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

2 Answers
May 29, 2017

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?

May 29, 2017

+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)