JEE Advance - Chemistry (2016 - Paper 2 Offline - No. 4)

The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is (are) :
The number of the nearest neighbours of an atom present in the topmost layer is 12
The efficiency of an atom packing is 74%
The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
The unit cell edge length is $$2 \sqrt 2$$ times the radius of the atom

Explanation

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

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

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