Answer:
Oxidation State of H in H₂SO₄ = +1
Oxidation State of S in H₂SO₄ = +6
Oxidation State of O in H₂SO₄ = -2
Step-by-step 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