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Close packing in crystals

In order to understand the packing of the constituent particles in a crystal, it is assumed that these particles are hard spheres of identical size (eg those of metal). The packing of these spheres takes place in such a way that they occupy the maximum available space and hence the crystal has maximum density. This type of packing is called close packing.

One Dimensional Packing

Spheres are arranged in a single straight row.

There is only one way of arranging spheres in a one dimensional close packed structure, that is to arrange them in a row and touching each.

In this arrangement, each sphere is in contact with two of its neighbours.

The number of nearest neighbours of a particle is called its coordination number.

Thus, in one dimensional close packed arrangement, the coordination number is 2.

Two Dimensional Packing

When the rows are combined touching each other, the crystal plane is obtained. The rows can be combined in two different ways

When the particles when placed in the adjacent rows, show a horizontal as well as vertical alignment and form squares. This type of packing is called square close packing (a)

When the particles in every next row are placed in the depressions between the particles of the first row. The particles in the third row will be vertically aligned with those in the first row. This type of packing gives a hexagonal pattern and is called hexagonal close packing (b).

Coordination number of square close packing is 4.

Coordination number of  hexagonal close packing is 6.

Three Dimensional Packing

In two dimensional packing, a more efficient packing is given by hexagonal close packing. In order to develop three dimensional close packing let us retain the hexagonal close packing in the first layer.

If the spheres in the second layer are just placed over the spheres in the first layer so that the spheres in the two layers are vertically aligned, its voids will come above the voids in the first layer. This is an inefficient way of filling the space and results in the formation of simple cubic lattice .

When the second layer is placed in such a way that its spheres find place in the ‘b voids of the first layer, the ‘c’ voids will be left unoccupied. Since under this arrangement no sphere can be placed in them, (c voids), i.e. only half the triangular voids in the first layer are occupied by spheres in the second layer (i.e. either b or c)

There are two alternative ways in which spheres in the third layer can be arranged over the second layer

When a third layer is placed over the second layer in such a way that the spheres cover the tetrahedral or ‘a’ voids; a three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically aligned (i.e. the third layer is directly above the first, the fourth above the second layer and so on) calling the first layer A and second layer as layer B, the arrangement is called ABAB …………. pattern or hexagonal close packing (hcp) as it has hexagonal symmetry.

Refer to the following video for hexagonal close packing

When a third layer is placed over the second layer in such a way that spheres cover the octahedral or ‘c’ voids, a layer different from layers A and B is produced. Let it be layer ‘C’. Continuing further a packing is obtained where the spheres in every fourth layer will be vertically aligned to the spheres present in the first layer. This pattern of stacking spheres is called ABCABC ……….. pattern or cubic close packing (ccp). It is similar to face centred cubic (fcc) packing as it has cubic symmetry.

Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed

When the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap an octahedral void is formed.

If the number of close packed spheres be N, then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N

In both hcp and ccp methods of stacking, a sphere is in contact with 6 other spheres in its own layer. It directly touches 3 spheres in the layer above and three spheres in the layer below. Thus sphere has 12 close neighbours.

The number of nearest neighbours in a packing is called coordination number. In close packing arrangement (hcp & ccp) each sphere has a coordination number of 12.

Interstitial Voids

In hcp as well as ccp only 74% of the available space is occupied by spheres.

The remaining space is vacant and constitutes interstitial voids or interstices or holes.

These are of two types: (a) Tetrahedral voids and (b) Octahedral voids

Tetrahedral voids

In close packing arrangement, each sphere in the second layer rests on the hollow (triangular void) in three touching spheres in the first layer.

The centres of theses four spheres are at the corners of a regular tetrahedral.

The vacant space between these four touching spheres is called tetrahedral void.

In a close packing, the number of tetrahedral void is double the number of spheres, so there are two tetrahedral voids for each sphere

Radius of the tetrahedral void relative to the radius of the sphere is 0.225

i.e

. 

In a multi layered close packed structure, there is a tetrahedral hole above and below each atom hence there is twice as many tetrahedral holes as there are in close packed atoms

Octahedral void

As already discussed the spheres in the second layer rest on the triangular voids in the first layer. However, one half of the triangular voids in the first layer are occupied by spheres in the second layer while the other half remains unoccupied.

The triangular voids ‘b’ in the first layer is overlapped by the triangular voids in the second layer.

