The decreasing order of rate of ${\mathrm{S}}_{\mathrm{N}}2$ reaction for given compounds is:

$\left(\mathrm{I}\right) {\mathrm{CH}}_3-\mathrm{Cl}$

$\left(\mathrm{IV}\right) {\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Cl}$

Following are the curves for nucleophilic substitution reaction.

The correct statement is : :

${\mathrm{S}}_{\mathrm{N}}1$ reaction undergoes through a carbocation intermediate as follows:

$\mathrm{R}-\mathrm{X}\left(\text{ aq. }\right)\stackrel{\text{ Slow }}{\rightleftharpoons }{\mathrm{R}}^{*}\left(\text{ aq. }\right)+{\mathrm{X}}^{-}\left(\text{aq. }\right)\underset{\mathrm{fast}}{\overset{{\mathrm{H}}_2\mathrm{O}}{\to }}\mathrm{ROH}(\mathrm{aq}.)+{\mathrm{H}}^{+}(\mathrm{aq}.)$

$[\mathrm{R}=\mathrm{t}-\mathrm{Bu},\text{ iso-Pr, }\mathrm{Et},\mathrm{Me}](\mathrm{X}=\mathrm{Cl},\mathrm{Br},\mathrm{I})$

The correct statements are,

(I) The decreasing order of rate of ${\mathrm{S}}_{\mathrm{N}}1$ reaction is $\mathrm{t}-\mathrm{BuX}>\mathrm{iso}-\mathrm{PrX}>\mathrm{EtX}>\mathrm{MeX}$.

(II) The decreasing order of ionization energy is $\mathrm{MeX}>\mathrm{EtX}>\mathrm{iso}-\mathrm{PrX}>\mathrm{t}-\mathrm{BuX}$.

(III) The decreasing order of energy of activation is $\mathrm{t}-\mathrm{BuX}>\text{ iso-PrX }>\mathrm{EtX}>\mathrm{MeX}$.