In order to analyze the close packing of constituent particles (ions, molecules or atoms) in a lattice, we get into the assumption that constituent particles (ions, molecules or atoms) are hard spheres of identical shape.
What is Close Packing?
The packing of constituent particles inside lattice in such a way that they occupy maximum available space in the lattice is known as Close Packing.
Close Packing is done in three ways, namely:
(a) One Dimensional Close Packing
(b) Two Dimensional Close Packing
(c) Three Dimensional Close Packing
(a) One Dimensional Close Packing
In one dimension close packing, the spheres are arranged in a row touching each other. In onedimensional close packing, each sphere is in direct contact with two of its neighbor spheres. The number of nearest spheres to a particle in a lattice is called Coordination Number. Since there are two spheres in direct contact with the one sphere, the coordination number of onedimensional close packing is 2.
(b) Two Dimensional Close Packing
Twodimensional close packing is done by stacking rows of hard spheres one above the other. This can be done in two ways:
• AAA type
• ABA type
AAA Type
The packing in which one sphere touches two spheres placed in two different rows one above and one below is called AAA type close packing. The coordination number of AAA type twodimensional close packing is 4. The AAA type is formed by placing onedimensional row directly one above the other in both horizontal and vertical directions. It is also called twodimensional square close packing as the rows of spheres when arranged in vertical and horizontal alignments form a square.
AB Type
The packing in which the spheres in the second row are located in the depressions of the first row. The ABA type close packing is formed by placing onedimensional row let’s say B type over the A type close packing and this series continues to form a two dimensional hexagonal. The coordination number of ABA Type packing is 6 as each sphere is in direct contact with 6 other spheres.
In ABA type closepacking we find triangular empty spaces called voids. These are of two types:
• Apex of triangle pointing upwards
• Apex of triangle pointing downwards
(c) Three Dimensional Close Packing
The formation of real lattices and structures take place through threedimensional close packing. They are formed by stacking twodimensional layers of spheres one above the other. This can be done by two ways:
Threedimensional close packing from twodimensional close packed layers
Formation of threedimensional close packing can be done by placing the second square closed packing exactly above the first one. In this close packing, the spheres are aligned properly in horizontally and vertically. Similarly, by placing more layers one above the other, we can obtain a simple cubic lattice. The unit cell of the simple cubic lattice is called the primitive cubic unit cell.
Threedimensional close packing from two dimensional hexagonal closepacked layers
Threedimensional close packing can be formed with the help of twodimensional hexagonal packed layers in two ways:
• Stacking the second layer over the first layer
• Stacking the third layer over the second layer
Stacking the second layer over the first layer
Suppose we take two hexagonal close packed layer ‘A’ and place it over the second layer B ( as both layers have different alignment of spheres) such that spheres of the second layer are placed in the depressions of the first layer. We observe that a tetrahedral void is formed when a sphere of the second layer is right above the void (empty space) of the first layer. Adding further we notice octahedral voids at the points where the triangular voids of the second layer are placed right triangular voids of the first one in such a way that triangular space doesn't overlap. Octahedral voids are bordered by six spheres.
If there are ‘N’ closed spheres, then:
• Number of Octahedral Voids equals to “N”
• Number of Tetrahedral Voids equals to “ 2N”
Stacking the third layer over the second layer
There are two possible ways of placing the third layer over the second layer:
(1) By Covering Tetrahedral Voids
(2) By Covering Octahedral Voids
(1) Covering Tetrahedral Voids
In this kind of threedimensional packing, the spheres of the third layer are aligned right above the spheres of the first layer. If we name the first layer as A and second layer as B, then the pattern will be ABAB… so far and so forth. The structure formed is also called hexagonal closepacked structure also known as HCP.
