Among the following, the best and the least `H^(Θ)` ion donor, respectively, are:

Options

A.
B.
C.
D.

Explanation

To determine the best and the least H⁻ ion donor among the given compounds, we will analyze the structures and their ability to donate hydride ions (H⁻).

### Step-by-Step Solution:

1. **Identify the Compounds**:
Let's denote the compounds as A, B, C, and D. We need to analyze their structures to understand their ion-donating capabilities.

2. **Analyze the Anions**:
- Compounds B and D are dianions (having two negative charges), while A and C are anions with one negative charge each.
- Generally, dianions are more stable due to resonance stabilization, which makes them more likely to donate H⁻ ions.

3. **Compare Compounds A and C**:
- Compound A has two hydrogen atoms (H) available for donation, while compound C has one phenyl group and one hydrogen atom.
- The presence of two hydrogen atoms in A suggests a greater tendency to lose H⁻ compared to C, which only has one hydrogen.

4. **Compare Compounds B and D**:
- Both B and D are dianions. However, D has two hydrogen atoms available for donation, while B has only one.
- Therefore, D would be more likely to donate H⁻ ions compared to B.

5. **Conclusion**:
- The best H⁻ ion donor is compound A because it has the highest number of hydrogen atoms available for donation.
- The least H⁻ ion donor is compound D, as it has the least stabilizing factors compared to the other dianion (B).

### Final Answer:
- **Best H⁻ ion donor**: A
- **Least H⁻ ion donor**: D