The number of bonds present in dimeric `AlCl_3 ` are

Options

A. 8
B. 3
C. 6
D. 4

Answer

To determine the number of bonds present in dimeric AlCl₃, we can follow these steps:

### Step 1: Understand the Dimeric Form
The dimeric form of AlCl₃ is represented as Al₂Cl₆. This means that two aluminum (Al) atoms and six chlorine (Cl) atoms are involved in the structure.

**Hint:** Remember that dimeric compounds consist of two identical units bonded together.

### Step 2: Analyze the Structure
In Al₂Cl₆, the aluminum atoms are connected to chlorine atoms. Each aluminum atom can form bonds with chlorine atoms.

**Hint:** Visualize the arrangement of atoms to better understand how they are bonded.

### Step 3: Count the Covalent Bonds
Each aluminum atom forms three covalent bonds with three chlorine atoms. Therefore, for two aluminum atoms, we have:
- 3 Al-Cl bonds from the first Al
- 3 Al-Cl bonds from the second Al

This gives us a total of 6 covalent bonds (3 + 3 = 6).

**Hint:** Focus on the bonds formed between aluminum and chlorine atoms.

### Step 4: Identify the Dative Bonds
In addition to the covalent bonds, there are dative (coordinate) bonds formed between the aluminum atoms. Each aluminum atom has a vacant p-orbital that can accept a lone pair of electrons from a chlorine atom.

In Al₂Cl₆, there are 2 dative bonds formed:
- One dative bond from a chlorine atom of one Al to the other Al.
- Another dative bond from a chlorine atom of the second Al to the first Al.

**Hint:** Dative bonds occur when one atom donates both electrons to form a bond.

### Step 5: Total Number of Bonds
Now, we can add the total number of covalent bonds and dative bonds:
- Total covalent bonds = 6
- Total dative bonds = 2

Thus, the total number of bonds in dimeric AlCl₃ is:
6 (covalent) + 2 (dative) = 8 bonds.

**Hint:** Always sum up the different types of bonds to get the total.

### Final Answer
The number of bonds present in dimeric AlCl₃ is **8**.

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