🧱 Comprehensive Solid State Chemistry Notes
Class XII - NCERT & Beyond
📌 1. Introduction to Solid State
🔍 Fundamental Characteristics
- Solids have definite shape, volume, and mass
- Particles are closely packed with strong intermolecular forces
- Minimal molecular motion - only vibrational motion
- High density compared to liquids and gases
- Incompressible nature
🏗️ Classification of Solids
| Classification Basis |
Types |
Examples |
Key Features |
| Arrangement |
Crystalline |
NaCl, Diamond, Quartz |
Regular arrangement, sharp melting point |
|
Amorphous |
Glass, Plastic, Rubber |
Irregular arrangement, gradual softening |
| Bonding |
Ionic |
NaCl, CsCl, ZnS |
Electrostatic forces |
|
Covalent |
Diamond, SiC, SiO2 |
Covalent bonds |
|
Metallic |
Cu, Fe, Au |
Metallic bonding |
|
Molecular |
Ice, I2, CO2 |
Van der Waals forces |
🔍 2. Detailed Comparison: Crystalline vs Amorphous Solids
| Property |
Crystalline Solid |
Amorphous Solid |
Examples |
| Structure |
Long-range order |
Short-range order only |
Diamond vs Glass |
| Melting Point |
Sharp, definite |
Gradual, over a range |
Ice melts at 0°C vs Glass softens gradually |
| Anisotropy |
Anisotropic |
Isotropic |
Calcite shows double refraction |
| Cleavage |
Distinct cleavage planes |
Irregular fracture |
Mica cleaves in sheets |
| Heat of Fusion |
Definite value |
No definite value |
Ice: 334 J/g |
| X-ray Diffraction |
Sharp, well-defined patterns |
Broad, diffuse patterns |
Salt vs Plastic |
🔬 Real-world Applications:
- Crystalline: Semiconductors (Si, Ge), Gems (Diamond, Ruby), Pharmaceuticals
- Amorphous: Glass fibers, Plastics, Amorphous metals for transformers
🔬 3. Detailed Classification of Crystalline Solids
3.1 Ionic Solids
Characteristics:
- Composed of cations and anions
- Held together by electrostatic forces
- High melting and boiling points
- Conduct electricity in molten state
- Hard but brittle
Examples with Properties:
| Compound |
Structure |
Melting Point (°C) |
Coordination Number |
| NaCl |
Rock Salt |
801 |
6:6 |
| CsCl |
Cesium Chloride |
645 |
8:8 |
| ZnS |
Zinc Blende |
1020 |
4:4 |
| CaF2 |
Fluorite |
1360 |
8:4 |
3.2 Molecular Solids
Sub-types:
- Non-polar molecules: Held by weak London forces (Ar, CH4)
- Polar molecules: Dipole-dipole interactions (HCl, SO2)
- Hydrogen-bonded: H-bonds (H2O, NH3)
Properties:
- Low melting and boiling points
- Soft and compressible
- Poor conductors of electricity
- Soluble in polar solvents
3.3 Covalent Network Solids
Examples and Properties:
| Compound |
Structure |
Hardness |
Electrical Conductivity |
| Diamond |
3D network of C-C bonds |
Hardest natural material |
Insulator |
| Graphite |
Layered structure |
