Solid State MCQ Quiz - Objective Question with Answer for Solid State - Download Free PDF

Last updated on Apr 26, 2025

Latest Solid State MCQ Objective Questions

Solid State Question 1:

Match the following types of unit cells with their correct properties:

Column I
(Unit Cell Type)
Column II
(Property)
A. Simple Cubic (SC) 1. Atomic radius = a/2, Z = 1
B. Body-Centered Cubic (BCC) 2.. Coordination number = 8, Packing efficiency = 68%
C. Face-Centered Cubic (FCC) 3. Atomic radius = a/2√2, Z = 2
  4.  Atomic radius = a/2√2, Z = 4

  1. A–3, B–1, C–2
  2. A–2, B–1, C–4
  3. A–2, B–3, C–4
  4. A–1, B–2, C–4

Answer (Detailed Solution Below)

Option 4 : A–1, B–2, C–4

Solid State Question 1 Detailed Solution

CONCEPT:

Unit Cell Geometry and Properties

  • SC (Simple Cubic)
    • Atomic radius = a / 2
    • Z (atoms per unit cell) = 1
    • Coordination number = 6
    • Packing efficiency = 52%
  • BCC (Body-Centered Cubic)
    • Atomic radius = (√3 / 4) × a
    • Z = 2
    • Coordination number = 8
    • Packing efficiency = 68%
  • FCC (Face-Centered Cubic)
    • Atomic radius = (√2 / 4) × a = a / (2√2)
    • Z = 4
    • Coordination number = 12
    • Packing efficiency = 74%

EXPLANATION:

  • Column I to Column II Matching:
    • A. SC → Atomic radius = a / 2, Z = 1 → Match: 1
    • B. BCC → Coordination number = 8, Packing efficiency = 68% → Match: 2
    • C. FCC → Atomic radius = a / (2√2), Z = 4 → Match: 4

Correct Matching: A–1, B–2, C–4

Solid State Question 2:

Calculate the number of atoms present in unit cell of an element having molar mass 190 g mol-1 and density 20 g cm-3.

[a3 ⋅ NA = 38 cm3 mol-1]

  1. 1
  2. 2
  3. 6
  4. 4
  5. 3

Answer (Detailed Solution Below)

Option 4 : 4

Solid State Question 2 Detailed Solution

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Explanation:

The number of atoms in the unit cell (Z) is given by:\( Z = \frac{d \cdot \text{a}^3 \cdot \text{N}_A}{M}\)

Calculation:

  • Given Data,
    • Molar mass (M) = 190 g/mol
    • Density (d) = \(20 g/cm^3\)
    • \(\text{a}^3 \cdot N_A = 38 \, \text{cm}^3 \cdot \,\text{mol}^{-1}\)
    • \(\text{N}_A = Avogadro's number = 6.022 \times 10^{23} \, \text{mol}^{-1}\)
  • Substitute the given values into the formula:

\(Z = \frac{20 \, \text{g/cm}^3 \cdot 38 \, \text{cm}^3/\text{mol}^{-1}}{190 \, \text{g/mol}} Z = \frac{760 \, \text{g/mol}}{190 \, \text{g/mol}} Z = 4\)

Conclusion:

The number of atoms present in the unit cell is: 4

Solid State Question 3:

Find the radius of an atom in fcc unit cell having edge length 393 pm.

  1. 196.51 pm
  2. 170.22 pm
  3. 78.63 pm
  4. 150.93 pm
  5. 138.93 pm

Answer (Detailed Solution Below)

Option 5 : 138.93 pm

Solid State Question 3 Detailed Solution

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Explanation:

  • Given:
    • Edge length of the fcc unit cell (a) = 393 pm
    • In a face-centered cubic (fcc) unit cell, the relation between the edge length (a) and the atomic radius (r) is given by:
  • Solving for the atomic radius (r):

                  
              
             
             
             

Conclusion:

Therefore, the radius of the atom in the fcc unit cell with an edge length of 393 pm is 138.93 pm..

