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Class 12 Chemistry · Chapter 11 Legacy

The Solid State

From salt crystals to silicon chips — the chemistry of ordered matter. With interactive packing fraction calculator, unit cell diagrams, defects, magnetic and electrical properties, and 20+ real-life examples.

⏱️ ~35 min read · 📚 7 sections · 🧮 1 interactive tool · 🎯 6 solved + 10 NCERT
7Sections
20+Real-life Examples
10Diagrams & Tables
1Calculator
18MCQs
The Big Picture

🗺️ How Everything Connects

Before diving in, here's the entire chapter on one page. How solids are classified, their structure, properties, and defects.

Classify Solids Crystal Lattice Packing Efficiency Density Calculations Defects in Crystals Magnetic & Electric SC / BCC / FCC Z × M / N_A a³ "How are atoms arranged in solids, and how does arrangement determine properties?" Structure → properties. Packing determines density, defects modify properties.
Key Takeaway

Solids are classified as crystalline (ordered, sharp MP) or amorphous (disordered, glassy). Crystals have repeating lattice + basis. Unit cells (primitive, BCC, FCC) determine packing efficiency. Density is calculated from unit cell parameters. Defects (point, line, impurity) affect properties. Magnetic and electrical properties arise from electron arrangement.

Section 01

Classification of Solids

11.1Crystalline vs Amorphous

Solids vs Liquids/Gases

Solid: State of matter with definite shape and volume. Particles are closely packed and held by strong intermolecular forces.

Two types: Crystalline and amorphous.

Crystalline Solids

Crystalline: Long-range order (atoms regularly arranged over long distances). Sharp melting point. Anisotropic (properties depend on direction). Examples: NaCl, diamond, quartz, metals, ice.

Properties:

  • Definite geometry (crystal faces at specific angles)
  • Sharp melting point
  • Anisotropic (refractive index, conductivity depend on direction)
  • Give XRD pattern (X-ray diffraction)
  • Cleave along definite planes
Amorphous Solids

Amorphous: Short-range order (no long-range arrangement). No sharp melting point — soften over a range. Isotropic. Examples: glass, rubber, plastics, amorphous silicon.

Properties:

  • Glass is a supercooled liquid (not truly solid)
  • Soften gradually on heating
  • Isotropic (properties same in all directions)
  • No sharp XRD pattern
  • Irregular cleavage (conchoidal fracture)
Classification of Crystalline Solids by Bonding
Type Particles Forces Properties Examples
Molecular Molecules Van der Waals / H-bonds Soft, low MP, non-conducting Ice, dry ice, iodine
Ionic Ions (+, −) Strong electrostatic (Coulomb) Hard, brittle, high MP, conduct in solution/melt NaCl, CaO, MgO
Covalent (network) Atoms (covalent) Covalent bonds Very hard, very high MP, non-conducting (except graphite) Diamond, quartz, SiC
Metallic Metal cations + sea of electrons Metallic bonds Malleable, ductile, conductive, high MP Fe, Cu, Al, Na
Real Life Connect
  • Quartz (SiO₂): Crystalline, sharp MP, used in watches, electronics
  • Glass: Amorphous, softens gradually, transparent
  • Diamond: Covalent network, hardest known natural material
  • Graphite: Covalent network (2D sheets), soft, conducts electricity
  • Ice: Molecular (H-bonds), low MP, less dense than water
  • NaCl: Ionic, brittle, conducts when dissolved
Key Takeaway

Crystalline: long-range order, sharp MP, anisotropic. Amorphous: short-range order, gradual softening, isotropic. By bonding: molecular (weak forces), ionic (electrostatic), covalent (network), metallic (sea of electrons).

Section 02

Crystal Lattice and Unit Cell

11.2Building Blocks of Crystals

Space Lattice and Unit Cell

Space lattice: Regular 3D arrangement of points representing the positions of atoms/ions/molecules in a crystal.

Lattice point: Each point in the space lattice. Represents identical environment.

Basis: The atom/molecule/ion placed at each lattice point.

Unit cell: Smallest repeating portion of the lattice that, when repeated in 3D, gives the entire crystal.

Lattice parameters: Edge lengths (a, b, c) and angles (α, β, γ).

