The Solid State: CBSE Class 12 Chemistry NCERT Guide

Classification of Solids: Crystalline vs. Amorphous

Solids are primarily classified into two categories based on the arrangement of their constituent particles (atoms, molecules, or ions): Crystalline and Amorphous.

Crystalline solids have a highly ordered, three-dimensional arrangement of particles that repeats throughout the entire crystal. This long-range order gives them sharp melting points, definite heats of fusion, and anisotropic properties (meaning their physical properties like electrical resistance or refractive index can be different when measured along different directions). Examples include sodium chloride (NaCl), quartz, and diamond.

Amorphous solids, on the other hand, have a disordered, random arrangement of particles, much like a supercooled liquid. They lack long-range order but may exhibit short-range order. This results in them melting over a range of temperatures and being isotropic (their physical properties are the same in all directions). Common examples are glass, rubber, and plastics. When cut, crystalline solids give clean, flat cleavages, whereas amorphous solids produce irregular fractures.

Crystal Lattices and Unit Cells

Crystal Lattice

A regular three-dimensional arrangement of points in space, representing the constituent particles (atoms, ions, or molecules) of a crystal.

Lattice Point

Each point in a crystal lattice that represents one constituent particle of the solid.

Unit Cell

The smallest repeating portion of a crystal lattice which, when repeated in three dimensions, generates the entire lattice. It is characterized by its edge lengths (a, b, c) and the angles between them (α, β, γ).

How to Calculate the Density of a Unit Cell

1. Step 1: Determine the number of atoms per unit cell (Z) — First, identify the type of cubic unit cell: Simple Cubic (Z=1), Body-Centred Cubic (BCC, Z=2), or Face-Centred Cubic (FCC, Z=4). This value represents the effective number of atoms belonging to that single unit cell.
2. Step 2: Calculate the mass of the unit cell — The mass of one unit cell is the number of atoms (Z) multiplied by the mass of a single atom (m). The mass of a single atom is its molar mass (M) divided by Avogadro's number (Nₐ). So, mass of unit cell = Z × (M / Nₐ).
3. Step 3: Calculate the volume of the unit cell — For a cubic unit cell with edge length 'a', the volume is simply a³. Be careful with units! Edge length is often given in picometers (pm) or Angstroms (Å), which must be converted to centimeters (cm) for density in g/cm³. (1 pm = 10⁻¹⁰ cm, 1 Å = 10⁻⁸ cm).
4. Step 4: Apply the Density Formula — Density (ρ) is mass divided by volume. Combining the steps above, we get the final formula: ρ = (Mass of unit cell) / (Volume of unit cell) = (Z × M) / (a³ × Nₐ) where Z = number of atoms, M = molar mass, a = edge length, and Nₐ = Avogadro's number (6.022 × 10²³ mol⁻¹).

Worked Examples

• Example 1: Density Calculation An element with molar mass 27 g/mol forms a cubic unit cell with an edge length of 405 pm. If its density is 2.7 g/cm³, what is the nature of the cubic unit cell (i.e., is it SC, BCC, or FCC)? Solution: Step 1: List the given values and convert units. M = 27 g/mol ρ = 2.7 g/cm³ a = 405 pm = 405 × 10⁻¹⁰ cm = 4.05 × 10⁻⁸ cm Nₐ = 6.022 × 10²³ mol⁻¹ Step 2: Rearrange the density formula to solve for Z. ρ = (Z × M) / (a³ × Nₐ) => Z = (ρ × a³ × Nₐ) / M Step 3: Substitute the values and calculate Z. Z = (2.7 g/cm³ × (4.05 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹) / (27 g/mol) Z = (2.7 × (66.43 × 10⁻²⁴) × 6.022 × 10²³) / 27 Z = (2.7 × 6.643 × 10⁻²³ × 6.022 × 10²³) / 27 Z = (108.0) / 27 Z ≈ 4 Final Answer: Since Z=4, the cubic unit cell is Face-Centred Cubic (FCC).
• Example 2: Formula of a Compound A compound is formed by two elements, X and Y. Atoms of element Y (as anions) make up the ccp (cubic close-packed) lattice, and those of element X (as cations) occupy all the octahedral voids. What is the formula of the compound? Solution: Step 1: Determine the number of Y atoms. The ccp lattice is equivalent to an FCC lattice. Let's assume the number of atoms of Y in the unit cell is 'n'. In an FCC lattice, the effective number of atoms is 4. So, n = 4. Step 2: Determine the number of X atoms. The number of octahedral voids in a ccp/fcc structure is equal to the number of atoms in the lattice. Number of octahedral voids = n = 4. Since atoms of X occupy all the octahedral voids, the number of X atoms in the unit cell is also 4. Step 3: Find the simplest ratio of X to Y. The ratio of atoms X : Y is 4 : 4, which simplifies to 1 : 1. Final Answer: The formula of the compound is XY.

