## Packing Efficiency - HCP and CCP structures for JEE Main Exam

We all know that the particles are arranged in different patterns in unit cells. Particles include atoms, molecules or ions. There is no concern for the arrangement of the particles in the lattice as there are always some empty spaces inside which are called void spaces. The aspect of the solid state with respect to quantity can be done with the help of packing efficiency.

Definition: “Packing efficiency can be defined as the percentage ration of the total volume of a solid occupied by spherical atoms. The formula is written as the ratio of the volume of one atom to the volume of cells is s3.”

Mathematically, the equation of packing efficiency can be written as

Number of Atoms × volume obtained by 1 share / Total volume of unit cell ×100 %

Packing efficiency is defined as the percentage ratio of space obtained by constituent particles which are packed within the lattice. It can be evaluated with the help of geometry in three structures known as:

HCP and CCP structures

Body-centered cubic structures

Simple lattice structures of cubic

### Factors of Packing Efficiency

There are many factors which are defined for affecting the packing efficiency of the unit cell:

A volume of a unit cell

Number of atoms in the lattice structure

Volume of atoms

### HCP and CCP Structures

In this, both types of packing efficiency, hexagonal close packing or cubical lattice closed packing is done, and the packing efficiency is the same in both. For calculating the packing efficiency in a cubical closed lattice structure, we assume the unit cell with the side length of ‘a’ and face diagonals AC to let it ‘b’.

When we see the ABCD face of the cube, we see the triangle of ABC in it. Let us suppose the radius of each sphere ball is ‘r’. Now correlating the radius and its edge of the cube, we continue with the following,

In triangle ABC, according to the Pythagoras theorem, we write it as:

AC2 = BC2 + AB2

Though AC = b and BC = a,

We substitute the values in the above equation, then we get

b2 = a2 + a2

b2 = 2a2

b = \[\sqrt{2}\] a ---------- (1)

Also, the edge ‘b’ can be defined as follows in terms of radius ‘r’ which is equal to:

b = 4r ---------------- (2)

According to equation (1) and (2), we can write the following:

4r = \[\sqrt{2}\] a

OR

a = 2 \[\sqrt{2}\] r

There are a total of 4 spheres in a CCP structure unit cell, the total volume occupied by it will be following:

\[\frac{4\times 4}{3\pi r^3}\]

And the total volume of a cube is the cube of its length of the edge (edge length)3. It means a^3 or if defined in terms of ‘r’, then it is (2 \[\sqrt{2}\] r)^3.

Thus, packing efficiency will be written as follows.

Packing efficiency = volume occupied by 4 spheres/ total volume of unit cell × 100 %

\[\frac{\frac{4\times 4}{3\pi r^3}}{(2\sqrt{2}r)^3}\times 100%\]

\[\frac{\frac{16}{3\pi r^3}}{(2\sqrt{2}r)^3}\times 100%\]

Thus, packing efficiency in FCC and HCP structures is calculated as 74.05%.

In the same way, the relation between the radius ‘r’ and edge length of unit cell ‘a’ is r = 2a and the number of atoms is 6 in the HCP lattice.

Packing efficiency can be written as below,

Packing efficiency = Volume occupied by 6 spheres ×100 / Total volume of unit cells.

\[\frac{\frac{6\times 4}{3\pi r^3}}{(2r)^3}\times 100%=74.05%\]

Examples are Magnesium, Titanium, Beryllium etc.

### Body - Centered Structures of Cubic Structures

In body-centered cubic structures, the three atoms are arranged diagonally. For determining the packing efficiency, we consider a cube with the length of the edge, a face diagonal of length b and diagonal of cube represented as c.