Soft, slippery |
Conductor |
| Silicon Carbide |
Diamond-like |
Very hard |
Semiconductor |
| Quartz (SiO2) |
3D silicate network |
Hard |
Insulator |
3.4 Metallic Solids
Electron Sea Model:
- Metal atoms lose valence electrons
- Electrons form a "sea" around metal cations
- Explains conductivity, malleability, ductility
Properties:
- Excellent electrical and thermal conductors
- Malleable and ductile
- Metallic luster
- Variable hardness and melting points
🧱 4. Crystal Lattices and Unit Cells
4.1 Fundamental Concepts
Crystal Lattice: A regular 3D arrangement of points in space
Unit Cell: The smallest repeating unit that generates the entire lattice
4.2 Unit Cell Parameters
- Edge lengths: a, b, c (in Angstroms or pm)
- Angles: α (between b and c), β (between a and c), γ (between a and b)
- Volume: V = abc√(1 + 2cosα cosβ cosγ - cos²α - cos²β - cos²γ)
4.3 Seven Crystal Systems
| Crystal System |
Edge Relations |
Angle Relations |
Examples |
| Cubic |
a = b = c |
α = β = γ = 90° |
NaCl, Diamond |
| Tetragonal |
a = b ≠ c |
α = β = γ = 90° |
TiO2, SnO2 |
| Orthorhombic |
a ≠ b ≠ c |
α = β = γ = 90° |
BaSO4, KNO3 |
| Hexagonal |
a = b ≠ c |
α = β = 90°, γ = 120° |
Graphite, ZnO |
| Monoclinic |
a ≠ b ≠ c |
α = γ = 90°, β ≠ 90° |
Gypsum, β-sulfur |
| Triclinic |
a ≠ b ≠ c |
α ≠ β ≠ γ ≠ 90° |
CuSO4·5H2O |
| Rhombohedral |
a = b = c |
α = β = γ ≠ 90° |
Calcite, Quartz |
📐 5. Types of Cubic Unit Cells
5.1 Simple Cubic (SC)
Structure:
● --- ● --- ●
| | |
● --- ● --- ●
| | |
● --- ● --- ●
Number of atoms per unit cell: 8 × (1/8) = 1
Coordination number: 6
Packing efficiency: 52.4%
5.2 Body-Centered Cubic (BCC)
Structure:
● --- ● --- ●
| \ | / |
● ---●● --- ●
| / | \ |
● --- ● --- ●
Number of atoms per unit cell: 8 × (1/8) + 1 × 1 = 2
Coordination number: 8
Packing efficiency: 68.0%
5.3 Face-Centered Cubic (FCC)
Structure:
● -◐- ● -◐- ●
◐ ◐ ◐ ◐
● -◐- ● -◐- ●
◐ ◐ ◐ ◐
● -◐- ● -◐- ●
Number of atoms per unit cell: 8 × (1/8) + 6 × (1/2) = 4
Coordination number: 12
Packing efficiency: 74.0%
5.4 Relationship between Edge Length and Atomic Radius
Simple Cubic: a = 2r
Body-Centered Cubic: a = (4r)/√3
Face-Centered Cubic: a = 2√2 r
📦 6. Close Packing in Solids
6.1 Two-Dimensional Packing
Square Close Packing:
- Spheres arranged in square arrays
- Each sphere touches 4 others
- Packing efficiency = 78.5%
Hexagonal Close Packing (2D):
- Most efficient 2D arrangement
- Each sphere touches 6 others
- Packing efficiency = 90.7%
6.2 Three-Dimensional Close Packing
Cubic Close Packing (CCP) or FCC
Layer sequence: A-B-C-A-B-C...
Examples: Cu, Ag, Au, Al
Coordination number: 12
Packing efficiency: 74%
Hexagonal Close Packing (HCP)
Layer sequence: A-B-A-B-A-B...