7 Crystal Systems
System a, b, c α, β, γ Example
Cubic a = b = c α = β = γ = 90° NaCl, Cu, Fe
Tetragonal a = b ≠ c α = β = γ = 90° White tin (Sn)
Orthorhombic a ≠ b ≠ c α = β = γ = 90° Rhombic sulfur
Monoclinic a ≠ b ≠ c α = γ = 90°, β ≠ 90° Monoclinic sulfur
Triclinic a ≠ b ≠ c α ≠ β ≠ γ CuSO₄·5H₂O
Hexagonal a = b ≠ c α = β = 90°, γ = 120° Graphite, ZnO
Rhombohedral a = b = c α = β = γ ≠ 90° Calcite, quartz

14 Bravais lattices: The 7 crystal systems combined with different centering (P, I, F, C) give 14 unique lattices.

Types of Unit Cells (Cubic)

1. Primitive (Simple Cubic, SC):

  • Lattice points only at corners
  • Atoms per unit cell: Z = 1 (each corner shared by 8 cells → 8 × 1/8 = 1)
  • Coordination number: 6
  • Edge length: a = 2r
  • Packing efficiency: 52.4%
  • Example: Polonium (α-Po)

2. Body-Centered Cubic (BCC):

  • Lattice points at corners + 1 at body center
  • Z = 2 (8 × 1/8 + 1 = 2)
  • Coordination number: 8
  • Edge length: a = 4r/√3 (or r = √3 a/4)
  • Packing efficiency: 68%
  • Example: Fe (α), Cr, W, Mo, V, Na, K

3. Face-Centered Cubic (FCC) — also called Cubic Close Packed (CCP):

  • Lattice points at corners + 1 at each face center
  • Z = 4 (8 × 1/8 + 6 × 1/2 = 4)
  • Coordination number: 12
  • Edge length: a = 2√2 r (or r = a/(2√2))
  • Packing efficiency: 74% (highest possible)
  • Example: Cu, Ag, Au, Al, Ni, Pt
V1 · Unit cell types — SC, BCC, FCC Simple Cubic (SC) Z = 1, PF = 52.4%, CN = 6 a = 2r Only corners Body-Centered (BCC) Z = 2, PF = 68%, CN = 8 a = 4r/√3 Corners + 1 center Face-Centered (FCC/CCP) Z = 4, PF = 74%, CN = 12 a = 2√2 r Corners + face centers

Packing Fraction Calculator

Choose unit cell type — see Z, coordination number, and packing efficiency.

Z (atoms/unit cell)
—
Coordination Number
—
Packing Efficiency
—
Relation (r, a)
—
Real Life Connect
  • NaCl: FCC arrangement of Cl⁻ with Na⁺ in octahedral holes → rock salt structure
  • Diamond: FCC of carbon + 4 tetrahedral carbons → extremely hard
  • Iron: BCC at low T (α-Fe), FCC at high T (γ-Fe) → different densities
  • Copper, silver, gold: FCC → best conductors, malleable
  • Hexagonal boron nitride: HCP structure → lubricant like graphite
Key Takeaway

Crystal = lattice + basis. 7 crystal systems + 14 Bravais lattices. Cubic unit cells: SC (Z=1, PF=52.4%), BCC (Z=2, PF=68%), FCC (Z=4, PF=74%). Packing efficiency: SC < BCC < FCC (max).

Section 03

Packing in Crystals

11.3How Atoms Fill Space

Close Packing

Close packing: Arrangement of spheres in a way that maximizes packing efficiency (fills space most densely).

1D: Atoms in a row — coordination number 2

2D: Atoms in a layer — square close packing (CN=4, PF=52.4%) or hexagonal close packing (CN=6, PF=60.4%)

3D: Layers stacked to give higher packing

3D Close Packing

1. Simple cubic (SC) packing: Layers directly on top → packing efficiency 52.4%

2. Hexagonal close packing (HCP): Layer B sits in depressions of layer A; layer C sits in depressions of B directly above A → ABAB... stacking. Packing efficiency 74%. Example: Mg, Zn, Ti.

3. Cubic close packing (CCP) = FCC: Layer B in depressions of A; layer C in depressions of B but not above A → ABCABC... stacking. Packing efficiency 74%. Example: Cu, Ag, Au, Al.

Void types:

  • Triangular void (in 2D): Between 3 spheres
  • Tetrahedral void: Formed by 4 spheres at corners of tetrahedron
  • Octahedral void: Formed by 6 spheres at vertices of octahedron

Number of voids:

  • Per atom in close packing: 1 octahedral void + 2 tetrahedral voids
  • So if there are N atoms, there are N octahedral voids and 2N tetrahedral voids.
V2 · FCC vs HCP — ABC vs AB stacking FCC (CCP) — ABCABC Cubic close packing Layer A Layer B Layer C CN = 12, PF = 74% HCP — ABABAB Hexagonal close packing Layer A Layer B Layer A (again) CN = 12, PF = 74% Both have 74% efficiency — most efficient packings in 3D FCC has ABC stacking; HCP has AB stacking. Both give same PF.
Key Takeaway

Close packing in 3D: SC (52.4%), BCC (68%), FCC/HCP (74% — most efficient). FCC = ABC stacking, HCP = AB stacking. Voids: tetrahedral (2 per atom) and octahedral (1 per atom).

