Which one of the following ionic species has the greatest proton affinity to form stable compound ?

Options

A. `HS^(-)`
B. `NH_(2)^(-)`
C. `F^(-)`
D. `I^(-)`

Explanation

To determine which ionic species has the greatest proton affinity to form a stable compound, we need to analyze the proton affinity of each species. The species we will consider are HS⁻, NH₂⁻, F⁻, and I⁻.

### Step-by-Step Solution:

1. **Understanding Proton Affinity**:
Proton affinity refers to the tendency of a species to accept a proton (H⁺). A higher proton affinity indicates a stronger base.

2. **Analyzing HS⁻ (Hydrosulfide Ion)**:
- HS⁻ has a negative charge that can be delocalized in the vacant 3d orbitals of sulfur.
- This delocalization decreases the negative charge density on sulfur, making it a weaker base.
- Therefore, HS⁻ has a lower proton affinity.

3. **Analyzing NH₂⁻ (Amide Ion)**:
- NH₂⁻ has a lone pair of electrons on nitrogen, which does not have vacant orbitals to delocalize the negative charge.
- Nitrogen can easily donate its lone pair to H⁺, making NH₂⁻ a strong base.
- Hence, NH₂⁻ has a high proton affinity.

4. **Analyzing F⁻ (Fluoride Ion)**:
- F⁻ has a high electronegativity, which means it holds onto its electrons tightly.
- It lacks vacant orbitals for delocalization, making it a weaker base compared to NH₂⁻.
- Thus, F⁻ has a lower proton affinity.

5. **Analyzing I⁻ (Iodide Ion)**:
- I⁻ has vacant 5p orbitals, allowing for the localization of the negative charge.
- This localization results in a very weak base, making I⁻ the poorest base among the species considered.
- Therefore, I⁻ has the lowest proton affinity.

6. **Conclusion**:
- After analyzing all four species, NH₂⁻ (Amide Ion) is determined to have the greatest proton affinity and is the strongest base among the options provided.

### Final Answer:
**NH₂⁻ (Amide Ion) has the greatest proton affinity to form a stable compound.**