Chapter 6 — Solid State Exercise Solutions
Solid State numericals require careful unit conversion. Always convert edge length from pm or Å to cm before substituting in the density formula.
Key Concepts Quick Recap
| Term | Key Fact |
|---|
| Crystalline solid | Long-range order; sharp melting point; anisotropic |
| Amorphous solid | Short-range order; no sharp melting point; isotropic |
| Unit cell | Smallest repeating unit of crystal lattice |
| SC (Z=1) | 1 atom per unit cell; packing efficiency 52.4% |
| BCC (Z=2) | 2 atoms per unit cell; packing efficiency 68% |
| FCC (Z=4) | 4 atoms per unit cell; packing efficiency 74% |
| Bragg equation | nλ = 2d sinθ — used to find inter-planar spacing d |
| Schottky defect | Ions missing from lattice → density decreases |
| Frenkel defect | Ion displaced to interstitial position → density unchanged |
| F-centre | Electron trapped in anion vacancy → crystal becomes coloured |
| Packing efficiency | Fraction of unit cell volume occupied by atoms |
Formula Recap
| Concept | Formula |
|---|
| Density of unit cell | ρ = ZM / (a³Nₐ) |
| Bragg equation | nλ = 2d sinθ |
| Packing efficiency | (Volume of atoms / Volume of unit cell) × 100 |
| Relation (FCC) | 4r = √2 a → r = a√2/4 |
| Relation (BCC) | 4r = √3 a → r = a√3/4 |
| Relation (SC) | 2r = a → r = a/2 |
I. Choose the Best Answer (MCQs)
1. An element crystallises as FCC. The number of atoms per unit cell is: (a) 1 (b) 2 (c) 4 (d) 6
Answer(c) 4 FCC: 8 corner atoms (×1/8) + 6 face atoms (×1/2) = 1 + 3 = 4 atoms per unit cell.
2. Which crystal defect decreases the density of a crystal? (a) Frenkel defect (b) Interstitial defect (c) Schottky defect (d) Metal excess defect
Answer(c) Schottky defect In Schottky defect, pairs of cation and anion vacancies are created. Mass decreases while volume stays same → density decreases.
3. The Bragg equation is nλ = 2d sinθ. What does 'd' represent? (a) Diameter of atom (b) Inter-planar spacing (c) Edge length (d) Wavelength
Answer(b) Inter-planar spacing 'd' is the distance between successive planes of atoms in the crystal lattice.
4. The packing efficiency of BCC is approximately: (a) 52% (b) 68% (c) 74% (d) 90%
Answer(b) 68% BCC packing efficiency = (2 × 4/3 πr³) / a³ × 100 = 68.04% ≈ 68%.
5. An ionic crystal shows colour when exposed to excess of one ion. This is due to: (a) Schottky defect (b) Frenkel defect (c) F-centre (metal excess defect) (d) Interstitial defect
Answer(c) F-centre NaCl becomes yellow in excess Na vapour — Na deposits and electrons occupy Cl⁻ vacancies (F-centres). These trapped electrons absorb visible light → colour.
6. In a BCC unit cell, the atoms touch along: (a) Face diagonal (b) Edge (c) Body diagonal (d) Corners only
Answer(c) Body diagonal In BCC, the body centre atom touches the 8 corner atoms along the body diagonal → 4r = √3 a.
7. CsCl has a body-centred cubic structure. The coordination number of Cs⁺ is: (a) 4 (b) 6 (c) 8 (d) 12
Answer(c) 8 In CsCl structure, each Cs⁺ is surrounded by 8 Cl⁻ at the corners of a cube → CN = 8. Note: this is NOT true BCC — CsCl has one formula unit per unit cell (Z = 1).
II. Short Answer Questions
Q1. Distinguish between crystalline and amorphous solids with two examples each.
Answer
| Property | Crystalline | Amorphous |
|---|
| Order | Long-range periodic order | Short-range order only |
| Melting point | Sharp, definite | Range of temperatures |
| Anisotropy | Anisotropic | Isotropic |
| Cleavage | Clean, regular | Irregular fracture |
| Examples | NaCl, diamond, quartz | Glass, rubber, plastics |
Q2. Calculate the number of atoms per unit cell in SC, BCC, and FCC.
