Single Choice

When $$MnO_2$$ is fused with KOH and $$O_2$$, what is the product formed and its colour?

AMnO - colourless
B$$KMnO_4$$ - purple
C$$K_2MnO_4$$ - dark green
Correct Answer
D$$MnO_3$$ - black

Solution

In alkaline medium, $$Hn$$ has a stable oxidation state +6. In the presence of $$KOH,{ MnO }_{ 2 }$$ reacts in the presence of atmospheric oxygen to form $${ K }_{ 2 }{ MnO }_{ 4 }$$ which is a green coloured product. $${ 2MnO }_{ 2 }+4KOH+{ O }_{ 2 }\longrightarrow 2{ K }_{ 2 }{ MnO }_{ 4 }+2{ H }_{ 2 }O$$

SIMILAR QUESTIONS

d - block and f - block Elements

Baeyer's reagent is:

d - block and f - block Elements

A solution of $$KMnO_{4}$$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on $$pH$$ of the solution. What different stages of the reduction do these represent and how are they carried out?

d - block and f - block Elements

These questions consists of two statements each, printed as assertion and reason, while answering these questions you are required to choose any one of the following responses. Assertion : $$KMnO_4$$ is purple in colour due to charge transits Reason : There is no electron present in dentine's of manganese in $$MnO_4$$.

d - block and f - block Elements

Heaviest atom among the following is:

d - block and f - block Elements

The color of $$KMnO_4$$ is due to :

d - block and f - block Elements

The colour of $$KMnO_4$$ disappears when oxalic acid is added to its solution in an acidic medium. This can be explained as :

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