Solid State Question 4:

The number of atoms in body centred and face centred cubic unit cell respectively are

  1. 2 and 3
  2. 4 and 3
  3. 1 and 2
  4. 4 and 6
  5. 2 and 4

Answer (Detailed Solution Below)

Option 5 : 2 and 4

Solid State Question 4 Detailed Solution

CONCEPT:

Number of Atoms in Unit Cells

  • A body-centered cubic (BCC) unit cell has atoms at each of the eight corners of a cube and one atom at the center of the cube.
  • A face-centered cubic (FCC) unit cell has atoms at each of the eight corners of a cube and one atom at the center of each of the six faces of the cube.

EXPLANATION:

  • For a body-centered cubic (BCC) unit cell:
    • The eight corner atoms contribute 1/8th of an atom each to the unit cell, so 8 * (1/8) = 1 atom from corners.
    • The single atom at the center contributes 1 atom.
    • Total number of atoms in a BCC unit cell = 1 (from corners) + 1 (from center) = 2 atoms.
  • For a face-centered cubic (FCC) unit cell:
    • The eight corner atoms contribute 1/8th of an atom each to the unit cell, so 8 * (1/8) = 1 atom from corners.
    • The six face atoms contribute 1/2 of an atom each to the unit cell, so 6 * (1/2) = 3 atoms from faces.
    • Total number of atoms in an FCC unit cell = 1 (from corners) + 3 (from faces) = 4 atoms.

Therefore, the number of atoms in body-centered and face-centered cubic unit cells are 2 and 4 respectively.

Solid State Question 5:

Which of the following has Frenkel defects?

  1. Sodium chloride
  2. Silver bromide
  3. Graphite
  4. Diamond
  5. Silver nitrate

Answer (Detailed Solution Below)

Option 2 : Silver bromide

Solid State Question 5 Detailed Solution

CONCEPT:

Frenkel Defects

  • Frenkel defects occur when an ion in a crystal lattice is displaced from its normal site to an interstitial site, leaving a vacancy at its original position.
  • These defects are common in ionic crystals where there is a significant size difference between the cations and anions.
  • Frenkel defects do not change the overall density of the crystal.

EXPLANATION:

The question provides several compounds. We need to analyze each one to see if they can exhibit a Frenkel defect:

  • - AgCl (Silver Chloride) - This compound has a significant size difference between Ag⁺ and Cl⁻ ions, making it a candidate for Frenkel defect.
  • - NaCl (Sodium Chloride) - Na⁺ and Cl⁻ ions are of similar size, so it does not exhibit a Frenkel defect.
  • - Graphite - Graphite is a covalent network solid and does not have cations and anions; hence, it cannot exhibit a Frenkel defect.
  • - AgBr (Silver Bromide) - Similar to AgCl, AgBr has a significant size difference between Ag⁺ and Br⁻ ions, making it another candidate for Frenkel defect.
  • Therefore, the correct answer is option 2) Silver bromide.

Therefore, Silver bromide (AgBr) exhibits Frenkel defects.

Top Solid State MCQ Objective Questions

The unit cell of a certain type of crystal is defined by three vectors a, b and c. The vectors are mutually perpendicular, but a ≠ b ≠ c. The crystal structure is

  1. Triclinic
  2. Tetragonal
  3. Orthorhombic
  4. Monoclinic

Answer (Detailed Solution Below)

Option 3 : Orthorhombic

Solid State Question 6 Detailed Solution

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Explanation:

If the atoms or atom groups in the solid are represented by points and the points are connected, the resulting lattice will consist of an orderly stacking of blocks or unit cells.

  • The orthorhombic unit cell is distinguished by three lines called axes of twofold symmetry about which the cell can be rotated by 180° without changing its appearance.
  • This characteristic requires that the angles between any two edges of the unit cell be right angles but the edges may be any length.

