Darkgreen manganate salts contain the $Mn{O_4}^{2-}$ion. The ion is stable in basic solution but disproportionates in acid to $Mn{O}_2\left(s\right)$ and $Mn{O_4}^{-}$.

(a) What is the oxidation state of$Mn$in $Mn{O_4}^{-}$,and $Mn{O}_2$?

(b)Write a balancedequation for thereaction of inacidic solution.1

The composition of dark green manganate salt is $Mn{O_4}^{2-}$ ions. This ion stays stable in the basic solution but shows disproportion reaction in acidic solution of $Mn{O_4}^{-}$ and $Mn{O}_2$ .

The loss of electrons during a chemical reaction which a molecule, atom or ion undergoes is known asoxidation.

The electrons are taken up by$Mn{O_4}^{2-}$as it is reduced to $Mn{O}_2$ in the presence of an acid. The reaction involves losing of two oxygen atoms and this produces water molecules with protons in solution.

(a)The oxidation state of$Mn$for $Mn{O_4}^{2-}$ is given by:

The overall charge is -2 .

The charge of each oxygen atom is also -2 .

The charge of is calculated as:

$$\begin{gathered} =-2-4(-2) \\ =-2+8 \\ =+6 \end{gathered}$$

The oxidation state of $Mn$for $Mn{O_4}^{-}$ is given by:

The overall charge is -1 .

The charge of each oxygen atom is also -2 .

The charge of is calculated as:

$$\begin{gathered} =-1-4(-2) \\ =-1+8 \\ =+7 \end{gathered}$$

The oxidation state of$Mn$for $Mn{O}_2$ is given by:

The overall charge is 0 .

The charge of each oxygen atom is also -2 .

The charge of is calculated as:

$$\begin{gathered} =0-2(-2) \\ =0+4 \\ =+4 \end{gathered}$$

c) The reaction for this is given by:

­$2e^{-}+Mn{{O_4}^{2-}}_{\left(aq\right)}+4{H^{+}}_{\left(aq\right)}\to Mn{O}_{2\left(s\right)}+2H_2{O}_{\left(l\right)}$