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Solid packing of crystals can be described by spherical packing where atoms, ions, and molecules have their centres on the lattice points in a crystal structure.
To explain the bonding and formations of metallic crystals, atoms are considered to be spherical. These spherical particles can be arranged in a variety of ways in packing solids. The arrangement of the spheres is tightly packed in structures in order to use the maximum amount of space possible.
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Key takeaways - Solid packing, Close packing, Hexagonal close packing, Square close packing, One dimension & two dimensions, Crystals, Lattice, Solid, Atom, Coordination number
Close Packing in One Dimension
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There aren't many ways of arranging spheres in a one-dimensional close-packed structure, other than to arrange them in a row so they touch each other.
In one dimensional close packing arrangement, every sphere is in contact with its two neighbours. The number of nearest neighbours of the particle is known as the coordination number. The coordination number is two in one dimensional close-packed arrangement.
Packing in One Dimension
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| Related Articles | ||
|---|---|---|
| Amorphous and Crystalline Solids | Imperfections in Solids | Magnetic Properties |
| Impurity Defects | Covalent Bond | Schottky Defect |
Close Packing in Two Dimension
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A two-dimensional close-packed structure can be made by placing the rows of close packed spheres side by side horizontally. This can be achieved in the following two ways -
Solid Packing: Square Close-Packing
In square close packing arrangement, every sphere is in contact with four of its neighbours, so the two-dimensional coordination number is 4. This particular type of packing in crystalline solids is known as square close packing in two dimensions. A square is formed when the centre of these four immediate neighbouring spheres are joined together. Hence, this type of packing is known as the square close packing in two dimensions and it occupies 52.4% of the space available. This arrangement is of AAAA type.
Square Close Packing in Two Dimensions
Solid Packing: Hexagonal Close-Packing
In hexagonal close packing arrangement, the free space is less and packing is more efficient than square close packing. The second row is placed on top of the first row in such that it's spheres fit in the depressions of the First row. So the two-dimensional coordination number is 6.We observe that if the centres of the six immediate neighbouring spheres are joined, a hexagon is formed. This type of packing in solids is known as hexagonal close packing in two dimensions.The centres of these six spheres are at the corners of a regular hexagon.It occupies 60.4 % of available space. This arrangement is of BABA type.
Hexagonal Close Packing
Close Packing in Three Dimension
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Layers of closed packed spheres in two dimensions, one on top of the other, can be used to create a three-dimensional packed structure.
This can be achieved in the following ways -
Three-Dimensional Close Packing from Two-Dimensional Square Layer
In this three dimensional close packing from two dimensional square layer arrangement, both the layers are placed together horizontally as well as vertically. The second layer is positioned over the first layer such that the spheres of the upper layer are exactly above those of the first one. So the two-dimensional coordination number is 6.
Three-dimensional close packing from two-dimensional square layer
Three-Dimensional Close Packing from Two-Dimensional Hexagonal Close-Packed Layer
In this arrangement, the structure is generated by placing layers one over the other.
A second layer similar to the bottom layer is positioned in such a way that the spheres of the second layer are positioned in the depressions of the first layer. in this type of close packing. If the first layer is referred to as 'A,' the second layer might be referred to as 'B,' since the spheres of the two levels are positioned differently. A tetrahedral void is generated if a sphere from the second layer is placed over the first layer's vacuum (or vice versa). At other times, the triangular voids in the second layer lie above the triangular voids in the first layer, thus their triangular forms do not overlap.
Things to Remember
- In a crystal lattice, particles such as ions, atoms, and molecules are filled with lattice points.
- The maximum possible crystal density is attained because of this arrangement.
- Crystal’s stability is related to close packing.
- When there is a maximum closed-packed structure, the minimum space that is unoccupied is left.
- To attain the highest closed-packed structure, the particles attempt to pack together as tightly as possible.
Also Read:
Sample Questions
Ques. Copper has the fcc crystal structure. Assuming an atomic radius of 130pm for copper atom (Cu = 63.54):
(a) What is the length of the unit cell of Cu?
(b) What is the volume of the unit cell?
(c) How many atoms belong to the unit cell?
(d) Find the density of Cu. [4 marks]
Ans - As we know
p = n × Mm / NA × a3
(a) for fcc structure
4r = √2 a
a = 2√2r =2√2 × 130 pm = 367.64 pm
(b) volume of unit cell = a3 = (367.64 × 10–10 cm)3 = 4.968 × 10–23 cm3
(c) n = 4
(d) p = 4 × 63.54 / 6.023 × 1023 × (3.67 × 10–8 cm3)3 = 8.54 gm / cm3
Ques- A compound is formed by atoms of elements A occupying the corners of the unit cell and an atom of element B present at the centre of the unit cell. Deduce the formula of the compound. [2 marks]
Ans - The description is of a BCC. For BCC formation, each atom at the corner is shared by 8 unit cells. The one atom at the centre completely belongs to the corresponding unit cell.
