What is the covalency of Al in `[AlCl(H_2O)_5]^(2+)` ?

Options

A. 3
B. 4
C. 5
D. 6

Answer

To determine the covalency of aluminum in the complex ion \([AlCl(H_2O)_5]^{2+}\), we need to analyze the number of bonds formed by the aluminum atom with its ligands.

### Step-by-Step Solution:

1. **Identify the Ligands**: In the complex \([AlCl(H_2O)_5]^{2+}\), we have two types of ligands:
- One chloride ion (Cl⁻)
- Five water molecules (H₂O)

2. **Count the Bonds**:
- Aluminum (Al) forms one bond with the chloride ion (Cl⁻).
- Aluminum also forms one bond with each of the five water molecules (H₂O).

3. **Calculate Total Bonds**:
- Total bonds from the chloride ion = 1
- Total bonds from the five water molecules = 5
- Therefore, the total number of bonds (covalency) formed by aluminum = 1 (from Cl) + 5 (from H₂O) = 6.

4. **Conclusion**: The covalency of aluminum in the complex \([AlCl(H_2O)_5]^{2+}\) is 6.

### Final Answer:
The covalency of Al in \([AlCl(H_2O)_5]^{2+}\) is **6**.

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