Hybridisation of Al in `AlCl_(3)` (monomeric from ltbgt above `800^(@) C) and Al_(2) Cl_(6)` (dimeric form below ` 400^(@)`C) respectively are

Options

A. `sp^(2) , sp^(3)`
B. `sp^(2) , sp^(2)`
C. `sp^(3) , sp^(3)`
D. `sp^(2), dsp^(2)`

Explanation

To determine the hybridization of aluminum in AlCl3 (monomeric form above 800°C) and Al2Cl6 (dimeric form below 400°C), we can follow these steps:

### Step 1: Analyze the Structure of AlCl3
- Aluminum (Al) has an atomic number of 13, with the electronic configuration:
- 1s² 2s² 2p⁶ 3s² 3p¹.
- In AlCl3, aluminum is bonded to three chlorine atoms.

### Step 2: Determine the Hybridization of Al in AlCl3
- In AlCl3, aluminum uses one 3s and two 3p orbitals to form three equivalent sp² hybrid orbitals.
- Each sp² hybrid orbital overlaps with the p orbital of chlorine to form sigma bonds.
- Therefore, the hybridization of aluminum in AlCl3 is **sp²**.

### Step 3: Analyze the Structure of Al2Cl6
- In Al2Cl6, two AlCl3 units dimerize, meaning that two aluminum atoms are involved.
- Each aluminum atom in Al2Cl6 forms four bonds: three with chlorine atoms and one with another aluminum atom.

### Step 4: Determine the Hybridization of Al in Al2Cl6
- In this dimeric form, aluminum uses one 3s orbital and three 3p orbitals to form four equivalent sp³ hybrid orbitals.
- Each sp³ hybrid orbital overlaps with the p orbital of chlorine and the other aluminum atom to form sigma bonds.
- Therefore, the hybridization of aluminum in Al2Cl6 is **sp³**.

### Conclusion
- The hybridization of aluminum in AlCl3 (monomeric form) is **sp²**.
- The hybridization of aluminum in Al2Cl6 (dimeric form) is **sp³**.

### Final Answer
- The hybridization of Al in AlCl3 is sp² and in Al2Cl6 is sp³.