Examples: Mg, Zn, Ti, Co
Coordination number: 12
Packing efficiency: 74%
6.3 Voids in Close Packing
| Type of Void |
Coordination Number |
Number per Sphere |
Size (radius ratio) |
| Tetrahedral |
4 |
2 |
0.225 |
| Octahedral |
6 |
1 |
0.414 |
📏 7. Density Calculations and Numerical Problems
7.1 Density Formula
ρ = (Z × M) / (a³ × NA)
Where:
ρ = density (g/cm³)
Z = number of atoms per unit cell
M = molar mass (g/mol)
a = edge length (cm)
NA = Avogadro's number = 6.022 × 10²³ mol⁻¹
7.2 Worked Examples
Example 1: Silver (FCC structure)
Given: Edge length = 4.07 × 10⁻⁸ cm, Molar mass = 107.87 g/mol
Solution:
For FCC: Z = 4
ρ = (4 × 107.87) / ((4.07 × 10⁻⁸)³ × 6.022 × 10²³)
ρ = 431.48 / (6.75 × 10⁻²³ × 6.022 × 10²³)
ρ = 431.48 / 40.65 = 10.6 g/cm³
Example 2: Iron (BCC structure)
Given: Density = 7.87 g/cm³, Molar mass = 55.85 g/mol
Find: Edge length
Solution:
For BCC: Z = 2
7.87 = (2 × 55.85) / (a³ × 6.022 × 10²³)
a³ = 111.7 / (7.87 × 6.022 × 10²³)
a³ = 2.355 × 10⁻²³ cm³
a = 2.87 × 10⁻⁸ cm
Example 3: Atomic Radius Calculation
Given: Copper (FCC), edge length = 3.61 × 10⁻⁸ cm
Find: Atomic radius
Solution:
For FCC: a = 2√2 r
r = a / (2√2) = 3.61 × 10⁻⁸ / (2√2)
r = 3.61 × 10⁻⁸ / 2.828 = 1.28 × 10⁻⁸ cm
7.3 Packing Efficiency Calculations
Packing Efficiency = (Volume occupied by atoms / Total volume) × 100%
Example 4: Packing Efficiency of BCC
Solution:
Number of atoms per unit cell = 2
Volume of atoms = 2 × (4/3)πr³
For BCC: a = 4r/√3, so r = a√3/4
Volume of atoms = 2 × (4/3)π(a√3/4)³ = πa³√3/8
Volume of unit cell = a³
Packing efficiency = (πa³√3/8) / a³ × 100% = π√3/8 × 100%
Packing efficiency = 68.0%
⚙️ 8. Imperfections in Solids (Crystal Defects)
8.1 Point Defects
8.1.1 Stoichiometric Defects
Vacancy Defect (Schottky Defect)
- Missing atoms from lattice sites
- Decreases density
- Common in ionic crystals with similar sized ions
- Examples: NaCl, KCl, CsCl
Normal: Na⁺ Cl⁻ Na⁺ Cl⁻
Defect: Na⁺ ⬜ Na⁺ ⬜ (⬜ = vacancy)
Interstitial Defect
- Extra atoms in interstitial sites
- Increases density
- Rare in ionic crystals
- More common in metals with small impurity atoms
Frenkel Defect
- Ion displaced from lattice site to interstitial site
- Density remains constant
- Common when cation is smaller than anion
- Examples: AgCl, AgBr, ZnS
Normal: Ag⁺ Cl⁻ Ag⁺ Cl⁻
Defect: ⬜ Cl⁻ Ag⁺ Cl⁻ (Ag⁺ in interstitial)
8.1.2 Non-Stoichiometric Defects
Metal Excess Defects
- F-Centers: Electrons trapped in anion vacancies
- Gives color to crystals
- Example: NaCl heated in Na vapor becomes yellow
- Metal Interstitials: Extra metal atoms in interstitial sites
Metal Deficiency Defects
- Missing metal atoms compensated by higher oxidation states
- Common in transition metal compounds
- Example: FeO (Fe0.95O) - some Fe²⁺ oxidized to Fe³⁺
- Creates p-type semiconductivity
8.2 Effects of Defects
| Defect Type |
Effect on Density |
Effect on Conductivity |
Other Effects |
| Vacancy |
Decreases |
May increase ionic conductivity |
Weakens crystal |
| Interstitial |
Increases |
May increase |
Distorts lattice |
| F-centers |
No change |
Increases electronic conductivity |
Imparts color |
| Metal deficiency |
Decreases |
Creates p-type conductivity |
Variable stoichiometry |
🔋 9. Electrical Properties of Solids
