Consider the following reactions
I. `AlH_(3)+H^(-)toAlH_(4)^(-)`
II. `H_(2)O+H^(-)toH_(2)+OH^(-)`
Select the correct statement based on these reactions.

Options

A. `H^(-)` is a lewis acid in I and lewis base in II
B. `H^(-)` is a lewis base in I and brosted base in II
C. `H^(-)` is a lewis acid in I and brosted acid in II
D. `H^(-)` is a lewis base in I and II

Answer

To analyze the given reactions and select the correct statement based on them, we will break down each reaction step by step.

### Step 1: Analyze the first reaction
**Reaction I:**
\[ \text{AlH}_3 + \text{H}^- \rightarrow \text{AlH}_4^- \]

1. **Identify the species involved:**
- \( \text{AlH}_3 \) (Aluminum hydride) contains aluminum, which is in group 13 and has 3 valence electrons.
- \( \text{H}^- \) (Hydride ion) has an extra electron, making it a strong reducing agent.

2. **Determine the electron configuration:**
- \( \text{AlH}_3 \) has 6 electrons around aluminum (3 from Al and 3 from H), making it electron-deficient and seeking more electrons to complete its octet.

3. **Identify the roles of the species:**
- \( \text{AlH}_3 \) acts as a Lewis acid because it can accept electrons.
- \( \text{H}^- \) acts as a Lewis base because it donates electrons.

4. **Conclusion for Reaction I:**
- The reaction shows that \( \text{H}^- \) donates electrons to \( \text{AlH}_3 \), forming \( \text{AlH}_4^- \).

### Step 2: Analyze the second reaction
**Reaction II:**
\[ \text{H}_2\text{O} + \text{H}^- \rightarrow \text{H}_2 + \text{OH}^- \]

1. **Identify the species involved:**
- \( \text{H}_2\text{O} \) (Water) can act as both an acid and a base.
- \( \text{H}^- \) (Hydride ion) is again present.

2. **Determine the roles of the species:**
- In this reaction, water donates a proton (H⁺) to \( \text{H}^- \), forming \( \text{H}_2 \) and \( \text{OH}^- \).
- Here, \( \text{H}_2\text{O} \) acts as a Bronsted acid (donating H⁺) and \( \text{H}^- \) acts as a Bronsted base (accepting H⁺).

3. **Conclusion for Reaction II:**
- The reaction shows that \( \text{H}^- \) accepts a proton from water, resulting in the formation of \( \text{H}_2 \) and \( \text{OH}^- \).

### Final Conclusion:
Based on the analysis:
- In Reaction I, \( \text{H}^- \) is a Lewis base.
- In Reaction II, \( \text{H}^- \) is a Bronsted base.

Thus, the correct statement based on these reactions is that \( \text{H}^- \) acts as a Lewis base in the first reaction and a Bronsted base in the second reaction.