JEE Challenger
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Calculate Packing Fraction of Face Centered Unit Cell Salt Structure

Consider the close-packed structure of a salt consisting of cation XX and anion YY depicted below (for clarity, only ions on a single face are displayed). Calculate the approximate packing fraction, given that packing fraction=packing efficiency100\text{packing fraction} = \frac{\text{packing efficiency}}{100}

Question Diagram 1

Options

A

0.74

B

0.63

Correct
C

0.52

D

0.48

Step-by-Step Solution

To calculate the packing fraction of the given salt structure, we analyze the arrangement of ions XX and YY in the cubic unit cell based on the provided figure.

1. Structure Analysis

From the diagram:

  • The anion YY spheres are positioned at the 8 corners of the cubic unit cell and touch each other along the edges.
  • The cation XX spheres are positioned at the 6 face-centers of the cubic unit cell and touch the surrounding YY spheres along the face diagonals.

2. Determination of Ionic Radii

Let aa be the edge length of the cubic unit cell, rYr_Y be the radius of anion YY, and rXr_X be the radius of cation XX.

  • Along the edge of the unit cell: Since adjacent YY ions touch each other along the edge: a=2rY  ⟹  rY=a2a = 2 r_Y \implies r_Y = \frac{a}{2}

  • Along the face diagonal of the unit cell: The face diagonal has a length of 2a\sqrt{2}a. Along the face diagonal, the central XX ion touches two corner YY ions: 2rY+2rX=2a2 r_Y + 2 r_X = \sqrt{2} a

    Substituting a=2rYa = 2 r_Y: 2rY+2rX=22rY2 r_Y + 2 r_X = 2\sqrt{2} r_Y rX=rY(2−1)r_X = r_Y (\sqrt{2} - 1)

3. Number of Ions per Unit Cell

  • Number of YY ions (ZYZ_Y): ZY=8 corners×18=1Z_Y = 8 \text{ corners} \times \frac{1}{8} = 1

  • Number of XX ions (ZXZ_X): ZX=6 face centers×12=3Z_X = 6 \text{ face centers} \times \frac{1}{2} = 3

4. Calculation of Packing Fraction

The total volume occupied by the ions in the unit cell is: Voccupied=ZY(43πrY3)+ZX(43πrX3)V_{\text{occupied}} = Z_Y \left(\frac{4}{3} \pi r_Y^3\right) + Z_X \left(\frac{4}{3} \pi r_X^3\right) Voccupied=43π[rY3+3rX3]V_{\text{occupied}} = \frac{4}{3} \pi \left[ r_Y^3 + 3 r_X^3 \right]

Substituting rX=rY(2−1)r_X = r_Y(\sqrt{2} - 1): Voccupied=43πrY3[1+3(2−1)3]V_{\text{occupied}} = \frac{4}{3} \pi r_Y^3 \left[ 1 + 3(\sqrt{2} - 1)^3 \right]

The total volume of the cubic unit cell is: Vcell=a3=(2rY)3=8rY3V_{\text{cell}} = a^3 = (2 r_Y)^3 = 8 r_Y^3

Thus, the packing fraction is given by: Packing Fraction=VoccupiedVcell=43πrY3[1+3(2−1)3]8rY3=π6[1+3(2−1)3]\text{Packing Fraction} = \frac{V_{\text{occupied}}}{V_{\text{cell}}} = \frac{\frac{4}{3} \pi r_Y^3 \left[ 1 + 3(\sqrt{2} - 1)^3 \right]}{8 r_Y^3} = \frac{\pi}{6} \left[ 1 + 3(\sqrt{2} - 1)^3 \right]

5. Numerical Computation

Substitute 2≈1.414\sqrt{2} \approx 1.414: 2−1≈0.414\sqrt{2} - 1 \approx 0.414 (2−1)3≈(0.414)3≈0.071(\sqrt{2} - 1)^3 \approx (0.414)^3 \approx 0.071 1+3(0.071)=1+0.213=1.2131 + 3(0.071) = 1 + 0.213 = 1.213

Now, calculating the packing fraction: Packing Fraction≈3.14166×1.213≈0.5236×1.213≈0.635≈0.63\text{Packing Fraction} \approx \frac{3.1416}{6} \times 1.213 \approx 0.5236 \times 1.213 \approx 0.635 \approx 0.63

Correct Answer: (B) 0.63