Question

The correct statement(s) for cubic close packed (ccp) three dimensional structure is(are):

• A. The number of the nearest neighbours of an atom present in the topmost layer is $12$
• B. The efficiency of atom packing is $74\%$
• C. The number of octahedral and tetrahedral voids per atom are $1$ and $2$, respectively
• D. The unit cell edge length is $2\sqrt{2}$ times the radius of the atom

Answer 1

Answer: (B), (C), (D)

The correct options are
B The efficiency of atom packing is $74\%$
C The number of octahedral and tetrahedral voids per atom are $1$ and $2$, respectively
D The unit cell edge length is $2\sqrt{2}$ times the radius of the atom

(B) the efficiency of atom packing is 74 $\%$

(C) the number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

(D) the unit cell edge length is $2\sqrt{2}$ times the radius of the atom

Solution:-

In the middle layer (B) the atom present at the center has 6 neighbouring atom in the same layer (B), 3 neighbouring atom in the top layer and 3 neighbouring atom in the bottom layer. Total 12 neighbouring atoms. The top-most layer will have 9 nearest neighbours.

Packing efficiency = $\frac{\text{Volume occupied by 4 spheres in the unit cell}\times 100}{\text{Total volume of unit cell}}\%$

Volume of 4 spheres = $4\times \left(\frac{4}{3}\right)\pi r^3$

Volume of unit cell = $a^3={\left(2\sqrt{2}r\right)}^3$

$\therefore$ Packing efficiency = $\frac{4\times \left(\frac{4}{3}\right)\pi r^3\times 100}{{\left(2\sqrt{2}r\right)}^3}\%=74\%$

In CCP,

Number of octahedral voids = 1 $\times$ no. of atom

Number of tetrahedral voids = 2 $\times$ no. of atom

Hence, the number of octahedral and tetrahedral voids per atom are 1 and 2, respectively.

For edge length-

$4r=\sqrt{2}a$, where ‘a’ is edge length of unit cell and ‘r’ is radius of atom.

$\Rightarrow a=2\sqrt{2}r$