The increasing order of reduction of alkyl halides with zinc and dilute HCl is

1. $\mathrm{R}-\mathrm{Cl}<\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Br}$

2. $\mathrm{R}-\mathrm{Cl}<\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{I}$

3. $\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{Cl}$

4. $\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Cl}$

HINT: Weaker bond of halogen with carbon gets easily reduced.

EXPLANATION (2).

The reactivity of halogens with alkane is ${\mathrm{F}}_{2}>{\mathrm{Cl}}_{2}>{\mathrm{Br}}_{2}>{\mathrm{I}}_{2}$. Hence, reduction of alkyl halide with Zn and dilute HCl follows reverse order i.e., $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}$. Further, the reactivity of this reduction increases as the strength of C-X bond decreases.