Amongst `OH^(-), CH_(3)COO^(-)` and `CH_(3)O^(-)` ions, nucleophilicity decreases in the order.................. .

Answer

To determine the order of nucleophilicity among the ions \( OH^- \), \( CH_3COO^- \), and \( CH_3O^- \), we need to analyze the stability of the negative charge on each ion. Nucleophilicity is inversely proportional to the stability of the negative charge: the less stable the negative charge, the more nucleophilic the species will be.

### Step-by-Step Solution:

1. **Identify the Ions**:
- The three ions we are comparing are \( OH^- \) (hydroxide ion), \( CH_3COO^- \) (acetate ion), and \( CH_3O^- \) (methoxide ion).

2. **Analyze the Stability of the Negative Charge**:
- **\( OH^- \)**: The negative charge is localized on the oxygen atom, which is relatively electronegative but not stabilized by resonance.
- **\( CH_3COO^- \)**: The negative charge is delocalized over two oxygen atoms due to resonance, making it more stable than \( OH^- \).
- **\( CH_3O^- \)**: Similar to \( OH^- \), the negative charge is localized on the oxygen atom, but it is less electronegative than \( OH^- \) due to the presence of the methyl group, which can donate electron density.

3. **Rank the Stability of the Negative Charges**:
- The stability of the negative charges can be ranked as follows:
- \( CH_3COO^- \) (most stable due to resonance)
- \( OH^- \) (less stable, localized negative charge)
- \( CH_3O^- \) (least stable, localized negative charge but less electronegative than \( OH^- \))

4. **Determine the Order of Nucleophilicity**:
- Since nucleophilicity is inversely proportional to stability:
- \( CH_3O^- \) > \( OH^- \) > \( CH_3COO^- \)

5. **Final Answer**:
- Therefore, the order of nucleophilicity decreases as follows:
\[ CH_3O^- > OH^- > CH_3COO^- \]