Section 04

Density Calculations

11.4From Unit Cell to Density

Density of a Unit Cell

Density (ρ) = mass of unit cell / volume of unit cell

Formula: ρ = Z × M / (N_A × a³)

Where:

  • Z = number of atoms per unit cell
  • M = molar mass (g/mol)
  • N_A = Avogadro's number = 6.022 × 10²³ mol⁻¹
  • a = edge length (cm) — a³ gives volume in cm³

For edge length in picometers (pm): 1 pm = 10⁻¹⁰ cm, so a³ (in pm³) × 10⁻³⁰ = a³ (cm³)

For edge length in Angstroms (Å): 1 Å = 10⁻⁸ cm, so a³ (in ų) × 10⁻²⁴ = a³ (cm³)

Common Unit Cell Relations
Unit Cell Z r and a relation a from r
Simple Cubic (SC) 1 a = 2r a = 2r
BCC 2 4r = √3 a → r = √3 a/4 a = 4r/√3 ≈ 2.31 r
FCC 4 4r = √2 a → r = a/(2√2) a = 2√2 r ≈ 2.83 r

Density Calculator

Enter unit cell parameters — calculate crystal density.

Volume a³ (cm³)
—
Density (g/cm³)
—
Real Life Connect
  • Iron (BCC): a ≈ 288 pm, M = 56 g/mol, Z = 2 → ρ ≈ 7.87 g/cm³
  • Copper (FCC): a ≈ 361 pm, M = 63.5 g/mol, Z = 4 → ρ ≈ 8.96 g/cm³
  • Silver (FCC): a ≈ 409 pm → ρ ≈ 10.5 g/cm³
  • Density calculation: Determines metal purity, alloy composition, mineral identification
  • X-ray crystallography: Uses density + unit cell to find Avogadro's number
Key Takeaway

Density = Z × M / (N_A × a³). SC: Z=1, a=2r. BCC: Z=2, a=4r/√3. FCC: Z=4, a=2√2 r. Higher Z or smaller a → higher density. Used to identify metals and determine Avogadro's number.

Section 05

Defects in Crystals

11.5Imperfections in Order

Types of Defects

1. Point defects (0D): At a single atomic position. Types:

  • Vacancy: Missing atom at a lattice site
  • Interstitial: Extra atom in interstitial space
  • Frenkel defect: Ion displaced from its site to interstitial site (cation moves). Example: AgCl, AgBr, ZnS. Density unchanged (same number of atoms).
  • Schottky defect: Equal number of cations and anions missing (pair of vacancies). Example: NaCl, KCl, CsCl. Density decreases.

2. Line defects (1D): Dislocations — edge or screw. Cause plastic deformation.

3. Interfacial defects (2D): Grain boundaries, twin boundaries, stacking faults.

4. Volume defects (3D): Voids, cracks, foreign inclusions.

Frenkel vs Schottky — Key Difference
Feature Frenkel Defect Schottky Defect
What happens Ion moves to interstitial site Cation + anion vacancies
Stoichiometry Preserved Preserved
Density Unchanged Decreases
Common in Large cation, small anion, low CN High CN, similar-sized ions
Examples AgCl, AgBr, ZnS, AgI NaCl, KCl, CsCl, KBr
Cause Large size difference High coordination, ionic
Impurity Defects

Substitutional impurity: Foreign atom replaces a host atom. Example: SrCl₂ in NaCl — Sr²⁺ replaces 2 Na⁺ (creates 1 cation vacancy per Sr²⁺).

Interstitial impurity: Small foreign atom in interstitial space. Example: C in Fe (steel).

Doping: Deliberate impurity addition in semiconductors.

Other Properties

Non-stoichiometric defects: Compounds with variable composition (e.g., FeO₁₋ₓ). Due to variable oxidation states of transition metals.

F-centers: Anionic vacancies filled with electrons → color (e.g., NaCl is yellow when F-centers present).

Real Life Connect
  • NaCl crystals: Schottky defects → slight non-stoichiometry, changes density
  • AgCl: Frenkel defects → used in photography (light sensitive)
  • Doped silicon: Impurity defect — critical for electronics
  • Steel: C in Fe interstitial defect → harder
  • Colored gemstones: Impurity defects (ruby = Cr in Al₂O₃, sapphire = Fe/Ti in Al₂O₃)
  • F-centers: Yellow NaCl (from electron in vacancy) — used in dosimeters
Key Takeaway

Point defects: Frenkel (ion to interstitial, density unchanged) vs Schottky (missing pairs, density decreases). Impurity defects: substitutional or interstitial. Non-stoichiometric compounds (variable composition). F-centers cause color. Used in electronics (doping) and materials science.