9.1 Classification Based on Conductivity
Conductivity (σ) = 1/Resistivity (ρ)
Units: S/m or Ω⁻¹m⁻¹
| Type |
Conductivity Range (S/m) |
Examples |
Band Gap (eV) |
| Conductors |
10⁴ - 10⁸ |
Cu, Ag, Au, Al |
0 (overlapping bands) |
| Semiconductors |
10⁻⁶ - 10⁴ |
Si, Ge, GaAs |
0.1 - 3.0 |
| Insulators |
< 10⁻⁸ |
Diamond, Glass, Ceramics |
> 3.0 |
9.2 Band Theory
Energy Band Diagrams:
CONDUCTOR:
Conduction Band ████████ (partially filled)
────────────────────────
Valence Band ████████ (overlaps)
SEMICONDUCTOR:
Conduction Band ░░░░░░░░ (small gap)
────── ~1eV ──────
Valence Band ████████ (filled)
INSULATOR:
Conduction Band ░░░░░░░░ (large gap)
────── >3eV ──────
Valence Band ████████ (filled)
9.3 Semiconductors
9.3.1 Intrinsic Semiconductors
- Pure semiconductors (Si, Ge)
- Conductivity increases with temperature
- Equal number of electrons and holes
- σ = σ₀e^(-Eg/2kT)
9.3.2 Extrinsic Semiconductors
n-Type (Negative)
- Doped with Group 15 elements (P, As, Sb)
- Donor impurities provide extra electrons
- Majority carriers: electrons
- Minority carriers: holes
Si - Si - Si
| | |
Si - P - Si (P has 5 valence electrons)
| /|\ |
Si - Si - Si (extra electron becomes mobile)
p-Type (Positive)
- Doped with Group 13 elements (B, Al, Ga)
- Acceptor impurities create holes
- Majority carriers: holes
- Minority carriers: electrons
Si - Si - Si
| | |
Si - B - Si (B has 3 valence electrons)
| /|\ |
Si - Si - Si (hole created, can accept electron)
9.4 Applications of Semiconductors
| Device |
Principle |
Application |
| p-n Junction Diode |
Rectification |
AC to DC conversion |
| Solar Cell |
Photovoltaic effect |
Solar energy conversion |
| LED |
Electroluminescence |
Light emission |
| Transistor |
Amplification |
Switching, amplification |
🧲 10. Magnetic Properties of Solids
10.1 Types of Magnetism
Diamagnetism
- All electrons paired
- Weakly repelled by magnetic field
- χ < 0 (negative susceptibility)
- Temperature independent
- Examples: NaCl, C₆H₆, H₂O
Paramagnetism
- Unpaired electrons present
- Weakly attracted to magnetic field
- χ > 0 (positive susceptibility)
- Follows Curie law: χ = C/T
- Examples: O₂, NO, transition metal salts
Ferromagnetism
- Parallel alignment of spins
- Strongly attracted to magnetic field
- Large positive susceptibility
- Shows hysteresis
- Examples: Fe, Co, Ni, CrO₂
Antiferromagnetism
- Antiparallel alignment of equal spins
- Net magnetic moment = 0
- Above Néel temperature becomes paramagnetic
- Examples: MnO, FeO, CoO, NiO
Ferrimagnetism
- Antiparallel alignment of unequal spins
- Net magnetic moment ≠ 0
- Shows ferromagnetic behavior
- Examples: Fe₃O₄, ferrites (MFe₂O₄)
10.2 Magnetic Susceptibility Values
| Substance |
Type |
χ (×10⁻⁶) |
Temperature Dependence |
| H₂O |
Diamagnetic |
-9.0 |
Independent |
| O₂ |
Paramagnetic |
+1900 |
χ ∝ 1/T |
| Fe |
Ferromagnetic |
+200,000 |
Complex |
| MnO |
Antiferromagnetic |
+650 |
Maximum at Tₙ |
📊 11. Advanced Calculations and Numerical Examples
11.1 Complex Density Problems
Problem 1: Mixed Crystal System
Question: A crystal has 75% FCC structure and 25% BCC structure. If the FCC portion has edge length 4.0 Å and BCC portion has edge length 3.5 Å, and the atomic mass is 60 g/mol, calculate the average density.
