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

CONCEPT:

Packing Efficiency in a Face-Centered Cubic (FCC) Lattice

• FCC (Face-Centered Cubic) is a common lattice structure where atoms are located at:
  • 8 corners (each contributes 1/8th to the unit cell)
  • 6 face centers (each contributes 1/2 to the unit cell)
• Total atoms per unit cell in FCC = 8 × (1/8) + 6 × (1/2) = 4 atoms
• Each atom is considered a hard sphere touching its nearest neighbors along the face diagonal.

EXPLANATION:

√2 × a = 4r → r = (√2 × a) / 4

Total volume of atoms = 4 × (4/3)π[(√2 × a)/4]³ = (16/3)π × (2√2 a³ / 64)

• In FCC, nearest neighbors touch along the face diagonal.
• Length of face diagonal = √2 × a (a = edge length)
• Along this diagonal, 4 radii fit (2 atoms touching):
• Volume of unit cell: a³
• Total volume occupied by atoms: 4 atoms × (4/3)πr³
• Substitute r = (√2 × a) / 4:
• Simplifying:

  = (16/3)π × (2√2 a³ / 64) = πa³ × (16 × 2√2) / (3 × 64)

  ≈ 0.7405 × a³

Hence, packing efficiency = volume occupied / volume of cube = 0.74

CONCEPT:

Common Structural Types in Solid State Chemistry

• Spinel Structure:
  • General formula: AB₂O₄
  • ‘A’ and ‘B’ are metal cations; oxygen forms a close-packed lattice.
  • ‘A’ occupies 1/8 of tetrahedral sites; ‘B’ occupies 1/2 of octahedral sites.
• Perovskite Structure:
  • General formula: ABX₃
  • ‘A’ is a large cation; ‘B’ is a smaller metal cation; ‘X’ is usually oxygen or a halide.
  • Common in many fluorides and oxides like KMnF₃, CaTiO₃

EXPLANATION:

• ZnFe₂O₄:
  • Fits the general formula AB₂O₄, where Zn²⁺ = A and Fe³⁺ = B
  • This confirms a spinel structure
• KMnF₃:
  • Fits the ABX₃ type formula with K⁺ = A, Mn³⁺ = B, and F^- = X
  • This confirms a perovskite structure

Therefore, the correct structural types are Spinel (for ZnFe₂O₄) and Perovskite (for KMnF₃)

CONCEPT:

Radius Ratio Rule and Coordination Geometry

• The radius ratio is the ratio of the cation radius (r⁺) to the anion radius (r^-).
• This ratio helps predict the coordination number and geometry of an ionic compound.
• Typical radius ratio ranges:
  • 0.15 – 0.22 → CN = 3 (Trigonal planar)
  • 0.22 – 0.41 → CN = 4 (Tetrahedral)
  • 0.41 – 0.73 → CN = 6 (Octahedral)
  • 0.73 – 1.00 → CN = 8 (Cubic)

EXPLANATION:

• Cation radius = 74 pm
• Anion radius = 170 pm
• Calculate the radius ratio:
  • r⁺/r^- = 74 / 170 ≈ 0.435
• Since 0.435 lies between 0.41 and 0.73:
  • The coordination number = 6
  • The geometry = Octahedral

Therefore, the correct answer is 6, Octahedral.

CONCEPT:

Quantum Confinement Effect in Nanomaterials

• Quantum confinement occurs when the size of a semiconductor crystal becomes comparable to or smaller than the exciton Bohr radius (typically in nanometers).
• This effect is prominent in quantum dots, which are nanoparticles with all three dimensions confined.
• Due to confinement, the motion of charge carriers (electrons and holes) is restricted, leading to quantization of energy levels.

EXPLANATION:

Sourced: https://www.researchgate.net/publication/306001639/figure/fig4/AS:393164155310108@1470749048148/Schematic-representation-of-the-quantum-confinement-effects-the-bandgap-or-HOMO-LUMO.png

• As the size of the quantum dot decreases:
  • The energy levels become more discrete.
  • More energy is required to excite an electron from the valence band to the conduction band.
• This leads to an increase in the band gap energy of the semiconductor.
• The material starts to absorb and emit light at shorter wavelengths (blue-shift) as the particle size decreases.

Therefore, due to quantum confinement, the band gap of semiconductors increases.

