(ii) The diagrammatic representation of the constituent particles (atoms, ions, or molecules) present in a crystal in a regular three-dimensional arrangement is called crystal lattice. A unit cell is the smallest three-dimensional portion of a crystal lattice. When repeated again and again in different directions, it generates the entire crystal lattice.
(iii) A void surrounded by 4 spheres is called a tetrahedral void and a void surrounded by 6 spheres is called an octahedral void. Figure 1 represents a tetrahedral void and figure 2 represents an octahedral void.
8. How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred
Answer:
(i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.
(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.
(iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.
9. Explain
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points. The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.
(ii) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle.
10. Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).
Answer:
(i) Simple cubic
In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be ‘a’ and the radius of each particle be r.
So, we can write: a = 2r
Now, volume of the cubic unit cell = a3 = (2r)3 = 8r3
We know that the number of particles per unit cell is 1.
Therefore, volume of the occupied unit cell = \(\frac43\pi r^3\)
Hence, packing efficiency = \(\frac{Volume\,of\,one \,particle}{Volume\,of\,cubic\,unit\,cell}\times 100 \%\)
(ii) Body-centred cubic
It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.
From ∆FED, we have:
Again, from ∆AFD, we have:
Let the radius of the atom be r.
Length of the body diagonal, c = 4π
Volume of the cube, a3 = \(\left(\frac{4r}{\sqrt3}\right)^3\)
A body-centred cubic lattice contains 2 atoms.
So, volume of the occupied cubic lattice = \(2 \times\frac43\pi r^3=\frac83\pi r^3\)
\(\therefore Packing \, efficiency=\frac{Volume\,occupied \,by\, two\,sphere\,in\, the\, unit\,cell}{Total\,volume\,of\,the\,unit\,cell}\times100\%\)
(iii) Face-centred cubic
Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b. From ∆ABC, we have:
Let r be the radius of the atom.
Now, from the figure, it can be observed that:
Now, volume of the cube, a3 = (2√2r)3
We know that the number of atoms per unit cell is 4.
So, the volume of the occupied unit cell = 4 × \(\frac43\pi r^3\), therefore
\(Packing \, efficiency=\frac{Volume\,occupied \,by\, four\,sphere\,in\, the\, unit\,cell}{Total\,volume\,of\,the\,unit\,cell}\times100\%\)
11. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver.
Answer:
It is given that the edge length, a = 4.077 × 10−8 cm
Density, d = 10.5 g cm−3
As the lattice is fcc type, the number of atoms per unit cell, z = 4
We also know that,
NA = 6.022 × 1023 mol−1
Using the relation:
= 107.13 gmol−1
Therefore, atomic mass of silver = 107.13 u
12. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer:
It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell = 8 × \(\frac18\) = 1
It is also given that the atoms of P are present at the body-centre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1
Hence, the formula of the compound is PQ. The coordination number of both P and Q is 8.
13. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm−3, calculate atomic radius of niobium using its atomic mass 93 u.
Answer:
It is given that the density of niobium, d = 8.55 g cm−3
Atomic mass, M = 93 gmol−1
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, NA = 6.022 × 1023 mol−1
Applying the relation:
= 3.612 × 10−23 cm3
So, a = 3.306 × 10−8 cm
For body-centred cubic unit cell:
= 1.432 × 10−8 cm
= 14.32 × 10−9 cm
= 14.32 nm