Section 06

Magnetic Properties

11.6Magnetic Behaviour of Solids

Types of Magnetism

1. Diamagnetic: All electrons paired. Weakly repelled by magnetic field. Examples: NaCl, benzene, H₂O.

2. Paramagnetic: Unpaired electrons. Weakly attracted by magnetic field. Examples: O₂, Cu²⁺ salts, TiO₂.

3. Ferromagnetic: Unpaired electrons aligned in domains → strong permanent magnetism. Examples: Fe, Co, Ni, Gd.

4. Antiferromagnetic: Adjacent spins antiparallel → net zero. Example: MnO.

5. Ferrimagnetic: Unequal antiparallel spins → net non-zero. Example: Fe₃O₄ (magnetite).

Property Diamagnetic Paramagnetic Ferromagnetic
Unpaired e⁻ None Present Present + aligned
Response Weakly repelled Weakly attracted Strongly attracted
Permeability μ < 1 μ > 1 (small) μ >> 1
Examples Diamond, H₂O TiO₂, CuSO₄ Fe, Co, Ni
Real Life Connect
  • Iron (ferromagnetic): Permanent magnets, magnetic storage, motors
  • MRI: Uses superconducting magnets (ferromagnetic coils)
  • Magnetite (Fe₃O₄, ferrimagnetic): Natural magnets (lodestone)
  • Antiferromagnetic MnO: Doesn't attract magnet — used in some magnetic devices
  • Hard drives: Ferromagnetic domains store data
Key Takeaway

Diamagnetic (all paired, weakly repelled), paramagnetic (unpaired, weakly attracted), ferromagnetic (strong permanent, aligned domains), antiferromagnetic (antiparallel, net zero), ferrimagnetic (unequal, net non-zero). Ferromagnetism → permanent magnets and data storage.

Section 07

Electrical Properties — Semiconductors

11.7Conductors, Insulators, Semiconductors

Band Theory

Valence band: Highest occupied band (full of electrons)

Conduction band: Lowest empty band (empty or partly filled)

Band gap (Eg): Energy difference between valence band and conduction band.

Classification:

  • Conductors: No band gap (or overlap) — electrons move freely. Examples: metals.
  • Insulators: Large band gap (>3 eV) — very few electrons in conduction band. Examples: diamond (Eg ~5.5 eV), quartz.
  • Semiconductors: Small band gap (~1 eV) — some electrons can be excited to conduction band. Examples: Si (1.1 eV), Ge (0.7 eV).
Intrinsic vs Extrinsic

Intrinsic semiconductor: Pure Si or Ge. Small conductivity due to thermal excitation (electron + hole pairs).

Extrinsic semiconductor: Doped with impurities.

n-type: Doped with pentavalent element (P, As, Sb) → extra electron → electron carriers. Donor doping.

p-type: Doped with trivalent element (B, Al, Ga) → missing electron (hole) → hole carriers. Acceptor doping.

Real Life Connect
  • Silicon: Foundation of modern electronics (chips, solar cells)
  • n-type and p-type semiconductors: Form p-n junctions (diodes, transistors, LED, solar cells)
  • Germanium: Early transistors (1947)
  • GaAs: High-speed electronics, LEDs, solar
  • Solar cells: Silicon p-n junction converts light to electricity
Key Takeaway

Band theory: conductors (no gap), semiconductors (~1 eV gap), insulators (>3 eV gap). Intrinsic (pure) vs extrinsic (doped). n-type (electron carriers, pentavalent dopant). p-type (hole carriers, trivalent dopant). Foundation of all electronics.

Practice

📝 Solved Examples

📋 2 Marks

Q1. Why does NaCl have a sharp melting point while glass softens gradually?

NaCl 1 Mark

NaCl is a crystalline solid with long-range order. All ionic bonds have the same energy → uniform melting requires same amount of energy throughout → sharp melting point (801°C).

Glass 1 Mark

Glass is amorphous — short-range order only. Different regions have different strengths → melt at different temperatures → gradual softening over a range.

Crystalline → sharp MP; amorphous → gradual softening.
📋 3 Marks

Q2. A compound forms an FCC crystal with edge length 4.07 Å. If density is 10.5 g/cm³, calculate the molar mass.

