JEE Advance - Chemistry (2023 - Paper 2 Online - No. 6)
Atoms of metals $\mathrm{x}, \mathrm{y}$, and $\mathrm{z}$ form face-centred cubic (fcc) unit cell of edge length $\mathrm{L}_{\mathrm{x}}$, body-centred cubic (bcc) unit cell of edge length $\mathrm{L}_y$, and simple cubic unit cell of edge length $\mathrm{L}_z$, respectively.
If $\mathrm{r}_{\mathrm{z}}=\frac{\sqrt{3}}{2} r_{\mathrm{y}} ; \mathrm{r}_{\mathrm{y}}=\frac{8}{\sqrt{3}} \mathrm{r}_{\mathrm{x}} ; M_{\mathrm{z}}=\frac{3}{2} M_{\mathrm{y}}$ and $M_{\mathrm{z}}=3 M_{\mathrm{x}}$, 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.
$\mathrm{r}_{\mathrm{x}}, \mathrm{r}_{\mathrm{y}}$, and $\mathrm{r}_{\mathrm{z}}$ are atomic radii of metals $\mathrm{x}, \mathrm{y}$, and $\mathrm{z}$, respectively.]
If $\mathrm{r}_{\mathrm{z}}=\frac{\sqrt{3}}{2} r_{\mathrm{y}} ; \mathrm{r}_{\mathrm{y}}=\frac{8}{\sqrt{3}} \mathrm{r}_{\mathrm{x}} ; M_{\mathrm{z}}=\frac{3}{2} M_{\mathrm{y}}$ and $M_{\mathrm{z}}=3 M_{\mathrm{x}}$, 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.
$\mathrm{r}_{\mathrm{x}}, \mathrm{r}_{\mathrm{y}}$, and $\mathrm{r}_{\mathrm{z}}$ are atomic radii of metals $\mathrm{x}, \mathrm{y}$, and $\mathrm{z}$, respectively.]
Packing efficiency of unit cell of $x>$ Packing efficiency of unit cell of $y>$ Packing efficiency of unit cell of $z$
$\mathrm{L}_{\mathrm{y}}>\mathrm{L}_{\mathrm{z}}$
$\mathrm{L}_{\mathrm{x}}>\mathrm{L}_{\mathrm{y}}$
Density of $x>$ Density of $y$
Explanation
1. Atomic radii relations :
Given,
$$ r_{y} = \frac{8}{\sqrt{3}} r_{x} \tag{1} $$ ......(1)
$$ r_{z} = \frac{\sqrt{3}}{2} r_{y} \tag{2} $$ ........(2)
Substituting Eq. (1) into Eq. (2), we get :
$$ r_{z} = \frac{\sqrt{3}}{2} \times \frac{8}{\sqrt{3}} r_{x} = 4 r_{x} \tag{3} $$ .......(3)
2. Edge lengths of the unit cells :
The edge lengths (L) of the unit cells for x, y, and z are given by :
$$ L_x = 2 \sqrt{2} r_x \quad \text{(for FCC)} \tag{4} $$ .......(4)
$$ L_y = 4 r_y = \frac{32}{\sqrt{3}} r_x \quad \text{(for BCC, using Eq. (1))} \tag{5} $$ .......(5)
$$ L_z = 2 r_z = 8 r_x \quad \text{(for simple cubic, using Eq. (3))} \tag{6} $$ .......(6)
3. Comparing the edge lengths :
From Eq. (4), (5) and (6), we find :
$$ L_y > L_z > L_x \tag{7} $$
This means Option B : $L_y > L_z$ is correct.
4. Packing Efficiencies :
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows :
- FCC :
$$ PE_{fcc} = \frac{4 \times \frac{4}{3} \pi r_x^3}{(L_x)^3} = \frac{\pi}{\sqrt{2}} \tag{8} $$
- BCC :
$$ PE_{bcc} = \frac{2 \times \frac{4}{3} \pi r_y^3}{(L_y)^3} = \frac{3 \pi \sqrt{3}}{32} \tag{9} $$
- Simple Cubic :
$$ PE_{sc} = \frac{1 \times \frac{4}{3} \pi r_z^3}{(L_z)^3} = \frac{\pi}{6} \tag{10} $$
Comparing these packing efficiencies, we find :
$$ PE_{fcc} > PE_{bcc} > PE_{sc} \tag{11} $$
This means Option A : Packing efficiency of unit cell of $x > y > z$ is correct.
5. Calculating Densities :
The densities of x and y are given by the formula $d = \frac{Z \cdot M}{N_a \cdot L^3}$, where $Z$ is the number of atoms in the unit cell, $M$ is the molar mass of the metal, $N_a$ is Avogadro's number, and $L$ is the edge length of the unit cell.
