In the compound KMnO4 potassium permanganate , the oxidation states of each element are as follows:1. Potassium K : The oxidation state of potassium in its compounds is always +1. So, in KMnO4, the oxidation state of K is +1.2. Manganese Mn : To determine the oxidation state of manganese, we need to consider the overall charge of the compound and the oxidation states of the other elements. Since KMnO4 is a neutral compound, the sum of the oxidation states of all elements must be equal to zero.3. Oxygen O : The oxidation state of oxygen in its compounds is usually -2 except in peroxides, where it is -1, and in OF2, where it is +2 . In KMnO4, the oxidation state of each oxygen atom is -2.Now, let's calculate the oxidation state of manganese Mn using the information above:Let x be the oxidation state of Mn. We have 1 potassium atom, 1 manganese atom, and 4 oxygen atoms in the compound. So, the equation for the sum of the oxidation states is: +1 + x + 4 -2 = 0Solving for x:x - 7 = 0x = +7Therefore, the oxidation state of manganese Mn in KMnO4 is +7.In summary, the oxidation states of each element in KMnO4 are:- Potassium K : +1- Manganese Mn : +7- Oxygen O : -2