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CBSE - XII Science - Chemistry - Coordination Compounds

How can we find the oxidation number of central atom in a coordination compound. Please explain with an example.

Asked by akanksha 23rd September 2013, 7:54 PM
Answered by Expert


The oxidation state is designated by Roman numerals (I, II, III, etc.) in the bracket at the end of the name of the complex without a gap between the two. 

Analyze the atomic composition of the molecule and determine to which group in the periodic table each atom belongs.

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.

Assign the oxidation number +1 for the group IA alkali metal atoms (for example, Na or K) if such atoms prresent in the molecule.
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.

Assign the oxidation number -1 for halogens (the group VIIA).
There are no such elements in our example.
Assign the oxidation number +1 for hydrogen.
Assign the oxidation number -2 for oxygen.
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."
In the example:
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.
In our example, 1 + A -8 = 0 or A - 7 = 0
Hence, A = +7
Answered by Expert 24th September 2013, 9:46 AM

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