The interstitial void, formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres centres of which occupy corners of a regular octahedron

In close packing, the number of octahedral voids is equal to the number of spheres. Thus, there is only one octahedral void associated with each sphere.

Radius of the octahedral void in relation to the radius of the sphere is 0.414 i.e.

Sizes of tetrahedral and octahedral voids

Derivation of the relationship between the radius (r) of the octahedral void and the radius (R) of the atoms in close packing. :-

A sphere into the octahedral void is shown in the diagram. A sphere above and a sphere below this small sphere have not been shown in the figure. ABC is a right angled triangle. The centre of void is A.

Applying Pythagoras theorem.

BC2 = AB2 + AC2

(2R)2 + (R + r)2 + (R + r)2 = 2(R + r)2

 4R2/2 = (R + r)2

 √2R = R +  r

 (√2) = (R + r)2

 r = √2R – R = (1.414 –1)R

 r = 0.414 R

Derivation of the relationship between radius (r) of the tetrahedral void and the radius (R) of the atoms in close packing:

To simplify calculations, a tetrahedral void may be represented in a cube as shown in the figure. In

which there spheres form the triangular base, the fourth lies at the top and the sphere occupies the tetrahedral void.

Let the length of the side of the cube = a

From right angled triangle ABC, face diagonal

AB = √AC2 + BC2 = √a2 + a2 = √2a

As spheres A and B are actually touching each other, face diagonal AB = 2R

∴ 2R = √2a or

R = 1/√2 a        ....(i)

Again from the right angled triangle ABD

AD √AB2 + BD2 = √(√2a)2 + a2 = √3a

But as small sphere (void) touches other spheres, evidently body diagonal AD = 2(R + r).

∴ 2(R + r) = √3a

 R + r = √3/2 a                ...(ii)

Dividing equation (ii) by equation (i)

R + r/R = √3 / 2 × a / a / √2 = √3 / √2

 1 + r/R = √3 / √2 = 1.225 => r/R = 1.225 – 1 = 0.225

 r = 0.225 R

Radius ratio in 1:1 or ab type structure

Structural Arrangement

Coordination number

Example

0.225 – 0.414

Tetrahedral

4

CuCl, CuBr, CuI, BaS, HgS

0.414 – 0.732

Octahedron

6

MgO, NaBr, CaS, MnO, KBr, CaO

0.732 – 1

Cubic

8

CsI, CsBr, TlBr, NH4Br

Solved Examples

Question

The two ions A+ and B- have radii 88 and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination number of A+.

Solution:

r+ / r– = 88/200 = 0.44

It lies in the range of 0.414 – 0.732

Hence, the coordination number of A+ = 6

__________________________________________

Question.

Br- ion forms a close packed structure. If the radius of Br- ions is 195 pm. Calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation A+ having a radius of 82 pm be slipped into the octahedral hole of the crystal A+Br-?

Solution:

Radius of the cations just filling into the tetrahedral hole  = Radius of the tetrahedral hole = 0.225´195  = 43.875 pm

For cation A+ with radius = 82 pm

Radius ratio r+ / r– = 82/195 = 0.4205

As it lies in the range 0.414 – 0.732, hence the cation A+ can be slipped into the octahedral hole of the crystal A+Br-.

_____________________________________________

Question 3:

Solution:

Maximum radius ratio in ionic crystals lies in the range 0.732 – 1 which corresponds to a coordination number of 8. Hence coordination number greater than 8 is not possible in ionic crystals.

Question 1: in one dimensional close packed arrangement, the coordination number is..

a. 1

b. 2

c. 3

d. 4

Question 2: If the number of close packed spheres be N, then:
The number of octahedral voids generated =

a. N/2

b. N

c. 2N

d. N2

Question 3: Coordination number of  atom in CuCl crystal is

a. 8

b. 6

c. 4

d. 3

Question 4:  If R is the radius of atom, then radius of tetrahedral void would be..

a. 0.225 R

b. 0.069 R

c. 1.864 R

d. 5.00 R

Question 5: Cubic close packing (ccp) has

a. ABABAB type pattern

b. AAAAA type pattern

c. AACABC type pattern

d. AABAAB type pattern

Q.1
Q.2
Q.3
Q.4
Q.5

a
b
c
a
c

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