Question

When \[Mn{O_2}\] is fused with \[KOH\] and \[{O_2}\], what is the product formed and its colour?
A. \[{\text{MnO - Colourless}}\]
B. \[KMn{O_4}{\text{ - Purple}}\]
C. \[{{\text{K}}_2}{\text{Mn}}{{\text{O}}_4}{\text{ - Dark green}}\]
D. \[{\text{Mn}}{{\text{O}}_3}{\text{ - Black}}\]

Answer 1

Hint:
1. As the reaction is carried out in presence of atmospheric oxygen, and the atmospheric oxygen is green in colour.
2. \[{H_2}O\]is colourless in nature.
3. For balancing the reaction, use 2 moles of \[Mn{O_2}\] and 4 moles of KOH and 1 mole of Oxygen.

Complete answer:
As per question we have to mix fused \[Mn{O_2}\] with \[KOH\] and \[{O_2}\],
In that case,the reaction will be,
\[2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O\]
Here, It is clearly observed that when, \[Mn{O_2}\] is fused with \[KOH\] and \[{O_2}\] we will get, \[2{K_2}Mn{O_4}\] and \[{H_2}O\]
Here, we know that the oxidation state of Mn is +6.
The above reaction is placed in presence of atmospheric oxygen,
The colour of atmospheric oxygen is green.
Hence, the obtained product i.e. \[2{K_2}Mn{O_4}\] will be green in nature.
We know that \[{H_2}O\] is colourless in nature.
So, the final obtained product obtaining after the reaction when \[Mn{O_2}\] is fused with \[KOH\] and \[{O_2}\] i.e. \[2{K_2}Mn{O_4}\] and \[{H_2}O\] will be green in colour.
Hence, right choice for this question will be \[{{\text{K}}_2}{\text{Mn}}{{\text{O}}_4}{\text{ - Dark green}}\].

Hence, Option C will be the correct choice for this answer.

Additional Information:
1. Fused is a process when two or more elements are joined into a single element.
2. Colour of \[Mn{O_2}\] is Brown/Black, and it is insoluble in water.
3. Colour of \[KOH\] is milky white.
4. Colour of atmospheric oxygen is green/ dark-green.

Note:
1. As the obtained water after reaction is colourless in nature, so, the colour of final obtained element will be same as the colour of atmospheric oxygen ,
2. Balance the number of moles of \[Mn{O_2}\],\[KOH\] and \[{O_2}\] for obtaining the balance equation.