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

Apr 15, 2015

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

$\stackrel{\textcolor{b l u e}{+ 1}}{L {i}_{2}}$ $\stackrel{\textcolor{b l u e}{\text{x}}}{S}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}}$

$2 \cdot \textcolor{b l u e}{\left(+ 1\right)} + x + 4 \cdot \textcolor{b l u e}{\left(- 2\right)} = 0$

$2 + x - 8 = 0 \implies x = 8 - 2 = + 6$

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

$L {i}_{2} S {O}_{4} \to 2 L {i}^{+} + S {O}_{4}^{2 -}$

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

$\stackrel{\textcolor{b l u e}{x}}{S}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}^{- 2}}$

$x + 4 \cdot \left(\textcolor{b l u e}{- 2}\right) = - 2 \implies x = - 2 + 8 = + 6$