Define oxidation and reduction in terms of oxidation number

1.	Define oxidation and reduction in terms of oxidation number
Answer» SOLUTION :Oxidation involves INCREASE in O.N while reduction involves decrease in O.N `Sn^(2+)+2HG^(2+)rarrSn^(4+)+Hg_(2)^(2+)` here `Sn^(2+)` gets oxidised to `Sn^(4+)` because its O.N increase from +2 while `HG^(2+)` gets reduced to `Hg_(2)^(2+)` because its O.N decreasefrom + 2 to +1

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