Question

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

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

Hint:

For a BCC structure (like CsCl), the ions touch along the body diagonal, so \(\sqrt{3}a = 2(r_+ + r_-)\). For FCC structure (like NaCl), the ions touch along the face diagonal, so \(\sqrt{2}a = 2(r_+ + r_-)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
CsCl has a body-centered cubic (BCC) structure. In the CsCl structure, the Cl\(^-\) ions form a simple cubic lattice, and the Cs\(^+\) ion occupies the body center. The ions touch along the body diagonal.
Step 2: Detailed Explanation:
In the CsCl structure, the relationship between the edge length (a) and the sum of the ionic radii (r\(_{Cs^+}\) + r\(_{Cl^-}\)) is given by: \[ \sqrt{3}a = 2(r_{Cs^+} + r_{Cl^-}) \] Given: \( r_{Cs^+} = 1.69 \) Å, \( r_{Cl^-} = 1.81 \) Å. Sum of radii = \( 1.69 + 1.81 = 3.50 \) Å. Thus, \( \sqrt{3}a = 2 \times 3.50 = 7.00 \) Å. So, \( a = \frac{7.00}{\sqrt{3}} = \frac{7.00}{1.732} \approx 4.04 \) Å.
Step 3: Final Answer:
The edge length of the unit cell is 4.04 Å, option (A).