The unit cell cube length for LiCl(just like NaCl structures) is `5.14Å`. Assuming anion-anion contact, the ionic radius for chloride ion is:

Options

A. 128.5 pm
B. 257 pm
C. 181.7 pm
D. None of the above

Answer

To solve the problem of finding the ionic radius for the chloride ion in LiCl, we will follow these steps:

### Step 1: Understand the Structure
LiCl has a face-centered cubic (FCC) structure similar to NaCl. In this structure, the chloride ions (Cl⁻) are arranged in an FCC lattice.

### Step 2: Identify the Unit Cell Length
The given unit cell length (edge length) for LiCl is \( a = 5.14 \, \text{Å} \).

### Step 3: Calculate the Face Diagonal
In an FCC structure, the face diagonal can be calculated using the formula:
\[ \text{Face Diagonal} = \sqrt{2} \cdot a \]
Substituting the value of \( a \):
\[ \text{Face Diagonal} = \sqrt{2} \cdot 5.14 \, \text{Å} \]

### Step 4: Relate Face Diagonal to Ionic Radii
In the FCC structure, the face diagonal is also equal to the sum of the diameters of the ions along that diagonal. Since there are two ions (one from each corner of the face), we have:
\[ \text{Face Diagonal} = 4r \]
where \( r \) is the ionic radius of the chloride ion.

### Step 5: Set Up the Equation
Equating the two expressions for the face diagonal:
\[ 4r = \sqrt{2} \cdot 5.14 \, \text{Å} \]

### Step 6: Solve for Ionic Radius \( r \)
Rearranging the equation to solve for \( r \):
\[ r = \frac{\sqrt{2} \cdot 5.14 \, \text{Å}}{4} \]

### Step 7: Calculate the Value
Calculating the value:
\[ r = \frac{1.414 \cdot 5.14}{4} \approx \frac{7.25}{4} \approx 1.8125 \, \text{Å} \]

### Step 8: Convert to Picometers
Since the options are given in picometers, we convert angstroms to picometers. We know that:
\[ 1 \, \text{Å} = 100 \, \text{pm} \]
Thus,
\[ r \approx 1.8125 \, \text{Å} \times 100 \, \text{pm/Å} \approx 181.25 \, \text{pm} \]

### Final Result
The ionic radius for the chloride ion is approximately \( 181.25 \, \text{pm} \).