Concept:

Miller Indices

• Miller indices are a notation system used to describe the orientation of a plane in a crystal lattice. These indices are based on the intercepts of the plane with the crystal axes.
• The Miller indices (h, k, l) are determined by the following steps:
  • Find the intercepts of the plane along the crystal axes. If the intercepts are not finite, the plane is parallel to the corresponding axis.
  • Take the reciprocal of the intercepts.
  • Clear the fractions to obtain the smallest integer values for h, k, and l.

Explanation:

• Given intercepts of the plane along the axes are:
  • Along the x-axis: 2a
  • Along the y-axis: 3b
  • Along the z-axis: 2c
• The Miller indices are found by taking the reciprocals of these intercepts:
  • 1 / (2a) = 1/2
  • 1 / (3b) = 1/3
  • 1 / (2c) = 1/2
• To eliminate fractions, multiply each reciprocal by 6 (the least common multiple of 2, 3, and 2):
  • 1/2 × 6 = 3
  • 1/3 × 6 = 2
  • 1/2 × 6 = 3
• Thus, the Miller indices for the plane are (3, 2, 3).

Hence, the correct answer is (3, 2, 3).

Concept:

• The material which is not a good conductor or a good insulator is called a semiconductor.
• For example: Silicon
• The charge carriers which are present in more quantity in a semiconductor compared to other particles are called the majority charge carrier.
• There are two types of semiconductors:

P-type semiconductor:

• The semiconductor having holes as majority charge carriers and electrons as a minority charge carrier is called as P-type semiconductor.

N-type semiconductor:

• The semiconductor having electrons as majority charge carriers and holes as a minority charge carrier are called as N-type semiconductor

Explanation:

• Fermi level in a semiconductor: It is that energy level in the energy-band-diagram of semiconductor for which the probability of occupancy (i.e., the presence of main current carriers electrons or holes) becomes half. 

From the above diagram, it is clear that the Fermi level is close to the conduction band edge.

• As Temperature increases then the fermi level moves towards the centre of forbidden gap irrespective of whether it is p-type or n-type.
  • For n-type material as the doping increases then fermi level moves towards the conduction band.
  • For p-type material as the doping increases then fermi level moves towards the valence band.
• If doping is very large, then the Fermi level may move into the conduction band.

Concept:

Solid state physics: The study of the physical properties of solids, including how atoms pack together to form structures such as crystals.

Crystallography: The science of determining the arrangement of atoms in solids. Understanding structures such as BCC lattices is fundamental to crystallography.

Materials Science: The understanding of how substances are packed plays a fundamental role in determing their properties. For example, a metal's packing efficiency can affect its hardness, melting point, and other physical properties. More tightly packed atoms generally lead to harder, denser materials.

Explanation:-

In BCC, the unit cell consists of one atom at each corner of the cube and one atom at the center of the cube. Since the corner atoms are shared among 8 adjacent cells, each unit cell gets 1/8 of each of the 8 corner atoms, so in total, it gets one full atom from the corners. Additionally, there's one complete atom at the center of the cube. So altogether, there are 2 atoms in a BCC unit cell.

The volume of a sphere (each atom can be considered a sphere here) is given by (4/3)πr³, where r refers to the radius of an atom.

The total volume occupied by 2 atoms in the BCC lattice cell would then be 2*(4/3)πr³.

b² = a² + a2
\(b= \sqrt{2a^2}=\sqrt{2}a\)

c² = a² + b²
\(c=\sqrt{a^2 + 2a^2}= \sqrt{3a^2}= \sqrt{3}a\)

c=4r
so, \(a=\frac{4r}{\sqrt3}\)

packing efficiency= volm of 2 sphere/ total volm of unit cell x100

\(P.E=\frac{ 2\times 4\pi r^3\times (\sqrt3)^3}{3 (4r)^3}\times 100\\ P.E=\frac{\sqrt(3)\pi}{8}\times 100\\ \)

P.E=68%
Hence, the packing efficiency for a BCC lattice is approximately 68%.

Conclusion:-

So, The packing efficiency (in %) of spheres for a body-centered cubic (bcc) is 68%

Concept:

→ The Miller index notation is a system used to describe the orientation of crystal planes in a lattice.

→ The notation consists of three integer numbers enclosed in square brackets [h k l], known as Miller indices, which represent the reciprocal intercepts of the plane with the crystallographic axes.

Explanation:

b represents y- axis and a represents x-axis and c parallel to this is y-axis, the miller index corresponding to c-axis will be infinity.

→ In the figure, the plane intersects the a-axis at 4 unit, the b-axis at 2 units, and is parallel to the c-axis which is at 0 unit.

