How can we find the oxidation number of central atom in a coordination compound. Please explain with an example.
Determining the oxidation number of manganese (Mn) in potassium permanganate, KMnO4, will be used as an example. Potassium (K) belongs to the group IA, manganese (Mn) is in group VIIB, while oxygen (O) is from the group VIA.
In our example, it is potassium (K).
Assign the oxidation number +2 for the group IIA alkaline earth atoms (for example, Ca or Mg).
There are no such elements in the molecule in our example.
In our example, there is oxygen (O) in the molecule KMnO4.
Multiply the oxidation number for each element by its quantity in the molecule. If the oxidation number cannot be derived from Steps 2 to 6 denote it as "A."
Potassium (K): +1 x 1 = 1
Oxygen (O): -2 x 4 = -8
Manganese (Mn): A x 1 = A
Add up all values obtained in Step 7 and make them equal to zero. Zero is the total charge of a molecule that must remain neutral.