The decreasing leaving group order of the following is:
i. `F^(o-)`, ii. `Cl^(o-)`, iii. `Br^(o-)`, Iv. `I^(o-)`

Options

A. `(i) gt (ii) gt (iii) gt (iv)`
B. `(iv) gt (iii) gt (ii) gt (i)`
C. `(ii) gt (i) gt (iii) gt (iv)`
D. `(ii) gt (i) gt (iv) gt (iii)`

Explanation

To determine the decreasing order of leaving group strength for the halides \( F^-, Cl^-, Br^-, \) and \( I^- \), we need to analyze the bond strength between carbon and each halogen. The strength of the leaving group is inversely related to the strength of the bond it forms with carbon; weaker bonds correspond to better leaving groups.

### Step-by-Step Solution:

1. **Identify the Halogens and Their Bonds with Carbon:**
- The halogens in question are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). We need to consider the carbon-halogen bonds: \( C-F, C-Cl, C-Br, \) and \( C-I \).

2. **Analyze the Bond Strength:**
- The bond strength decreases as we move down the group in the periodic table. This is due to the increasing atomic size and the decreasing electronegativity of the halogens.
- The bond strengths are approximately as follows:
- \( C-F \) bond is the strongest.
- \( C-Cl \) bond is weaker than \( C-F \).
- \( C-Br \) bond is weaker than \( C-Cl \).
- \( C-I \) bond is the weakest.

3. **Determine the Leaving Group Strength:**
- Since a good leaving group is one that can easily leave, the weaker the bond with carbon, the better the leaving group.
- Therefore, the order of leaving group strength from strongest to weakest is:
- Iodide \( (I^-) \) is the best leaving group because it forms the weakest bond with carbon.
- Bromide \( (Br^-) \) is next.
- Chloride \( (Cl^-) \) follows.
- Fluoride \( (F^-) \) is the weakest leaving group due to the strong bond with carbon.

4. **Write the Final Order:**
- Thus, the decreasing order of leaving group strength is:
\[ I^- > Br^- > Cl^- > F^- \]

### Final Answer:
The decreasing order of leaving group strength is \( I^- > Br^- > Cl^- > F^- \).

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