 Chemistry Close Packed Structure Packing Efficiency, Calculations Involving Unit Cell Dimensions
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### Topic to be covered

• Close Packed Structures :
• Formula of a Compound and Number of Voids Filled :
• Packing Efficiency
• Packing Efficiency in hcp and ccp Structures :
• Efficiency of Packing in Body-Centred Cubic Structures :
• Packing Efficiency in Simple Cubic Lattice :
• Calculations Involving Unit Cell Dimensions :

### 𝐂𝐥𝐨𝐬𝐞 𝐏𝐚𝐜𝐤𝐞𝐝 𝐒𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐞𝐬 : In solids constituent particles are closed-packed, leaving the minimum vacant space. Constituent particles are considered as identical hard spheres.

(𝐢) 𝐂𝐥𝐨𝐬𝐞 𝐏𝐚𝐜𝐤𝐢𝐧𝐠 𝐢𝐧 𝐎𝐧𝐞 𝐃𝐢𝐦𝐞𝐧𝐬𝐢𝐨𝐧 (𝟏 𝐃) : There is only one way of arranging spheres in one dimensional. We can arrange them in a row and touching each other (Fig 1.13).

The number of nearest neighbours of a particle is called its co-ordination number. Thus in 1-D close packed arrangement the co-ordination number is 2.
color{purple}(✓✓)color{purple} " DEFINITION ALERT"
The number of nearest neighbouring particles around a specific particle in a given crystalline substance is called as coordination number of that crystalline substance

(𝐢𝐢) 𝐂𝐥𝐨𝐬𝐞 𝐏𝐚𝐜𝐤𝐢𝐧𝐠 𝐢𝐧 𝐓𝐰𝐨 𝐃𝐢𝐦𝐞𝐧𝐬𝐢𝐨𝐧𝐬 (𝟐 𝐃) : Two dimensional close packed structure can be generated in two different ways.

(a) One row may be placed over other row in a way that the spheres of the one row are exactly above those of the other row and doing so we get A A A..... type of arrangement as shown in fig 1.14 (a).

=> In this each sphere is in contact with 4 of its neighbours.

Therefore, its co-ordination number is 4. It is also called text(square close packing in two dimensions).

(b) In this second row may be placed above the first one in a staggered manner such that its sphere fit in the depression of the first row. This is ABAB..... type arrangement.

In this arrangement there is less free space. Each sphere is in contact with 6 neighbouring spheres.

Therefore, its co-ordination number is 6. This is also called text(two dimensional hexagonal close-packing) (Fig 1.14 b).

There are some voids (empty spaces) in this arrangement. These are triangular in shape. These are of two types i.e. upward triangular voids and downwards triangular voids.

(𝐢𝐢𝐢) 𝐂𝐥𝐨𝐬𝐞 𝐏𝐚𝐜𝐤𝐢𝐧𝐠 𝐢𝐧 𝐓𝐡𝐫𝐞𝐞 𝐃𝐢𝐦𝐞𝐧𝐬𝐢𝐨𝐧𝐬 (𝟑𝐃) : These are obtained by stacking two dimensional layers one above the other. Types of 3-D close-packed structure obtained is :

(a) Three Dimensional Close Packing from Two Dimensional Square Closed Packed Layers : The second layer is placed over the first layer such that the spheres of the upper layer are exactly above those of the first layer (Fig 1.15). This is A A A...... type pattern. This is simple cubic lattice and its unit cell is the primitive cubic unit cell (Fig. 1.9).

(b) Three Dimensional Close Packing from Two Dimensional Hexagonal Close Packed Layers : This is done in following ways :

(A) Placing Second Layer over the First Layer : Second layer is placed above the first layer in a way that it covers the depressions of the first layers. It is observed that all triangular voids are not covered (Fig 1.16).

Wherever a sphere of the second layer is above the void of the first layer (or vice-versa), a void is formed called text(tetrahedral void) because a tetrahedron is formed when the centres of these four spheres are joined. It is marked as T in Fig 1.16. See Fig 1.17 also.

At some places the triangular voids in the second layer are above the triangular voids in the first layer and voids formed by this are called text(octahedral voids) because these voids are surrounded by six sphere. This is marked as O in Fig 1.16. See Fig 1.17 also.

The number of these voids depends upon the number of close packed spheres.

Let the number of close packed spheres be N, then :

The number of octahedral voids generated = N

The number of tetrahedral voids generated = 2N

(B) Placing Third Layer over the Second Layer : In this case, there are two possibilities.

