Assign oxidation number 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 ?
Answers
Answer:
(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
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Answer:
Answer:
(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