Carbon is in the lowest oxidation state in

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(A) To determine the oxidation state of carbon in each compound:$1$. In $CH_4$,let the oxidation state of $C$ be $x$. Then $x + 4(+1) = 0$,so $x = -4$

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$CH_4$

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(A) To determine the oxidation state of carbon in each compound:$1$. In $CH_4$,let the oxidation state of $C$ be $x$. Then $x + 4(+1) = 0$,so $x = -4$.$2$. In $CCl_4$,$x + 4(-1) = 0$,so $x = +4$.$3$. In $CF_4$,$x + 4(-1) = 0$,so $x = +4$.$4$. In $CO_2$,$x + 2(-2) = 0$,so $x = +4$.Comparing these values,the lowest oxidation state is $-4$,which corresponds to $CH_4$.

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