Question

Which of the following expressions is correct in case of a C_SCl unit cell (edge length, a) ?

• A. \(r_{cs^+}+r_{cl^-}=a\)
• B. \(r_{cs^+}+r_{cl^-}=\frac{a}{\sqrt2}\)
• C. \(r_{cs^+}+r_{cl^-}=\frac{\sqrt3}{2}a\)
• D. \(r_{cs^+}+r_{cl^-}=\frac{a}{2}\)

Hint:

Use geometric relationships in unit cells to relate ionic radii and edge length

Answer 1

The Correct Option is C

To determine the correct expression for the sum of ionic radii in a CsCl unit cell, we need to understand the structure and geometry of the CsCl crystal lattice. 

1. Cesium chloride (CsCl) has a simple cubic structure with chloride ions (\(Cl^-\)) forming the main cube and cesium ions (\(Cs^+\)) located in the center of the cube.
2. The chloride ion lies at each corner of the cube, and the cesium ion is located at the body center. This arrangement forms a body-centered cubic (BCC) lattice when looked from the perspective of the CsCl itself.
3. In this type of lattice, the body diagonal of the cube goes through the central cesium ion and two opposite chloride ions at the corners of the cube.
4. The formula for the edge length \(a\) of a cube can be connected to the body diagonal using the geometry of the cube. The body diagonal of a cubic unit cell, \(d\), can be given by:

\(d = \sqrt{3} \times a\)

1. In a CsCl unit cell, this body diagonal is also equal to the sum of the diameters of one cesium ion and two chloride ions, expressed as:

\(d = 2(r_{cs^+} + r_{cl^-})\)

1. Substituting the expression for the body diagonal, we get the following relationship:

\(2(r_{cs^+} + r_{cl^-}) = \sqrt{3} \times a\)

1. Simplifying gives us the expression for the sum of the ionic radii:

\(r_{cs^+} + r_{cl^-} = \frac{\sqrt{3}}{2}a\)

This matches the correct answer provided. Therefore, the correct expression is:

\(r_{cs^+} + r_{cl^-} = \frac{\sqrt{3}}{2}a\)

The other options are incorrect because they do not accurately represent the relationship between the ionic radii and the edge length in a CsCl unit cell structure.