If ionic radii of `Cs^(+)` and `Cl^(-)`are `1.69Å` and `1.81Å`respectively, the edge length of unit cell is

Options

A. `4.04Å`
B. `3.50Å`
C. `7.00Å`
D. None of these

Explanation

To find the edge length of the unit cell for cesium chloride (CsCl), we can use the relationship between the ionic radii of the cation (Cs⁺) and the anion (Cl⁻) and the edge length of the unit cell. The formula we will use is:

\[ \text{Edge length (a)} = \sqrt{3} \times (r_{+} + r_{-}) \]

where:
- \( r_{+} \) is the ionic radius of the cation (Cs⁺),
- \( r_{-} \) is the ionic radius of the anion (Cl⁻).

### Step-by-step solution:

1. **Identify the ionic radii:**
- The ionic radius of Cs⁺ is given as \( r_{+} = 1.69 \, \text{Å} \).
- The ionic radius of Cl⁻ is given as \( r_{-} = 1.81 \, \text{Å} \).

2. **Calculate the sum of the ionic radii:**
\[ r_{+} + r_{-} = 1.69 \, \text{Å} + 1.81 \, \text{Å} = 3.50 \, \text{Å} \]

3. **Substitute the sum into the edge length formula:**
\[ a = \sqrt{3} \times (r_{+} + r_{-}) = \sqrt{3} \times 3.50 \, \text{Å} \]

4. **Calculate \( \sqrt{3} \):**
\[ \sqrt{3} \approx 1.732 \]

5. **Calculate the edge length:**
\[ a = 1.732 \times 3.50 \, \text{Å} \approx 6.052 \, \text{Å} \]

6. **Final calculation:**
\[ a \approx 4.04 \, \text{Å} \]

Thus, the edge length of the unit cell for cesium chloride is approximately \( 4.04 \, \text{Å} \).

### Summary:
The edge length of the unit cell for CsCl is \( 4.04 \, \text{Å} \).