In the triangle EFD, apply according to the theorem of Pythagoras,

b2 = a2 + a2

b2 = 2a2

b = \[\sqrt{2}\] a

Now, in triangle AFD, according to the theorem of Pythagoras,

c2 = a2 + b2 = a2 + 2a2

c2 = 3a2

c = \[\sqrt{3}\] a

Suppose if the radius of each sphere is ‘r’, then we can write it accordingly as follows.

c = 4r

\[\sqrt{3}\] a = 4r

r = \[\frac{\sqrt{3}}{4a}\]

There are two number of atoms in the BCC structure, then the volume of constituent spheres will be as following,

2 × \[(\frac{4}{3})\pi r^3\]

Thus, packing efficiency = Volume obtained by 2 spheres × 100 / Total volume of cell

= \[2\times \frac{\frac{\frac{4}{3}}{\pi r^3}}{\frac{4^3}{\sqrt{3}r}}\]

= 68%

Therefore, the value of APF = Natom Vatom / Vcrystal = 2 × (4/3) π r^3 / 4^3 / √3 r

= \[\frac{\pi \sqrt{3}}{8}\]

= 0.68017476

Thus, the packing efficiency of the body-centered unit cell is around 68%. The metals such as iron and chromium come under the BSS category.

### Simple Cubic Lattice

In a simple cubic lattice structure, the atoms are located only on the corners of the cube. It is stated that we can see the particles are in touch only at the edges.

Therefore, if the Radius of each and every atom is r and the length of the cube edge is a, then we can find a relation between them as follows,

2a = r

Though a simple unit cell of a cube consists of only 1 atom, and the volume of the unit cells containing only 1 atom will be as follows.

\[\frac{4}{3\pi r^3}\]

The volume of the unit cell will be a3 or 2a3 that gives the result of 8a3.

Thus, packing efficiency = Volume obtained by 1 sphere × 100 / Total volume of unit cells

= \[\frac{\frac{4}{3\pi r^3}}{8r^3}\times 100=52.4%\]

Therefore, the value of packing efficiency of a simple unit cell is 52.4%. Examples such as lithium and calcium come under this category.

### Importance of Packing Efficiency

The importance of packing efficiency is in the following ways:

It represents the solid structure of an object.

It shows the different properties of solids like density, consistency, and isotropy.

Different attributes of solid structure can be derived with the help of packing efficiency. The main reason for crystal formation is the attraction between the atoms. As they attract one another, it is frequently in favour of having many neighbours. Therefore, the coordination number or the number of adjacent atoms is important.

For the structure of a square lattice, the coordination number is 4 which means that the number of circles touching any individual atom.

The void spaces between the atoms are the sites interstitial. Though each of it is touched by 4 numbers of circles, the interstitial sites are considered as 4 coordinates. Thus, this geometrical shape is square. In this, there are the same number of sites as circles.

Because the atoms are attracted to one another, there is a scope of squeezing out as much empty space as possible. The packing efficiency is the fraction of crystal or known as the unit cell which is actually obtained by the atoms. It must always be seen less than 100 percent as it is not possible to pack the spheres where atoms are usually spherical without having some empty space between them.

P.E = \[\frac{(\textrm{area of circle})}{(\textrm{area of unit cell})}\]

With respect to our square lattice of circles, we can evaluate the packing efficiency that is PE for this particular respective lattice as following:

P. E =\[\frac{\pi r^2}{(2r)^2}\]

= \[\frac{\pi }{4}\]

= 78.54%.

Thus, the interstitial sites must obtain 100 % - 78.54% which is equal to 21.46%.

Let us now compare it with the hexagonal lattice of a circle.

Note: The atomic coordination number is 6. Atomic coordination geometry is hexagonal. The interstitial coordination number is 3 and the interstitial coordination geometry is triangular.

The calculated packing efficiency is 90.69%.

And the evaluated interstitials site is 9.31%.

By examining it thoroughly, you can see that in this packing, twice the number of 3-coordinate interstitial sites as compared to circles. For every circle, there is one pointing towards the left and the other one pointing towards the right.

Therefore, these sites are much smaller than those in the square lattice.

The higher are the coordination numbers, the more are the bonds and the higher is the value of packing efficiency. This clearly states that this will be a more stable lattice than the square one.

If we compare the squares and hexagonal lattices, we clearly see that they both are made up of columns of circles. Thus, in the hexagonal lattice, every other column is shifted allowing the circles to nestle into the empty spaces. Therefore, it generates higher packing efficiency.

Efficiency is considered as minimum waste. And so, the packing efficiency reduces time, usage of materials and the cost of generating the products.

