Question

\[C{{l}_{2}}\] reacts with \[C{{S}_{2}}\] in presence of \[{{I}_{2}}\] catalyst to form
(A) \[CHC{{l}_{3}}\]
(B) \[CC{{l}_{4}}\]
(C) \[{{C}_{2}}{{H}_{5}}Cl\]
(D) \[{{C}_{2}}{{H}_{6}}\]

Answer 1

Hint: Interaction between polar and nonpolar molecules is due to London dispersion force or induced dipole. In this type of interaction, at any particular time, both molecules acquire a temporary dipole.

Complete Step by Step Answer:
A bond between two chlorine atoms to form chlorine is a covalent bond as both chlorine atoms are non-metal. Now covalent bonds are of two types, one is polar, and the other is non-polar. Chlorine is non-polar as there is no electronegativity difference between two chlorine atoms.

Also, carbon disulfide is non-polar as both are non-metal however, there is a small electronegativity difference between carbon and sulphur but sulphur bonded with carbon with two double bonds and thus, both sides dipole cancel out and the molecule becomes non-polar.

When \[C{{l}_{2}}\] reacts with \[C{{S}_{2}}\]in the presence of \[{{I}_{2}}\]catalyst then the negative temporary positive pole of chlorine molecule will interact with the temporary negative pole of carbon disulfide and the negative temporary pole of chlorine molecule will interact with the positive temporary pole of carbon disulfide in the presence of iodine catalyst to give carbon tetrachloride such as

Thus, the correct option is B.

Note: In this reaction, iodine catalyst has been used as iodine is majorly used in the process of conversion of unstable reaction or compound to a stable reaction or stable compound. Carbon tetrachloride is more stable than carbon disulfide. Carbon tetrachloride can also be prepared when chlorine reacts with methane.