(𝐈) 𝐂𝐨𝐯𝐞𝐫𝐢𝐧𝐠 𝐓𝐞𝐭𝐫𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : Tetrahedral voids of the second layer are covered by the spheres of third layer. So, the spheres of third layer are exactly aligned with those of the first layer. So, we get ABAB...... pattern. This structure is called text[hexagonal close packed (hcp)] structure (Fig 1.18). e.g. Zn and Mg.

(𝐈𝐈) 𝐂𝐨𝐯𝐞𝐫𝐢𝐧𝐠 𝐎𝐜𝐭𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : In this case, third layer is placed in such a way that its spheres cover the octahedral voids. In this way, the spheres of the third layer are not aligned with spheres of either first or third layer. So, ABCABC......... pattern is obtained. This structure is called text[cubic close packed (ccp)] or text[face-centred cubic (fcc)] structure. e.g Ag and Cu.

Note : (x) Both these types of packing are highly efficient.

(y) 74% space is filled.

(z) Each sphere is in contact with twelve spheres. So, co-ordination number is 12.

### Formula of a Compound and Number of Voids Filled :

In ionic solids the bigger ions (anions) from the close packed structure and the smaller ions (usually cations) occupy the tetrahedral or octahedral voids according to its size. All octahedral or tetrahedral voids are not occupied and the fraction of occupancy of tetrahedral and octahedral voids depends upon the chemical formula of the compound.
Q 2605780668 A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound? Solution:

The ccp lattice is formed by the element Y. The number of octahedral voids generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1 : 1 ratio. Therefore, the formula of the compound is XY.
Q 2615780669 Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B? Solution:

The number of tetrahedral voids formed is equal to twice the number of atoms of element B and only 2/3rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3) : 1 or 4 : 3 and the formula of the compound is A_4B_3.

### Locating Tetrahedral and Octahedral Voids

We know that close packed structures have both tetrahedral and octahedral voids. Let us take 𝐜𝐜𝐩 (or 𝐟𝐜𝐜) structure and locate these voids in it.

(𝐚) 𝐋𝐨𝐜𝐚𝐭𝐢𝐧𝐠 𝐓𝐞𝐭𝐫𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : Let us consider a unit cell of 𝐜𝐜𝐩 or 𝐟𝐜𝐜 lattice [Fig. 1(a)]. The unit cell is divided
into eight small cubes. Each small cube has atoms at alternate corners [Fig. 1(a)]. In all, each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron. Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total. Each of the eight small cubes have one void in one unit cell of 𝐜𝐜𝐩 structure. We know that 𝐜𝐜𝐩 structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms. (𝐛) 𝐋𝐨𝐜𝐚𝐭𝐢𝐧𝐠 𝐎𝐜𝐭𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : Let us again consider a unit cell of 𝐜𝐜𝐩 or 𝐟𝐜𝐜 lattice [Fig. 2(a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube.

Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. 2(b)]. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centre) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent

unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs
to a particular unit cell. Thus in cubic close packed structure:

Octahedral void at the body-centre of the cube = 1

12 octahedral voids located at each edge and shared between four unit cells  = 12 xx 1/4 = 3

therefore Total number of octahedral voids = 4

We know that in 𝐜𝐜𝐩 structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number.

### Packing Efficiency :

It is the percentage of total space filled by the particles. Let us calculate the packing efficiency in different types of structures.

### Packing Efficiency in hcp and ccp Structures : Both types of close packing (hcp and ccp) are equally efficient. Let us calculate the efficiency of packing in ccp structure. In Fig. 1.20, let the unit cell edge length be ‘a’ and face diagonal AC = b.

In Delta ABC,

AC^2 = b^2 = BC^2 +AB^2

 = a^2 + a^2 = 2a^2 or b = sqrt2a

If r is the radius of the sphere, we find

b = 4 r = sqrt2a or a = (4 r)/sqrt2 = 2 sqrt2 r (we can also write r = a/(2 sqrt2))

We know, that each unit cell in ccp structure has effectively 4 spheres. Total volume of four spheres is equal to 4xx (4/3) pir^3 and volume of the cube is a^3 or (2 sqrt2 r)^3

Therefore, text(Packing efficiency) = (text(Volume occupied by four spheres in the unit cell) xx 100) /text(Total volume of the unit cell) %

 = (4xx (4/3) pi r^3 xx 100)/(2 sqrt2r)^3 %

 = ((16/3) pi r^3 xx 100)/(16 sqrt2 r^3) % = 74%

### Efficiency of Packing in Body-Centred Cubic Structures : From Fig.1.21, it is clear that the atom at the centre will be in touch with the other two atoms diagonally arranged.

In Delta EFD,

b^2 = a^2 + a^2 = 2a^2

b = sqrt2a

Now in Delta AFD,

c^2 = a^2 + b^2 = a^2 + 2a^2 = 3a^2

c = sqrt3a

The length of the body diagonal c is equal to 4r, where r is the radius of the sphere (atom), as all the three spheres along the diagonal touch each other.