→ Intercept value along a, b and c axes are (4, 2, \(\propto \))

→ Now, take the reciprocal of these intercepts:

\(\frac{1}{4},\frac{1}{2},\frac{1}{\propto }\) ⇒ \(\frac{1}{4},\frac{1}{2},0\)

Multiply all of these with highest integer i.e., 4 we will get

[1,2,0]

→ This notation indicates that the plane is perpendicular to the c-axis and contains the direction of the vector [1, 2, 0] in the crystal lattice.

Conclusion:
The correct answer is option 2.

The Correct answer is 6,6.

Key Points

• The coordination number refers to the number of ions of opposite charge that surround an ion in a crystal lattice.
• In an NaCl lattice, each Na+ ion is surrounded by 6 Cl- ions.
• Similarly, each Cl- ion is surrounded by 6 Na+ ions.
• This arrangement is known as the face-centered cubic (FCC) lattice.
• The high coordination number ensures the lattice is highly stable and forms a strong ionic bond.
• This type of structure is also known as a rock-salt structure.

 Additional Information

• Face-Centered Cubic (FCC) Lattice
  • In an FCC lattice, atoms are located at each of the corners and the centers of all the cubic faces.
  • This arrangement is highly efficient and is observed in many metallic crystals and ionic compounds.
• Rock-Salt Structure
  • The rock-salt structure is a specific type of FCC lattice where each ion is surrounded by six ions of the opposite charge.
  • This structure is common in alkali halides like NaCl, KCl, and others.

Concept:

We can use Bragg's law to solve this problem:

nλ = 2d sinθ

where n is the order of the reflection (which is 1 in this case, since it is the first order reflection), λ is the wavelength of the X-ray, d is the distance between the crystal planes, and θ is the angle of incidence of the X-ray.

For a cubic primitive crystal, the distance between adjacent (111) planes is given by:

d = \(\frac{a}{\sqrt{3}}\)

where a is the length of the unit cell.

Explanation:

Substituting this into Bragg's law and solving for a, we get:

a =\(\frac{ d\times\sqrt{3}}{sin\theta }\)

a = \( \frac{\lambda}{2\times sin\theta }\times\sqrt{3}\)

a = \( \frac{173pm}{2\times sin30^{o}} \times\sqrt{3}\)

a ≈ 299.636 pm

Conclusion:
Therefore, the length of the unit cell is closest to 300 pm.

Explanation:- 

• CsCl forms a BCC-packed structure, where Eight Cs⁺ ions are bonded to Eight Cl^- ions.
• CaF₂ has a tetrahedron FCC-packed or Closed packed structure, this lattice structure is also known as the Fluorite structure. In which Four Ca²⁺ ions are coordinated with Eight F^- ions.

Important Points 

• In Cesium Chloride the Cesium ion is at the body center and the chloride ions are present at the corners.
• In Calcium Fluoride the  4 Calcium ions are linked with 8 Fluoride ions.
• The radius ratio of Cs/ Cl is 0.93 and that of Ca/ F is 0.73.

Concept:

• Miller Indices are imaginary mathematical representation. 
• It represents the set of parallel crystallographic planes passing using the numeric values (h, k, l)
• If the intercept of the plane at imaginary three coordinate axes is known, miller indices can be calculated by simply taking the reciprocal of intercept values and converting them to integers.
• In Miller Index notation, if a plane has 0 value with respect to certain axis, it means it is parallel to that axis.

Explanation.

The given planes are parallel to b-axis, that means the miller index corresponding to b-axis will be 0.

To get the miller indices values for other two axes, first define an origin for different planes and find the intercept values.

For plane (i)

Intercept value along a, b and c axes are (2, 0, 2)

Corresponding Miller Indices value is

= \((\frac{1}{2}0 \frac{1}{2}) \)  

= \((101)\) 

For plane (ii)

Intercept value along a, b and c axes are (4, 0, 2)

Corresponding Miller Indices value is 

= \((\frac{1}{4}0 \frac{1}{2}) \)  

= \((102)\)                

Conclusion:

Hence, the Miller indices of the planes parallel to the b axis and intersecting the a and c axis as given are (i) 101, (ii) 102                  

Concept:

• Plutonium(Pu) is crystallizing in the monoclinic lattice which can be compared to the parallelepiped.
• the volume of unit cell with monoclinic lattice can be calculated as : \(Volume = abcsin\beta \). Here, a, b, c are the lattice constants and \(\beta \) is the angle greater than 90°.
• the density of unit cell is calculated by dividing the total mass of atoms present in the unit cell and the volume of unit cell and its formula is written as:\(Density = {\frac{M\times N}{N_A\times Volume\;of\;unit\;cell}} \;\;\;-(1)\)

             where,

  \(M = molar\;mass\;of\; each\;atom \)