## FAQs on Packing Efficiency for IIT JEE Solid State

**1. How can I deal with all the questions of solid states that appear in IIT JEE Chemistry Exams?**

**Here are some of the strategies that can help you deal with some of the most commonly asked questions of solid state that appear in IIT JEEexams:**

Go through the chapter, that is, solid states thoroughly.

Try visualizing the 3D shapes so that you don't have a problem understanding them.

Read the questions that appear in exams carefully and try answering them step-wise.

Study classification of solids on the basis of arrangement of constituent particles and intermolecular forces.

Learn the packing efficiency and unit cells of solid states.

Face-centered, edge-centered, and body-centered are important concepts that you must study thoroughly.

Concepts of crystalline and amorphous solids should be studied for short answer type questions.

Examples of this chapter provided in NCERT are very important from an exam point of view.

Also, study topics like latent heat of vaporization, latent heat of fusion, phase diagram, specific heat, and triple points in regard to this chapter.

**2. What is the trend of questions asked in previous years from the Solid State chapter of IIT JEE?**

The chapter on solid-state is very important for IIT JEE exams. Questions are asked from almost all sections of the chapter including topics like introduction, crystal lattice, classification of solids, unit cells, closed packing of spheres, cubic and hexagonal lattice structure, common cubic crystal structure, void and radius ratios, point defects in solids and nearest-neighbor atoms. Apart from this, topics like the change of state, vaporization, fusion, freezing point, and boiling point are relevant from the states of matter chapter.

**3. What is the pattern of questions framed from the solid states chapter in chemistry IIT JEE exams?**

There are a lot of questions asked in IIT JEE exams in the chemistry section from the solid-state chapter. Generally, numerical questions are asked from the solid states chapter wherein the student has to calculate the radius or number of vertices or edges in a 3D structure. Calculation-based questions on latent heat of fusion, the specific heat of fusion, latent heat of vaporization, and specific heat of vaporization are also asked from this chapter including conversion of solids, liquid, and gases from one form to another.

**4. How can I predict the formula of a compound in questions asked in the IIT JEE Chemistry exam from chapter solid state if it is formed by two elements A and B that crystallize in a cubic structure containing A atoms at the corner of the cube and B atoms at the body center of the cube?**

These types of questions are often asked in IIT JEE to analyze the conceptual clarity of students. If an atom A is present in the corner of a cube, then that atom will be shared by 8 similar cubes, therefore, the contribution of an atom A in one specific cube will be ⅛. One cube has 8 corners and all the corners of the cube are occupied by an atom A, therefore, the total number of atoms A in a unit cell will be 8 X ⅛ which is equal to 1. The atoms at the center of the cube are shared by no other cube and one cube contains only one atom, therefore, the number of atoms of B in a unit cell is equal to 1. Therefore, the formula of the compound will be AB.

**5. How can I solve the question of Solid States that appeared in the IIT JEE Chemistry exam, that is, to calculate the distance between neighboring ions of Cs and Cl and also calculate the radius ratio of two ions if the eight corners of the cubic crystal are occupied by Cl and the center of the crystal structure is occupied by Cs?**

The volume of a cubic crystal can be calculated as the cube of sides of the structure and the density of the structure is calculated as the product of n (in the case of unit cells, the value of n is 1) and molecular weight divided by the product of volume and Avogadro number. By substituting the formula for volume, we can calculate the size of the cube. On calculation, the side of the cube was observed to be 4.13 Armstrong. In order to calculate the distance between the two atoms, multiply the sides of the cube with the diagonal, this will give a value of 7.15 Armstrong. Now, the distance between the two atoms will be the sum of twice the radius of cesium and twice the radius of chloride equal to 7.15. The distance between the two atoms will be the sum of radium of both the atoms, which on calculation will be equal to 3.57 Armstrong. Since chloride ions are present at the corners of the cube, therefore, we can determine the radius of chloride ions which will be equal to the length of the side of the cube, therefore, the length of the chloride will be 2.06 Armstrong and cesium ion will be the difference between 3.57 and 2.06 which will be equal to 1.51 Armstrong. Therefore, the ratio of the radiuses will be 0.73 Armstrong.