The `S_(N)2` reactivity order for halides is :

Options

A. `R-F gt R-Cl gt R-Br gt R-I`
B. `R-I gt R -Br gt R-Cl gt R-F`
C. `R-Br gt R-I gt R-Cl gt R-F`
D. `R-C gt R- Br gt R- F gt R-I`

Explanation

To determine the `S_(N)2` reactivity order for halides, we need to analyze the stability of the leaving groups formed when halides participate in the `S_(N)2` reaction. The order of reactivity is influenced by the ability of the halide to act as a leaving group, which is related to the stability of the corresponding anion formed after the leaving group departs.

### Step-by-Step Solution:

1. **Identify the Halides**: The halides we are considering are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I).

2. **Understand Leaving Group Ability**: The ability of a halide to act as a leaving group in an `S_(N)2` reaction is crucial. A better leaving group will increase the rate of the reaction.

3. **Electronegativity and Stability of Anions**: The stability of the anion formed after the halide leaves is important. The more electronegative an atom is, the less stable the negative charge is on it. This means:
- Fluorine (F) is the most electronegative and thus forms a less stable anion (F⁻).
- Iodine (I) is the least electronegative and forms a more stable anion (I⁻).

4. **Order of Leaving Group Ability**: Based on electronegativity and the stability of the corresponding anions, the leaving group ability increases down the group:
- I⁻ > Br⁻ > Cl⁻ > F⁻
This means that I⁻ is the best leaving group and F⁻ is the worst.

5. **Reactivity Order for `S_(N)2` Reactions**: Since better leaving groups correlate with higher reactivity in `S_(N)2` reactions, we can establish the reactivity order for halides:
- R-I > R-Br > R-Cl > R-F

6. **Final Answer**: Therefore, the `S_(N)2` reactivity order for halides is:
- R-I > R-Br > R-Cl > R-F