The sum of the oxidation numbers of all the carbons in $C_{6} H_{5} C H O$ is :

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-4

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Solution

The correct option is D -4
oxidation state of $C$ in $C_{6} H_{5} C H O$ is
$6 x + 5 + x + 1 + (- 2) = 0$
$7 x + 6 - 2 = 0$
$7 x + 4 = 0$
$7 x = - 4$
as per the given question sum of oxidation states of all C atoms is sum therefore answer becomes $- 4$

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