0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\mathrm{+IV}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.