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

Last updated on Jul 2, 2025

Latest Solid State MCQ Objective Questions

Solid State Question 1:

The donor band close to _______ and acceptor band close to _______ are found in n and p type semiconductors respectively. 

  1. conduction band. conduction band 
  2. conduction band. valence band 
  3. Valence band, conduction band  
  4. Valence band. valence band 

Answer (Detailed Solution Below)

Option 2 : conduction band. valence band 

Solid State Question 1 Detailed Solution

CONCEPT:

Donor Band and Acceptor Band in Semiconductors

  • Semiconductors can be classified into n-type and p-type based on the type of doping.
  • In an n-type semiconductor:
    • Donor atoms are introduced into the material.
    • These donor atoms add extra electrons, which occupy energy levels just below the conduction band.
  • In a p-type semiconductor:
    • Acceptor atoms are introduced into the material.
    • These acceptor atoms create energy levels just above the valence band, where electrons can be captured, leaving holes in the valence band.

EXPLANATION:

  • The donor band is an energy level close to the conduction band in n-type semiconductors. This is because donor atoms contribute additional electrons, which are easily excited to the conduction band.
  • The acceptor band is an energy level close to the valence band in p-type semiconductors. Acceptor atoms create holes in the valence band, allowing electrons to move in and create electrical conduction.
  • Thus, the donor band is close to the conduction band, and the acceptor band is close to the valence band.

Therefore, the correct answer is Conduction band, Valence band.

Solid State Question 2:

If interplanar distance is 4 A, the x-rays of wavelength 4 A are diffracted and is known as first order reflection. Then calculate the angle. 

  1. 30° 
  2. 45° 
  3. 60° 
  4. 90°

Answer (Detailed Solution Below)

Option 1 : 30° 

Solid State Question 2 Detailed Solution

CONCEPT:

Bragg's Law for X-ray Diffraction

  • Bragg's Law is used to determine the angles at which X-rays are diffracted by a crystal lattice.
  • The equation for Bragg's Law is:

    nλ = 2d sinθ

  • Where:
    • n: Order of reflection (1 for first order reflection).
    • λ: Wavelength of X-rays.
    • d: Interplanar distance in the crystal.
    • θ: Angle of diffraction.

EXPLANATION:

  • Given data:
    • Interplanar distance (d) = 4 Å.
    • Wavelength of X-rays (λ) = 4 Å.
    • Order of reflection (n) = 1 (for first order reflection).
  • Substituting into Bragg's Law:

    nλ = 2d sinθ

    1 × 4 = 2 × 4 × sinθ

    4 = 8 × sinθ

    sinθ = 4 / 8 = 0.5

  • Using the inverse sine function:

    θ = sin-1(0.5)

    θ = 30°

Therefore, the angle of diffraction (θ) is 30°.

Solid State Question 3:

The total number of elements present in FCC and BCC crystal systems, respectively are: 

  1. 2 and 3
  2. 3 and 4
  3. 4 and 2
  4. 2 and 4

Answer (Detailed Solution Below)

Option 3 : 4 and 2

Solid State Question 3 Detailed Solution

CONCEPT:

Total Number of Atoms in FCC and BCC Unit Cells

  • FCC (Face-Centered Cubic) and BCC (Body-Centered Cubic) are two types of crystal structures commonly found in solid materials.
  • The total number of atoms in a unit cell can be calculated based on the arrangement of atoms in the cell:
    • Atoms at the corners of the unit cell contribute 1/8 of an atom per corner (as each corner atom is shared among eight adjacent unit cells).
    • Atoms located at the center of faces of the unit cell contribute 1/2 of an atom per face (as each face atom is shared between two adjacent unit cells).
    • Atoms at the center of the unit cell contribute fully (1 atom).

EXPLANATION:

  • For the FCC structure:
    • There are 8 corner atoms, contributing 8 × (1/8) = 1 atom.
    • There are 6 face-centered atoms, contributing 6 × (1/2) = 3 atoms.
    • Total number of atoms = 1 + 3 = 4 atoms.
  • For the BCC structure:
    • There are 8 corner atoms, contributing 8 × (1/8) = 1 atom.
    • There is 1 atom at the center of the unit cell, contributing fully (1 atom).
    • Total number of atoms = 1 + 1 = 2 atoms.
  • Therefore, the total number of atoms in FCC and BCC unit cells are 4 and 2, respectively.

Correct Answer: Option 3 (4 and 2)

Solid State Question 4:

Metallic crystal structure of calcium is:  

  1. Face centered cubic 
  2. Hexagonal close packing  
  3. Simple cubic  
  4. Body centered cubic  

Answer (Detailed Solution Below)

Option 1 : Face centered cubic 

Solid State Question 4 Detailed Solution

CONCEPT:

Metallic Crystal Structure

  • Metallic elements crystallize into various structures, which are arrangements of atoms in space.
  • The most common metallic crystal structures are:
    • Face-Centered Cubic (FCC)
    • Body-Centered Cubic (BCC)
    • Hexagonal Close Packing (HCP)
    • Simple Cubic (SC) is rare for metals.
  • The type of metallic crystal structure depends on the element and its atomic properties.

EXPLANATION:

  • The metallic crystal structure of calcium is face-centered cubic (FCC). Specifically, it has an FCC lattice structure with an edge length of 0.556 nm

So the correct answer is FCC.

Solid State Question 5:

Which of the following metal hydrides has the crystal structure of rutile? 

  1. MgH2
  2. LiH
  3. NaH
  4. BaH2

Answer (Detailed Solution Below)

Option 1 : MgH2

Solid State Question 5 Detailed Solution

CONCEPT:

Rutile Crystal Structure

  • The rutile crystal structure is a tetragonal structure commonly observed in compounds like titanium dioxide (TiO2).
  • This structure consists of metal cations (e.g., Ti4+ in TiO2) surrounded by oxygen anions (or hydride ions in metal hydrides) in a specific arrangement.
  • The coordination number of the metal cation in the rutile structure is 6, meaning it is surrounded by 6 anions in an octahedral geometry.

EXPLANATION:


Rutile structure

MgH2

  • MgH2 has the rutile crystal structure. This is because:
    • Magnesium (Mg2+) forms a coordination number of 6 with hydride ions (H-).
    • This results in a tetragonal rutile-like arrangement, similar to TiO2 but with H- ions replacing O2-.
  • Other compounds:
    • LiH and NaH crystallize in a rock salt structure (not rutile).
    • BaH2 adopts an orthorhombic structure (not rutile).

Therefore, the correct answer is Option 1: MgH2.

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.

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

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

Effective no. of

lattice points

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:

The distance between the centers of two nearest tetrahedral voids in the lattice and also the minimum distance between two tetrahedral voids is ..

In fcc (face centered cubic), tetrahedral voids are located on the body diagonal at a distance of  from the corner. Together they form a smaller cube of edge length.

Therefore, distance between centres of two nearest tetrahedral voids in the lattice is also .

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:

 

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:

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.

  • 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-  tetrahedral voids- B
  2. hcp lattice-  tetrahedral voids- B
  3. hcp lattice-  tetrahedral voids- A
  4. hcp lattice-  tetrahedral voids- A

Answer (Detailed Solution Below)

Option 3 : hcp lattice-  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

 (hcp)

 (6)

 1

⇒ 2, 3

So,  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 =  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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