Formula ½ Mark

ρ = Z·M / (N_A · a³)

Convert a to cm ½ Mark

a = 4.07 Å = 4.07 × 10⁻⁸ cm

a³ = (4.07 × 10⁻⁸)³ = 6.74 × 10⁻²³ cm³

Substitute 1 Mark

FCC → Z = 4, N_A = 6.022 × 10²³

10.5 = (4 × M) / (6.022 × 10²³ × 6.74 × 10⁻²³)

Solve 1 Mark

M = (10.5 × 6.022 × 10²³ × 6.74 × 10⁻²³) / 4

= (10.5 × 6.022 × 6.74) / 4 ≈ 106.8 g/mol

This matches silver (M_Ag = 108 g/mol) ✓

M ≈ 107 g/mol → Silver (Ag)
📋 3 Marks

Q3. Calculate the packing efficiency of a simple cubic unit cell.

Setup ½ Mark

In SC, atoms are at 8 corners. Z = 1 (8 × 1/8). Edge length a = 2r (atoms touch along edge).

Volume of atom ½ Mark

Volume of 1 atom = (4/3)πr³

Total volume of atoms = Z × (4/3)πr³ = 1 × (4/3)πr³

Volume of unit cell ½ Mark

V_cell = a³ = (2r)³ = 8r³

Packing efficiency 1 Mark

PF = (Volume of atoms) / (Volume of cell) × 100

= [(4/3)πr³] / (8r³) × 100

= (π/6) × 100

≈ 52.4%

Interpretation ½ Mark

SC has the lowest packing efficiency of the three cubic structures. HCP/FCC has 74% (highest).

PF ≈ 52.4%
📋 5 Marks

Q4. (a) What is a Schottky defect? (b) How does it affect density? (c) Which crystals show this defect? (d) Compare with Frenkel defect.

(a) Schottky defect 1 Mark

Equal numbers of cations and anions are missing from the crystal lattice, creating paired vacancies. This maintains the electrical neutrality of the crystal.

(b) Effect on density 1 Mark

Since atoms/ions are removed from the lattice, mass decreases while volume stays nearly the same → density decreases.

(c) Examples 1 Mark

Common in ionic compounds with high coordination number and similar-sized ions:

  • NaCl (both ions similar size, CN=6)
  • KCl, KBr, CsCl
  • AgBr (partially)
(d) Compare with Frenkel 2 Marks
Feature Schottky Frenkel
What Missing ion pairs Ion to interstitial site
Density Decreases Unchanged
Stoichiometry Preserved Preserved
Occurs in NaCl, KCl (high CN, similar size) AgCl, ZnS (large size difference)
Cause High coordination number Large cation/small anion mismatch
Schottky: vacancy pairs, density ↓. Frenkel: interstitial, density unchanged.
📋 5 Marks

Q5. Calculate the density of copper. Copper crystallizes in FCC with a = 3.61 Å, M = 63.5 g/mol.

Given values ½ Mark

FCC → Z = 4

a = 3.61 Å = 3.61 × 10⁻⁸ cm

M = 63.5 g/mol

N_A = 6.022 × 10²³

a³ 1 Mark

a³ = (3.61 × 10⁻⁸)³ = 47.04 × 10⁻²⁴ cm³

= 4.704 × 10⁻²³ cm³

Density formula 1 Mark

ρ = Z × M / (N_A × a³)

= 4 × 63.5 / (6.022 × 10²³ × 4.704 × 10⁻²³)

Calculate 1½ Marks

Numerator: 4 × 63.5 = 254

Denominator: 6.022 × 4.704 = 28.32

ρ = 254 / 28.32 ≈ 8.97 g/cm³

Check with literature 1 Mark

Accepted density of Cu ≈ 8.96 g/cm³. Our calculation gives 8.97 — excellent agreement!

This method is used to determine unknown crystal parameters or identify materials from their crystal data.

ρ ≈ 8.97 g/cm³ (matches Copper)
📋 5 Marks

Q6. What are the differences between n-type and p-type semiconductors?

Definition 1 Mark

n-type: Semiconductor doped with pentavalent impurity (P, As, Sb) → extra electrons → electron carriers.

p-type: Semiconductor doped with trivalent impurity (B, Al, Ga) → electron holes → hole carriers.

Majority carriers 1 Mark

n-type: majority carriers = electrons, minority = holes

p-type: majority carriers = holes, minority = electrons

Dopant 1 Mark

n-type: donor atom (gives extra electron). E.g., P has 5 valence e⁻ — 4 form bonds with Si, 1 free.

p-type: acceptor atom (creates hole). E.g., B has 3 valence e⁻ — needs 1 more electron → creates hole.

