STATEMENT-1: In `LiAlH_(4),Al` is `sp^(3)` hybridised .
STATEMENT-2: `LiAlH_(4)` is a good reducing agent .
STATEMENT-3: `LiAlH_(4)` complex is unstable in water .

Options

A. TTT
B. TTF
C. TFF
D. FTT

Answer

To analyze the three statements regarding lithium aluminum hydride (LiAlH₄), we will evaluate each statement step by step.

### Step 1: Evaluate Statement 1
**Statement 1:** In LiAlH₄, Al is sp³ hybridized.

- **Hybridization of Aluminum in LiAlH₄:**
- Aluminum (Al) has 3 valence electrons.
- In LiAlH₄, Al forms four bonds with four hydrogen atoms.
- To determine the hybridization, we can use the formula:
\[ n = \frac{1}{2} \left( \text{Number of valence electrons in central atom} + \text{Number of univalent atoms} + \text{Charge} \right) \]
- For Al:
- Valence electrons = 3
- Univalent atoms (H) = 4
- Charge = -1 (since AlH₄⁻)
- Plugging in the values:
\[ n = \frac{1}{2} \left( 3 + 4 - 1 \right) = \frac{1}{2} \times 6 = 3 \]
- Since \( n = 4 \), the hybridization is sp³.

**Conclusion for Statement 1:** True.

### Step 2: Evaluate Statement 2
**Statement 2:** LiAlH₄ is a good reducing agent.

- **Reducing Agent Properties:**
- LiAlH₄ is known to reduce various functional groups.
- It can reduce aldehydes, ketones, and carboxylic acids to their respective alcohols.
- Due to its ability to donate hydride ions (H⁻), it acts as a strong reducing agent.

**Conclusion for Statement 2:** True.

### Step 3: Evaluate Statement 3
**Statement 3:** LiAlH₄ complex is unstable in water.

- **Reactivity with Water:**
- When LiAlH₄ comes into contact with water, it reacts vigorously.
- The reaction produces lithium hydroxide (LiOH), aluminum hydroxide (Al(OH)₃), and hydrogen gas (H₂).
- This reaction indicates that LiAlH₄ is indeed unstable in water.

**Conclusion for Statement 3:** True.

### Final Conclusion
All three statements are true. Therefore, the overall answer is that all statements regarding LiAlH₄ are correct.

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