Therefore, total number of atoms of A present=18x8 =1
Total number of atoms of B present=1
so, A:B=1:1 stating the formula of the compound is AB.
Ques- The density of CaO is 3.35 gm/cm3. The oxide crystallises in one of the cubic systems with an edge length of 4.80 Å. How many Ca++ ions and O–2 ions belong to each unit cell, and which type of cubic system is present? [3 marks]
Ans - From equation
r(density) = 3.35 gm/cm3
a = 4.80 Å
Mm of CaO = (40 + 16) gm = 56 gm CaO
Q r = where n = no. of molecules per unit cell
p = n × Mm / a3 × NA
∴ n = 3.35 × (4.8 ×10–8)3 × 6.023 × 1023 / 56 = 3.98
or n ≈ 4
So, 4-molecules of CaO are present in 1 unit cell
So, no. of Ca+ + ion = 4
No. of O– – ion = 4
So, the cubic system is fcc type.
Ques - A metal crystallises into two cubic system-face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0Å respectively. Calculate the ratio of densities of fcc and bcc. [3 marks]
Ans - fcc unit cell length = 3.5Å
bcc unit cell length = 3.0Å
Density in fcc = n1 × atomic weight / V1 × Avogadro number
Density in bcc = n2 × atomic weight / V2 × Avogadro number
Dfcc / Dbcc = n1 / n2 × V2 / V1
n1 for fcc = 4; Also V1 = a3 = (3.5 × 10–8)3
n2 for bcc = 2; Also V2 = a3 = (3.0 × 10–8)3
= 1.259
Ques- Are CCP Cubic Close Packed and FCC Face Centred Cube the Same thing? [2 marks]
Ans - NO, CCP and FCC are different things. Both define the structure of a compound.
FCC is a form of lattice that does not have definable properties like volume or density and consists of only points. Whereas, CCP is a form of crystal structure made of edges and planes and its properties like volume, mass, density and length can be determined.
Ques- Distinguish between Hexagonal close packing and Cubic close packing. [3 marks]
Ans - The difference are:
| Hexagonal close packing | Cubic close packing |
|---|---|
| The spheres can be arranged so that they fit into the depression in such a way that the 3rd layer is placed directly over the first layer | The 3rd layer may be placed over the 2nd layer in such a way that all the spheres of the 3rd layer fit in octahedral voids |
| The tetrahedral voids of the 2nd layer are covered by the spheres of the 3rd layer | The octahedral voids of the 2nd layer are covered by the spheres of the 3rd layer |
| Hexagonal close packed (hcp) arrangement is also known as abab arrangement | Cubic close packed (ccp) arrangement is also known as abcabcabc arrangement |
Ques - What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure? [2 marks]
Ans - The number of adjacent or nearest neighbours that surround a particle in a crystal is called the coordination number of that particle.
In the bcc structure formation the entire corner atom touches the atom that occupies the body centre. Therefore, every atom is surrounded by 8 of its adjacent neighbours and the coordination number is 8.
Ques- Calculate the number of atoms in a fcc unit cell. [3 marks]
Ans - Through the following method we can calculate the atomic number in a fcc unit cell:
- Identical atoms present at every corner as well as in the centre of every face.
- The atom in the centre of the face is shared by two unit cells as shown in image below.
- Every corner atom is shared by 8 unit cells and makes 1/8 contribution to the unit cell.
Number of atoms in a fcc unit cell = (Nc / 8) + (Nf / 2)
= (8 / 8) + (6 / 2)
= 1+3 = 4
Ques - Differentiate crystalline solids and amorphous solids. [5 marks]
Ans - The differences are:
| Crystalline solids | Amorphous solids |
|---|---|
| Long range arrangement | Short or random range arrangement |
| Definite shape | Indefinite or Irregular shape |
| Sharp melting point | Low melting point |
| Example - Diamonds, NaCl | Example - Glass, Plastic |
Ques- NH4Cl crystallises in a body centred cubic lattice, with a unit cell distance of 387 pm. Calculate (a) the distance between the oppositely charged ions in the lattice, and (b) the radius of the NH4+ ion if the radius of the Cl- ion is 181 pm.? [2 marks]
Ans - In a body centred cubic lattice oppositely charged ions touch each other along the cross - diagonal of the cube. Hence, we can write,
2r+ 2r– = √3a, r+ + r– = √3a / 2 = √3 / 2 (387 pm) = 335.15 pm
(b) Now, since
r– = 181 pm
We haver+ = (335.15 – 181) pm = 154.15 pm.
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