Fermi level 1 Mark

n-type: Fermi level closer to conduction band (electrons easier to excite)

p-type: Fermi level closer to valence band

Applications 1 Mark

n-type + p-type → p-n junction (diode)

Used in: LEDs, solar cells, transistors, integrated circuits, rectifiers

Real devices often use p-n-p or n-p-n configurations (transistors).

n-type: electron carriers (donor dopant). p-type: hole carriers (acceptor dopant). Together form p-n junctions.
NCERT Textbook

📚 NCERT Exercise Questions (Legacy)

Selected exercise questions from the LEGACY NCERT Chapter 1 (The Solid State). Removed from current syllabus but useful for competitive exams.

NCERT 1.1
Define the term 'amorphous'. Give a few examples of amorphous solids.
Show Solution

Amorphous solid: A solid in which constituent particles (atoms, ions, molecules) are arranged in a short-range order but lack long-range order. They do not have definite geometry, have no sharp melting point, and are isotropic.

Examples: Glass, rubber, plastics, amorphous silicon, pitch (tar).

NCERT 1.3
Distinguish between (a) hexagonal close packing and cubic close packing, (b) crystal lattice and unit cell.
Show Solution

(a) HCP vs CCP:

HCP: ABABAB... stacking of layers. Coordination number 12, packing efficiency 74%. Examples: Mg, Zn, Ti.

CCP (= FCC): ABCABC... stacking. Coordination number 12, packing efficiency 74%. Examples: Cu, Ag, Au, Al.

Both have the same packing efficiency, only the layer arrangement differs.

(b) Crystal lattice vs unit cell:

Crystal lattice: The 3D arrangement of lattice points (all equivalent) that represents the overall structure.

Unit cell: The smallest portion of the lattice that, when repeated in all 3D, generates the entire crystal. Contains all the structural information.

NCERT 1.5
How many lattice points are there in one unit cell of each of the following: (a) FCC, (b) BCC, (c) Simple cubic?
Show Solution

(a) FCC: 8 corners (each 1/8) + 6 face centers (each 1/2) = 8(1/8) + 6(1/2) = 1 + 3 = 4 lattice points

(b) BCC: 8 corners (each 1/8) + 1 body center (whole) = 1 + 1 = 2 lattice points

(c) Simple cubic: 8 corners (each 1/8) = 1 lattice point

NCERT 1.7
What is meant by the term 'coordination number'? What is the coordination number of atoms in (a) FCC, (b) BCC, (c) simple cubic?
Show Solution

Coordination number: Number of nearest neighbour atoms surrounding a given atom in the crystal.

(a) FCC = 12 (each atom has 12 nearest neighbors)

(b) BCC = 8 (each atom has 8 nearest neighbors)

(c) Simple cubic = 6 (each atom has 6 nearest neighbors)

NCERT 1.9
Niobium crystallises in a body-centred cubic structure. If density is 8.55 g/cm³, calculate the atomic radius of niobium (molar mass = 93 g/mol).
Show Solution

Given: BCC → Z = 2, M = 93 g/mol, ρ = 8.55 g/cm³

Formula: a³ = Z × M / (N_A × ρ)

= 2 × 93 / (6.022 × 10²³ × 8.55)

= 186 / (5.148 × 10²⁴)

= 3.61 × 10⁻²³ cm³

a = (3.61 × 10⁻²³)^(1/3) = 3.31 × 10⁻⁸ cm = 331 pm

For BCC: r = √3 a / 4

r = (1.732 × 331) / 4 = 143 pm

Atomic radius of niobium ≈ 143 pm

NCERT 1.11
Silver crystallises in FCC lattice. If edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g/cm³, calculate the atomic mass of silver.
Show Solution

Given: FCC → Z = 4, a = 4.07 × 10⁻⁸ cm, ρ = 10.5 g/cm³

Formula: M = ρ × N_A × a³ / Z

a³ = (4.07 × 10⁻⁸)³ = 6.74 × 10⁻²³ cm³

M = (10.5 × 6.022 × 10²³ × 6.74 × 10⁻²³) / 4

= (10.5 × 6.022 × 6.74) / 4

= 426.2 / 4 = 106.6 g/mol

Atomic mass of Ag ≈ 107 g/mol ✓ (literature: 107.87)

NCERT 1.13
Explain the following terms with suitable examples: (a) Frenkel defect, (b) Schottky defect, (c) F-centre.
Show Solution

(a) Frenkel defect: When an ion (usually a cation) leaves its normal lattice site and occupies an interstitial site, creating a vacancy + interstitial pair. Stoichiometry preserved. Density unchanged. Common in AgCl, AgBr, AgI, ZnS. Caused by large size difference between cation and anion.

(b) Schottky defect: Equal number of cations and anions are missing from lattice, creating vacancy pairs. Stoichiometry preserved. Density decreases. Common in NaCl, KCl, KBr, CsCl. Caused by high coordination number and similar-sized ions.

