Question

What is the oxidation number of the sulfur atom in `Li_2SO_4`?

Answer 1

We has `S^(VI+)`

Explanation:

Sulfur in sulfate expresses it maximum oxidation state, its Group Number, `+VI`.

How? Well, as you know the sum of the oxidation numbers MUST equal the charge on the ion..........

And so `S_"oxidation number"+4xxO_"oxidation number"=-2`

Now oxygen generally assumes an oxidation number of `-II` in its compounds, and certainly it does so here.........

`S_"oxidation number"+4xx(-2)=-2`

Add `8` to both sides of the equation..........

`S_"oxidation number"+cancel8+ cancel(4xx(-2))=-2+8`

`S_"oxidation number"=+6=+VI`..........

What about `"thiosulfate"`, `S_2O_3^(2-)`? What are the oxidation numbers of sulfur here?

Answer 2

`+6`

Explanation:

Charge of `Li_2SO_4` is balanced

The charge of `Li` is `+1` (because it is a `1^(st)` group element)

There are `2` `Li` atoms therefore the charge of `Li_2` is `+2` and `SO_4` must have a total charge of `-2` (to balance the charge)

Oxygen "in this case (because it is not a peroxide)" has a charge of `-2` (if it was a peroxide "`(O_2)^(2-)`" it would have had a charge of `-1`). Because there is `4` oxygen atoms, the total charge of oxygen will be `4*(-2)=color(red)(-8)`

Sulfur must have a charge that will make( `color(red)(-8)+` its charge)`=-2` ("charge of `SO_4`")

Therefore the charge of sulfur `=-2+8=color(blue)(+6)`