`S_(N)1` reaction underoges through carbocation intermediate as follows:
`R-X(aq.) overset("Slow") hArr R^(+)(aq.)+`
`X^(-) (aq.) underset ("fast") overset(H_(2)O)to ROH(aq.)+H^(+)(aq.)`
[R=t-Bu, iso-Pr, Me] `(X=Cl, Br, I)`
The correct statements are
I. The decreasing order of rate of `S_(N)1` reaction is `t-BuXgtiso-PrXgtEtXgtMeX`
II. The decreasing order of ionisation energy is `MeXgtEtXgtiso-PrXgt t-BuX`
III. The decreasing order of energy of activation is `t-BuX gt iso- PrXgtEtXgtMeX`

Options

A. I & II are correct
B. I & III are correct
C. II and III are correct
D. I,II & III are correct

Explanation

To solve the problem regarding the SN1 reaction and the stability of carbocations, we will analyze each statement provided in the question step by step.

### Step 1: Analyze the Rate of SN1 Reaction
The rate of an SN1 reaction is primarily determined by the stability of the carbocation formed during the reaction. The order of stability for carbocations is as follows:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (0°)

Given the carbocations:
- **t-Butyl (tert-butyl)**: 3° carbocation
- **Isopropyl**: 2° carbocation
- **Ethyl**: 1° carbocation
- **Methyl**: 0° carbocation

Thus, the order of rate of SN1 reaction based on stability is:
- **t-BuX > iso-PrX > EtX > MeX**

### Conclusion for Statement I:
The statement "The decreasing order of rate of SN1 reaction is t-BuX > iso-PrX > EtX > MeX" is **correct**.

### Step 2: Analyze Ionization Energy
Ionization energy is inversely related to the stability of the carbocation. The more stable the carbocation, the lower the ionization energy required to form it. Therefore, the order of ionization energy will be the reverse of the stability order:
- **Methyl > Ethyl > Isopropyl > t-Butyl**

### Conclusion for Statement II:
The statement "The decreasing order of ionization energy is MeX > EtX > iso-PrX > t-BuX" is **correct**.

### Step 3: Analyze Energy of Activation
The activation energy for the formation of a carbocation is also inversely related to its stability. A more stable carbocation will have a lower activation energy. Therefore, the order of activation energy will also be the reverse of the stability order:
- **Methyl > Ethyl > Isopropyl > t-Butyl**

### Conclusion for Statement III:
The statement "The decreasing order of energy of activation is t-BuX > iso-PrX > EtX > MeX" is **incorrect**. The correct order should be:
- **MeX > EtX > iso-PrX > t-BuX**

### Final Conclusion:
- Statement I: Correct
- Statement II: Correct
- Statement III: Incorrect

Thus, the correct statements are I and II.