Therefore, sqrt3a = 4r

a = (4 r)/sqrt3

Also, we can write r = sqrt3/4 a

In this type of structure, total number of atoms is 2 and their volume is 2xx (4/3) pi r^3

Volume of the cube, a^3 will be equal to (4/sqrt3 r)^3 or a^3 = (4/ sqrt3 r)^3

Therefore, text (Packing efficiency ) = (text(Volume occupied by two spheres in the unit cell) xx 100)/text(Total volume of the unit cell) %

 = (2xx (4/3) pir^3 xx100)/([4/sqrt3 r]^3) %

 = ((8/3) pi r^3 xx 100)/(64//(3 sqrt3)r^3) % = 68 %

### Packing Efficiency in Simple Cubic Lattice : In a simple cubic lattice the atoms are located only on the corners of the cube. The particles touch each other along the edge (Fig. 1.22). Thus, the edge length or side of the cube ‘a’, and the radius of each particle, r are related as

a = 2r

The volume of the cubic unit cell  = a^3 = (2r)^3 = 8r^3

Since a simple cubic unit cell contains only 1 atom,

The volume of the occupied space  = 4/3 pi r^3

therefore text(Packing efficiency) = text(Volume of one atom)/text(Volume of cubic unit cell) xx100%

(4/3 pi r^3)/(8 r^3) xx 100 = pi/6 xx 100

 = 52.36 % = 52.4 %

Thus, we may conclude that ccp and hcp structures have maximum packing efficiency.

### Calculations Involving Unit Cell Dimensions :

Unit cell dimensions can be used to calculate the volume of the unit cell.

And if density is known, we can calculate the mass of the atoms in the unit cell. And knowing the mass of a single atom, Avogadro constant can be determined.

Let a be the edge length of the unit cell of a cubic crystal determined by X-ray diffraction.

d be the density of the solid.

M be the molar mass.

In case of cubic crystal :

Volume of a unit cell = a^3

Mass of the unit cell = number of atoms in unit cell × mass of each atom = z × m

(Here z is the number of atoms present in one unit cell and m is the mass of a single atom.)

Mass of an atom present in the unit cell :

m = M/N_A \ \ \ \ (M text(is molar mass))

Therefore, density of the unit cell = text(mass of unit cell)/text(volume of unit cell)

= ( z * m)/a^3 = ( z * M)/(a^3 * N_A)  or d = (z M)/(a^3 N_A)

Note : The density of the unit cell is the same as the density of the substance.
Q 2635880762 An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g//cm^3. How many atoms are present in 208 g of the element? Solution:

Volume of the unit cell  = (288 text(pm) )^3

 = (288xx10^(-12) m) = (288xx10^(-10) cm)^3

 = 2.39xx10^(-23) cm^3

Volume of 208 g of the element

 = text(mass)/text(density) = (208 g)/(7.2 g cm^(-3)) = 28.88 cm^(3)

Number of unit cells in this volume

 = (28.88 cm^3)/((2.39xx10^(-23) cm^3)/text( unit cell)) = 12.08xx10^(23)  unit cells

Since each bcc cubic unit cell contains 2 atoms, therefore, the total number
of atoms in 208 g = 2 text((atoms/unit cell)) × 12.08 × 10^(23) unit cells
= 24.16×10^(23) atoms
Q 2645880763 X-ray diffraction studies show that copper crystallises in an fcc unit cell with cell edge of 3.608×10^(-8) cm. In a separate experiment, copper is determined to have a density of 8.92 g//cm^3, calculate the atomic mass of copper. Solution:

In case of fcc lattice, number of atoms per unit cell, z = 4 atoms

Therefore, M = ( d xx N_A xxa^3)/z

 = (8.92 g cm^3 xx 6.022xx10^(23) text(atom) mol^(-1) xx (3.608xx10^(-8) cm)^3)/text(4 atoms)

 = 63.1 g//mol

Atomic mass of copper = 63.1u
Q 2655880764 Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u). Solution:

Since the lattice is ccp, the number of silver atoms per unit cell  = z = 4

Molar mass of silver  = 107.9 g mol^(-1) = 107.9 xx 10^(-3) kg mol^(-1)

Edge length of unit cell  = a = 408.6 pm = 408.6xx10^(-12) m

Density d = ( z * M)/(a^3 * N_A)

 = (4xx (107.9xx10^9-3) kg mol^(-1))/((408.6xx10^(-12) m)^3 (6.022xx10^(23) mol^(-1))) = 10.5xx10^3 kg m^(-3)

 = 10.5 g cm^(-3) 