Find the oxidation number of elements in each case :

$C$ in $C H_{4}, C_{2} H_{6}, C_{3} H_{8}, C_{2} H_{4}, C_{2} H_{2}, H_{2} C_{2} O_{4}$ and $C O_{2}$ .

- 4, - 3, - \frac{8}{3}, - 2, - 1, + 3, + 4

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- 3, - 3, - \frac{8}{3}, - 2, - 2, + 3, + 4

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- 2, - 3, - 8, - 2, - 1, + 3, + 4

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None of the above

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Solution

The correct option is A $- 4, - 3, - \frac{8}{3}, - 2, - 1, + 3, + 4$
For compounds containing $C$ atoms, the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}}$
Here, $n_{O}, n_{H}$ and $n_{C}$ represents the number of oxygen, hydrogen and carbon atoms respectively.
For $C H_{4}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{0 - 4}{1} = - 4$
For $C_{2} H_{6}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{0 - 6}{2} = - 3$
For $C_{3} H_{8}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{0 - 8}{3} = - 8 / 3$
For $C_{2} H_{4}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{0 - 4}{2} = - 2$
For $C_{2} H_{2}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{0 - 2}{2} = - 1$
For $H_{2} C_{2} O_{4}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{2 \times 4 - 2}{2} = + 3$
For $C O_{2}$ , the oxidation number of $C$ atom $= \frac{2 n_{O} - n_{H}}{n_{C}} = \frac{2 \times 2 - 0}{1} = + 4$
Thus, the oxidation number of $C$ in $C H_{4}, C_{2} H_{6}, C_{3} H_{8}, C_{2} H_{4}, C_{2} H_{2}, H_{2} C_{2} O_{4}$ and $C O_{2}$ is $- 4, - 3, - \frac{8}{3}, - 2, - 1, + 3$ and $+ 4$ respectively.

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Find the oxidation number of elements in each case :C in CH4,C2H6,C3H8,C2H4,C2H2,H2C2O4 and CO2.

Find the oxidation number of elements in each case :

$C$ in $C H_{4}, C_{2} H_{6}, C_{3} H_{8}, C_{2} H_{4}, C_{2} H_{2}, H_{2} C_{2} O_{4}$ and $C O_{2}$ .