When $M n O_{2}$ is fused with KOH, a colored compound is formed. The product and its color is:
A. $K_{2} M n O_{4}$ , purple green
B. $K M n O_{4}$ , purple
C. $M n_{2} O_{3}$ , Brown
D. $M n_{3} O_{4}$ , Black

Answer

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Hint: As it is given that, the reactants are

M n 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

M n O_{2}

and 4 moles of KOH react with each other or the formation of colored products.

Complete step by step answer:
When

M n 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:

2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n 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,

M n O_{2}

fused with KOH and formed 2 moles of

K_{2} M n 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

K N 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.