In which of the following compounds is the oxidation state of $C$ highest?

H C O O H

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H C H O

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C H_{3} O H

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C H_{4}

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Solution

The correct option is A $H C O O H$
Let $x$ be the oxidation state of carbon.
Since, the overall charge on the complex is $0$ , the sum of oxidation states of all elements in it should be equal to $0$ .
a. $C H_{4} : x + 4 = 0 \to x = - 4$
b. $C H_{3} O H : x + 4 - 2 = 0 \Rightarrow x = - 2$
c. $H C O O H : x + 2 - 4 = 0 \Rightarrow x = 2$
d. $C_{6} H_{12} O_{6} : 6 x + 12 - 12 = 0 \Rightarrow x = 0$
Hence, the oxidation state of carbon is highest in $H C O O H$ .

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