(a) For HCP geometry Coordination number = 12
(b) For HCP geometry no. Of atoms per unit cell (corner)
= 12 (corner) (inside the body) ∗ 1 = 6
(c) For HCP geometry packing efficiency = 74%
(2) Covering Octahedral Voids
In this kind of packing the third layer, spheres are not placed with either of the second layer or first layer. If we name the first layer as A, second as B and then the third layer will be C (as it is now a different layer) then the pattern will be ABCABC… The structure formed is also called cubic closed packed (ccp) or facecentred packed cubic structure (fcc). For Example metals like copper and iron crystallize in the structure.
The coordination number in both cases will be 12 as each sphere in the structure is in direct contact with 12 other spheres. The packing is highly efficient and around 74% of the crystal is completely occupied.
ABC Type of Close Packing
The cubic close packing has face centred (fcc) unit cell.
(a) For CCP geometry coordination number = 12
(b) For CCP geometry number of atoms per unit cell = 4 (as calculated before)
(c) For CCP geometry packing efficiency = 74% (as calculated before) In the close packing of spheres, certain hollows are left vacant. These holes or voids in the crystals are called interstitial sites or interstitial voids.
Two important interstitial sites are
(i) Triangular
(ii) Tetrahedral
(iii) Octahedral
(iv) Cubical void
• The number of octahedral voids present in a lattice = The number of closepacked particles.
• The number of tetrahedral voids produced is twice this number.
• In ionic solids, the bigger ions i.e. anions forms closepacked structure and the smaller ions i.e. cations occupy the voids.
• If the latter ion is small then tetrahedral voids are occupied, if bigger, then octahedral voids are occupied.
• In a compound, the fraction of octahedral or tetrahedral voids that are occupied depends upon the chemical formula of the compound.
• Close packed structures have both tetrahedral and octahedral voids.
With the help of geometry and attributes of unit cells, we can easily evaluate the volume of the unit cell. With the volume and mass of atoms, we can also evaluate the density of the unit cell. A crystal lattice is represented in terms of unit cells; we can determine the density of crystal lattices by evaluating the density of unit cells.
Volume Density
Formula volume density of metal:
ρ_{v} = mass per unit cell/Volume per unit cell
The density of Unit Cell
A unit cell is a threedimensional structure occupying one, two or more atoms. With the help of the dimensions of unit cells, we can evaluate the density of the unit cell. To do so let’s consider a unit cell of edge length ‘a’, therefore the volume of the cell will be ‘a3’. Also, density is defined as the ratio of the mass of the unit cell and volume of the unit cell.
The density of Unit Cell = mass of unit cell/Volume of unit cell......(1)
Mass of unit cell varies with number of atoms “n” and mass of a single atom “m”. Mathematically mass of unit cell is the product of number of atoms “n” and mass of one atom “m” i.e.
Mass of Unit Cell = n × m
Also from quantitative aspect of atoms, mass of one atom can be written in terms of Avogadro Number ( NA ) and molar mass of atom ( M), that is,
Volume of Unit Cell = a^{3}
Placing the required values in equation 1 we get
Therefore if we know molar mass of atom “M”, number of atoms “n”, the edge length of unit cell “a” we can evaluate the density of unit cell.
Example 1
The density of a facecentred unit cell is 6.23 g cm^{3}. Given the atomic mass of a single atom is 60, evaluate the edge length of the unit cell. (Take value of N_{A} = 6.022 × 10^{23})
As the unit cell is a facecentred unit cell, a number of atoms will be 4. We have
n = 4 , M = 60 , N_{A} = 6.022 × 10^{23} and d = 6.23 g cm^{3}
We know that
After evaluating we get the value of edge length of the cube to be 4 × 10^{8} cm.
Example 2
KBr or potassium bromide has density 2.75 g cm^{3}. The edge length of its unit cell is 654 pm. Prove that KBr depicts facecentred cubic structure.
We have edge length of unit cell = 654 pm = 6.54 × 10^{8} cm
Therefore volume of the cell = (6.54 ×10^{8})^{3} cm^{3}
Molar Mass of Potassium Bromide ( KBr) is 119 g/mol
Density of KBr = 2.75 g cm^{3}
We know that
Evaluating we get value of n to be 3.09 ≈ 4
Since a number of atoms is 4 we can clearly say that KBr is a facecentred cubic structure.
