QJEEAdv16Chbb5d: JEE Advanced 2016 Chemistry - Class 12 JEEAdv16Chbb5d Solution: The CORRECT statement(s) for cubic close packed (ccp) three…

Coordination number cannot be 12, for any atom in the topmost layer, as there is no layer above it. Thus, each atom is in contact with six atoms in the same layer and three atoms from the layer below it.

For cubic close packing, we have

Packing fraction = ${{Volume\,of\,four\,spheres\,in\,the\,unit\,cell} \over {Total\,volume\,of\,the\,unit\,cell}}$

$= 4 \times {{(4/3)\pi {r^3}} \over {16\sqrt 2 {r^3}}} = {\pi \over {3\sqrt 2 }} = 0.74 = 74\%$

In fcc unit cell, the effective number of atoms is

One atom at each corner = 8 corner atoms $\times {1 \over 8} = 1$

Atoms at each of the six face centres :

6 face centered atoms $\times {1 \over 2} = 3$ .

Number of octahedral voids = 4.

Number of tetrahedral voids = 8.

Therefore, per atom, there is one octahedral void and two tetrahedral voids.

In fcc (or ccp), the unit edge length is given by

$\sqrt 2 a = 4r \Rightarrow a = {{4r} \over {\sqrt 2 }} = 2\sqrt 2 r$