balance MnO4- + I- gives MnO2 +I2 in acidic Medium by ion electron methods#class11 #jee #neet
balance MnO4- + I- gives MnO2 +I2 in acidic Medium by ion electron methods#class11 #jee #neet

Potassium permanganate (KMnO4) contain one potassium atom, one manganese atom and four oxygen atoms.

There are two methods to find the oxidation number of an atom in a molecule or ion and we are using both methods to find oxidation number of KMnO4.

Oxidation number of manganese is important to decide to whether KMnO4 can be oxidized or reduced to other compounds.

In algebra method, we are going to use known oxidation numbers in the molecule to find out unknown oxidation number of an atom in the same molecule.

Oxidation number of K + 4 × (oxidation number of oxygen) + oxidation number of Mn = 0

Let’s take oxidation number of Manganese atom as x.

The Oxidation number of manganese in KMnO4 is +7.

First, we have to draw the best lewis structure of KMnO4.

Then, we can find the Oxidation number due to bonds and Oxidation number due to charge using the lewis structure.

Oxidation number due to bonds: there are three double bonds and one single bond around manganese with oxygen atoms. Oxygen is more electronegative than manganese. Therefore, electrons in the bonds are attracted towards the oxygen atoms. Manganese gives seven electrons for these bonds. So, manganese atom losses 7 electrons and get (+7) oxidation number due to bonds.

Oxidation number due to charge: in KMnO4 molecule manganese atom has no charge. So, oxidation number due to charge is zero.

Overall oxidation number = Oxidation number due to bonds + Oxidation number due to charge

Overall oxidation number of Mn in KMnO4 = +7

Because manganese is at +7 oxidation state, that manganese atom can be reduced to lower oxidation states such as +2 or +4. Therefore, KMnO4 can act as a oxidizing agent.

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