Dark green manganate salts contain the$\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$ion. The ion is stable in basic solution but disproportionates in acid to$\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$ and $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}$ .

(a) What is the oxidation state of$Mn$ in role="math" localid="1663416208358" $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$,role="math" localid="1663416222031" $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}$ , and role="math" localid="1663416239015" $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}$?

(b) Write a balanced equation for the reaction of$\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$in acidic 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 $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$as it is reduced to role="math" localid="1663417345770" $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}$ 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:

­$\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{+}\mathbf{}\mathit{M}\mathit{n}{{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}}_{\left(aq\right)}\mathbf{}\mathbf{+}\mathbf{4}{{\mathit{H}}^{\mathbf{+}}}_{\left(aq\right)}\mathbf{}\mathbf{\to }\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}\left(s\right)}\mathbf{}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\left(l\right)}$