Solution:
For FCC: Z₁ = 4, a₁ = 4.0 × 10⁻⁸ cm
ρ₁ = (4 × 60) / ((4.0 × 10⁻⁸)³ × 6.022 × 10²³) = 6.22 g/cm³
For BCC: Z₂ = 2, a₂ = 3.5 × 10⁻⁸ cm
ρ₂ = (2 × 60) / ((3.5 × 10⁻⁸)³ × 6.022 × 10²³) = 4.45 g/cm³
Average density = 0.75 × 6.22 + 0.25 × 4.45 = 5.78 g/cm³
Problem 2: Defect Concentration
Question: In a crystal of AgBr, 1 in every 10⁶ Ag⁺ ions is missing from its lattice site (Schottky defect). If the edge length is 5.77 Å, calculate the number of defects per unit cell.
Solution:
For rock salt structure: Z = 4 formula units
Number of Ag⁺ ions per unit cell = 4
Defect concentration = 1/10⁶
Number of defects per unit cell = 4 × (1/10⁶) = 4 × 10⁻⁶ defects/unit cell
11.2 Semiconductor Calculations
Problem 3: Doping Concentration
Question: Silicon (density 2.33 g/cm³, atomic mass 28.1 g/mol) is doped with 1 ppm phosphorus. Calculate the number of donor atoms per cm³.
Solution:
Number of Si atoms per cm³ = (2.33 × 6.022 × 10²³) / 28.1 = 4.99 × 10²² atoms/cm³
Donor concentration = 1 ppm = 1 × 10⁻⁶
Number of donor atoms = 4.99 × 10²² × 1 × 10⁻⁶ = 4.99 × 10¹⁶ atoms/cm³
🔍 12. Important Ionic Structures
12.1 Rock Salt Structure (NaCl type)
Characteristics:
• FCC arrangement of Cl⁻ ions
• Na⁺ ions in octahedral voids
• Coordination number: 6:6
• Examples: NaCl, KCl, MgO, CaO
Radius ratio: 0.414 - 0.732
12.2 Cesium Chloride Structure (CsCl type)
Characteristics:
• Simple cubic arrangement of Cl⁻ ions
• Cs⁺ ions at body center
• Coordination number: 8:8
• Examples: CsCl, CsBr, CsI
Radius ratio: 0.732 - 1.000
12.3 Zinc Blende Structure (ZnS type)
Characteristics:
• FCC arrangement of S²⁻ ions
• Zn²⁺ ions in alternate tetrahedral voids
• Coordination number: 4:4
• Examples: ZnS, CuCl, GaAs
Radius ratio: 0.225 - 0.414
12.4 Fluorite Structure (CaF₂ type)
Characteristics:
• FCC arrangement of Ca²⁺ ions
• F⁻ ions in all tetrahedral voids
• Coordination number: 8:4
• Examples: CaF₂, BaF₂, SrF₂
Radius ratio: > 0.732
📈 13. Phase Diagrams and Polymorphism
13.1 Polymorphism
Carbon Allotropes:
- Diamond: sp³ hybridized, tetrahedral, hardest natural material
- Graphite: sp² hybridized, layered, conductor
- Fullerenes: sp² hybridized, cage-like (C₆₀, C₇₀)
- Graphene: Single layer graphite, excellent conductor
13.2 Silica Polymorphs
| Polymorph |
Structure |
Density (g/cm³) |
Stability Range |
| α-Quartz |
Hexagonal |
2.65 |
< 573°C |
| β-Quartz |
Hexagonal |
2.53 |
573-870°C |
| Tridymite |
Hexagonal |
2.26 |
870-1470°C |
| Cristobalite |
Cubic |
2.32 |
1470-1713°C |
🎯 30 Single Choice Questions (SCQs)
Test your knowledge with these comprehensive questions!
