Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length L_x, body-centred cubic (bcc) unit cell of edge length L_y, and simple cubic unit cell of edge length L_z, respectively. If r_z = $\frac{\sqrt{3}}{2}$ r_y; r_y = 8√3 r_x; M_z = $\frac{3}{2}$ M_y and M_z = 3M_x, then the correct statement(s) is(are): Given: M_x, M_y, and M_z are molar masses of metals x, y, and z, respectively. r_x, r_y, and r_z are atomic radii of metals x, y, and z, respectively

Correct Options:
• Packing efficiency of unit cell of x > Packing efficiency of unit cell of y > Packing efficiency of unit cell of z.
• L_y > L_z.
• Density of x > Density of y.

Step-by-step Explanation:

1. Analysis of Packing Efficiency:
• Metal x forms a face-centred cubic (fcc) unit cell.
The packing efficiency of an fcc unit cell is approximately 74% (or $\frac{\pi }{3\sqrt{2}}$).
• Metal y forms a body-centred cubic (bcc) unit cell.
The packing efficiency of a bcc unit cell is approximately 68% (or $\frac{\sqrt{3}\pi }{8}$).
• Metal z forms a simple cubic (sc) unit cell.
The packing efficiency of a simple cubic unit cell is approximately 52.4% (or $\frac{\pi }{6}$).

Comparing these values: 74% > 68% > 52.4%
Thus, Packing efficiency of x > Packing efficiency of y > Packing efficiency of z is correct.

2. Comparison of Edge Lengths L_y and L_z:
For a bcc unit cell of metal y (edge length L_y and atomic radius r_y):

$\sqrt{3}{L}_y=4r_y\Rightarrow r_y=\frac{\sqrt{3}}{4}{L}_y$

For a simple cubic unit cell of metal z (edge length L_z and atomic radius r_z):

$L_z=2r_z\Rightarrow r_z=\frac{L_z}{2}$

We are given the relationship between r_z and r_y:

$r_z=\frac{\sqrt{3}}{2}{r}_y$

Substituting the expressions for r_z and r_y in terms of edge lengths:

$\frac{L_z}{2}=\frac{\sqrt{3}}{2}\times \left(\frac{\sqrt{3}}{4}{L}_y\right)$

$\frac{L_z}{2}=\frac{3}{8}{L}_y\Rightarrow L_z=\frac{3}{4}{L}_y$

Since L_z = 0.75 L_y, it clearly follows that L_y > L_z. Thus, this statement is correct.

3. Comparison of Densities of x and y:
Density of a crystal lattice is given by:

$d=\frac{Z\times M}{N_A\times L^3}$

For metal x (fcc unit cell, Z_x = 4):
Relation between edge length L_x and radius r_x:

$\sqrt{2}{L}_x=4r_x\Rightarrow L_x=2\sqrt{2}{r}_x$

$L_x^3=16\sqrt{2}{r}_x^3$

Density of x:

$d_x=\frac{4M_x}{N_A\times 16\sqrt{2}{r}_x^3}=\frac{M_x}{4\sqrt{2}{N}_A{r}_x^3}$

For metal y (bcc unit cell, Z_y = 2):
Relation between edge length L_y and radius r_y:

$L_y=\frac{4}{\sqrt{3}}{r}_y\Rightarrow L_y^3=\frac{64}{3\sqrt{3}}{r}_y^3$

Density of y:

$d_y=\frac{2M_y}{N_A\times \frac{64}{3\sqrt{3}}{r}_y^3}=\frac{3\sqrt{3}{M}_y}{32N_A{r}_y^3}$

Given relations:
• ${\mathrm{Maccess\_z}}_{}=3M_x$ and $M_z=\frac{3}{2}{M}_y\Rightarrow 3M_x=\frac{3}{2}{M}_y\Rightarrow M_x=\frac{1}{2}{M}_y$
• $r_y=8\sqrt{3}{r}_x\Rightarrow r_x=\frac{r_y}{8\sqrt{3}}$

Substituting M_x and r_x into the density formula for d_x:

$d_x=\frac{\left(\frac{M_y}{2}\right)}{4\sqrt{2}{N}_A{\left(\frac{r_y}{8\sqrt{3}}\right)}^3}=\frac{192\sqrt{3}{M}_y}{\sqrt{2}{N}_A{r}_y^3}$

Comparing d_x and d_y:

$\frac{d_x}{d_y}=\frac{\left(\frac{192\sqrt{3}{M}_y}{\sqrt{2}{N}_A{r}_y^3}\right)}{\left(\frac{3\sqrt{3}{M}_y}{32N_A{r}_y^3}\right)}=\frac{192\times 32}{3\sqrt{2}}=1024\sqrt{2}>1$

Therefore, Density of x > Density of y is correct.