Test: Solid State - 3 - Chemistry MCQ

To calculate the Miller indices of crystal planes that intersect the crystal axes at the coordinates (2a, -3b, -3c), follow these steps:

• Identify the intercepts: The plane cuts the x-axis at 2a, the y-axis at -3b, and the z-axis at -3c.
• Convert intercepts to fractions:
  • For the x-axis: 1/(2a) = 1/2
  • For the y-axis: 1/(-3b) = -1/3
  • For the z-axis: 1/(-3c) = -1/3
• Take the reciprocals: This results in:
  • x = 1/2
  • y = -1/3
  • z = -1/3
• Clear the fractions: To do this, multiply through by the least common multiple (LCM) of the denominators, which is 6:
  • 1/2 × 6 = 3
  • -1/3 × 6 = -2
  • -1/3 × 6 = -2
• Resulting Miller indices: Therefore, the final indices are (3, -2, -2).

These Miller indices represent the orientation of the crystal plane in relation to the axes. They are essential for understanding the properties of the crystal structure.

- Calcium fluoride (CaF\(_2\)) crystallizes in the fluorite structure.
- In this structure:
- Calcium ions (Ca\(^{2+}\)) are at the center of a cube surrounded by 8 fluoride ions (F\(^-\)). Thus, the coordination number for Ca\(^{2+}\) is 8.
- Fluoride ions (F\(^-\)) are surrounded by 4 calcium ions (Ca\(^{2+}\)). Thus, the coordination number for F\(^-\) is 4.
- Therefore, the coordination numbers for the cation and anion are 8 and 4, respectively. This corresponds to option D: 8, 4.

The atomic radius can be calculated using the molecular weight and density of the element. For an element with a body-centred cubic unit cell, the formula to estimate the atomic radius (r) is:

• Density (ρ) = Mass/Volume
• The volume of the unit cell (V) can be expressed as: V = a³, where a is the edge length.
• The mass of the unit cell (M) can be derived from the molecular weight.

Using these relationships, we can find the atomic radius as follows:

• Calculate the volume of the unit cell using the density and the molecular weight.
• Determine the edge length (a) from the volume.
• The atomic radius (r) can then be calculated using the formula: r = a/2.

In this case:

• Molecular weight = 137.3 g/mol
• Density = 3.62 g/cm^–3

After performing the calculations, the estimated atomic radius is approximately 2.2 × 10^–8 cm.

The (111) plane in a face-centred cubic (FCC) structure is significant for reflection. This plane is known for its high atomic density, leading to strong diffraction effects. Here are some key points regarding its reflection characteristics:

• High Density: The (111) plane is densely packed with atoms, which enhances the likelihood of allowed reflections.
• Symmetry: The symmetry of FCC lattices allows the (111) planes to efficiently reflect X-rays and other forms of radiation.
• Common Usage: In materials science, the (111) plane is often examined because it provides clear insights into the material's properties.
• Other Planes: While (100) and (110) planes are also present in FCC structures, the (111) plane exhibits the most pronounced reflection characteristics.

Overall, the (111) plane is the most notable for showing allowed reflections in FCC structures, making it crucial for various applications in crystallography and materials analysis.

The crystal lattices can be distinguished based on their diffraction patterns:

• Crystal A: Shows diffraction from the (200) and (111) planes, but not from the (110) plane.
• Crystal B: Diffraction occurs from the (110) and (100) planes, but not from the (111) plane.

From the diffraction data:

• Crystal A corresponds to a body-centred cubic (BCC) structure.
• Crystal B corresponds to a face-centred cubic (FCC) structure.

When considering the options:

• A: BCC for A and FCC for B.
• B: FCC for A and BCC for B.
• C: Both A and B are BCC.
• D: Both A and B are FCC.

Based on the analysis, the most accurate classification is:

• Option B: A is FCC and B is BCC.

The percentage of voids in a Body-Centred Cubic (BCC) lattice is calculated based on the arrangement of atoms within the structure.