- For x (FCC), $Z=4$ :
$$ d_x = \frac{4 M_x}{N_a (L_x)^3} \tag{12} $$
- For y (BCC), $Z=2$ :
$$ d_y = \frac{2 M_y}{N_a (L_y)^3} \tag{13} $$
$M_z=\frac{3}{2} M_y$ and $M_z=3 M_x$. This implies that $M_x=\frac{M_y}{2}$. Thus, $2M_x=M_y$, we can write :
$$ \frac{d_x}{d_y} = \frac{2 M_x}{M_y} \times \left(\frac{L_y}{L_x}\right)^3 \tag{14} $$ .......(7)
Substituting the given relation $2 M_x = M_y$ into Eq. (7), we get :
$$ \frac{d_x}{d_y} = 2 \times 1 \times \left(\frac{32/3}{2 \sqrt{2}}\right)^3 \tag{15} $$ ..........(8)
After simplifying Eq. (8), we find $d_x > d_y$.
This means Option D : Density of $x > y$ is correct.
So, the correct answers are Options A, B, and D.
Given,
$$ r_{y} = \frac{8}{\sqrt{3}} r_{x} \tag{1} $$ ......(1)
$$ r_{z} = \frac{\sqrt{3}}{2} r_{y} \tag{2} $$ ........(2)
Substituting Eq. (1) into Eq. (2), we get :
$$ r_{z} = \frac{\sqrt{3}}{2} \times \frac{8}{\sqrt{3}} r_{x} = 4 r_{x} \tag{3} $$ .......(3)
2. Edge lengths of the unit cells :
The edge lengths (L) of the unit cells for x, y, and z are given by :
$$ L_x = 2 \sqrt{2} r_x \quad \text{(for FCC)} \tag{4} $$ .......(4)
$$ L_y = 4 r_y = \frac{32}{\sqrt{3}} r_x \quad \text{(for BCC, using Eq. (1))} \tag{5} $$ .......(5)
$$ L_z = 2 r_z = 8 r_x \quad \text{(for simple cubic, using Eq. (3))} \tag{6} $$ .......(6)
3. Comparing the edge lengths :
From Eq. (4), (5) and (6), we find :
$$ L_y > L_z > L_x \tag{7} $$
This means Option B : $L_y > L_z$ is correct.
4. Packing Efficiencies :
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows :
- FCC :
$$ PE_{fcc} = \frac{4 \times \frac{4}{3} \pi r_x^3}{(L_x)^3} = \frac{\pi}{\sqrt{2}} \tag{8} $$
- BCC :
$$ PE_{bcc} = \frac{2 \times \frac{4}{3} \pi r_y^3}{(L_y)^3} = \frac{3 \pi \sqrt{3}}{32} \tag{9} $$
- Simple Cubic :
$$ PE_{sc} = \frac{1 \times \frac{4}{3} \pi r_z^3}{(L_z)^3} = \frac{\pi}{6} \tag{10} $$
Comparing these packing efficiencies, we find :
$$ PE_{fcc} > PE_{bcc} > PE_{sc} \tag{11} $$
This means Option A : Packing efficiency of unit cell of $x > y > z$ is correct.
5. Calculating Densities :
The densities of x and y are given by the formula $d = \frac{Z \cdot M}{N_a \cdot L^3}$, where $Z$ is the number of atoms in the unit cell, $M$ is the molar mass of the metal, $N_a$ is Avogadro's number, and $L$ is the edge length of the unit cell.
- For x (FCC), $Z=4$ :
$$ d_x = \frac{4 M_x}{N_a (L_x)^3} \tag{12} $$
- For y (BCC), $Z=2$ :
$$ d_y = \frac{2 M_y}{N_a (L_y)^3} \tag{13} $$
$M_z=\frac{3}{2} M_y$ and $M_z=3 M_x$. This implies that $M_x=\frac{M_y}{2}$. Thus, $2M_x=M_y$, we can write :
$$ \frac{d_x}{d_y} = \frac{2 M_x}{M_y} \times \left(\frac{L_y}{L_x}\right)^3 \tag{14} $$ .......(7)
Substituting the given relation $2 M_x = M_y$ into Eq. (7), we get :
$$ \frac{d_x}{d_y} = 2 \times 1 \times \left(\frac{32/3}{2 \sqrt{2}}\right)^3 \tag{15} $$ ..........(8)
After simplifying Eq. (8), we find $d_x > d_y$.
This means Option D : Density of $x > y$ is correct.
So, the correct answers are Options A, B, and D.
Comments (0)
