\( \mathrm{S}_{\mathrm{N}} 1 \) reaction undergoes through a carbocation intermediate as follows...
0\( \mathrm{S}_{\mathrm{N}} 1 \) reaction undergoes through a carbocation intermediate as follows:
\[
\begin{array}{l}
\left.\left.\mathrm{R}-\mathrm{X}(\text { aq. }) \stackrel{\text { Slow }}{\rightleftharpoons} \mathrm{R}^{+} \text {(aq. }\right)+\mathrm{X}^{-} \text {(aq. }\right) \\
\stackrel{\text { fast }}{\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow}} \mathrm{ROH}(\text { aq. })+\mathrm{H}^{+} \text {(aq.) } \\
\end{array}
\]
\( \mathrm{R}=\mathrm{t}-\mathrm{Bu} \), iso \( -\operatorname{Pr}, \mathrm{Et}, \mathrm{Me}](\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I}) \)
The correct statement(s) is/are:
I. The decreasing order of rate of \( \mathrm{S}_{\mathrm{N}} 1 \) reaction is \( \mathrm{t}-\mathrm{BuX} \) iso- \( \operatorname{Pr} \mathrm{X}\mathrm{MeX} \).
II. The decreasing order of ionisation energy is \( \mathrm{MeX}\mathrm{EtX} \) \( \mathrm{t}-\mathrm{BuX} \).
III. The decreasing order of energy of activation is \( \mathrm{t}-\mathrm{BuX} \) \( \mathrm{EtX}\mathrm{MeX} \).
(a) I \& II are correct
(b) I \& III are correct
(c) II \& III are correct
(d) I, II \& III are correct
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