Question

Question #a3715

Answer 1

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

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

`overbrace([Ar]3d^5 4s^2)^(stackrel(color(blue)(0))(Mn)) -> overbrace([Ar] 3d^5 4s^0)^(stackrel(color(blue)(+2))(Mn))`

Oxidation states are hypothetical charges from assuming full ionization, and manganese is a metal, to a good approximation, by ionizing manganese twice from the `4s` orbital, we generate `bb("Mn"^(2+))` as the more common oxidation state.

`"Mn"^(3+)` is not as favorable, because of the `3d` orbitals being lower in energy than the `4s` orbitals.

`overbrace([Ar]3d^5 4s^2)^(stackrel(color(blue)(0))(Mn)) -> overbrace([Ar] 3d^(color(red)(4)) 4s^0)^(stackrel(color(red)(+3))(Mn))`

The incoming atom would have to be particularly electronegative to allow a `+3` oxidation state to occur (such as in `"Mn"_2"O"_3(s)`), and if it is too electronegative, it could give rise to a `+4` (such as in `"MnO"_2(s)`) or `+7` (such as in `"MnO"_4^(-)(aq)`) oxidation state instead, which are also common.