What is the common oxidation state of manganese metal?

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 $V I I +$. Subvalent oxidation states of $- I I$ 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 $M {n}^{2 +}$ ions. So $M n \left(I I +\right)$ compounds and complexes are common. This has a ${d}^{5}$ ground state, and as a consequence, $M n \left(I I +\right)$ 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 $\text{Mn}$ ion should define the most stable oxidation state.

You can write the valence configuration as ($Z = 25$)$\boldsymbol{\text{^"[*]}}$:

$\underline{\text{ "" }}$
$\text{ 4s}$ ($- \text{7.84 eV}$)

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" }}$
$\text{ "" "" "" "" ""3d orbitals}$ ($- \text{11.4 eV}$)

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 $4 s$ electrons, going from an electron configuration of $\left[A r\right] 3 {d}^{5} 4 {s}^{2}$ for $\stackrel{\textcolor{b l u e}{0}}{\text{Mn}}$ to $\left[A r\right] 3 {d}^{5}$ for $\stackrel{\textcolor{b l u e}{+ 2}}{\text{Mn}}$.

So, the $+ 2$ state is most stable.

$\boldsymbol{\text{^"[*]}}$: http://media.pearsoncmg.com/bc/bc_0media_chem/adv_chem/pdf/11054_appB_ts.pdf