Question

What is the oxidation number of the sulfur atom in Li2SO4 ? A) +4 B) +6 C) +2 D) -2

Answer 1

The answer is B) +6.

Since you're dealing with a neutral compound, the sum of the oxidation numbers of all the atoms that form said compound must be zero.

Since lithium is a group I alkali metal, its oxidation number will be +1. Oxygen, on the other hand, will have an oxidation number equal to -2. This means that you get

`stackrel(color(blue)(+1))(Li_2)` `stackrel(color(blue)("x"))(S)` `stackrel(color(blue)(-2))(O_4)`

`2 * color(blue)((+1)) + x + 4 * color(blue)((-2)) = 0`

`2 + x - 8 = 0 => x = 8 -2 = +6`

Alternatively, you can reason that, since lithium sulfate is an ionic compound, you can isolate the ions to get

`Li_2SO_4 -> 2Li^(+) + SO_4^(2-)`

Since sulfur is a part of the sulfate anion, which has a -2 net charge, you'll get

`stackrel(color(blue)(x))(S)` `stackrel(color(blue)(-2))(O_4^(-2))`

`x + 4 * (color(blue)(-2)) = -2 => x = -2 + 8 = +6`