# Clarification on Oxidation Number and Ionic Charge

I am confused on the differentiation between ionic charge and oxidation number (state). For example and simplicity, $\ce{NaCl}$ is an ionic compound so it forms $\ce{Na+}$ and $\ce{Cl-}$. Are these ($1+$ for $\ce{Na}$ and $1-$ for $\ce{Cl}$) both the oxidation states and the ionic charges? Or is the net of the compound ($\ce{NaCl}$) the ionic charge?

Additionally, when you have something such as $\ce{C2O4^2-}$ in this case a covalent compound , carbon's is $3+$. Is this carbon's oxidation state? What then is the net charge of the compound ($2-$) called?

For simple mononuclear ions in ionic compounds, as you have in $\ce{NaCl, Al2O3, ZnS}$ or $\ce{GaN}$, the oxidation state of each element will always equal its ionic charge: $\ce{Na+}$ has $\mathrm{+I}$, $\ce{Al^3+}$ $\mathrm{+III}$, $\ce{Zn^2+}$ $\mathrm{+II}$, $\ce{Ga^3+}$ $\mathrm{+III}$, $\ce{Cl-}$ $\mathrm{-I}$, $\ce{O^2-}$ $\mathrm{-II}$, $\ce{S^2-}$ $\mathrm{-II}$ and $\ce{N^3-}$ $\mathrm{-III}$.
Something like the oxalate anion $\ce{C2O4^2-}$ is a multinuclear anion. Its ionic charge is $2-$, as evident by the superscript. However, you cannot always determine the constituent elements’ oxidation states a priori — the only thing you do know is that the sum of the oxidation states must equal the charge number. In oxalate, carbon has $\mathrm{+III}$ as you correctly mentioned and oxygen has $\mathrm{-II}$, as it should be of no surprise to you. Added up, this leaves us with:
$$2 \times (+3) + 4 \times (-2) = +6 + -8 = -2$$
And $2-$ is the ionic charge as we already know. This rule goes both ways, so you can use the (known) ionic charge of a multinuclear ion to determine an element’s oxidation state if the other oxidation states are known: in $\ce{SO3^2-}$ oxygen has $\mathrm{-II}$, so sulphur can only have $\mathrm{+IV}$.