Q1. Which of the following has the highest packing efficiency?
(a) Simple cubic
(b) Body-centered cubic
(c) Face-centered cubic
(d) End-centered cubic
Q2. The number of atoms per unit cell in BCC structure is:
(a) 1
(b) 2
(c) 4
(d) 8
Q3. In CsCl structure, the coordination number is:
(a) 4:4
(b) 6:6
(c) 8:8
(d) 12:12
Q4. Frenkel defect is shown by:
(a) NaCl
(b) KCl
(c) AgCl
(d) CsCl
Q5. The relationship between edge length and atomic radius in FCC is:
(a) a = 2r
(b) a = 4r/√3
(c) a = 2√2r
(d) a = 4r/√2
Q6. Which defect does not change the density of the crystal?
(a) Vacancy defect
(b) Interstitial defect
(c) Frenkel defect
(d) Schottky defect
Q7. The coordination number in HCP structure is:
(a) 6
(b) 8
(c) 12
(d) 14
Q8. n-type semiconductors are formed by doping silicon with:
(a) Boron
(b) Aluminum
(c) Phosphorus
(d) Gallium
Q9. The number of tetrahedral voids per sphere in close packing is:
(a) 1
(b) 2
(c) 4
(d) 6
Q10. Which of the following is paramagnetic?
(a) NaCl
(b) H₂O
(c) O₂
(d) C₆H₆
Q11. The density of a cubic crystal with edge length 'a', molar mass 'M', and 'Z' atoms per unit cell is:
(a) ZM/(a³NA)
(b) ZM/(aNA)
(c) M/(Za³NA)
(d) ZMNA/a³
Q12. In zinc blende structure, zinc ions occupy:
(a) All octahedral voids
(b) All tetrahedral voids
(c) Alternate tetrahedral voids
(d) Alternate octahedral voids
Q13. F-centers are:
(a) Electrons trapped in cation vacancies
(b) Electrons trapped in anion vacancies
(c) Holes trapped in anion vacancies
(d) Extra cations in interstitial sites
Q14. The packing efficiency of simple cubic structure is:
(a) 52.4%
(b) 68.0%
(c) 74.0%
(d) 90.7%
Q15. Which crystal system has all angles equal to 90° but all edges different?
(a) Cubic
(b) Tetragonal
(c) Orthorhombic
(d) Monoclinic
Q16. Antiferromagnetism is exhibited by:
(a) Fe₃O₄
(b) MnO
(c) Fe
(d) NaCl
Q17. The band gap of silicon is approximately:
(a) 0.1 eV
(b) 1.1 eV
(c) 3.0 eV
(d) 5.0 eV
Q18. In fluorite structure (CaF₂), the coordination numbers of Ca²⁺ and F⁻ are:
(a) 6:6
(b) 8:4
(c) 4:8
(d) 4:4
Q19. The number of octahedral voids per sphere in close packing is:
(a) 1
(b) 2
(c) 4
(d) 6
Q20. Schottky defect decreases the density because:
(a) Cations and anions leave crystal
(b) Cations leave crystal
(c) Anions leave crystal
(d) Interstitial atoms are formed
Q21. A compound AB has rock salt structure. If edge length is 580 pm and radius of A⁺ is 140 pm, the radius of B⁻ is:
(a) 140 pm
(b) 150 pm
(c) 160 pm
(d) 170 pm
Q22. The magnetic moment of Fe³⁺ (3d⁵) in Bohr magnetons is:
(a) 3.87
(b) 4.90
(c) 5.92
(d) 6.93
Q23. Which of the following is an example of ferrimagnetic substance?