(c) F-centre: An anionic vacancy filled with one or more electrons (F = color center, from German "Farbe"). These trapped electrons absorb visible light → give color to the crystal. Example: NaCl heated in Na vapor becomes yellow because Cl⁻ vacancies trap electrons. Alkali metal halides with F-centers are colored.

NCERT 1.15
Explain the following with suitable examples: (a) ferromagnetism, (b) paramagnetism, (c) diamagnetism.
Show Solution

(a) Ferromagnetism: Strong permanent magnetism due to aligned magnetic moments in domains. Examples: Fe, Co, Ni, Gd. These metals are strongly attracted to magnets and can be permanently magnetized.

(b) Paramagnetism: Weak attraction due to unpaired electrons. No long-range alignment (thermal energy disrupts). Examples: O₂, CuSO₄, TiO₂, some transition metal complexes.

(c) Diamagnetism: Weak repulsion due to all paired electrons. No net magnetic moment per atom. Examples: NaCl, benzene, H₂O, most organic compounds.

NCERT 1.17
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Show Solution

Answer: When a solid (especially an ionic solid) is heated:

Schottky defect can arise — some cations and anions leave their lattice sites, creating vacancy pairs. This is more common at high temperatures because thermal energy allows ions to escape from their sites.

Property affected: Density decreases (mass lost, volume nearly constant).

Also, if the defect involves trapped electrons (F-centers), the crystal may become colored.

Conductivity may also increase due to increased ion mobility in the defective crystal.

NCERT 1.19
What is a semiconductor? Describe the two main types of semiconductors with examples.
Show Solution

Semiconductor: A material with electrical conductivity between that of conductors and insulators. Band gap ~1 eV. Conductivity increases with temperature (unlike metals).

Two main types:

(a) Intrinsic (pure) semiconductors: Pure Si or Ge. Small conductivity due to thermal excitation of electrons across the band gap. Both electrons and holes contribute. Conductivity is low but increases with T.

(b) Extrinsic (doped) semiconductors:

• n-type: Doped with pentavalent element (P, As, Sb). Extra electron per dopant atom → electron-rich → electrons are majority carriers. Examples: Si doped with P.

• p-type: Doped with trivalent element (B, Al, Ga). Creates a hole per dopant atom → hole-rich → holes are majority carriers. Examples: Si doped with B.

n-type + p-type → p-n junction → basis of diodes, transistors, LEDs, solar cells.

NCERT 1.22
Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (a) What is the length of the side of the unit cell? (b) How many unit cells are there in 1.00 cm³ of aluminium?
Show Solution

Given: CCP = FCC. r = 125 pm. M = 27 g/mol.

(a) Side length:

For FCC: a = 2√2 r = 2 × 1.414 × 125 = 353.5 pm ≈ 354 pm

(b) Number of unit cells in 1 cm³:

a = 354 pm = 3.54 × 10⁻⁸ cm

Volume of unit cell = a³ = (3.54 × 10⁻⁸)³ = 4.44 × 10⁻²³ cm³

Number of cells in 1 cm³ = 1 / (4.44 × 10⁻²³) = 2.25 × 10²² unit cells

Quick Reference

📐 Formula Sheet

Chapter 11 — All Formulas on One Page

Density of Unit Cell
ρ = Z × M / (N_A × a³)
SC Parameters
Z=1, a=2r, PF=52.4%, CN=6
BCC Parameters
Z=2, a=4r/√3, PF=68%, CN=8
FCC Parameters
Z=4, a=2√2 r, PF=74%, CN=12
Radius from Edge (BCC)
r = √3 a / 4
Radius from Edge (FCC)
r = a / (2√2)
Radius from Edge (SC)
r = a / 2
Edge from Density
a³ = Z M / (N_A ρ)
Avogadro's Number
N_A = Z M / (ρ a³)
Volume Conversions
1 pm = 10⁻¹⁰ cm; 1 Å = 10⁻⁸ cm
Packing Efficiency
PF = (V_atoms / V_cell) × 100%
Unit Cell Volume
V = a³ (for cubic)

Key Values & Unit Cell Data

N_A 6.022 × 10²³ mol⁻¹ Avogadro's number
1 pm 10⁻¹⁰ cm = 10⁻¹² m Picometer
1 Å 10⁻⁸ cm = 10⁻¹⁰ m Angstrom
SC efficiency 52.4% Simple cubic
BCC efficiency 68% Body-centered
FCC efficiency 74% Face-centered (max)
Diamond band gap ~5.5 eV Insulator
Si band gap ~1.1 eV Semiconductor
Ge band gap ~0.7 eV Semiconductor

