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

In the CaCl structure:

• Ca²⁺ ions are located at the body center.
• Cl⁻ ions are present at the corners of the cubic unit cell.

Geometric Considerations:

The relationship between the ionic radii and edge length \( a \) is derived as follows:

• The body diagonal of the cubic unit cell passes through the center of the unit cell, connecting opposite corners.
• The length of the body diagonal is given by \( \sqrt{3}a \), where \( a \) is the edge length of the cube.
• In the CaCl structure, the body diagonal is equal to the sum of the diameters of the cation and anion:

\[ \text{Body diagonal} = 2(r_{\text{Ca}^{2+}} + r_{\text{Cl}^-}) \]

Relationship:

Equating the body diagonal to \( \sqrt{3}a \):

\[ 2(r_{\text{Ca}^{2+}} + r_{\text{Cl}^-}) = \sqrt{3}a \]

Dividing by 2:

\[ r_{\text{Ca}^{2+}} + r_{\text{Cl}^-} = \frac{\sqrt{3}a}{2} \]

Final Formula:

The relationship between the ionic radii and edge length \( a \) is:

\[ r_{\text{Ca}^{2+}} + r_{\text{Cl}^-} = \frac{\sqrt{3}a}{2} \]