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(A) $S_{N}2$ reactivity depends on two main factors: steric hindrance at the $\alpha$-carbon and the strength of the $C-X$ bond.$1$. Steric hindrance:

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$III > I > IV > II$

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(A) $S_{N}2$ reactivity depends on two main factors: steric hindrance at the $\alpha$-carbon and the strength of the $C-X$ bond.$1$. Steric hindrance: Primary alkyl halides $(1^{\circ})$ are more reactive than secondary alkyl halides $(2^{\circ})$ due to less steric crowding. Thus,$1$-substituted compounds $(I, III)$ are more reactive than $2$-substituted compounds $(II, IV)$.$2$. Leaving group ability: The $C-I$ bond is weaker and longer than the $C-Cl$ bond,making $I^-$ a better leaving group than $Cl^-$. Therefore,iodides are more reactive than chlorides.Combining these factors,the order of reactivity is:$1$-Iodopropane $(III)$ > $1$-Chloropropane $(I)$ > $2$-Iodopropane $(IV)$ > $2$-Chloropropane $(II)$Therefore,the correct order is $III > I > IV > II$.

Arrange the following in the correct order of reactivity towards $S_{N}2$ reaction:
$I$: $1$-Chloropropane
$II$: $2$-Chloropropane
$III$: $1$-Iodopropane
$IV$: $2$-Iodopropane

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Arrange the following in the correct order of reactivity towards $S_{N}2$ reaction:$I$: $1$-Chloropropane$II$: $2$-Chloropropane$III$: $1$-Iodopropane$IV$: $2$-Iodopropane