CsCl has bcc structure with `Cs^(+)` at the centre and `Cl^(-)` ion at each corner. If `r_(Cs^(+))` is `1.69Å` and `r_(Cl^(-))` is `1.81Å` what is the edge length of the cube?

Options

A. `3.50Å`
B. `3.80Å`
C. `4.04Å`
D. `4.50Å`

Answer

To find the edge length of the CsCl unit cell with a body-centered cubic (BCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- In a BCC structure, the cesium ion (Cs⁺) is located at the body center, while the chloride ions (Cl⁻) are located at each corner of the cube.

2. **Identify the Radii**:
- The radius of the cesium ion (r(Cs⁺)) is given as 1.69 Å.
- The radius of the chloride ion (r(Cl⁻)) is given as 1.81 Å.

3. **Determine the Relationship Between Edge Length and Ion Radii**:
- In a BCC structure, the body diagonal of the cube is equal to the sum of the diameters of the ions that touch along that diagonal.
- The body diagonal (d) can be expressed in terms of the edge length (a) of the cube:
\[ d = \sqrt{3}a \]
- The body diagonal also equals the sum of the diameters of the ions:
\[ d = 2r(Cs⁺) + 2r(Cl⁻) \]

4. **Set Up the Equation**:
- From the above relationships, we can equate the two expressions for the body diagonal:
\[ \sqrt{3}a = 2r(Cs⁺) + 2r(Cl⁻) \]

5. **Substitute the Values**:
- Substitute the given radii into the equation:
\[ \sqrt{3}a = 2(1.69) + 2(1.81) \]
- Calculate the right side:
\[ \sqrt{3}a = 3.38 + 3.62 = 7.00 \, \text{Å} \]

6. **Solve for Edge Length (a)**:
- Now, solve for a:
\[ a = \frac{7.00}{\sqrt{3}} \]
- Calculate the value:
\[ a \approx \frac{7.00}{1.732} \approx 4.04 \, \text{Å} \]

### Final Answer:
The edge length of the cube is approximately **4.04 Å**.

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