Exam Traps and Important Points

Here are key areas to focus on for your exams:

• Value of Z: Memorize the effective number of atoms (Z) for each unit cell: Simple Cubic (Z=1), BCC (Z=2), and FCC/CCP (Z=4). A mistake here will make your entire calculation incorrect.
• Unit Conversion is CRITICAL: Density calculations almost always require converting the edge length from picometers (pm) or Angstroms (Å) to centimeters (cm). A common mistake is using the wrong conversion factor. Remember: 1 pm = 10⁻¹⁰ cm and 1 Å = 10⁻⁸ cm. Double-check this step!
• Voids: Understand the relationship between the number of atoms (N) in a close-packed structure and the number of voids. Number of Octahedral Voids = N; Number of Tetrahedral Voids = 2N. Problems on formula determination depend heavily on this.
• Defects: Clearly distinguish between Schottky defect (missing ions, decreases density) and Frenkel defect (displaced ion, density remains same). Questions often test this difference.

Practice Questions with Solutions

• Q: Silver crystallizes in an FCC lattice. If the edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g/cm³, calculate the atomic mass of silver. A: Step 1: Identify the knowns for an FCC lattice. For FCC, Z = 4. We are given a = 4.07 × 10⁻⁸ cm, ρ = 10.5 g/cm³, and Nₐ = 6.022 × 10²³ mol⁻¹. Step 2: Rearrange the density formula to solve for molar mass (M). ρ = (Z × M) / (a³ × Nₐ) => M = (ρ × a³ × Nₐ) / Z. Step 3: Substitute the values and calculate. M = (10.5 g/cm³ × (4.07 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹) / 4. M = (10.5 × 67.42 × 10⁻²⁴ × 6.022 × 10²³) / 4. M = 426.3 / 4 ≈ 106.57 g/mol. Final answer: The atomic mass of silver is approximately 106.57 u.
• Q: What is the difference between Ferromagnetism and Antiferromagnetism? A: Step 1: Define Ferromagnetism. In ferromagnetic substances, the magnetic moments of the domains (groups of ions) are spontaneously aligned in the same direction, even in the absence of an external magnetic field. This results in a strong attraction to magnetic fields. Example: Iron (Fe). Step 2: Define Antiferromagnetism. In antiferromagnetic substances, the magnetic moments of adjacent domains are aligned in opposite directions, cancelling each other out. This results in a net magnetic moment of zero. Example: Manganese(II) oxide (MnO). Final answer: The key difference is the alignment of magnetic moments: parallel and reinforcing in ferromagnetism, leading to strong magnetism, versus anti-parallel and cancelling in antiferromagnetism, leading to zero net magnetism.
• Q: Niobium crystallizes in a body-centred cubic (BCC) structure. If the density is 8.55 g/cm³, calculate the atomic radius of niobium given its atomic mass is 93 u. A: Step 1: Find the edge length 'a' using the density formula. For BCC, Z = 2. ρ = (Z × M) / (a³ × Nₐ). Rearranging, a³ = (Z × M) / (ρ × Nₐ) = (2 × 93 g/mol) / (8.55 g/cm³ × 6.022 × 10²³ mol⁻¹) = 186 / (5.149 × 10²⁴) ≈ 36.12 × 10⁻²⁴ cm³. So, a = (36.12 × 10⁻²⁴)¹/³ cm ≈ 3.30 × 10⁻⁸ cm. Step 2: Relate edge length 'a' to atomic radius 'r' for a BCC structure. For BCC, the atoms touch along the body diagonal. The relationship is √3a = 4r. So, r = (√3 × a) / 4. Step 3: Calculate the radius 'r'. r = (1.732 × 3.30 × 10⁻⁸ cm) / 4 ≈ 1.43 × 10⁻⁸ cm. Final answer: The atomic radius of niobium is approximately 1.43 × 10⁻⁸ cm or 143 pm.
• Q: What type of stoichiometric defect is shown by (i) ZnS and (ii) AgBr? A: Step 1: Analyze the structure of ZnS. Zinc sulfide (ZnS) has a large difference in the size of the cation (Zn²⁺) and the anion (S²⁻). The smaller Zn²⁺ ion can easily fit into interstitial sites. Step 2: Identify the defect in ZnS. Due to the size difference, ZnS shows the Frenkel defect, where the smaller cation is dislocated from its normal lattice site to an interstitial site. This does not change the density of the solid. Step 3: Analyze the structure of AgBr. Silver bromide (AgBr) is a unique case where the size difference between Ag⁺ and Br⁻ is intermediate. This allows for both types of defects to occur. Final answer: (i) ZnS shows the Frenkel defect. (ii) AgBr is interesting because it shows both Schottky and Frenkel defects.

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