F9 Tapesh 29-1-2021 Swati D014

Important Points

There are 7 types of crystal systems:

Crystal System

Angles between Axis

Unit Cell Dimensions

Cubic

α = β = γ = 90°

a = b = c

Tetragonal

α = β = γ=90°

a = b ≠ c

Orthorhombic

α = β = γ= 90°

a ≠ b ≠ c

Rhombohedral

α = β = γ ≠ 90°

a = b = c

Hexagonal

α = β = 90°, γ = 120°

a = b ≠ c

Monoclinic

α = γ = 90°, β ≠ 90°

a ≠ b ≠ c

Triclinic

α ≠ β ≠ γ

a ≠ b ≠ c

The radius of the largest sphere which fits properly at the edge of a body centered cubic unit cell is (Edge length is represented by ‘a’)

  1. 0.134 a
  2. 0.027 a
  3. 0.047a
  4. 0.067 a

Answer (Detailed Solution Below)

Option 4 : 0.067 a

Solid State Question 7 Detailed Solution

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Concept:

  • The body-centered cubic unit cell has atoms at each of the eight corners of a cube plus one atom in the center of the cube.
  • Each of the corner atoms is the corner of another cube so the corner atoms are shared among eight unit cells.

Calculation:

For body-centered cubic bcc structure,

Radius, (R) = (a×√3)/4    or a×√3 = 4×R  --- (1)

Where, a = edge length.

According to the question, the structure of a cubic unit cell can be shown as follows:

∴ a = 2(R + r)

On substituting the value of R from Eq. (1) and Eq (2), we get

\(\frac{a}{2}=\frac{√{3}}{4}a+r\)

\(r=\frac{a}{2}=\frac{√{3}}{4}a=2a-\frac{√{3a}}{4}\)

\(r=\frac{a\left( 2-√{3} \right)}{4}\)

r = 0.067 a.

The ratio of number of atoms present in a simple cubic, body centred cubic and face centred cubic structure are, respectively:

  1. 8 : 1 : 6
  2. 1 : 2 : 4
  3. 4 : 2 : 1
  4. 4 : 2 : 3

Answer (Detailed Solution Below)

Option 2 : 1 : 2 : 4

Solid State Question 8 Detailed Solution

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Explanation:

Unit Cell

Coordination No.

No. of Atoms Per Unit Cell

Atomic packing factor

Simple Unit Cell

6

1

52%

Body-centred Cubic

8

2

68%

Face-centred Cubic

12

4

74%

Hexagonal Closest Packed

12

6

74%

 

Diagram

GATE ME 2009 Images-Q15

GATE ME 2009 Images-Q15.1

GATE ME 2009 Images-Q15.2

Effective no. of

lattice points

\(\Rightarrow \frac{1}{8} \times 8 = 1\)

\(\frac{1}{8} \times 8 + 1 = 2\)

\(\frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4\)

Diamond is a ________.

  1. Metallic Crystal
  2. Covalent Crystal
  3. Ionic Crystal
  4. Molecular Crystal

Answer (Detailed Solution Below)

Option 2 : Covalent Crystal

Solid State Question 9 Detailed Solution

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The correct answer is Covalent Crystal.
CONCEPT:

The crystal lattice is known as the regular arrangement of constituent particles such as atoms, ions, or molecules of a crystalline solid in 3-D space.

Van Der Waals Crystal / Molecular Crystal.

  • A solid that consists of a lattice array of molecules such as hydrogen, methane, or more organic compounds bound by Van Der Waals forces or hydrogen bonds are known as Van Der Waals Crystal.
  • The general properties of the Van der Waals crystal are soft, low melting point, and a poor conductor of heat and electricity. 

Ionic crystals.

  • Lattice points which occupied by charged particles (ions) and are held together by columbic forces. 
  • Size and the relative number of each ion determine the crystal structure. 

Covalent Crystals. 

  • Lattice points are occupied by neutral atoms. In which atoms are held together by covalent bonds. 
  • These crystals are hard solids and poor conductors of electricity. 
  • Covalent Solids are solids in which the constituent particles are atoms and interparticle forces are strong covalent bonds.
    Examples - Diamond, Quartz, SiO2.

