Which of the following has the highest nucleophilicity ?
`F^(-)`
`OH^(-)`
`CH_(3)^(-)`
`NH_(2)^(-)`

Options

A. `F^(-)`
B. `OH^(-)`
C. `CH_(3)^(-)`
D. `NH_(2)^(-)`

Explanation

To determine which of the given anions has the highest nucleophilicity, we can analyze the properties of each species based on their electronegativity and the ability to donate electrons.

### Step-by-Step Solution:

1. **Understand Nucleophilicity**: Nucleophilicity refers to the ability of a species to donate a pair of electrons to form a new covalent bond. Nucleophiles are electron-rich species and are often Lewis bases.

2. **List the Given Anions**: The anions provided are:
- \( F^- \) (Fluoride)
- \( OH^- \) (Hydroxide)
- \( CH_3^- \) (Methyl)
- \( NH_2^- \) (Amide)

3. **Consider Electronegativity**: Electronegativity is the tendency of an atom to attract electrons. The higher the electronegativity, the less likely the atom will donate its electrons. The electronegativity order for the elements in question is:
- Fluorine (F) > Oxygen (O) > Nitrogen (N) > Carbon (C)

4. **Analyze Each Anion**:
- **\( F^- \)**: Fluorine is highly electronegative, making it a poor nucleophile because it holds onto its electrons tightly and is less willing to donate them.
- **\( OH^- \)**: Oxygen is also electronegative but less so than fluorine. It can donate electrons but not as readily as less electronegative species.
- **\( NH_2^- \)**: Nitrogen is less electronegative than oxygen and fluorine, so it can donate its electrons more easily than \( OH^- \).
- **\( CH_3^- \)**: Carbon is the least electronegative of the four, making \( CH_3^- \) the most willing to donate its electrons and thus the strongest nucleophile.

5. **Conclusion**: Based on the analysis, \( CH_3^- \) (Methyl) has the highest nucleophilicity among the given anions.

### Final Answer:
The species with the highest nucleophilicity is \( CH_3^- \).