Question

What is the oxidation state of S in each anion? SO₄^2-, SO₃^2- , HSO₃^- , HSO₄^-,

Answer 1

Answer: a) SO4^^-2 =   x + (-2 x 4) =-2 [ where x = the oxidation state of the element to be found,

                                                             - 2 = is oxidation state of oxygen in sulphate ion , it is multiplied by 4 since since                                                                    there are 4 oxygen atoms

                                                                  the whole equation is equated to -2, since valency of sulphate ion is -2

                              on solving this equation , x- 8 = -2 ===> x = -2 +8  ===> x = 6 (which is oxidation state of S in

                                                                                                                                             sulphate ion )

similarly in b) SO3^-2 = oxidation state of S is , x = -2 +6[ got from x + -2 x3 = -2 ]

                                 oxidation state of S in SO3^-2 = +4

                   c) HSO3^-1 = 1+ x + (-2 x3) =-1 [ where valency of H=1 ] =====> x = 6 [ which is oxidation state of sulphur is bisulphite ]

     d) HSO4^-1 = 1+x +(-2 x 4) =-1 ==> x = 6 [which is oxidation state of sulphur in bisulphate ion }