Identify the correct increasing order of the actual oxidation state of the underlined atom in the following species: O F ₂ , Pb O ₂ , Cr O ₅ , K Mn O ₄

Step-by-step solution

Let us calculate the actual oxidation state of the underlined atom in each compound. In OF ₂ : Fluorine is more electronegative than oxygen and always has an oxidation state of -1 . Let the oxidation state of O be x . x + 2(-1) = 0 x = +2 . In PbO ₂ : It is a normal oxide (lead dioxide), not a peroxide. Let the oxidation state of Pb be y . y + 2(-2) = 0 y = +4 . In CrO ₅ : Calculating algebraically gives z + 5(-2) = 0 z = +10 , which is impossible since chromium belongs to group 6 and its maximum oxidation state is