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 πr3 

 

1965_cubic unit cell.JPG
 

Volume of the cube = a3= (2r)3 = 8r3

 

 Fraction of the space occupied = 1/3πr3 / 8r3 = 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 πr3) 

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

AD = 4r

1129_face centred unit cell.JPG 

 

 

But from the right angled triangle ACD

 

AD = √AC2 + DC2 = √a2 + a2= √2a

 

√2a = 4r or a = 4/√2 r

 

 volume of cube = (4/√2 r)3 = 64 / 2√2 r3 

 

percentage of space occupied by sphere

 

= volume of sphere / volume of cube × 100 = 16/3 πr3 / 64 /2√2 r3  × 100 = 74%

 

In body centred cubic unit cell

 

          1457_body centred cubic unit cell.JPG

 

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 = √AB2 + BC2 = √a2 + a2 = √2a

 

In right angled triangle ACD =

 

AD = √AC2 + CD2 = √2a2 + a2 = √3a

 

 

 √3a = 4r

 

a = 4r  / √3

 

 Volume of the unit cell = a3 = (4r / √3)3 = 64r3 / 3√3

 

No. of spheres in bcc = 2

 

 volume of 2 spheres = 2 × 4/3πr3

 

 percentage of space occupied by spheres

 

= volume of sphere / volume of cube × 100 = 8/3 πr3 × 100 / 64r3 / 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%

 

                              1252_hexagonal cubic packing.JPG

 

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 2nd layer B.

 

The centre sphere and the spheres of 2nd 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 = √PR2 – QR2 = √4r2 – 4r2 / 3

 

h1 = √8r2 / 3 = 2 √2/3 r

 

 

 h = 2h1 = 4 √2/3 r

 

Now, volume of hexagon = area of base x height

 

= 6 × √3 / 4 a2 × 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 × 4/3πr3

 

 Percentage of space occupied by sphere = 6 × 4/3 πr3 / 6 × √3/4 × 4r2 × 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 10-24cm3 and density of the element is 7.20gm/cm3, 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 cm3 × 7.2 gm cm3

 

                        = 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 × 1024 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 πrA3

 

                        Total volume of B atoms = 4 × 4/3 πrA3  4 × 4/3 π(0.414rA)3

 

                        Since rB/rA as B is in octahedral void of A

 

                        Volume of HCP = 24√2rA3

 

                        Packing fraction = 6 × 4/3 πrA3  + 4 × 4/3 π (0.414rA)3 / 24√2rA3 = 0.7756

 

                        Void fraction = 1-0.7756 = 0.2244          

 

Illustration 13. Lithium iodide crystal has a face-centred 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 face-centres. These ions will touch each other along the face diagonal.

 

                                           1865_face diagonal.JPG

 

                           Face diagonal AB = 4r1 = √a2 + a2 = √2a

 

                          r1 = √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) dbcc / dccp = 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 (edge-length = a), rc + ra = √3/2 a

 

For NaCl unit cell (edge-length =), rc + ra = l/2

 

The void space in a b.c.c. unit cell is 0.68

 

The void space % in a face-centered unit cell is 26%

 

Solution:          (A), (B), (C), (D). In bcc structure are r+ + r = √3/2 a

 

                        Hence, for CsCl, rC + ra = √3/2 a

 

                         (A) is correct

 

                        Since, NaCl crystalise in fcc structure

 

                         2rC + 2ra = edge length of the unit cell

 

                        Hence, rC + ra = 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?

Related Resources
Answers to Assignment Problems

ANSWERS TO ASSIGNMENT PROBLEMS Subjective: Level -...

Simple Cubic Lattice

SIMPLE CUBIC LATTICE There are eight atoms at the...

Interstitial Voids

INTERSTITIAL VOIDS In hcp as well as ccp only 74%...

Space Lattice or Crystal Lattice

SPACE LATTICE OR CRYSTAL LATTICE All crystals...

Locating Tetrahedral and Octahedral Voids

LOCATING TETRAHEDRAL AND OCTAHEDRAL VOIDS The...

Objective Questions Part I

Objective Questions Part I Objective: Prob 1. If...

Answers to Miscellaneous Exercises Part I

ANSWER TO MISCELL ANEOUS EXERCISES Exercise 1: One...

Structure of Ionic Compounds

STRUCTURE OF IONIC COMPOUNDS Simple ionic...

Imperfections in Solids and defects in Crystals

IMPERFECTIONS IN SOLIDS: DEFECTS IN CRYSTALS...

Sizes of Tetrahedral and Octahedral Voids

SIZES OF TETRAHEDRAL AND OCTAHEDRAL VOIDS (i)...

Structure Determination by X Rays

STRUCTURE DETERMINATION BY X – RAYS The...

Answers to Questions Part I

ANSWERS TO EXERCISES Exercise 1: One Exercise 2:...

Assignment Problems

ASSIGNMENT PROBLEMS Subjective: Level – 0 1....

Solved Problems

SOLVED PROBLEMS Subjective: Board Type Questions...

Classification of Solids

INTRODUCTION A solid is defined as that form of...

Objective Questions Level I

Objective: Level – I 1. If the coordination...

Non Stoichiometric Defects

NON – STOICHIOMETRIC DEFECTS If as a result...

Consequences of Schottky Defects

CONSEQUENCES OF SCHOTTKY DEFECT (a) As the number...

Calculation Involving Unit Cell Dimensions

CALCULATION INVOLVING UNIT CELL DIMENSIONS From...

Effect of Temperature and Pressure on Crystal Structure

Effect of temperature on crystal structure...

Miscellaneous Exercises Part I

MISCELLANEOUS EXERCISES Exercise 1: How many atoms...

Questions Level I

Level - I 1. In a face centred lattice of X and Y,...

Properties of Solids

PROPERTIES OF SOLIDS The three main properties of...

Bravais Lattices

BRAVAIS LATTICES The French crystallographer...

Ionic Compound of Type AB2

IONIC COMPOUND OF THE TYPE AB 2 Calcium fluoride...

Close Packing in Crystals

CLOSE PACKING IN CRYSTALS In order to understand...