Ionic species are stailised by the dispersal of charge. Which of the following carboxylate ion is the most stable ?

Options

A. `CH_(3) - overset(O)overset(||)(C) - O^(-)`
B. `Cl - CH_(2) - overset(O)overset(||)(C) - O^(-)`
C. `F - CH_(2) - overset(O)overset(||)(C) - O^(-)`
D.

Explanation

To determine which carboxylate ion is the most stable, we need to analyze the resonance stabilization and the effect of substituents on the stability of the carboxylate ions. Here’s a step-by-step solution:

### Step 1: Understand the Structure of Carboxylate Ions
Carboxylate ions have the general structure RCOO⁻, where R is a hydrocarbon group. The negative charge resides on one of the oxygen atoms, and this charge can be delocalized through resonance.

**Hint:** Remember that resonance can stabilize a negative charge by spreading it over multiple atoms.

### Step 2: Identify the Electron-Withdrawing Groups
Electron-withdrawing groups (EWGs) stabilize negative charges by pulling electron density away from the negatively charged atom. The more effective the EWG, the more stable the carboxylate ion will be.

**Hint:** Look for groups that are known to withdraw electrons, such as halogens or nitro groups.

### Step 3: Analyze Each Option
1. **Option A:** Contains a methyl group (−CH₃), which is an electron-donating group. This will destabilize the carboxylate ion.

2. **Option B:** Contains a chlorine atom (−Cl), which is an electron-withdrawing group (−I effect). This will provide some stabilization to the carboxylate ion.

3. **Option C:** Contains a fluorine atom (−F), which is more electronegative than chlorine and thus has a stronger −I effect. This will provide greater stabilization than chlorine.

4. **Option D:** Contains two fluorine atoms (−F). The presence of two electronegative fluorine atoms will enhance the −I effect even more, leading to the highest stabilization of the carboxylate ion.

**Hint:** Compare the strength of the electron-withdrawing effects of the groups present in each option.

### Step 4: Compare the Stabilization Effects
- **Methyl group (A)** destabilizes the ion.
- **Chlorine (B)** provides moderate stabilization.
- **Fluorine (C)** provides greater stabilization than chlorine.
- **Two fluorines (D)** provide the greatest stabilization due to the cumulative −I effect.

**Hint:** The more electron-withdrawing groups present, the more stable the carboxylate ion will be.

### Step 5: Conclusion
Based on the analysis, the carboxylate ion in **Option D** is the most stable due to the presence of two fluorine atoms, which exert a strong electron-withdrawing effect.

**Final Answer:** **Option D is the correct answer.**