  \(N = number\;of\;atoms\;in\;the\;unit\;cell\)

  \(N_A= avogadro\;number \)​

Explanation:

 Before using the density formula, we must find the volume of   the unit cell

 \(volume \;of\; the\; unit\; cell = abc sin\beta \) 

 given,

 \(a=620\;pm=6.2 \times 10^{-10}m\)

 \(b=480\;pm=4.8\times10^{-10}m\)

 \(c= 1100\;pm=11\times10^{-10}m\)

 \(sin108 ^0=0.98\)

putting these values gives,

 \(Volume = (6.2\times10^{-10}m)\times(4.8\times10^{-10}m)\times(11\times10^{-10}m)\times0.98\)

  \(=3.208 \times10^{-28} \;m^3\)

Now, density of the unit cell can be calculated using formula in equation (1), Given that \(M =244\;gmol^{-1},\;N=16,\;N_A=6.022 \times10^{23}\)

\(density=\frac{244gmol^{-1}\times 16}{6.022\times 10^{23}mol^{-1}\times 3.208\times 10^{-28}m^3}\)

 \(=2.021\times10^{-5}\;gm^{-3}\)

 \(= 20.21\; gcm^{-3} \)

Conclusion:

The density of the monoclinic unit cell in which plutonium is crystallizing is 20.23 gcm^-3

CONCEPT:

Properties of ZnS (Zinc Sulfide)

• ZnS shows both cubic and hexagonal structures:
  • ZnS can exist in two distinct crystal structures: cubic (sphalerite) and hexagonal (wurtzite). Therefore, Statement I is correct.
• Sphalerite exhibits the ZnS structure:
  • Sphalerite is the cubic form of ZnS, where Zn and S are tetrahedrally coordinated. This structure is commonly referred to as the ZnS structure. Therefore, Statement II is correct.
• ZnS is a semiconductor:
  • Zinc sulfide is a semiconductor with a wide band gap. It is used in optoelectronic devices such as phosphors and LEDs. Therefore, Statement III is correct.
• Precipitation of ZnS from an acidic solution:
  • This statement is incorrect because ZnS precipitates in basic solutions when H₂S is passed through, but not in acidic solutions. Therefore, Statement IV is incorrect.

CONCLUSION:

• Statements I, II, and III are correct, while Statement IV is incorrect. Thus, the correct option is the one that includes I, II, and III only.

Concept:

• A lattice point is a point at the intersection of two or more grid lines in a regularly spaced array of points, which is a point lattice.
• The lattice is a 2D representation of a 3D crystal on the plane of a paper. It is constructed by using certain points known as lattice points and lattice lines.
• The unit cell is a smaller part of a crystal that is unique for a particular crystal substance.
• In crystallography, miller indices form a notation system for lattice planes in crystal lattices. In particular, a family of lattice planes of a given Bravais lattice is determined by three integers h, k, and ℓ, the Miller indices.

​Explanation:

• In crystallography, x-rays are allowed to be passed through a crystal in a series of parallel beams so that they diffract from different sets of parallel miller planes.
• The allowed cubic planes for a BCC (body-centered cubic) lattice in crystallography depend upon miller indices (h, k, ℓ) as, that the sum of h, k, and ℓ, must be even (\({\rm{h + k + l\; = \;even}}\)).
• Now, for the (100) plane, the value of h, k, and ℓ is 1, 0, and 0. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 1 + 0 + 0}}\)

\({\rm{ = 1}}\)

\({\rm{ = odd}}\)

• For the (111) plane, the value of h, k, and ℓ is 1, 1, and 1. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 1 + 1 + 1}}\)

\({\rm{ = 3}}\)

​\({\rm{ = odd}}\)

• For the (200) plane, the value of h, k, and ℓ is 2, 0, and 0. The sum of h, k, and ℓ is an even number and this cubic plane is allowed for powder diffraction.

\({\rm{h + k + l = 2 + 0 + 0}}\)

\({\rm{ = 2}}\)

\({\rm{ = even}}\)

• For the (210) plane, the value of h, k, and ℓ is 2, 1, and 0. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 2 + 1 + 0}}\)

\({\rm{ = 3}}\)

​\({\rm{ = odd}}\)

Conclusion:

• Hence, the Miller indices of the second allowed reflection in the powder diffraction pattern of iron would be (200)