Question

Assign oxidation number to the underlined elements in each of the following species:
Ni( Co)4​

Answer 1

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

Explanation:

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Answer 2

Explanation:

the answer is the

x=+6 is the answer