• The BCC lattice structure consists of one atom at each corner of a cube and one atom in the centre.
• This arrangement leads to a certain amount of empty space, or voids, between the atoms.
• The percentage of void space can be determined using the formula:
• Void percentage = (Volume of voids / Total volume) × 100%
• In the case of a BCC lattice, the void percentage is approximately 32%.
• Understanding the void percentage is crucial for applications in materials science, as it affects the properties of metals and alloys.

The second order diffraction from the (100) planes of a cubic crystal can be understood in relation to its diffraction patterns.

• The diffraction from the (100) planes is characterised by specific orders.
• In this context, the second order diffraction refers to the condition where the path difference between waves leads to constructive interference.
• For cubic crystals, this process is intricately linked to the Miller indices of the planes involved.
• The (200) plane represents a doubling of the spacing compared to the (100) plane.
• This means that the first order diffraction from the (200) plane corresponds to the same physical condition as the second order from the (100) plane.

Thus, understanding these relationships is crucial for interpreting diffraction patterns in crystallography.

8:8 coordination refers to a specific arrangement of ions in a crystal structure, where each ion is surrounded by eight others. The following crystals exhibit this coordination:

• NH₄Cl - Ammonium chloride features a structure that allows for 8:8 coordination.
• AlFe - This compound also shows a similar coordination pattern.
• NH₄Br - Like ammonium chloride, ammonium bromide supports 8:8 coordination.

However, while MnO is a common compound, it does not exhibit 8:8 coordination.

⟶ NH₄I and MgO crystallizes in rock salt structure (NaCl(NaCl Type)Type) where NH⁺⁴ and Mg⁺² occupies all octahedral voids and I⁻ and O²⁻ forms FCC lattice having coordination number 6:6

⟶ MnO also have 6:6 coordination as Mn⁺² forms FCC lattice and O²⁻ occupies all tetrahedral voids. 

 (Inverse of NaCl structure)

⟶ ZnS have 4:4 coordination as Zn⁺ occupies alternate tetrahedral voids and O²⁻ forms FCC lattice.

4:4 coordination refers to a structural arrangement where each atom is surrounded by four others in a tetrahedral configuration. In the context of the mentioned crystals, here's a breakdown of their coordination:

• HgS (Cinnabar): This crystal exhibits a 4:4 coordination. Each mercury (Hg) atom is surrounded by four sulphur (S) atoms.
• NH4F (Ammonium Fluoride): This compound does not display 4:4 coordination. The coordination number is typically less than four.
• SiC (Silicon Carbide): Silicon carbide also showcases 4:4 coordination, with each silicon (Si) atom bonded to four carbon (C) atoms.
• NaCl (Sodium Chloride): This crystal structure is based on a 3:3 coordination model, where each sodium (Na) ion is surrounded by three chloride (Cl) ions.

In summary, the crystals that demonstrate 4:4 coordination are HgS and SiC.

Statement A: In a CsCl crystal structure, the coordination number for each type of ion is 8. This means each ion is surrounded by eight other ions, which is typical for this type of ionic compound.

Statement B: A metal that exhibits a body-centred cubic (bcc) structure does not have a coordination number of 12. Instead, it has a coordination number of 8, where each atom is surrounded by eight others.

Statement C: In ionic crystals, a unit cell shares some of its ions with adjacent unit cells. This sharing is essential for forming the overall structure of the crystal.

Statement D: The unit cell length for NaCl is 552 pm. This measurement indicates the distance between the ions in the crystal lattice.

Defects in Solids can be understood through the following points:

• The Frenkel defect occurs when an atom or ion is displaced from its normal position, creating a vacancy. This defect is more likely to happen when there is a small size difference between cations and anions.

• It is classified as a dislocation defect because it involves disruption in the crystal lattice structure.

• The formation of F-centres happens when an electron gets trapped in the lattice, leading to colour changes in the material.

• In contrast, Schottky defects do affect the physical properties of solids, including density and electrical conductivity.

