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Calculate the oxidation number of all the atoms in the following compounds and ions : CO_2,SiO_2,PbSO_4,ClO_4^(-) |
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Answer» SOLUTION :`CO_2` : OXIDATION NUMBER of each oxygen atom = -2 Oxidation number of carbon (say x) `{:(x,-2),(C,O_2):}` `x+(-2xx2) =0 " or " x = +4` `:.` Oxidation number of C = +4 , O = -2 `SiO_2` : Oxidation number of each oxygen atom = -2 Oxidation number of silicon (say x) `{:(x,-2),(Si,O_2):}` `x +(-2xx2) " or "x =+4` `PbSO_4` : Oxidation number of each oxygen =-2 Oxidation number of lead (Pb) = +2 Oxidation number of SULPHUR (say x) `{:(+2,x,-2),(Pb,S,O_4):}` Oxidation number of Pb = +2, S = +6 , O = - 2 `ClO_4^(-)` : Oxidation number of chlorine (say x) `[{:(x,-2),(CL,O_4):}]` `x+(-2xx4)=-1 " or "x=+7` `:.` Oxidation number of O = 2 , Cl = +7 |
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