(a) Fe
(b) Fe₃O₄
(c) MnO
(d) NaCl
Q24. The coordination number in simple cubic structure is:
(a) 4
(b) 6
(c) 8
(d) 12
Q25. If 200 pm is the edge length of FCC unit cell, the radius of atom is:
(a) 50 pm
(b) 70.7 pm
(c) 100 pm
(d) 141.4 pm
Q26. The number of nearest neighbors in BCC structure is:
(a) 6
(b) 8
(c) 12
(d) 14
Q27. p-type semiconductors have:
(a) Excess electrons
(b) Excess holes
(c) Equal electrons and holes
(d) No charge carriers
Q28. A metal crystallizes in FCC structure. If edge length is 400 pm and density is 8.0 g/cm³, the atomic mass is:
(a) 38.6 g/mol
(b) 51.5 g/mol
(c) 77.2 g/mol
(d) 102.9 g/mol
Q29. Which defect is responsible for the color of alkali halide crystals?
(a) Schottky defect
(b) Frenkel defect
(c) F-center
(d) Interstitial defect
Q30. The ratio of octahedral to tetrahedral voids in close packed structure is:
(a) 1:1
(b) 1:2
(c) 2:1
(d) 1:4
📝 ANSWER KEY
Q1: (c)
Q2: (b)
Q3: (c)
Q4: (c)
Q5: (c)
Q6: (c)
Q7: (c)
Q8: (c)
Q9: (b)
Q10: (c)
Q11: (a)
Q12: (c)
Q13: (b)
Q14: (a)
Q15: (c)
Q16: (b)
Q17: (b)
Q18: (b)
Q19: (a)
Q20: (a)
Q21: (c)
Q22: (c)
Q23: (b)
Q24: (b)
Q25: (b)
Q26: (b)
Q27: (b)
Q28: (c)
Q29: (c)
Q30: (b)
📚 DETAILED EXPLANATIONS
Q1. (c) Face-centered cubic has 74% packing efficiency, highest among cubic structures.
Q2. (b) BCC: 8×(1/8) + 1×1 = 2 atoms per unit cell.
Q3. (c) In CsCl structure, both ions have coordination number 8.
Q4. (c) AgCl shows Frenkel defect as Ag⁺ is small and can fit in interstitial sites.
Q5. (c) In FCC, face diagonal = 4r and face diagonal = a√2, so a = 2√2r.
Q6. (c) Frenkel defect: ion moves from lattice to interstitial site, density unchanged.
Q7. (c) HCP has coordination number 12, same as FCC.
Q8. (c) Phosphorus (Group 15) has 5 valence electrons, provides extra electron.
Q9. (b) For every sphere in close packing, there are 2 tetrahedral voids.
Q10. (c) O₂ has unpaired electrons in π* orbitals, making it paramagnetic.
Q11. (a) Density formula: ρ = ZM/(a³NA).
Q12. (c) In zinc blende, Zn²⁺ occupies alternate tetrahedral voids (4 out of 8).
Q13. (b) F-centers are electrons trapped in anion vacancies, giving color to crystals.
Q14. (a) Simple cubic packing efficiency = π/6 × 100% = 52.4%.
Q15. (c) Orthorhombic: a ≠ b ≠ c, α = β = γ = 90°.
Q16. (b) MnO shows antiferromagnetism with opposing magnetic moments.
Q17. (b) Silicon has band gap of 1.1 eV, making it a semiconductor.
Q18. (b) In CaF₂: Ca²⁺ (coordination 8), F⁻ (coordination 4).
Q19. (a) One octahedral void per sphere in close packing.
Q20. (a) Schottky defect: equal cations and anions leave, decreasing density.
Q21. (c) a = 2(r₊ + r₋), so 580 = 2(140 + r₋), r₋ = 160 pm.
Q22. (c) μ = √[n(n+2)] = √[5×7] = 5.92 BM for Fe³⁺.
