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CCP or FCC has two lattice point corner as well as face centred:

Suppose ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube As there are 4 sphere in FCC unit cell

(1) Relation between radius (r) and side (a)
In FCC, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere AC = 4r
But from the right angled triangle ABC:
(2)  Effective no. of atoms per unit cell (Z):
(3) Volume of four spheres (atoms):
(4) Volume of unit cube:
(5) Packing efficiency (PE): Percentage of space occupied by sphere
(6) Percentage Voids: 100- PE= 26 %
(7) Density of FCC(CCP):
(8) Coordination Numbers:

(A) Tetrahedral voids: The FCC/CCP unit cell has eight tetrahedral voids per unit cell. Just below every corner of the unit cell, there is one tetrahedral void. As there are eight corners, there are eight tetrahedral voids.

(B) Octahedral voids: In an FCC/CCP unit cell, there are four octahedral voids. They are present at all the edge centres and at the body centre. The contribution of the edge centre is 1/4
Hence, total number of octahedral voids:
 Rank (Z) = 4,
 Number of tetrahedral voids = 8 and
 Number of tetrahedral voids = 2 × Z
Number of tetrahedral voids in close packing = 2 × eff. no. of spheres.
Hence, there are two Tetrahedral Voids per sphere in closed packing arrangements.
 Z = 4
Number of octahedral voids = 4 
Number of octahedral voids = Z
There is exactly one OV per sphere in close packing.

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