What is the oxidation number of Carbon in $N a_{2} C_{2} O_{4}$ $Na_2C_2O_4$ ?

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What is the oxidation number of Carbon in $N a_{2} C_{2} O_{4}$ $Na_2C_2O_4$ ?

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Answer 1 · 2016-06-23T19:27:16

Each carbon in oxalate ion has an oxidation number of $+ I I I$ $+III$ .

Explanation:

The sum of the oxidation numbers of the constituent atoms equals the charge on the ion. Oxygen normally has an oxidation number of $- I I$ $-II$ in its compounds, and it certainly does here.

Thus $2 \times C_{Oxidation number} + 4 \times (- 2) = - 2$ $2xxC_"Oxidation number"+4xx(-2)=-2$ (Of course, $- 2$ $-2$ was the charge on the ion.) It does not take too much algebra to determine that $C_{Oxidation number} = + I I I$ $C_"Oxidation number"=+III$ . Carbon here is almost fully oxidized.

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What is the oxidation number of Carbon in $N a_{2} C_{2} O_{4}$ $Na_2C_2O_4$ ?

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What is the oxidation number of Carbon in Na_2C_2O_4Na2C2O4Na_2C_2O_4?

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What is the oxidation number of Carbon in $N a_{2} C_{2} O_{4}$ $Na_2C_2O_4$ ?