Question

Question 8.1 Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O

Class XI Redox Reactions Page 272

Answer 1

(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
     
                         

Answer 2

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