Question

When $Mn{O_2}$ is fused with KOH, a colored compound is formed. The product and its color is:
A. ${K_2}Mn{O_4}$ , purple green
B. $KMn{O_4}$ , purple
C. $M{n_2}{O_3}$ , Brown
D. $M{n_3}{O_4}$ , Black

Answer 1

Hint: As it is given that, the reactants are $Mn{O_2}$ and KOH. Here both react with each other and form new colored compounds. This reaction occurs in the presence of air and it forms a compound which has color. In this reaction, 2 moles of $Mn{O_2}$ and 4 moles of KOH react with each other or the formation of colored products.

Complete step by step answer:
When $Mn{O_2}$ reacts with KOH, it forms a purple green colored product. The reaction is taking place in the presence of air.
Now we see the reaction given as below:
 $2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O$
Here Mn has +6 oxidation state and when it reacts with an alkaline compound in the presence of atmospheric oxygen to form purple green colored products.
So that, $Mn{O_2}$ fused with KOH and formed 2 moles of ${K_2}Mn{O_4}$ which is in purple green color and 2 moles of water formed.

Hence option (A) is the correct answer.

Note: Here if the reaction being in the process in the presence of $KN{O_3}$ as an oxidizing agent, it gives a dark green compound. Here we remember both reactions in the presence of air and potassium nitrate. The product formed in different colors because here Mn is a transition element which is colorful metal. The equation must be balanced in this reaction. We remember the number of moles in reactants and the number of moles formed in product.