# Chemistry Questions and Answers – Solid State – Crystal Lattices and Unit Cells

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This set of Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Solid State – Crystal Lattices and Unit Cells”.

1. Which of the following is regarded as the ‘repeatable entity’ of a 3D crystal structure?
a) Unit cell
b) Lattice
c) Crystal
d) Bravais Index

Explanation: Unit cell is the smallest entity of a crystal lattice which, when repeated in space (3 dimensions) generates the entire crystal lattice. Lattice comprises of the unit cells which hold all the particles in a particular arrangement in 3 dimensions. Crystal is a piece of homogenous solid and Bravais indices are used to define planes in crystal lattices in the hexagonal system.

2. Which of the following unit cells has constituent particles occupying the corner positions only?
a) Body-centered cell
b) Primitive cell
c) Face centered cell
d) End-centered cell

Explanation: According to classification of unit cells, a primitive unit cell is one which has all constituent particles located at its corners. BCC has one particle present at the center including the corners. FCC has an individual cell shared between the faces of adjacent cells. End centered cells have cells present at centers of two opposite faces.

3. What is the coordination number of a body-centered unit cell?
a) 6
b) 12
c) 8
d) 4

Explanation: Coordination number of a unit cell is defined as the number of atoms/ions that surround the central atom/ion. In the case of BCC, the central particle is surrounded by 8 particles hence, 8.
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4. Which of the following arrangements of particles does a simple cubic lattice follow?
a) ABAB
b) AABB
c) ABCABC
d) AAA

Explanation: Simple cubic lattice results from 3D close packing from 2D square-packed layers. When one 2D layer is placed on top of the other, the corresponding spheres of the second layer are exactly on top of the first one. Since both have the same, exact arrangement it is AAA type.

5. If a crystal lattice has 6 closed-pack spheres, what the number of tetrahedral voids in the lattice?
a) 12
b) 6
c) 36
d) 3

Explanation: For a crystal lattice, if there are N close-packed spheres the number of tetrahedral voids are 2N and number octahedral voids are N. For N=6, number of tetrahedral voids = 2 × 6 = 12.

6. Which of the following possess anisotropic nature within their structure?
a) Hair wax
b) Snowflakes
c) Polythene
d) Crystal glass

Explanation: Crystalline solids possess anisotropic nature within their structure. Anisotropy is the directional dependence of a property. Meaning, a property within the crystal structure will have different values when measured in different directions. Snowflake is a crystalline solid whereas the rest are amorphous solids.

7. Identify the dimensional relation for the unit cell illustrated below.

a) a = b = c
b) a = b ≠ c
c) a ≠ b ≠ c
d) a ≠ b = c

Explanation: The given figure represents an orthorhombic unit cell. Experimentally, it is determined that for orthorhombic unit cells a ≠ b ≠ c. All sides are unequal. It results from extension of cube along two pairs of orthogonal sides by two distinct factors.

8. 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 center of the unit cell. Deduce the formula of the compound.
a) AB2
b) AB3
c) AB4
d) AB

Explanation: The description is of a BCC. For BCC, each atom at the corner is shared by 8 unit cells. One atom at the center wholly belongs to the corresponding unit cell.
Therefore, total number of atoms of A present=$$\frac{1}{8}$$ x 8=1
Total number of atoms of B present=1
Therefore, A:B=1:1 implying the formula of the compound is AB.

9. Atoms of element X form a BCC and atoms of element Y occupy 3/4th of the tetrahedral voids. What is the formula of the compound?
a) X2Y3
b) X3Y2
c) X3Y4
d) X4Y3

Explanation: The number of tetrahedral voids form is equal to twice the number of atoms of element X. Number of atoms of Y is 3/4th the number of tetrahedral voids i.e. 3/2 times the number of atoms of X. Therefore, the ratio of numbers of atoms of X and Y are 2:3, hence X2Y3.

10. What is the total volume of the particles present in a body centered unit cell?
a) 8πr3
b) $$\frac{8}{3}$$πr3
c) $$\frac{16}{3}$$πr3
d) $$\frac{32}{3}$$πr3

Explanation: Since particles are assumed to be spheres and volume of one sphere is $$\frac{4}{3}$$πr3, total volume of all particles in BCC = 2 x $$\frac{4}{3}$$πr3=$$\frac{8}{3}$$πr3 since a BCC has 2 particles per cell.

11. If the aluminum unit cell exhibits face-centered behavior then how many unit cells are present in 54g of aluminum?
a) 1.2042 x 1024
b) 5.575 x 1021
c) 3.011 x 1023
d) 2.4088 x 1024

Explanation: Atomic mass of Al = 27g/mole (contains 6.022 x 1023 Al atoms)
Since it exhibits FCC, there are 4 Al atoms/unit cell.
If 27g Al contains 6.022 x 1023 Al atoms then 54g Al contains 1.2044 x 1024atoms.
Thus, if 1 unit cell contains 4 Al atoms then number of unit cells containing 1.2044 x 1024 atoms=(1.2044 x 1024 x 1)/4 = 3.011× 1023 unit cells.

12. What is the radius of a metal atom if it crystallizes with body-centered lattice having a unit cell edge of 333 Pico meter?
a) 1538.06 pm
b) 769.03 pm
c) 288.38 pm
d) 144.19 pm

Explanation: For body-centered unit cells, the relation between radius of a particle ‘r’ and edge length of unit cell ‘a’ is given as $$\frac{\sqrt{3}}{4}$$a=r
On substituting the values we get r = $$\frac{\sqrt{3}}{4}$$ x 333 pm = 144.19 pm is the radius of the metal atom.

13. How many parameters are used to characterize a unit cell?
a) Six
b) Three
c) Two
d) Nine

Explanation: A unit cell is characterized by six parameters i.e. the three common edge lengths a, b, c and three angles between the edges that are α, β, γ. These are referred to as inter-axial lengths and angles, respectively. The position of a unit cell can be determined by fractional coordinates along the cell edges.

14. What is each point (position of particle) in a crystal lattice termed as?
a) Lattice index
b) Lattice point
c) Lattice lines
d) Lattice spot

Explanation: Each point of the particle’s position is referred to as ‘lattice point’ or ‘lattice site’. Every lattice point represents one constituent particle which may be an atom, ion or molecule.

15. If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.
a) 2923 kg/m3
b) 5846 kg/m3
c) 8768 kg/m3
d) 1750 kg/m3

Explanation:
Given,
Edge length (a) = 500 pm = 500 x 10-12 m
Atomic mass (M) = 110 g/mole = 110 x 10-3 kg/mole
Avogadro’s number (NA) = 6.022 x 1023/mole
z = 4 atoms/cell
The density, d of a metal is given as d=$$\frac{zM}{a^3N_A}$$
On substitution, d=$$\frac{4 \times 110 \times 10^{-3}}{(500 \times 10^{-12})^3 \times 6.022 \times 10^{23}}$$=5846 kg/m3.

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