The way \(Li_2\) displaces a \(Fe\) in the lattice of \(FeS_2\) puts the validity of oxidation numbers into question. The formula \(FeS_2\) is arrived at after the considering the number of neighbors the basic unit member of the lattice has. In this case, the basic unit member containing \(Fe\) has six \(S\) around it, two of these are connected to another unit member. Six \(S\) are shared by three other unit members. This make on average, two \(S\) for each \(Fe\), and thus the formula \(FeS_2\). When one \(Fe\) is replaced, half of the two \(Fe\) out of six in total is removed. From the diagram repeated below, \(Fe\)'s oxidation state is reduced by one, from \(+6\) to \(+5\).
But the formula,
\(Li_2FeS_2 \)
indicates that the oxidation state of \(Fe\) has reduced by two. In the diagram, the oxidation state of \(Fe\) is obtained by counting the number of ionic bonds it has.
And chemistry is more fun than stamp collecting.