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 will follow these steps:

### Step 1: Identify the oxidation state of aluminum (Al)
We start by letting the oxidation state of aluminum be \(X\). The complex consists of one chloride ion (Cl\(^-\)) and five water molecules (H\(_2\)O), which are neutral.

The overall charge of the complex is +2. Therefore, we can set up the following equation:

\[ X + (-1) + 0 \cdot 5 = +2 \]

This simplifies to:

\[ X - 1 = 2 \]

Solving for \(X\):

\[ X = 2 + 1 = 3 \]

Thus, the oxidation state of aluminum in this complex is +3.

### Step 2: Determine the coordination number
The covalency of an atom in a coordination complex is defined as the number of atoms or groups directly bonded to it. In this case, we need to count the number of ligands surrounding the aluminum ion.

- The complex contains one chloride ion (Cl\(^-\)), which acts as a ligand.
- There are five water molecules (H\(_2\)O), which also act as ligands.

### Step 3: Count the total number of ligands
Now, we count the total number of ligands around the aluminum ion:

- 1 chloride ion
- 5 water molecules

Adding these together gives:

\[ 1 + 5 = 6 \]

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

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

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