What is the oxidation number of carbon in Na2C2O4?

Oct 25, 2015

${C}^{\text{+3}}$

Explanation:

You need to do a little algebra on this one.

You have the substance

$N {a}_{2} {C}_{2} {O}_{4}$ = zero oxidation state

Converting this into numbers based on their known oxidation states, you will have:

Na = 2 (+1) = +2
C = 2 (x) = 2x, where x is the unknown
O = 4 (-2) = -8

Hence,

(+2) + 2x + (-8) = 0

2x + (-6) = 0

2x = +6

x = +3

Therefore, the oxidation state of one Carbon atom in the substance $N {a}_{2} {C}_{2} {O}_{4}$ is +3.

Oct 25, 2015

The oxidation number of each carbon atom in $\text{Na"_2"C"_2"O"_4}$ is +3.

Explanation:

$\text{Na"_2"C"_2"O"_4}$ is the compound sodium oxalate.

The sum of the oxidation numbers of the elements in a compound is zero.

The oxidation number for sodium is pretty much always $+ 1$.

Since there are two sodium atoms, the total oxidation number for sodium is $+ 2$.

The oxidation number for oxygen is $- 2$, except in peroxides. Oxalate is not a peroxide, so the oxidation number here is still $- 2$.

Since there are four oxygen atoms, the total oxidation number for the oxygen atoms is $- 8$.

The sum of the oxidation numbers for sodium and oxygen is $+ 2 - 8 = - 6$. Therefore, the total oxidation number for carbon must be $+ 6$ in order for the sum of the oxidation numbers to equal zero.

Divide $+ 6$ by two to get the oxidation number of each carbon atom, which is $\boldsymbol{+ 3}$.

Oxidation numbers for all elements in the compound sodium oxalate:

$\stackrel{2 \times + 1}{{\text{Na"_2)" "stackrel(2 xx +3)("C"_2)" "stackrel(4 xx -2)("O}}_{4}}$

$2 \times \left(+ 1\right) + 2 \times \left(+ 3\right) + 4 \times \left(- 2\right) = 0$