What is the common oxidation state of manganese metal?

2 Answers
Oct 17, 2016

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 0 to VII+. Subvalent oxidation states of -II and -I are known. Manganese has 7 valence electrons, and thus many possible oxidation states.

In water, in common with most of the transition metals, manganese commonly forms Mn^(2+) ions. So Mn(II+) compounds and complexes are common. This has a d^5 ground state, and as a consequence, Mn(II+) are VERY palely coloured, i.e. almost colourless.

Oct 17, 2016

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 "Mn" ion should define the most stable oxidation state.

You can write the valence configuration as (Z = 25)bb(""^"[*]"):

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 4s electrons, going from an electron configuration of [Ar] 3d^5 4s^2 for stackrel(color(blue)(0))"Mn" to [Ar] 3d^5 for stackrel(color(blue)(+2))"Mn".

So, the +2 state is most stable.

bb(""^"[*]"): http://media.pearsoncmg.com/bc/bc_0media_chem/adv_chem/pdf/11054_appB_ts.pdf