There are rules to assigning conventional oxidation numbers to some elements. For those unspecified, you can solve them. This is how you solve it.
*S8. Since this is in elemental form, its oxidation number is assigned as 0.
*H2S. H is assigned with +1. Since the compound is neutral, the overall charge is 0. So,
2(+1) + x = 0
x = -2
The charge of S here is -2.
*SO₂. O is assigned with (-2). Using the same procedure,
x + 2(-2) = 0
x = +2
The charge of S here is +2.
*H₂SO₃.
2(+1) + x + 3(-2) = 0
x = +4.
The charge of S here is +4.
*K₂SO₄. K is assigned with +1.
2(+1) + x + 4(-2) = 0
x = +6
The charge of S here is +6.
<em>The S with the highest oxidation number is the one in K₂SO₄.</em>
*S8. Since this is in elemental form, its oxidation number is assigned as 0.
*H2S. H is assigned with +1. Since the compound is neutral, the overall charge is 0. So,
2(+1) + x = 0
x = -2
The charge of S here is -2.
*SO₂. O is assigned with (-2). Using the same procedure,
x + 2(-2) = 0
x = +2
The charge of S here is +2.
*H₂SO₃.
2(+1) + x + 3(-2) = 0
x = +4.
The charge of S here is +4.
*K₂SO₄. K is assigned with +1.
2(+1) + x + 4(-2) = 0
x = +6
The charge of S here is +6.
<em>The S with the highest oxidation number is the one in K₂SO₄.</em>
