What is the change in oxidation number of carbon in the following reaction?
$C H_{4} (g) + 4 C l_{2} (g) \to C C l_{4} (l) + 4 H C l (g)$

0 to -4

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-4 to +4

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0 to +4

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+4 to +4

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Solution

The correct option is B -4 to +4
(I) $C H_{4}$
$\Rightarrow x + 4 = 0$
$\Rightarrow x = - 4$
Oxidation number of C in $C H_{4}$ is −4.

(II) $C C l_{4}$
$\Rightarrow x - 4 = 0$
$\Rightarrow x = + 4$
Oxidation number of C in $C C l_{4}$ is +4.
Therefore, change in oxidation state of carbon is from −4 to +4.

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