The structure of each ionic compound is determined by stoichiometry and ion sizes. Larger cations can fit into cubic or octahedral holes. In tetrahedral holes, smaller cations can be accommodated. If we examine an array of anions in the form of cubic tight packing, the diameters of the tetrahedral and octahedral holes will vary. As a result, the cations will only occupy the voids if there is adequate room for them.
The Radius Ratio can be used to determine whether the ions will be able to retain the cations. The shape of the unit cell is also determined by the ion’s coordination number in the crystal structure. For a particular coordination number, there is a limiting value of the ratio of cation radius to anion radius, i.e. r+ / r–. The ionic structure becomes unstable if the value of the ratio r+ / r– is smaller than the predicted value.
Hence, the radius ratio is defined as the ratio of a smaller ionic radius (cation) to a larger ionic radius (anion), and is given by,
ρ = r+ / r–
where,
ρ is the radius ratio,
r+ is the radius of cation, and
r– is the radius of the anion.
Given below are the limiting values of r+ / r– and their coordination numbers.
Examples are B_{2}O_{3}, ZnS, NaCl, CsCl, MgO, CuCl
Some properties are:
Problem 1: If a solid “X^{+}Y^{–}” has a structure similar to NaCl and the radius of anion is 250 pm. Then, find the ideal radius of the cation in the structure. Also, state your reason, is it possible to fit a cation Z+ of radius 180 pm in the tetrahedral site of the structure(X^{+}Y^{–})?
Solution:
Given that, Structure X^{+}Y^{–}
Radius of anion = 250 pm = 2.5 A° [1 picometer = 0.01A°]
Limiting ratio = 0.414 [refer to the above table]
If the X^{+}Y^{–} structure is comparable to that of the Na^{+}Cl^{– }ion, then six Cl^{–} ions will surround the Na^{+} and vice versa.
Thus, octahedral void is occupied.
Radius ration ρ = r+ / r–
r+ = 0.414 × 2.5
= 1.035 A°
Now, the limiting ratio for tetrahedral site is 0.225 [refer to table]
As a result, r+ / r– = 0.225
r+ = 0.225 × 2.5
= 0.5625 A° or 56.25 pm
Thus, for a tetrahedral site, the optimum radius for the cation in the given structure is 56.25 pm.
We know that the radius of Z+ is 180 pm. This implies that Z+‘s radius is substantially bigger than 56.25 pm.
Thus, cation Z+ cannot be accommodated in a tetrahedral location.
Problem 2: Predict the coordination number of Cs+ ion and the structure of CsCl if r_{cs+} = 1.69 A° and r_{Cl–} = 1.81A°
Given that:
r_{cs+} = 1.69 A°
r_{Cl– }= 1.81A°
Calculation:
Radius ration ρ = r+ / r–
= 1.69 / 1.81
= 0.9337 A°
As 0.9337 > 0.732, hence the coordination number is 8 and geometry of CsCl is cubic.
Problem 3: Predict the coordination number of Na+ ion and the structure of NaCl crystal if r_{Na+} = 0.95 A° and r_{Cl–} = 1.81A°
Given that:
r_{Na+} =0.95A°
r_{Cl–} = 1.81A°
Radius ration ρ = r+ / r–
= 0.95 / 1.81
= 0.5248 A°
As 0.5248 lies between (0.414 to 0.732), hence the coordination number is 6 and geometry of NaCl is Octahedral.
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One Shot: Properties of Solid State Video  56:45 min 
1. What are close packed structures? 
2. What are the types of close packed structures? 
3. How do close packed structures determine the physical properties of materials? 
4. How are close packed structures formed? 
5. What are some examples of materials with close packed structures? 
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One Shot: Properties of Solid State Video  56:45 min 

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