EXPLANATION: From the above concept, it is clear that a diamond is a Covalent Crystal.

An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centers of two nearest tetrahedral voids in the lattice is:

  1. √2a 
  2. a
  3. a/2
  4. 3a/2

Answer (Detailed Solution Below)

Option 3 : a/2

Solid State Question 10 Detailed Solution

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Concept:

12.04.2019 Shift 1 Synergy JEE Mains D74

The distance between the centers of two nearest tetrahedral voids in the lattice and also the minimum distance between two tetrahedral voids is \(\frac{a}{2}\)..

In fcc (face centered cubic), tetrahedral voids are located on the body diagonal at a distance of \(\frac{{\sqrt {3a} }}{4}\) from the corner. Together they form a smaller cube of edge length\(\frac{a}{2}\).

Therefore, distance between centres of two nearest tetrahedral voids in the lattice is also \(\frac{a}{2}\).

Face-centered cubic (fcc) refers to a crystal structure consisting of an atom at each cube corner and an atom in the center of each cube face. It is a close-packed plane in which on each face of the cube atoms are assumed to touch along face diagonals.

Which primitive unit cell has unequal edge lengths (a ≠ b ≠ c) and all axial angles different from 90°?

  1. Hexagonal
  2. Monoclinic
  3. Tetragonal
  4. Triclinic

Answer (Detailed Solution Below)

Option 4 : Triclinic

Solid State Question 11 Detailed Solution

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Concept:

Triclinic primitive unit cell has dimensions as, a ≠ b ≠ c and α ≠ β ≠ 90°.

Among the seven basic or primitive crystalline systems, the triclinic system is most unsymmetrical, the triclinic. In other cases, edge length and axial angles are given as follows:

Hexagonal: a = b ≠ c and α = β = 90°, γ = 120°

Monoclinic: a ≠ b ≠ c and α = γ = 90°, β ≠ 90°

Tetragonal: a ≠ b ≠ c and α = β = γ = 90°

  • The parameters of all other crystal systems are given below:

F3 Vinanti Teaching 27.04.23 D1 
F4 Vinanti Teaching 02.05.23 D1

NaCl crystals appear yellow due to:

  1. Schottky defect
  2. F - centres
  3. Frenkel defect
  4. Interstitials

Answer (Detailed Solution Below)

Option 2 : F - centres

Solid State Question 12 Detailed Solution

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Concept:

Stoichiometric Defects:

  • The Stoichiometric defects are those in which the imperfections are such that the ratio between the cations and anions remains the same as represented by the chemical formula.
  • Types of Stoichiometric defects are:-Schottky defect and Frenkel defect.

Non-Stoichiometric Defect:

  • The Non- Stoichiometric defects are those in which the imperfections are such that the ratio between the cations and anions differs from the ideal chemical formula.
  • Here, the balance of + and - charges are maintained either by having extra electrons or +ve charge.
  • Its types include:- Metal excess (i. due to anion vacancy and ii. due to interstitial cations) and Metal Deficiency.

The classification of defects in solids is given below:

F1 Shraddha Utkarsh 02.02.2021 D23

Explanation:

Metal excess defect due to anionic vacancies:

  • Alkali halide like NaCl and KCl show this type of defect.
  • When crystals of NaCl is heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal.
  • The Cl ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl.
  • This happens by loss of an electron by sodium atoms to form Na+ ions.
  • The released electrons diffuse into the crystal and occupy anionic sites.

F1 Shraddha Utkarsh 02.02.2021 D24

  • As a result, the crystal now has an excess of sodium.
  • The anionic sites occupied by unpaired electrons are called F-centres.
  • They impart a yellow colour to the crystals of NaCl.
  • The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.
  • Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).

Hence, NaCl crystals appear yellow due to F - centres.

Additional Information

 Schottky Defect:

  • In order to maintain electrical neutrality, the number of missing cations and anions are equal.
  • It decreases the density of the substance.
  • The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes.
  • For example, NaCl, KCl, CsCl and AgBr.