📖 Notation Key

Z Number of atoms per unit cell
a, b, c Unit cell edge lengths
α, β, γ Unit cell angles
r Atomic radius
M Molar mass (g/mol)
N_A Avogadro's number
ρ Density (g/cm³)
PF Packing fraction / efficiency
SC Simple cubic
BCC Body-centered cubic
FCC Face-centered cubic
HCP Hexagonal close packed
CN Coordination number
E_g Band gap energy
One-Page Revision

📌 Chapter Summary

Everything You Need to Remember

  1. Crystalline solids: Long-range order, sharp MP, anisotropic. Amorphous: Short-range order, gradual softening, isotropic.
  2. By bonding: Molecular (weak), ionic (electrostatic), covalent network (very strong), metallic (sea of electrons).
  3. Crystal lattice: Regular 3D arrangement. Unit cell = smallest repeating unit. 7 crystal systems, 14 Bravais lattices.
  4. Cubic unit cells: SC (Z=1, CN=6, PF=52.4%), BCC (Z=2, CN=8, PF=68%), FCC (Z=4, CN=12, PF=74%).
  5. Relations: SC a=2r; BCC a=4r/√3; FCC a=2√2 r.
  6. Close packing: FCC = ABCABC stacking. HCP = ABABAB stacking. Both 74% efficiency.
  7. Voids: N octahedral voids + 2N tetrahedral voids per N atoms.
  8. Density formula: ρ = Z·M / (N_A·a³). Rearrange to find M, a, or N_A.
  9. Point defects: Frenkel (ion to interstitial, density unchanged), Schottky (missing ion pairs, density decreases).
  10. Impurity defects: Substitutional (Sr²⁺ in NaCl) or interstitial (C in Fe).
  11. F-centers: Anionic vacancies with electrons → color.
  12. Magnetic types: Diamagnetic (weakly repelled), paramagnetic (unpaired, weakly attracted), ferromagnetic (Fe, Co, Ni — strong), antiferromagnetic (net zero, MnO), ferrimagnetic (Fe₃O₄).
  13. Band theory: Conductors (no gap), semiconductors (~1 eV), insulators (>3 eV).
  14. n-type: Doped with pentavalent (P) → electron carriers. p-type: Doped with trivalent (B) → hole carriers.
  15. Applications: Electronics (Si semiconductors), solar cells, LEDs, gemstones (impurity defects), magnets (ferromagnetism).
Exam Preparation

🎯 Common Exam Questions (Legacy)

2-Mark Questions

  1. Distinguish between crystalline and amorphous solids.
  2. What is a unit cell? What are its parameters?
  3. Calculate the number of atoms in a BCC unit cell.
  4. What is meant by coordination number? Give the CN of an atom in FCC.
  5. Define Schottky and Frenkel defects with examples.
  6. What is a semiconductor? Give two examples.
  7. Distinguish between diamagnetic and paramagnetic substances.

3-Mark Questions

  1. Classify crystalline solids by bonding, giving examples.
  2. Calculate the packing efficiency of FCC.
  3. Derive the relationship between radius and edge length for BCC.
  4. What are F-centers? How do they cause color in crystals?
  5. Explain intrinsic vs extrinsic semiconductors.
  6. Describe the different types of point defects in crystals.
  7. Give the difference between n-type and p-type semiconductors.

5-Mark Questions

  1. Describe the various types of unit cells (SC, BCC, FCC) with their parameters.
  2. Derive the formula for the density of a unit cell. Use it to find the atomic mass of an element.
  3. Discuss the different types of magnetic properties with examples.
  4. Compare Frenkel and Schottky defects in detail.
  5. Explain close packing in 3D. Distinguish between FCC and HCP.
  6. Discuss the band theory of solids and its classification (conductors, semiconductors, insulators).
Test Yourself

✍️ MCQ Practice

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🎯 Assertion–Reason

(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.

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Quick Revision

⚡ Formula Flashcards

All key formulas at one place — perfect for quick revision before competitive exams.

Density Formula
ρ = Z·M / (N_A·a³)
SC Parameters
Z=1, a=2r, PF=52.4%, CN=6
BCC Parameters
Z=2, a=4r/√3, PF=68%, CN=8
FCC Parameters
Z=4, a=2√2 r, PF=74%, CN=12
Radius (BCC)
r = √3 a / 4
Radius (FCC)
r = a / (2√2)
Frenkel Defect
Ion to interstitial (density same)
Schottky Defect
Missing ion pairs (density ↓)
FCC Stacking
ABCABC...
HCP Stacking
ABABAB...
n-type Doping
Pentavalent (P) → electron carriers
p-type Doping
Trivalent (B) → hole carriers