oxidation number of Carbon in Na2C2O4?

Each Na is +1, and each O is -2, for a total of (+2) + (-8) = -6. Therefore the two carbons between them must have an oxidation state of +6, so each one is in the oxidation state +3.
This is actually fairly easy to see from the oxalate ion’s Lewis structure too. Each carbon is double-bonded to one oxygen and single-bonded to another, so it has formally “lost” 2 + 1 = 3 valence electrons to the more electronegative oxygen.

Na almost always has an oxidation number of +1, while O almost always has an oxidation number of -2.
So, now you want an oxidation number of carbon that will cause the molecule as a whole to have an oxidation number of zero.
Check it out:
+1 * 2 Na atoms + -2 * 4 O atoms + X * 2 C atoms = 0, or
1 * 2 + -2 * 4 + 2X = 0 Solve for x, and you will have the oxidation number.
In the end, you should get X= +3
+3 is your oxidation number.

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