Simple Cubic (SC):
Corner atoms: 8 × (1/8) = 1 Total Z = 1
Body-Centred Cubic (BCC):
Corner: 8 × (1/8) = 1
Body centre: 1 × 1 = 1 Total Z = 2
Face-Centred Cubic (FCC):
Corner: 8 × (1/8) = 1
Face centres: 6 × (1/2) = 3 Total Z = 4
Q3. What is the difference between Schottky and Frenkel defects?
Answer
| Feature | Schottky Defect | Frenkel Defect |
|---|
| Nature | Both cation and anion vacancies | Ion displaced to interstitial site |
| Effect on density | Density decreases | Density unchanged |
| Crystal type | Ionic with similar sized ions (NaCl, KCl) | Ionic with large size difference (AgBr, ZnS) |
| Electrical conductivity | Slightly increased | Slightly increased |
Q4. What is packing efficiency? Compare SC, BCC, and FCC.
Answer
Packing efficiency = (volume occupied by atoms / total volume of unit cell) × 100
| Crystal | Z | Atom volume | Packing efficiency |
|---|
| SC | 1 | 1 × 4/3πr³ | 52.4% |
| BCC | 2 | 2 × 4/3πr³ | 68% |
| FCC | 4 | 4 × 4/3πr³ | 74% |
FCC (and HCP) give the closest packing — used by many metals (Cu, Al, Ag, Au).
Worked Examples
Example 1 — Density of a unit cell
An element (molar mass 60 g/mol) crystallises in FCC structure with edge length 400 pm. Find its density.
Given: Z = 4 (FCC), M = 60 g/mol, a = 400 pm = 4×10⁻⁸ cm
Nₐ = 6.022×10²³ mol⁻¹
ρ = ZM / (a³ × Nₐ)
= (4 × 60) / ((4×10⁻⁸)³ × 6.022×10²³)
= 240 / (6.4×10⁻²³ × 6.022×10²³)
= 240 / 38.54
≈ 6.23 g cm⁻³
Example 2 — Edge length from density
An FCC metal (M = 108 g/mol, ρ = 10.5 g/cm³). Find edge length.
ρ = ZM / (a³ × Nₐ) → a³ = ZM / (ρ × Nₐ)
a³ = (4 × 108) / (10.5 × 6.022×10²³)
= 432 / (6.323×10²⁴)
= 6.83×10⁻²³ cm³
a = (6.83×10⁻²³)^(1/3) = 4.08×10⁻⁸ cm = 408 pm
(This matches silver: Z=47, M=108 g/mol)
Example 3 — Packing efficiency of FCC
In FCC: atoms touch along face diagonal
4r = √2 a → r = a√2/4
Volume of 4 atoms = 4 × (4/3)πr³ = 4 × (4/3)π(a√2/4)³
= 4 × (4/3)π × (a³ × 2√2/64)
= π a³ √2 / 6
Packing efficiency = (π√2/6) / 1 × 100 = 74.05% ≈ 74%
Example 4 — Bragg equation
X-rays of wavelength 154 pm hit a crystal. Reflection at θ = 45°. Find d (n=1).
nλ = 2d sinθ
1 × 154 = 2 × d × sin 45°
154 = 2 × d × 0.7071
d = 154 / 1.4142 = 108.9 pm ≈ 109 pm
Example 5 — Ionic crystal radius ratio
Radius ratio (r⁺/r⁻) determines coordination number and structure:
r⁺/r⁻ CN Structure
< 0.155 2 Linear
0.155–0.225 3 Triangular planar
0.225–0.414 4 Tetrahedral (ZnS type)
0.414–0.732 6 Octahedral (NaCl type)
0.732–1.000 8 Cubic (CsCl type)
Common Mistakes
- Using wrong Z value (SC=1, BCC=2, FCC=4).
- Not converting edge length from pm to cm (1 pm = 10⁻¹⁰ m = 10⁻⁸ cm).
- Forgetting to cube the edge length in the density formula.
- Confusing Schottky (density decreases) with Frenkel (density unchanged).
- Using radius in density formula instead of edge length.
Exam Checklist
- Crystal type and Z are identified first.
- Edge length is converted to cm before use in formula.
- All steps shown: formula → substitution → calculation → unit.
- Defect type is matched to its effect on density.
- Final answer has correct unit (g cm⁻³ for density, pm for length).
