I always get confused with oxidation number rules for transition metals. Can someone show the full working for manganese in potassium permanganate?
Use the rule that the sum of all oxidation numbers in a neutral compound equals zero. In KMnO4, potassium (an alkali metal) is always +1, and oxygen is usually -2. Let the oxidation number of manganese be x. Now write the sum: (+1 for K) + (x for Mn) + (4 times -2 for the four oxygens) = 0. That gives 1 + x - 8 = 0, so x - 7 = 0, which means x = +7. So manganese is in the +7 oxidation state in KMnO4, its highest possible. This is why KMnO4 is such a strong oxidising agent, manganese readily drops from +7 to lower states like +2. Always fix the known elements first (K and O here), then solve for the unknown.
Sign in as a tutor to answer this doubt.