Question

Atoms of metals x, y, and z form face-centered cubic (fcc) unit cells of edge length L_x, body-centered cubic (bcc) unit cells of edge length L_y, and simple cubic unit cells of edge length L_z, respectively. If r_z =$\sqrt{\frac{3}{2}}$r_y ; r_y = $\sqrt{\frac{8}{3}}$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.]

• A. Packing efficiency of a unit cell of x> Packing efficiency of a unit cell of y > Packing efficiency of a unit cell of z
• B. L_y > L_z
• C. L_x > L_y
• D. Density of x > Density of y

Answer 1

The Correct Option is A, B, D

The correct options are (A),(B) and (D)
Given, $r_z=\frac{\sqrt3}{2}r_y$ and $r_y=\frac{8}{\sqrt3}r_x$
$\therefore\, r_z=\frac{\sqrt3}{2}\times\frac{8}{\sqrt3}r_x=4r_x$
$M_z=\frac{3}{2}M_y$
$M_z=3M_x$
$\therefore M_y=2M_x$
Packing efficiency FCC>BCC>SC
Packing efficiency unit cell x>y>z
In the FCC unit cell, $\sqrt2L_x=4r_x$
$\Rightarrow L_x=2\sqrt2 r_x$
$\therefore \sqrt3l_y=4r_y\Rightarrow L_y=\frac{4}{\sqrt3}r_y=\frac{4}{\sqrt3}\times\frac{8}{\sqrt3}r_x=\frac{32}{3}r_x$
$L_y=\frac{32}{3}r_x$
$\therefore\,l_z=2r_z$$=2\times4r_x=8r_x$
$l_x=2\sqrt2r_x$ and $l_y=\frac{32}{3}r_x$
$\therefore l_z=8r_x$
$\therefore\,l_y>l_z>l_x$
density of  x = $d_x=\frac{zM_x}{(L_x)^3N_A}=\frac{4\times M_X}{(2\sqrt2)r_x)^3\times N_A}=\frac{M_X}{4\sqrt2 r^3xN_A}$
density of y = $d_y=\frac{zM_y}{(L_y)^3N_A}= \frac{2\times2M_X}{(\frac{32}{3}R_X)^3N_A}=\frac{108M_X}{32768r^3_xN_A}$
$\therefore$ density of x > density of y