PACKING FRACTIONS
Both of the above patterns of packing (i.e. hcp & ccp) though different in form are equally efficient. They occupy the maximum possible space which is about 74% of the available volume. Hence they are called closest packing.
In addition to the above two types of arrangements a third type of arrangement found in metals is body centred cubic (bcc) in which space occupied is about 68%.
CALCULATION OF THE SPACE OCCUPIED
In simple cubic unit cell
Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.
As sphere are touching each other
Therefore a = 2r
No. of spheres per unit cell = 1/8 × 8 = 1
Volume of the sphere = 4/3 πr^{3}


Volume of the cube = a^{3}= (2r)^{3} = 8r^{3}
∴ Fraction of the space occupied = 1/3πr^{3} / 8r^{3} = 0.524
∴ % occupied = 52.4 %
In face centred unit cell
Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube
As there are 4 sphere in fcc unit cell
∴ Volume of four spheres = 4 (4/3 πr^{3})
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


But from the right angled triangle ACD
AD = √AC^{2} + DC^{2} = √a^{2} + a^{2}= √2a
√2a = 4r or a = 4/√2 r
∴ volume of cube = (4/√2 r)^{3} = 64 / 2√2 r^{3}
percentage of space occupied by sphere
= volume of sphere / volume of cube × 100 = 16/3 πr^{3} / 64 /2√2 r^{3} × 100 = 74%
In body centred cubic unit cell
Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube
As the sphere at the centre touches the sphere at the corner. Therefore body diagonal
AD = 4r
Face diagonal AC = √AB^{2} + BC^{2} = √a^{2} + a^{2} = √2a
In right angled triangle ACD =
AD = √AC^{2} + CD^{2} = √2a^{2} + a^{2} = √3a
∴ Volume of the unit cell = a^{3} = (4r / √3)^{3} = 64r3 / 3√3
No. of spheres in bcc = 2
∴ volume of 2 spheres = 2 × 4/3πr^{3}
∴ percentage of space occupied by spheres
= volume of sphere / volume of cube × 100 = 8/3 πr^{3} × 100 / 64r^{3} / 3√3 = 8/3 × 22/7 × 3√3/64 = 68%
Illustration 10. Show by simple calculation that the percentage of space occupied by spheres in hexagonal cubic packing (hcp) is 74%
Solution: Let the edge of hexagonal base =a
And the height of hexagon = h
And radius of sphere = r
The centre sphere of the first layer lies exactly over the void of 2^{nd} layer B.
The centre sphere and the spheres of 2^{nd} layer B are in touch
So, In Δ PQR (an equilateral triangle)
PR = 2r, Draw QS tangent at points
∴ In Δ QRS ∠QRS = 30°, SR = r
Cos30° = SR/QR
QR = r / √3/2 = 2r / √3
∴ PQ = √PR^{2 }– QR^{2} = √4r^{2} – 4r^{2} / 3
h^{1} = √8r^{2} / 3 = 2 √2/3 r
∴ h = 2h^{1} = 4 √2/3 r
Now, volume of hexagon = area of base x height
= 6 × √3 / 4 a^{2} × h => 6 × √3/4 (2r)^{2} × 4 √2/3 r
[∴ Area of hexagonal can be divided into six equilateral triangle with side 2r)
No. of sphere in hcp = 12 × 1/6 + 1/2 × 2 + 3
= 2+1+3 = 6
∴ Volume of spheres = 6 × 4/3πr^{3}
∴ Percentage of space occupied by sphere = 6 × 4/3 πr^{3} / 6 × √3/4 × 4r^{2} × 4√2/3 r × 100 = 74%
Illustration 11. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10^{24}cm^{3} and density of the element is 7.20gm/cm^{3}, calculate no. of atoms present in 200gm of the element.
Solution: Number of atoms contributed in one unit cell
= one atom from the eight corners + one atom from the two face diagonals
= 1+1 = 2 atoms
Mass of one unit vell = volume × its density
= 24 × 10^{–24} cm^{3} × 7.2 gm cm^{3}
= 172.8 × 10^{–24} gm
∴ 172.8 10^{–24} gm is the mass of one – unit cell i.e., 2 atoms
∴ 200 gm is the mass = 2 × 200 / 172.8 × 10^{–24} atoms = 2.3148 × 10^{24} atoms
Illustration 12. Calculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed
by A
Solution: Total volume of A atom = 6 × 4 / 3 πr_{A}^{3}
Total volume of B atoms = 4 × 4/3 πr_{A}^{3} 4 × 4/3 π(0.414r_{A})^{3}
Since r_{B}/r_{A} as B is in octahedral void of A
Volume of HCP = 24√2r_{A}^{3}
Packing fraction = 6 × 4/3 πr_{A}^{3} + 4 × 4/3 π (0.414r_{A})^{3} / 24√2r_{A}^{3} = 0.7756
Void fraction = 10.7756 = 0.2244
Illustration 13. Lithium iodide crystal has a facecentred cubic unit cell. If the edge length of the unit cell is 620 pm, determine the ionic radius of I^{} ion.
Solution: As LiI has fcc arrangement, I^{} ions will occupy the corners and facecentres. These ions will touch each other along the face diagonal.
∴ Face diagonal AB = 4r^{–}_{1} = √a^{2 }+ a^{2} = √2a
∴ r_{1} = √2a / 4 = 1.414 × 620 pm / 4 = 219.17 pm
Illustration 14. When heated above 916°C, iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is:
(A) 1.089 (B) 0.918
(C) 0.725 (D) 1.231
Solution: (B) d_{bcc} / d_{ccp} = packing efficiency of bcc / packing efficiency of ccp = 67.92/74.02 = 0.918
Illustration 15. Select the correct statements:
(A) For CsCl unit cell (edgelength = a), r_{c} + r_{a} = √3/2 a
For NaCl unit cell (edgelength =), r_{c} + r_{a} = l/2
The void space in a b.c.c. unit cell is 0.68
The void space % in a facecentered unit cell is 26%
Solution: (A), (B), (C), (D). In bcc structure are r_{+} + r_{–} = √3/2 a
Hence, for CsCl, r_{C} + r_{a} = √3/2 a
∴ (A) is correct
Since, NaCl crystalise in fcc structure
∴ 2r_{C} + 2r_{a} = edge length of the unit cell
Hence, r_{C} + r_{a} = l/2
∴ (B) is correct
Since packing fraction of a bcc unit cell is 0.68
∴ void space = 1–0.68 = 0.32
∴ (C) is incorrect
In fcc unit cell PF = 74%
∴ VF = 100–74 = 26%
∴ (D) is correct
Exercise 4.
(a) What is the empty fraction in case of S.C., B.C.C and F.C.C?
(b) Which is most packed structure S.C., B.C.C or F.C.C?