Answer :

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Ferric Sulphate Fe2(SO4)3=Fe2S3O12 Oxidation no. of iron is 3 & oxygen is -2 So, Fe charge =2*3=6 O charge=12*-2=-24 Fe+O=6-24=-18 To make charge 0, -18+S3=0 => S3=18 S=6 Ferrous Sulphte, FeSO4 Oxidation no. og Fe is 2 and of O is -2 So, Fe charge=2 O charge=4*-2=-8 F+O=2-8=-6 As in ferric sulphate -6+S=0 S=6
Ferric Sulphate :  Iron (III) sulphate    valency of iron is 3.
   Chemical formula:      Fe₂  (SO₄)₃        Ions:  Fe³⁺    and  SO₄²⁻

oxidation number of  Fe is +3   
oxidation number of  SO₄  is  -2
 Oxidation number of oxygen (in general) is  - 2
         hence,  - 2 = S + 4 * -2            =>    S = 6

   oxidation number of Sulphur in Ferric Sulphate   6
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Ferrous sulphate  or  Iron (II) suplhate  : chemical formula:  Fe  SO₄
 Here Iron has valency - oxidation number 2.  as  the ions are:  Fe²⁺    SO₄²⁻

Hence the oxidation number of  Sulphur will be 6 like in the case of  Ferric sulphate.  It will be  6.

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