Comparison of Edge Lengths Packing Efficiency and Density for Cubic Unit Cells

To determine the correct statements, we analyze each property step-by-step.

1. Packing Efficiency (Option A)

The packing efficiency of a cubic unit cell depends only on its crystal lattice type:

• Metal x forms a face-centred cubic (fcc) unit cell: $\text{Packing Efficiency}_x = \frac{\pi}{3\sqrt{2}} \approx 74\%$
• Metal y forms a body-centred cubic (bcc) unit cell: $\text{Packing Efficiency}_y = \frac{\pi\sqrt{3}}{8} \approx 68\%$
• Metal z forms a simple cubic (sc) unit cell: $\text{Packing Efficiency}_z = \frac{\pi}{6} \approx 52.4\%$

Comparing these values: $\text{Packing efficiency of } x > \text{Packing efficiency of } y > \text{Packing efficiency of } z$

Thus, Option A is correct.

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2. Edge Length Comparison (Options B and C)

The relationship between unit cell edge length ($L$) and atomic radius ($r$) for each unit cell is:

• For fcc (metal $x$): $\sqrt{2} L_x = 4 r_x \implies L_x = 2\sqrt{2} r_x$
• For bcc (metal $y$): $\sqrt{3} L_y = 4 r_y \implies L_y = \frac{4}{\sqrt{3}} r_y$
• For simple cubic (metal $z$): $L_z = 2 r_z$

Given the relations between radii: $r_y = \frac{8}{\sqrt{3}} r_x$ $r_z = \frac{\sqrt{3}}{2} r_y = \frac{\sqrt{3}}{2} \left( \frac{8}{\sqrt{3}} r_x \right) = 4 r_x$

Now, express all edge lengths in terms of $r_x$:

• $L_x = 2\sqrt{2} r_x \approx 2.83 r_x$
• $L_y = \frac{4}{\sqrt{3}} \left( \frac{8}{\sqrt{3}} r_x \right) = \frac{32}{3} r_x \approx 10.67 r_x$
• $L_z = 2(4 r_x) = 8 r_x$

Comparing the edge lengths: $L_y \left(\frac{32}{3} r_x\right) > L_z \left(8 r_x\right) > L_x \left(2\sqrt{2} r_x\right)$

• $L_y > L_z$ is true $\implies$ Option B is correct.
• $L_x > L_y$ is false $\implies$ Option C is incorrect.

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3. Density Comparison (Option D)

The density ($\rho$) of a unit cell is given by: $\rho = \frac{Z \cdot M}{N_A \cdot L^3}$

Given the relations between molar masses: $M_z = 3 M_x \implies M_x = \frac{M_z}{3}$ $M_z = \frac{3}{2} M_y \implies M_y = \frac{2}{3} M_z = 2 M_x$

For the respective unit cells:

• Metal x (fcc, $Z_x = 4$): $\rho_x = \frac{4 M_x}{N_A \cdot L_x^3} = \frac{4 M_x}{N_A \left(2\sqrt{2} r_x\right)^3} = \frac{4 M_x}{N_A \left(16\sqrt{2} r_x^3\right)} = \frac{M_x}{4\sqrt{2} N_A r_x^3}$

• Metal y (bcc, $Z_y = 2$): $\rho_y = \frac{2 M_y}{N_A \cdot L_y^3} = \frac{2(2 M_x)}{N_A \left(\frac{32}{3} r_x\right)^3} = \frac{4 M_x}{N_A \left(\frac{32768}{27} r_x^3\right)} = \frac{27 M_x}{8192 N_A r_x^3}$

Comparing $\rho_x$ and $\rho_y$: $\frac{\rho_x}{\rho_y} = \frac{\frac{1}{4\sqrt{2}}}{\frac{27}{8192}} = \frac{8192}{108\sqrt{2}} = \frac{2048}{27\sqrt{2}} \approx 53.6 > 1$

Hence, $\rho_x > \rho_y$.

Thus, Option D is correct.

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Correct Options:

(A), (B), and (D)