Oxidation number of carbon in " $C_{3} O_{2}$ " and " $M g_{2} C_{3}$ " are respectively :

- \frac{4}{3}, + \frac{4}{3}

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+ \frac{4}{3}, - \frac{4}{3}

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- \frac{2}{3}, + \frac{2}{3}

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+ \frac{2}{3}, - \frac{2}{3}

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Solution

The correct option is B $+ \frac{4}{3}, - \frac{4}{3}$
Let oxidation state of $C$ in $C_{3} O_{2}$ be $x$ .
Since the compound is neutral, sum of oxidation states of all elements = 0
$3 x + 2 (- 2) = 0$
or, $x = \frac{4}{3}$
Oxidation state of $C$ in $M g_{2} C_{3}$ is $y$ .
Since the compound is neutral, sum of oxidation states of all elements = 0
$2 \times 2 + 3 \times y = 0$
or, $y = - \frac{4}{3}$

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