The correct order of decreasing ionic nature of lead dihalides is:

Options

A. `PbF_(2) gt PbCl_(2) gt PbBr_(2) gt PbI_(2)`
B. `PbF_(2) gt PbBr_(2) gt PbCl_(2) gt PbI_(2)`
C. `PbF_(2) lt PbCl_(2) gt PbBr_(2) lt PbI_(2)`
D. `PbI_(2) lt PbBr_(2) lt PbCl_(2) lt PbF_(2)`

Answer

To determine the correct order of decreasing ionic nature of lead dihalides (PbF2, PbCl2, PbBr2, and PbI2), we need to analyze the ionic and covalent character of these compounds based on the size of the anions involved. Here’s a step-by-step solution:

### Step 1: Identify the Compounds
The lead dihalides in question are:
- PbF2 (Lead(II) fluoride)
- PbCl2 (Lead(II) chloride)
- PbBr2 (Lead(II) bromide)
- PbI2 (Lead(II) iodide)

### Step 2: Understand Ionic vs. Covalent Character
Ionic character is influenced by the size of the anions. As the size of the anions increases, the covalent character increases, and consequently, the ionic character decreases. This is due to the increased polarizability of larger anions, which leads to greater sharing of electron density between the lead cation and the anion.

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

From the periodic table, we know that:
- Fluoride (F⁻) is the smallest anion.
- Chloride (Cl⁻) is larger than fluoride.
- Bromide (Br⁻) is larger than chloride.
- Iodide (I⁻) is the largest anion.

### Step 4: Determine the Order of Ionic Nature
Since smaller anions have higher ionic character, we can arrange the lead dihalides in order of decreasing ionic nature:
1. PbF2 (highest ionic character due to smallest anion F⁻)
2. PbCl2 (next highest, Cl⁻ is larger than F⁻)
3. PbBr2 (Br⁻ is larger than Cl⁻)
4. PbI2 (lowest ionic character due to largest anion I⁻)

### Final Order
Thus, the correct order of decreasing ionic nature of lead dihalides is:
**PbF2 > PbCl2 > PbBr2 > PbI2**