Frenkel Defect:

  • The smaller ion (usually cation) is dislocated from its normal site to an interstitial site.
  • It creates a vacancy defect at its the original site and an interstitial defect at its new location.
  • Frenkel defect is also called a dislocation defect.
  • It does not change the density of the solid.
  • Frenkel defect is shown by the ionic substance in which there is a large difference in the size of ions.
  • For example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and Ag+ ions.

Interstitial Defect:

  • When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect
  • This defect increases the density of the substance.
  • Ionic solids must always maintain electrical neutrality. 

A compound of formula A2 B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms?

  1. hcp lattice- \(A,\frac{2}{3}\) tetrahedral voids- B
  2. hcp lattice- \(A,\frac{1}{3}\) tetrahedral voids- B
  3. hcp lattice- \(B,\frac{1}{3}\) tetrahedral voids- A
  4. hcp lattice- \(B,\frac{2}{3}\) tetrahedral voids- A

Answer (Detailed Solution Below)

Option 3 : hcp lattice- \(B,\frac{1}{3}\) tetrahedral voids- A

Solid State Question 13 Detailed Solution

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Concept:

The total effective number of atoms in hcp unit lattice = Number of octahedral voids in hcp = 6

∴ Number of tetrahedral voids (TV) in hcp 

= 2 × Number of atoms in hcp lattice 

= 2 × 6 = 12

As the formula of the lattice is A2B3

Suppose, A B

\(\left( {\frac{1}{3} \times TV} \right)\) (hcp)

\(⇒ {\rm{\;}}\frac{1}{3} \times 12\) (6)

\(⇒ \frac{2}{3}\) 1

⇒ 2, 3

So, \(A = \frac{1}{3}\) tetrahedral voids, B = hcp lattice.

Which of the following statements best describes the characteristics of a crystalline solid?

  1. Easily experiences geometric deformation
  2. There is no precise melting point
  3. Has an uneven 3-dimensional layouts
  4. Changes abruptly from solid to liquid when heated

Answer (Detailed Solution Below)

Option 4 : Changes abruptly from solid to liquid when heated

Solid State Question 14 Detailed Solution

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The correct answer is Changes abruptly from solid to liquid when heated. Key Points

  • Crystalline solids have a highly ordered and repeating atomic arrangement, which gives them their characteristic geometric shapes.
  • They have a precise melting point, which is the temperature at which the ordered atomic arrangement breaks down and the solid transitions to a liquid state.
  • The 3-dimensional layout of a crystalline solid is highly regular and symmetrical, with atoms or molecules arranged in a repeating pattern.
  • When heated, a crystalline solid undergoes an abrupt phase transition from solid to liquid, without passing through an intermediate liquid crystal phase.

Additional Information

  • Crystalline solids are resistant to geometric deformation due to their highly ordered atomic arrangement.
  • Crystalline solids have a precise melting point, which is a characteristic feature of their structure.
  • The layout of a crystalline solid is highly regular and symmetrical, with no unevenness in the arrangement of atoms or molecules.
  • The key term 'crystalline solid' refers to a type of solid material with a highly ordered atomic structure, which gives it unique physical and chemical properties.

Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

  1. A2BO4
  2. AB2O4
  3. A2B2O
  4. A4B2O

Answer (Detailed Solution Below)

Option 2 : AB2O4

Solid State Question 15 Detailed Solution

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Concept:

Atom A:

The number of atoms in octahedral void = 4 atoms

Half of the octahedral voids = \(\frac{4}{2} = 2\) atoms

Number of A atoms = 2 atoms

Atom B:

Number of atoms in CCP lattice structure = 4 atoms

Number of B atoms = 4 atoms

Atom Oxygen:

Number of atoms in all tetrahedral voids = 8

Number of oxygen atoms = 8 atoms

Thus, the compound formed is: A2B4O8 ⇒ AB2O4
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