Question
When $$Mn{O_2}$$ is fused with $$KOH,$$ a coloured compound is formed, the product and its colour is:
A.
$${K_2}Mn{O_4},\,{\text{purple}}\,{\text{green}}$$
B.
$$KMn{O_4},\,\,{\text{purple}}$$
C.
$$M{n_2}{O_3},\,\,{\text{brown}}$$
D.
$$M{n_3}{O_4},\,{\text{black}}$$
Answer :
$${K_2}Mn{O_4},\,{\text{purple}}\,{\text{green}}$$
Solution :
Stable oxidation state of $$Mn$$ in alkaline medium is$${{ + 6}}{{.}}$$
So, $$Mn{O_2}$$ is oxidised to $${K_2}Mn{O_4}$$ (purple green) by atmospheric oxygen in $$KOH$$ medium.
$$2Mn{O_2} + 4KOH + {O_2} \to \mathop {2{K_2}Mn{O_4}}\limits_{\left( {{\text{Purple}}} \right)} + 2{H_2}O$$