In KMn04 molecule, the oxidation states of the elements potassium (K), manganese (Mn) and oxygen (O) are respectively

In the neutral compound potassium permanganate (KMnO4), the sum of the oxidation states of all atoms must equal zero [1]. Potassium (K) is an alkali metal from Group 1, which consistently exhibits an oxidation state of +1 in its compounds [1]. Oxygen (O) typically has an oxidation state of -2, except in peroxides or superoxides; in KMnO4, it follows the standard -2 state [1]. With four oxygen atoms, the total contribution from oxygen is -8 [1]. To maintain electrical neutrality, the calculation is (+1) + x + 4(-2) = 0, where x represents the oxidation state of manganese (Mn). Solving this equation (1 + x - 8 = 0) yields x = +7 [1]. Therefore, the oxidation states for K, Mn, and O are +1, +7, and -2, respectively, making KMnO4 a powerful oxidizing agent.

Sources

1. [1] https://pubchem.ncbi.nlm.nih.gov/compound/Potassium-Permanganate