Cesium and chloride ions are in contact along the body diagonal in a body-centred cubic lattice. The edge lenth of the unit cell is 350 pm and `Cs^+` has a radius of 133pm. Hence, the radius of `Cl^-` ion is approximately

Options

A. 170
B. 133
C. 180
D. 150

Answer

To find the radius of the chloride ion (Cl⁻) in a body-centered cubic (BCC) lattice where cesium ions (Cs⁺) and chloride ions are in contact along the body diagonal, we can follow these steps:

### Step 1: Understand the relationship between the body diagonal and the unit cell edge length
In a body-centered cubic lattice, the body diagonal (d) can be expressed in terms of the edge length (a) of the unit cell:
\[ d = \sqrt{3} \cdot a \]

### Step 2: Calculate the length of the body diagonal
Given that the edge length \( a = 350 \, \text{pm} \):
\[ d = \sqrt{3} \cdot 350 \, \text{pm} \]
\[ d \approx 1.732 \cdot 350 \, \text{pm} \]
\[ d \approx 605.8 \, \text{pm} \]

### Step 3: Set up the equation for the body diagonal
In a BCC lattice, the body diagonal is equal to the sum of the radii of the cation (Cs⁺) and the anion (Cl⁻):
\[ d = 2r_{Cs^+} + 2r_{Cl^-} \]
Where:
- \( r_{Cs^+} = 133 \, \text{pm} \) (radius of cesium ion)
- \( r_{Cl^-} \) is the radius of chloride ion we need to find.

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 605.8 \, \text{pm} = 2(133 \, \text{pm}) + 2r_{Cl^-} \]

### Step 5: Simplify the equation
Calculating the left side:
\[ 605.8 \, \text{pm} = 266 \, \text{pm} + 2r_{Cl^-} \]

### Step 6: Solve for the radius of the chloride ion
Rearranging the equation to isolate \( r_{Cl^-} \):
\[ 2r_{Cl^-} = 605.8 \, \text{pm} - 266 \, \text{pm} \]
\[ 2r_{Cl^-} = 339.8 \, \text{pm} \]
\[ r_{Cl^-} = \frac{339.8 \, \text{pm}}{2} \]
\[ r_{Cl^-} \approx 169.9 \, \text{pm} \]

### Final Answer
The radius of the chloride ion \( r_{Cl^-} \) is approximately **170 pm**.

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