What is the common oxidation state of manganese metal?
2 Answers
Good question. How do you determine the most stable oxidation state?
Explanation:
Manganese metal has a very large redox manifold, and can form compounds with metal oxidation states from
In water, in common with most of the transition metals, manganese commonly forms
You can determine the most stable oxidation state by identifying the ground-state electron configuration.
Since an oxidation state is defined as the hypothetical charge of a pure ion, the ground-state configuration for the corresponding
You can write the valence configuration as (
#ul(" "" ")#
#" 4s"# (#-"7.84 eV"# )
#ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))#
#underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")#
#" "" "" "" "" ""3d orbitals"# (#-"11.4 eV"# )
This is said to be most stable because
- as many electrons are unpaired as possible (coulombic repulsions from paired electrons are minimized), being distributed in accordance with Hund's rule.
- they all occupy the non-highest-energy valence orbitals.
(We don't worry about the core orbitals because they are already low in energy and fully-occupied, thus not participating in bonding.)
In this state, manganese has lost its two valence
So, the