
For $Mn{O_2} + KOH + {O_2}\xrightarrow{\Delta }2{K_2}Mn{O_4} + 4{H_2}O$
If true enter $1$ else $0$.
Answer
568.8k+ views
Hint: It is of the $ + 6$ oxidation state of the permanganate ion. In a strongly basic solution, the permanganate (VII) is reduced to the $ + 6$ oxidation state. It produces a green color.
Complete step by step answer:
In compounds, manganese remains with a number of different oxidation states. It can be easily raised to the $ + 2$ state. For example, by reaction with hydrochloric acid $\left( {HCl} \right)$ to form manganous chloride, $MnC{l_2}$. Manganese can also be found in the $ + 3$ (manganic) oxidation state. But this oxidation state is unstable. It usually goes back to the $ + 2$ oxidation state. Both of these manganous and manganic ions can form acidic solutions. Manganese is actually found largely in the $ + 4$ oxidation state in manganese dioxide $\left( {Mn{O_2}} \right)$. The $ + 4$ oxidation state is amphoteric, which means in the $ + 4$ oxidation state manganese can either donate or accept electrons in any chemical reactions. Manganese also exists in $ + 6$ and $ + 7$ oxidation states. The $ + 6$ oxidation state is found largely in the manganate ion $\left( {MnO_4^ - } \right)$ and the $ + 7$ oxidation state is found largely in the permanganate ion $\left( {MnO_4^ - } \right)$. These two ions are stable in basic solutions.
In a strongly basic solution, we can also see that the permanganate (VII) is reduced to the (green) $ + 6$ oxidation state of the manganate ion $\left( {MnO_2^{4 - }} \right)$:
$Mn{O^{4 - }} + {e^ - } \to MnO_2^{4 - }$
Here, $KOH$ is the basic medium. So, when $Mn{O_2}$ reacts with $KOH$ in the presence of ${O_2}$, potassium permanganate $\left( {{K_2}Mn{O_4}} \right)$ (green colored) is obtained. The required reaction is:
\[2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O\]
So, it is true.
Note: Manganese can be found in abundance in nature. We should remember that manganese can also be found in $ + 1$ oxidation state (in the case of complex cyanide). Also, we should remember that manganese is not very stable in its $ + 5$ oxidation state in basic solution.
Complete step by step answer:
In compounds, manganese remains with a number of different oxidation states. It can be easily raised to the $ + 2$ state. For example, by reaction with hydrochloric acid $\left( {HCl} \right)$ to form manganous chloride, $MnC{l_2}$. Manganese can also be found in the $ + 3$ (manganic) oxidation state. But this oxidation state is unstable. It usually goes back to the $ + 2$ oxidation state. Both of these manganous and manganic ions can form acidic solutions. Manganese is actually found largely in the $ + 4$ oxidation state in manganese dioxide $\left( {Mn{O_2}} \right)$. The $ + 4$ oxidation state is amphoteric, which means in the $ + 4$ oxidation state manganese can either donate or accept electrons in any chemical reactions. Manganese also exists in $ + 6$ and $ + 7$ oxidation states. The $ + 6$ oxidation state is found largely in the manganate ion $\left( {MnO_4^ - } \right)$ and the $ + 7$ oxidation state is found largely in the permanganate ion $\left( {MnO_4^ - } \right)$. These two ions are stable in basic solutions.
In a strongly basic solution, we can also see that the permanganate (VII) is reduced to the (green) $ + 6$ oxidation state of the manganate ion $\left( {MnO_2^{4 - }} \right)$:
$Mn{O^{4 - }} + {e^ - } \to MnO_2^{4 - }$
Here, $KOH$ is the basic medium. So, when $Mn{O_2}$ reacts with $KOH$ in the presence of ${O_2}$, potassium permanganate $\left( {{K_2}Mn{O_4}} \right)$ (green colored) is obtained. The required reaction is:
\[2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O\]
So, it is true.
Note: Manganese can be found in abundance in nature. We should remember that manganese can also be found in $ + 1$ oxidation state (in the case of complex cyanide). Also, we should remember that manganese is not very stable in its $ + 5$ oxidation state in basic solution.
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