Question

Suppose we carry out the reaction of Iodine on ethylene in presence of UV light such that the obtained product is symmetrical. Which of the following options are accurate?

• A. The oxidation state of Iodine increases.
• B. The oxidation state of Carbon increases.
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is B The oxidation state of Carbon increases.

Ethylene undergoes reduction with iodine to form 1,2-Diiodoethane
$C{H}_2=C{H}_2+I_2\to I-C{H}_2-C{H}_2-I$
(Ethylene) (Iodine) (1,2-Diiodoethane)

Now that we have figured the product, check the validity of each of the given options. Clearly Iodine is more electronegative than Carbon. So the electrons in the $C-I$ bond will be closer to the Iodine atom. Hence, Iodine will be assigned a -1 oxidation state. Overall, the molecule is neutral. In ethylene, each Carbon atom has -2 oxidation state while in 1,2-Diiodoethane, each Carbon has -1 oxidation state.

The oxidation state of Carbon increases.