Understanding Intermolecular Forces, Unit Cells, and Crystal Defects

Posted on Mar 16, 2025 in Chemistry

Intermolecular Forces in Molecular Crystals

Q: What are the various types of intermolecular forces of attraction that hold molecules of molecular crystals together?

A:

  1. Weak dipole-dipole interactions in polar molecules such as solid HCl, H2O, SO2, etc., which possess a permanent dipole moment.
  2. Very weak dispersion or London forces in non-polar molecules such as solid CH4, H2, etc. These forces are also involved in monoatomic solids like argon, neon, etc.
  3. Intermolecular hydrogen bonds in solids such as H2O (ice), NH3, HF, etc.

Understanding Unit Cells

Q: Explain the term: Unit cell

A:

  1. The smallest repeating structural unit of a crystalline solid is called a unit cell.
  2. When the unit cells are stacked together to generate the crystal, each unit cell shares its faces, edges, and corners with neighboring unit cells.
  3. The geometric shape of a unit cell is the same as that of the macroscopic crystal. For example, if the crystal has a cubic shape, the unit cell will also have its constituent particles arranged to form a tiny cube.
  4. The dimensions of a unit cell along the three axes are denoted by the symbols a, b, and c. The angles between these axes are represented by the symbols α, β, and γ.

Determining Crystal Lattice Type

Q: A unit cell of a metal has an edge length of 288 pm & a density of 7.86 g/cm³. Determine the type of the crystal lattice.

A:

Given: a = 288 pm = 2.88 × 10-8 cm, M = 56 g/mol, density (ρ) = 7.86 g/cm³, n = ?

Formula: Density (ρ) = Mn / a³NA

Calculation: n = 7.86 × (2.88)³ × 6.022 × 1023 / 56 ≈ 2.018

Nature of Cubic Unit Cell

Q: An element with a molar mass of 27 g/mol forms a cubic unit cell with an edge length of 405 pm. If the density of the element is 2.7 g/cm³, what is the nature of the cubic unit cell?

A:

Given: a = 405 pm = 4.05 × 10-8 cm, M = 27 g/mol, Density = 2.7 g/cm³, n = ?

Formula: Density (ρ) = Mn / a³NA

Calculation: n = 2.7 × (4.05)³ × 6.022 × 1023 / 27 ≈ 4.00

Packing Efficiency of BCC Structure

Q: Calculate the packing efficiency of a metal crystal that has a BCC structure.

A:

Packing efficiency of a metal crystal in a body-centered cubic lattice:

Step 1: Radius of sphere (particle):

In a BCC unit cell, particles occupy the corners, and one particle is at the center of the cube. The particle at the center of the cube touches two corner particles along the diagonal of the cube. To obtain the radius of the particle (sphere), the Pythagorean theorem is applied.

For triangle FED, ∠FED = 90°

FD² = FE² + ED² = a² + a² = 2a² (because FE = ED = a) …(1)

For triangle AFD, ∠ADF = 90°

AF² = AD² + FD²… (2)

Substitution of equation (1) into (2) yields:

AF² = a² + 2a² = 3a² (because AD = a) or AF = √3 a …(3)

From the figure, AF = 4r

Substitution for AF from equation (3) gives √3a = 4r hence, r = (√3 / 4) × a …(4)

Step 2: Volume of sphere:

Volume of sphere particle = (4/3) × π × r³

Substitution for r from equation (4) gives:

Volume of one particle = (4/3)π × [(√3 / 4)a]³ = (4/3)π × (√3)³ / 64 a³ = (√3πa³) / 16

Step 3: Total volume of particles:

A BCC unit cell contains 2 particles. Hence, the volume occupied by particles in a BCC unit cell = 2 × (√3πa³) / 16 = (√3πa³) / 8…(5)

Step 4: Packing efficiency:

Packing efficiency = (Volume occupied by particles in unit cell / Total volume of unit cell) × 100

= (√3πa³ / 8a³) × 100 ≈ 68%

Thus, 68% of the total volume in a BCC unit lattice is occupied by atoms, and 32% is empty space or void volume.

Vacancy Defect Explained

Q: What is a vacancy defect? Explain with a suitable diagram.

A:

  1. During the crystallization of a solid, a particle is missing from its regular site in the crystal lattice.
  2. The missing particle creates a vacancy in the lattice structure. Thus, some of the lattice sites are vacant because of missing particles, as shown in the figure. Such a crystal is said to have a vacancy defect.
  3. The vacancy defect can also develop when the substance is heated.
  4. Due to the absence of particles, the mass of the substance decreases. However, the volume remains unchanged. As a result, the density of the substance decreases.