Which of the following has maximum covalent character?

Options

A. `LiI`
B. `LiF`
C. `LiCl`
D. `LiBr.`

Answer

To determine which compound has the maximum covalent character among the given options (Lithium Iodide, Lithium Fluoride, Lithium Chloride, and Lithium Bromide), we can use Fajans' rule. Here’s a step-by-step solution:

### Step 1: Understand Fajans' Rule
Fajans' rule states that the covalent character of an ionic compound increases with:
- A smaller cation size (higher charge density)
- A larger anion size (higher polarizability)

### Step 2: Identify the Cation
In all the given compounds, the cation is Lithium (Li⁺). Since the cation is the same in all cases, we will focus on the anions.

### Step 3: Compare the Anions
The anions in the compounds are:
- Lithium Iodide (I⁻)
- Lithium Fluoride (F⁻)
- Lithium Chloride (Cl⁻)
- Lithium Bromide (Br⁻)

### Step 4: Determine the Size of the Anions
The size of the halogen anions increases as we move down the group in the periodic table:
- F⁻ < Cl⁻ < Br⁻ < I⁻

This means that I⁻ is the largest anion among the options provided.

### Step 5: Apply Fajans' Rule
Since the cation (Li⁺) is the same in all cases, the covalent character will increase with the size of the anion. Therefore:
- I⁻ (largest) will have the highest covalent character.
- F⁻ (smallest) will have the lowest covalent character.

### Conclusion
Based on the analysis, Lithium Iodide (LiI) has the maximum covalent character among the given options.

### Final Answer
**Lithium Iodide (LiI)** has the maximum covalent character.

---