What’s the easiest way for finding the oxidation and reduction numbe

1.	What’s the easiest way for finding the oxidation and reduction numbers of an element in a compound in a chemical equation?
Answer» There are different rules on finding the oxidation number for different compound. These are as follows 1.oxidation number of element is zero i.e. either mono-atomic or di atomic e.g. Na, H₂, Cl₂ ,Al etc. 2. Oxidation number of ion is charge present on it e.g Al⁺⁺⁺ means oxidation number of aluminium is 3. 3. Sum of oxidation number of neutral compound is always zero. 4. Oxidation number of alkali metal is +1 and alkaline earth metal is +2 . In oxygen generally oxidation state is -2 but contain limitation on peroxide and super oxide. 5 Oxidation of halogen is generally -1 but with Chlorine and Bromine contain exception when they formed the compound with Florine. 6. Hydrogen also give the oxidation state of +1 but some exception with metal hydride.

What’s the easiest way for finding the oxidation and reduction numbers of an element in a compound in a chemical equation?

Answer»

There are different rules on finding the oxidation number for different compound. These are as follows

1.oxidation number of element is zero i.e. either mono-atomic or di atomic e.g. Na, H₂, Cl₂ ,Al etc.

2. Oxidation number of ion is charge present on it e.g Al⁺⁺⁺ means oxidation number of aluminium is 3.

3. Sum of oxidation number of neutral compound is always zero.

4. Oxidation number of alkali metal is +1 and alkaline earth metal is +2 . In oxygen generally oxidation state is -2 but contain limitation on peroxide and super oxide.

5 Oxidation of halogen is generally -1 but with Chlorine and Bromine contain exception when they formed the compound with Florine.

6. Hydrogen also give the oxidation state of +1 but some exception with metal hydride.

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