Question

# Consider the compounds$\left( I \right){\text{ }}R - I{\text{ }}\left( {II} \right){\text{ }}R - Br{\text{ }}\left( {III} \right){\text{ }}R - Cl{\text{ }}\left( {IV} \right){\text{ }}R - F$. The rate of $S{N_2}$reaction is $A. I > II > III > IV \\ B. II > I > III > IV \\ C. I = II = III = IV \\ D.{\text{}}IV > III > II > I \\$

Hint: First we must know that the $S{N_2}$ stands for Nucleophilic Substitution, Second Order reaction. Types of Alkyl halide, solvent, nucleophilicity, leaving group, etc are some factors that affect the rate of $S{N_2}$ reaction.

There are various factors that may affect the $S{N_2}$ reaction. Leaving the group is one the important factor that impacts the $S{N_2}$ reaction.
The leaving group is almost always excluded with a full negative charge, so those leaving groups are best that can best stabilize an anion. $^ - OTos{,^ - }OMs{,^ - }Br{,^ - }I{,^ - }Cl{,^ - }NH2{,^ - }OH{,^ - }F,{\text{ }}etc.$are some examples of leaving group.
In the $S{N_2}$ mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.
The reactivity of the halogen group decreases with an increase in the $C - X$ bond energy. Bond energy of $C - F$ is maximum and so fluorides are least reactive. On the other hand, the bond energy of $C - I$ is minimum and hence, iodides are most reactive.
Thus, Alkyl halides follow the reactivity sequence, $R - F{\text{ }} < {\text{ }}R - Cl{\text{ }} < {\text{ }}R - Br{\text{ }} < {\text{ }}R - I$
$S{N_2}$ reactions are particularly sensitive to crowding or steric factors. $S{N_2}$ reactions get greatly affected by steric hindrance at the site of reaction. So, the order of reactivity of alkyl halides in $S{N_2}$ reactions is as follows:
The tertiary alkyl halides have high steric factors or are so crowded that they do not generally react by an $S{N_2}$ mechanism.