In h2so4, the oxidation number of h is , that of s is , and that of o is .

Oxidation State of H in H₂SO₄  = +1

Oxidation State of S in H₂SO₄  = +6

Oxidation State of O in H₂SO₄  = -2

Explanation:

Oxidation number:

oxidation number is an apparent charge of an atom in a compound.

the total charge on a compound is zero

So for H₂SO₄ total charge = 0

Hydrogen donate an electron in H₂SO₄ and have +1 charge that is oxidation number

***so to calculate oxidation number of  Sulfur and Oxygen in H₂SO₄

oxygen have 2- state

O.N. of H = 1+

O.N. of O = 2-

O.N. of S = x

H₂SO₄ =0

put values

(+1)2 + S + (-2)4=0

+2 + S -8 =0

S = 0-2 + 8

S = +6

Now to find oxidation number of Oxygen

O.N. of H = 1+

O.N. of O = x

O.N. of S = +6

H₂SO₄ =0

put values

(+1)2 + (+6) + (O)4=0

+2 + 6 - (O) 4 =0

(O)4 = 0-2 - 8

(O)4 = -8

O = -8/4

O = -2

***Now to find oxidation number of Oxygen

O.N. of H = 1+

O.N. of O = x

O.N. of S = +6

H₂SO₄ =0

put values

(H)2 + (+6) + (-2)4=0

(H)2 + 6 - 8 =0

(H)2 = 0 - 6 + 8

(H)2 = +2

H = +2/2

H = +1

the answer is 75.0%.