Question

The oxidation number of Mn in the product of alkaline oxidative fusion of $Mn{{O}_{2}}$ is:

Answer 1

Hint: Reaction that takes place in the alkaline oxidative fusion of $Mn{{O}_{2}}$ is,\[2Mn{{O}_{2}}+4KOH+{{O}_{2}}\to {{K}_{2}}Mn{{O}_{4}}+2{{H}_{2}}O\], Here the product of alkaline oxidative fusion will be ${{K}_{2}}Mn{{O}_{4}}$ or $Mn{{O}_{4}}^{2-}$. The two negative charges given here it not the charge on the oxygen atom it is the charge of the whole element,${{K}_{2}}Mn{{O}_{4}}$.

Complete answer:
From your chemistry lessons you have learned about the oxidation number and what is alkaline oxidative fusion. Oxidation number is the imaginary charge which is present in an atom in its combined state because of the real charge or electronegativity difference present on a monatomic ion.
In alkaline oxidative fusion, alkaline refers to the basic condition like use of KOH or NaOH and oxidative states that in a reaction there must be an increase in oxidation number and therefore oxidation has taken place and fusion means the process of combination.
Hence it is a redox reaction in which compounds undergo fusion with themselves only, in the presence of alkali or bases.
In the question we have given that we have to find the alkaline oxidative fusion of $Mn{{O}_{2}}$ in the presence of an alkaline solution of KOH.
\[4KOH+2Mn{{O}_{2}}+{{O}_{2}}\to 2{{K}_{2}}Mn{{O}_{4}}+2{{H}_{2}}O\]
In the above reaction there is a fusion of $Mn{{O}_{2}}$ to ${{K}_{2}}Mn{{O}_{4}}$ or $Mn{{O}_{4}}^{2-}$
Now let us find the oxidation state of Mn in $Mn{{O}_{2}}$ of the reactant, Here Oxygen is a negatively charged ligand, we have to find the oxidation no. of Mn so let it be 'X And there no charge on the complex so it is neutral, then
So, the oxidation state of Mn in $Mn{{O}_{2}}$ will be,
$X + 2 \times (-2) = 0$
$X= +4$
Here $Mn $ exists in $+4$ state.
Now let us find the oxidation state of Mn in the product ${{[Mn{{O}_{4}}]}^{2-}}$ , Here oxygen is the negatively charged ligand (having 2- charge on it) and the whole complex is anionic because it have two negative charge on it so, let the oxidation state of Mn will be 'X'
So, the oxidation state Mn in ${{[Mn{{O}_{4}}]}^{2-}}$ will be,
$X + 4 \times (-2) = -2$
$X – 8 = -2$
$X = +6$
Here, Mn exists in +6 state.

Thus the oxidation state on Mn in the product of alkaline oxidative fusion of $Mn{{O}_{2}}$ is +6.

Note:
In the above reaction the oxidation state of Mn changes from +4 to +6 oxidation state. We can write ${{K}_{2}}Mn{{O}_{4}}$ as ${{[Mn{{O}_{4}}]}^{-2}}$ because if you break the complex K (potassium) will have +2 charge on it and rest will have negative charge on it (-2) and behave as an anionic complex. The oxidation of Mn will happen only in the presence of KOH.