# What is the oxidation number of Carbon in Na_2C_2O_4?

Each carbon in oxalate ion has an oxidation number of $+ I I I$.
The sum of the oxidation numbers of the constituent atoms equals the charge on the ion. Oxygen normally has an oxidation number of $- I I$ in its compounds, and it certainly does here.
Thus $2 \times {C}_{\text{Oxidation number}} + 4 \times \left(- 2\right) = - 2$ (Of course, $- 2$ was the charge on the ion.) It does not take too much algebra to determine that ${C}_{\text{Oxidation number}} = + I I I$. Carbon here is almost fully oxidized.