The $4 s$ orbital of manganese is higher in energy, and so, the two $4 s$ electrons get ionized first.

${\overbrace{\left[A r\right] 3 {d}^{5} 4 {s}^{2}}}^{\stackrel{\textcolor{b l u e}{0}}{M n}} \to {\overbrace{\left[A r\right] 3 {d}^{5} 4 {s}^{0}}}^{\stackrel{\textcolor{b l u e}{+ 2}}{M n}}$

Oxidation states are hypothetical charges from assuming full ionization, and manganese is a metal, to a good approximation, by ionizing manganese twice from the $4 s$ orbital, we generate $\boldsymbol{{\text{Mn}}^{2 +}}$ as the more common oxidation state.

${\text{Mn}}^{3 +}$ is not as favorable, because of the $3 d$ orbitals being lower in energy than the $4 s$ orbitals.

${\overbrace{\left[A r\right] 3 {d}^{5} 4 {s}^{2}}}^{\stackrel{\textcolor{b l u e}{0}}{M n}} \to {\overbrace{\left[A r\right] 3 {d}^{\textcolor{red}{4}} 4 {s}^{0}}}^{\stackrel{\textcolor{red}{+ 3}}{M n}}$

The incoming atom would have to be particularly electronegative to allow a $+ 3$ oxidation state to occur (such as in ${\text{Mn"_2"O}}_{3} \left(s\right)$), and if it is too electronegative, it could give rise to a $+ 4$ (such as in ${\text{MnO}}_{2} \left(s\right)$) or $+ 7$ (such as in ${\text{MnO}}_{4}^{-} \left(a q\right)$) oxidation state instead, which are also common.