Question

Percentage of free space in cubic close packed structure and in body centered packed structure are respectively
A. 30% and 26%
B. 26% and 32%
C. 48% and 48%
D. 32% and 26%

Answer 1

Any crystal lattice of a compound contains some free spaces, and packed spaces in which the atoms are arranged in a certain sequence. A unit cell is the smallest part of any crystal lattice. The crystals are made of many unit cells.

Formula used: Packing efficiency = packing fraction $\times$ 100
Packing fraction = $\dfrac{Z\times volume\,occupied\,by\,atom\,in\,unit\,cell}{Volume\,of\,unit\,cell}$

Complete answer: Packing efficiency of any crystal is the capacity which can be filled with atoms.
In a close packed structure the unit cell is FCC, that has particles on corners as well as the centre of all faces. The Z= 4 here, so, according to the packing efficiency formula, we will calculate its packing efficiency, then subtract this efficiency from 100 to obtain the vacant space.
So, P.E. of CCP = $\dfrac{Z\times volume\,occupied\,by\,atom\,in\,unit\,cell}{Volume\,of\,unit\,cell}$ $\times$ 100
P.E. of CCP = $\dfrac{4\times {}^{4}/{}_{3}\pi {{r}^{3}}}{{{(2\sqrt{2}r)}^{3}}}$ $\times$100
P.E. of CCP = 0.74$\times$100
P.E. of CCP = 74%
So, vacant space = 100-74
Vacant space = 26 %
Now, for body centered packed structure, the unit cell is Z = 2. It has particles on corners as well as one at the centre of the cube. So, according to the packing efficiency formula, we will calculate its packing efficiency, then subtract this efficiency from 100 to obtain the vacant space.
So, P.E. of BCC = $\dfrac{Z\times volume\,occupied\,by\,atom\,in\,unit\,cell}{Volume\,of\,unit\,cell}$ $\times$ 100
P.E. of BCC = $\dfrac{2\times {}^{4}/{}_{3}\pi {{r}^{3}}}{{{\left( \dfrac{4r}{\sqrt{3}} \right)}^{3}}}$ $\times$100
P.E. of BCC = 0.68 $\times$100
P.E. of BCC = 68%
So, vacant space = 100-68
Vacant space = 32 %
Hence, the vacant space in cubic close packed structure and in body centered packed structure are 26% and 32% respectively.

So, option B is correct.

Note: The volume of unit cell is calculated by $\dfrac{4}{3}\pi {{r}^{3}}$and the volume of the unit cell is calculated by $2\sqrt{2}r$ , and for BCC volume of unit cell by $\left( \dfrac{4r}{\sqrt{3}} \right)$, both of which are obtained by making a structural diagram of CCP and BCC.