Question

Ratio of densities of two crystals formed when the same metal crystallizes once in bcc and ccp structure will be:
(A) 1:2
(B) $\sqrt{2}:\sqrt{3}$
(C) $2\sqrt{2}:3\sqrt{3}$
(D) $3\sqrt{6}:8$

Answer 1

We know that any crystal lattice is made up of very large unit cells and every lattice point is occupied by one constituent particle (atom, molecular, or ion). We must consider three types of cubic unit cells and for simplicity assume that the constituent particle is an atom, which is a primitive cubic unit cell, body-centered cubic unit cell, and face-centered cubic unit cell.

Complete step by step solution:
A Body-centred cubic (bcc) unit cell with an atom at each of its corners and also one atom at its body center. An atom at the body center can be seen that wholly belongs to the unit cell in which it is present.
Hence, in a bcc unit cell:
(i) $8~corners\times \dfrac{1}{8}per~~corner~atom=8\times \dfrac{1}{8}$ = 1 atom
(ii) 1 body centre atom = $1\times 1$ = 1 atom
Therefore, the total number of atoms per unit cell for bcc, Z = 2 atoms
A cubic close packing (ccp) or Face-centred cubic (fcc) which consists of atoms at all the corners and the center of all the faces of the cube.
Hence, in the FCC unit cell,
(a) $8~corners\times \dfrac{1}{8}per~~corner~atom=8\times \dfrac{1}{8}$ = 1 atom
(b) $6\text{ }face-centred\text{ }atoms\times \dfrac{1}{2}atoms\text{ }per\text{ }unit\text{ }cell=6\times \dfrac{1}{2}$= 3 atoms
Therefore, the total number of atoms per unit cell = 4 atoms
The molar mass of an element (M) = $\dfrac{d\times {{N}_{A}}\times {{a}^{3}}}{Z}$
Density (d) = $\dfrac{ZM}{{{N}_{A}}{{a}^{3}}}$
For bcc, the value of Z = 2, and edge length ‘b’ = $\dfrac{4r}{\sqrt{3}}$
For ccp or fcc, the value of Z = 4, and edge length ‘a’ = $\dfrac{4r}{\sqrt{2}}$
Densities ratio of bcc and ccp,
$\dfrac{{{d}_{bcc}}}{{{d}_{ccp}}}=\dfrac{{{Z}_{bcc}}}{{{Z}_{ccp}}}\times \dfrac{{{a}_{ccp}}}{{{a}_{bcc}}}=\dfrac{2}{4}\times {{(\dfrac{4r}{\sqrt{2}})}^{3}}\times {{(\dfrac{\sqrt{3}}{4r})}^{3}}=\dfrac{3\sqrt{3}}{4\sqrt{2}}$
On multiplying the above value of denominator and numerator by $\sqrt{2}$ , then
$\dfrac{{{d}_{bcc}}}{{{d}_{cpp}}}=\dfrac{3\sqrt{6}}{8}$
Hence, the ratio of densities of two crystals formed when the same metal crystallizes once in bcc and ccp structure will be $3\sqrt{6}:8$

So, the correct answer is option D.

Note: A cubic crystal is made up of a crystal lattice or a space lattice structure. A repeating arrangement of a space lattice is made up of unit cells. The most basic structure of a crystalline solid is a unit cell. The density of the unit cell is said to be the density of the cubic crystal itself.