Q23. (b) Fe₃O₄ (magnetite) is ferrimagnetic with unequal opposing spins.
Q24. (b) Simple cubic coordination number = 6 (face-touching neighbors).
Q25. (b) r = a/(2√2) = 200/(2×1.414) = 70.7 pm.
Q26. (b) BCC has 8 nearest neighbors (body diagonal distance).
Q27. (b) p-type: acceptor impurities create excess holes.
Q28. (c) M = ρa³NA/Z = 8.0×(4×10⁻⁸)³×6.022×10²³/4 = 77.2 g/mol.
Q29. (c) F-centers (color centers) absorb visible light, giving color.
Q30. (b) Octahedral:Tetrahedral = 1:2 in close packing.
📚 Additional Numerical Problems
Numerical Problem 1: Crystal Density
Question: Tungsten crystallizes in BCC structure with edge length 3.16 Å. If density is 19.3 g/cm³, calculate Avogadro's number.
Solution:
For BCC: Z = 2, Atomic mass of W = 183.84 g/mol
ρ = ZM/(a³NA)
NA = ZM/(ρa³) = (2 × 183.84)/(19.3 × (3.16 × 10⁻⁸)³)
NA = 367.68/(19.3 × 3.16 × 10⁻²³)
NA = 6.02 × 10²³ mol⁻¹
Numerical Problem 2: Ionic Radius Calculation
Question: In NaCl structure, edge length is 564 pm. If radius of Cl⁻ is 181 pm, calculate the radius of Na⁺.
Solution:
In rock salt structure: a = 2(r+ + r-)
564 = 2(rNa+ + 181)
282 = rNa+ + 181
rNa+ = 101 pm
Numerical Problem 3: Defect Calculation
Question: A crystal of AgBr has density 6.47 g/cm³ while theoretical density is 6.50 g/cm³. Calculate the percentage of Schottky defects.
Solution:
Percentage defect = [(ρtheoretical - ρactual)/ρtheoretical] × 100%
Percentage defect = [(6.50 - 6.47)/6.50] × 100%
Percentage defect = 0.46%
Numerical Problem 4: Packing Efficiency
Question: Calculate the packing efficiency of a crystal in which atoms are arranged in simple cubic pattern.
Solution:
In simple cubic: a = 2r, Volume of unit cell = a³ = 8r³
Number of atoms per unit cell = 1
Volume of atoms = 1 × (4/3)πr³ = (4/3)πr³
Packing efficiency = [(4/3)πr³]/8r³ × 100% = π/6 × 100%
Packing efficiency = 52.36%
Numerical Problem 5: Semiconductor Doping
Question: Pure silicon has 5 × 10²² atoms/cm³. If it's doped with 1 ppm boron, calculate the hole concentration at room temperature.
Solution:
Boron concentration = 5 × 10²² × 1 × 10⁻⁶ = 5 × 10¹⁶ atoms/cm³
Each boron atom creates one hole
Hole concentration = 5 × 10¹⁶ holes/cm³
🎯 Key Formulas Summary
💡 Important Points for Board Exams
🔥 Must Remember Facts:
- Highest packing efficiency: FCC and HCP (74%)
- F-centers: Responsible for color in alkali halides
- Frenkel defect: Density unchanged, common in AgCl
- Schottky defect: Density decreases, common in NaCl
- n-type: Group 15 dopants (P, As, Sb)
- p-type: Group 13 dopants (B, Al, Ga)
- Ferromagnetism: Fe, Co, Ni
- Antiferromagnetism: MnO, FeO, CoO
🎓 Exam Strategy Tips:
- Always identify the crystal structure first
- Remember the relationship between edge length and atomic radius
- Practice numerical problems daily
- Understand the concept behind each formula
- Draw diagrams for better understanding
- Connect defects with their effects on properties
🏆 "Crystallography reveals the hidden order in matter's chaos"
Master the Solid State - Master Chemistry!