Question

Find the oxidation number of mn in the product of alkaline oxidative fusion of mno2

Answer 1

Answer: +6

Explanation:

The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is +6.

2MnO2+4KOH+O2→2K2MnO4+2H2O

In K2MnO4 the oxidation state of Mn is +6.

Answer 2

The oxidation number of Mn in the product of the alkaline oxidative fusion of $MnO_2$ is +6.

Given:

The reaction is an alkaline oxidative fusion of $MnO_2$.

To Find:

The oxidation number of Mn in the product in the reaction.

Solution:

The reaction of alkaline oxidative fusion of $MnO_2$ is as follows:

$2MnO_2+4KOH+O_2= 2K_2MnO_4+2H_2O$

Here the product of Mn is $K_2MnO_4$.

Let us assume the oxidation number of Mn in  $K_2MnO_4$ is x.

So, 2×1+x+(-2)4=0

x= 6

Hence, The oxidation number of Mn in the product of the alkaline oxidative fusion of $MnO_2$ is +6.

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