Question

In the preparation of ${\text{KMn}}{{\text{O}}_{\text{4}}}$, pyrolusite (${\text{Mn}}{{\text{O}}_{\text{2}}}$ ) is first converted to potassium manganate (${{\text{K}}_2}{\text{Mn}}{{\text{O}}_{\text{4}}}$). In this conversion, the oxidation state of manganese changes from
A.$ + 1{\text{ to + 3}}$
B.$ + 2{\text{ to + 4}}$
C.$ + 3{\text{ to + 5}}$
D.$ + 4{\text{ to + 6}}$

Answer 1

Hint:On oxidation, the oxidation number is found to be increased. On reduction, the oxidation number is reduced. Potassium permanganate is prepared from pyrolusite ore. We know that pyrolusite can be converted to potassium permanganate by oxidizing it.
Complete step by step answer:
We know that the starting material is pyrolusite ore. The crushed pyrolusite ore is taken and is fused with alkali metal hydroxide like KOH in the presence of air to give potassium permanganate (${{\text{K}}_2}{\text{Mn}}{{\text{O}}_{\text{4}}}$).
First, pyrolusite ore is converted to potassium permanganate
${\text{2Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4KOH + }}{{\text{O}}_{\text{2}}} \to {\text{2}}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}}$
The color of potassium permanganate is dark green.
The pyrolusite can also be treated with an oxidizing agent like ${\text{KN}}{{\text{O}}_{\text{3}}}$ to give potassium permanganate.
Now let us find out the change in oxidation number of manganese in this primary reaction.
In ${\text{Mn}}{{\text{O}}_{\text{2}}}$, the oxidation number on manganese is $x - 2(2) = 0$, x is the oxidation number of Mn. The oxidation number of Oxygen atoms is -2. There is no charge on ${\text{Mn}}{{\text{O}}_{\text{2}}}$. Therefore, $x = 4$
In ${{\text{K}}_2}{\text{Mn}}{{\text{O}}_{\text{4}}}$, the oxidation number of K is 1. The oxidation number of oxygen is $ - 2$. Here, ${\text{2(1) + x + 4( - 2) = 0}}$, since the overall charge is 0.
${\text{ - 6 + x = 0}}$, ${\text{x = + 6}}$
Here, we could see that the oxidation number is changing from $x = 4$ in ${\text{Mn}}{{\text{O}}_{\text{2}}}$to x=+6 in ${{\text{K}}_2}{\text{Mn}}{{\text{O}}_{\text{4}}}$.
Since there was an increase in the oxidation number, we could confirm the reaction as oxidation.
Therefore, the correct answer is $ + 4{\text{ to + 6}}$

Thus, the correct option is D.

Note: After the pyrolusite is converted to potassium permanganate by oxidation, it is then converted to potassium manganate. On electrolytic oxidation of the permanganate ion formed, we get manganate ion. Now, let us look at the change in the oxidation number of Mn in ${\text{Mn}}{{\text{O}}_{\text{2}}}$ to ${\text{KMn}}{{\text{O}}_{\text{4}}}$ . We know the oxidation number of Mn in ${\text{Mn}}{{\text{O}}_{\text{2}}}$is $ + 4$. $1 + x + 4( - 2) = 0$, ${\text{x = + 7}}$