The unit cell of a mineral features a cubic close packed (ccp) arrangement of oxygen atoms. Within this structure:

• Aluminium ions occupy a fraction m of the available octahedral holes.
• Magnesium ions fill a fraction n of the tetrahedral holes.

Understanding the proportions of m and n is essential for determining the mineral's composition and properties. Here are the key points:

• The ccp structure consists of a specific number of octahedral and tetrahedral holes.
• Each unit cell has octahedral holes that can accommodate cations like aluminium.
• It also contains tetrahedral holes suitable for smaller cations like magnesium.

By analysing the arrangement and occupancy of these holes, we can derive the values of m and n, which are crucial for understanding the mineral's overall structure.

Cubic close packed (ccp) structure has several key characteristics:

• The number of nearest neighbours for an atom in the topmost layer is 12.

• The packing efficiency of atoms in this structure is 74%.

• For each atom, there are 1 octahedral void and 2 tetrahedral voids.

• The unit cell edge length is 2√2 times the radius of the atom.

Statement A: In an anti-fluorite structure, the anions form a face-centred cubic (FCC) lattice, while the cations occupy all the tetrahedral voids.

Statement B: When the radius of the cations and anions is 0.2 Å and 0.95 Å respectively, the coordination number of the cation in the crystal is 4.

Statement C: In a Frenkel defect, an atom or ion is transferred from a lattice site to an interstitial position.

Statement D: The density of a crystal does not always increase due to a substitution impurity defect.

• Statement A is correct.
• Statement B is incorrect as the coordination number is influenced by the relative sizes of the ions.
• Statement C is correct.
• Statement D is incorrect; density may decrease depending on the mass and size of the substituting impurity.

Improved Solutions

• Cs⁺ ions are surrounded by eight neighbouring ions in the second nearest position.

• In a two-dimensional hexagonal close packed layer, each sphere has six voids around it.

• If cations and anions have radii of 0.3 Å and 0.4 Å, respectively, the coordination number of the cation in the crystal is 6.

• In AgCl, the silver ion moves from its lattice position to an interstitial site; this type of defect is known as a Frenkel defect.

To calculate the number of silver atoms on a given surface area, follow these steps:

• Determine the volume of silver: Using its density, we find the volume of silver that corresponds to the area of interest.
• Density of silver: The density is 10.5 g/cm^–3. This means that 1 cm³ of silver weighs 10.5 grams.
• Convert area to volume: For a surface area of 10^–12 m², we can calculate the volume of a thin layer of silver (assuming a thickness, for example, of 1 nm or 10^–9 m).
• Calculate mass: Use the formula: mass = density × volume. Convert the volume to cm³ for compatibility with the density unit.
• Find the number of moles: Use the formula: moles = mass / atomic weight. The atomic weight of silver is 108 g/mol.
• Calculate the number of atoms: Multiply the number of moles by Avogadro's number (approximately 6.022 × 10²³ atoms/mol) to get the total number of silver atoms.

Through this method, the value of x in the scientific notation Y × 10^x can be derived, yielding the answer as 7.

In a ZnS (Blend) structure, the number of particles present in an octahedral void is zero.

Here's a brief explanation:

• In the ZnS structure, which is known as zinc blende, the arrangement of atoms plays a crucial role.
• The octahedral voids refer to specific spaces between atoms where additional particles can potentially reside.
• However, in the case of zinc blende, these voids do not contain any additional particles.
• This structure is primarily composed of zinc and sulphur ions, forming a stable lattice without any octahedral inclusions.

The hcp unit cell (hexagonal close-packed) has a specific arrangement that creates space within its structure. The amount of empty space is crucial in understanding how atoms are packed in materials.

• The hcp structure consists of atoms closely packed in a hexagonal arrangement.
• In this configuration, not all spaces are occupied by atoms, leading to empty spaces.
• The total empty space in an hcp unit cell is approximately 26%.
• This empty space affects the material's properties, such as density and strength.

Understanding the empty space in the hcp unit cell is essential for applications in materials science and engineering, as it can influence how materials behave under various conditions.