Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : $\text{C₃H₇–I undergoes}{S}_{\mathrm{N}}2\text{reaction faster than C₃H₇–Cl.}$

Reason (R) : Iodine is a better leaving group because of its large size.
In the light of the above statements, choose the correct answer from the options given below:

Let us analyze the given statements one by one:

Assertion (A): $C_3{H}_7-I$ undergoes $S_{N}2$ reaction faster than $C_3{H}_7-\mathrm{Cl}$.
In nucleophilic substitution reactions (both $S_{N}1$ and $S_{N}2$ ), the rate of the reaction depends heavily on the leaving group ability of the halide ion. Iodine ($I^{-}$) is a much better leaving group than chlorine (${\mathrm{Cl}}^{-}$). Therefore, propyl iodide ($C_3{H}_7-I$) reacts faster than propyl chloride ($C_3{H}_7-\mathrm{Cl}$). Thus, Assertion (A) is true.

Reason (R): Iodine is a better leaving group because of its large size.
As we go down the halogen group (from F to I), the size of the halide ion increases. The iodide ion ($I^{-}$) has a larger ionic radius compared to the chloride ion (${\mathrm{Cl}}^{-}$). Because of its large size, the negative charge is distributed over a larger volume, making it more stable and a weaker base. Additionally, the carbon-iodine ($C-I$) bond is weaker and has a lower bond dissociation enthalpy than the carbon-chlorine ($C-\mathrm{Cl}$) bond, making it easier to break during the reaction. Thus, Reason (R) is true.

Since iodine is a better leaving group due to its larger size, the activation energy for the nucleophilic attack is lower for the iodide, which directly explains why $C_3{H}_7-I$ reacts faster than $C_3{H}_7-\mathrm{Cl}$. Therefore, Reason (R) is the correct explanation of Assertion (A).