What is the oxidation number of carbon in the salt sodium oxalate, Na_2^(+)C_2O_4^(2-)?

Aug 1, 2017

$3$. See below.

Explanation:

We assign oxidation states to atoms in a molecule by basically using one premise: the more electronegative atom gets all the bonding electrons.

Here we are considering the oxalate ion:

We know that oxygen is more electronegative than carbon and it is typically assigned an oxidation number of $\text{-2}$.

So
$\stackrel{\textcolor{red}{2 x}}{\text{C}}$ ${\stackrel{\textcolor{b l u e}{4 \left(- 2\right)}}{\text{O}}}_{2}$

As the whole complex ion has a charge of $- 2$

$\therefore$ color(red)(2x)+color(blue)(4(-2)=color(green)(-2)

$\textcolor{w h i t e}{344}$$\textcolor{red}{2 x} - \textcolor{b l u e}{8} = \textcolor{g r e e n}{- 2}$

$\implies$ $\textcolor{red}{2 x} = \textcolor{g r e e n}{6}$

i.e. $\textcolor{w h i t e}{34}$ $x = 3$

Aug 1, 2017

We have $C \left(+ I I I\right)$........

Explanation:

By definition, the oxidation state is the charge left on the central atom, when all the bonding atoms are removed with the charge (the electrons) devolved to the most electronegative atom.....

We gots ""^(-)O(O=)C-C(=O)O^(-). The oxidation number of oxygen is usually $- I I$ and it is here. The sum of the oxidation numbers equals the charge on the ion.....here $- 2$.

And so $2 \times {C}_{\text{oxidation number}} + 4 \times \left(- 2\right) = - 2$

And $2 \times {C}_{\text{oxidation number}} = + 6$

And ${C}_{\text{oxidation number}} = + I I I$ if you will forgive me mixing Roman and Arabic numerals.........of course you will.

Another way we could look at this is to split the $C - C$ bond in oxalate ion to give 2xx""^(-)O(O=)C*, $C \left(+ I I I\right)$ as required. And thus the carbon in oxalic acid is almost fully oxidized.

Aug 1, 2017

Oxygen is one of the most electronegative elements after flourine so it shows 2 as oxidation state for most compounds [exception - hydrogen peroxide $H 2 O 2$]
SO IN $C 2 O {4}^{2 -}$ Oxidation state of oxygen must be -2 . net charge on compound is (-2) .
OXIDATION STATE OF CARBON IS $\frac{6}{2}$ = 3.