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

May 29, 2017

We has ${S}^{V I +}$

#### Explanation:

Sulfur in sulfate expresses it maximum oxidation state, its Group Number, $+ V I$.

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

And so ${S}_{\text{oxidation number"+4xxO_"oxidation number}} = - 2$

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

${S}_{\text{oxidation number}} + 4 \times \left(- 2\right) = - 2$

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

${S}_{\text{oxidation number}} + \cancel{8} + \cancel{4 \times \left(- 2\right)} = - 2 + 8$

${S}_{\text{oxidation number}} = + 6 = + V I$..........

What about $\text{thiosulfate}$, ${S}_{2} {O}_{3}^{2 -}$? What are the oxidation numbers of sulfur here?

May 29, 2017

$+ 6$

#### Explanation:

Charge of $L {i}_{2} S {O}_{4}$ is balanced

The charge of $L i$ is $+ 1$ (because it is a ${1}^{s t}$ group element)

There are $2$ $L i$ atoms therefore the charge of $L {i}_{2}$ is $+ 2$ and $S {O}_{4}$ 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 "${\left({O}_{2}\right)}^{2 -}$" it would have had a charge of $- 1$). Because there is $4$ oxygen atoms, the total charge of oxygen will be $4 \cdot \left(- 2\right) = \textcolor{red}{- 8}$

Sulfur must have a charge that will make( $\textcolor{red}{- 8} +$ its charge)$= - 2$ ("charge of $S {O}_{4}$")

Therefore the charge of sulfur $= - 2 + 8 = \textcolor{b l u e}{+ 6}$