Question

Assign the oxidation number of a) S in Na 2 S 2 O 3 and Na 2 S 4 O 6 b) Mn in K 2 MnO 4 c) Cr in Cr2 O7 2- d) B in NaBH 4

Answer 1

Explanation:

a) let the oxidation number of P is x 

we know,

  O.N of Na = +1

  O.N of H = +1

  O.N of O = -2

              now, 1(+1) +2(+1) + x +4(-2) =0

                      1 + 2 + x -8 = 0

                      x = +5

hence, the oxidation number of P = +5 

(b) let oxidation  number of S = x

 now, 1(+1) +1(+1) +x + 4(-2) =0

          1 + 1 + x -8 = 0

                x = 6 

hence, oxidation number of S = +6

(C) H₄P₂O7

 let o.s of P = X

 4(+1) + 2x + 7(-2) = 0

  4 + 2x - 14 = 0

  x = + 5

hence, o.s of P = +5

(d) k₂MnO₄ 

let o.s of Mn is x

 2(+1) + x + 4(-2) = 0

 2 + x -8 = 0

  x = +6 

hence, o.s of Mn = +6

(e) CaO2

let o.s of O is x

 +2 + 2x = 0

    x = -1 

hence, o.s of  O = -1

(F) NaBH₄ 

let o.s of B is x 

+1 + x + 4(-1)  =0 [HERE,o.s of H is -1 due to ionic compound ]

   x = +3

(g) H₂S₂O₇

 let the o.s of S is x

 2 + 2x + 7(-2) = 0

 x = +6

(h) KAl(SO₄)₂.12H₂O

lety the o.s of S is x 

 +1 + 3 + 2x +8(-2) + 12(+2 -2) =0

  2x -12 = 0

   x = +6

hence, o.s of S IS +6