IIT-JEE & NEET | Class 12 Chemistry
Solid State
Master crystalline structures, defects, packing efficiency, and ionic models — the cornerstone chapter for competitive chemistry.
⚛ 12 Core Topics
🎯 40+ Practice Questions
📜 PYQ Collection
01
🔷 Types of Solids
Classification of matter in solid state — the foundation of this chapter
📌 Definition
In solids, constituent particles (atoms, ions, molecules) are held together by strong interparticle forces. They have definite shape, volume, and are rigid due to closely packed particles with very little free movement.
Crystalline vs Amorphous Solids
| Property | Crystalline Solids | Amorphous Solids |
|---|---|---|
| Arrangement | Regular, long-range order | Irregular, short-range order |
| Melting Point | Sharp, definite | Gradual softening (range) |
| Geometry | Definite geometry | No definite geometry |
| Anisotropy | Anisotropic | Isotropic |
| Cleavage | Clean cleavage planes | Irregular surfaces |
| Nature | True solid | Pseudo solid / Supercooled liquid |
| Examples | NaCl, ZnS, Quartz | Glass, Rubber, Plastics, Gel |
🔥 Key Point — Anisotropy
Crystalline solids are Anisotropic — they show different physical properties in different directions. Amorphous are Isotropic (same in all directions). This is a frequently tested concept in both JEE and NEET!
Classification by Bonding
🔗 Molecular Solids
Binding: van der Waals / H-bond / Dipole
- Non-polar: Ar, CCl₄, I₂ — very soft, very low mp
- Polar: SO₂, HCl — soft, low-moderate mp
- H-bonded: H₂O(ice), NH₃ — volatile
⚡ Ionic Solids
Binding: Electrostatic (Coulombic) forces
- Hard and brittle
- High melting point
- Conduct electricity only in molten/dissolved state
- Examples: NaCl, MgO, ZnS
🔴 Covalent / Network Solids
Binding: Covalent bonds
- Very hard (except graphite)
- Very high melting point
- Non-conductors (except graphite)
- Examples: Diamond, SiC, Quartz, Graphite
🥈 Metallic Solids
Binding: Metallic bond (electron sea)
- Hard to soft (Na to Fe)
- Good electrical & thermal conductors
- Lustrous, malleable, ductile
- Examples: Fe, Al, Cu, Mg
🧠 Memory Trick
"My Iron Can Melt" → Molecular, Ionic, Covalent, Metallic — the four types of crystalline solids in order of increasing melting point (generally)!
02
🔬 Crystal Lattice & Unit Cell
Space lattices, Bravais lattices, and the seven crystal systems
📌 Definitions
Crystal Lattice: Regular 3D arrangement of points (lattice points) showing positions of particles.Unit Cell: The smallest repeating unit of a crystal lattice. When stacked in 3D, it generates the entire crystal.
Seven Crystal Systems
| Crystal System | Axial Lengths | Axial Angles | Bravais Lattices | Examples |
|---|---|---|---|---|
| Cubic | a = b = c | α = β = γ = 90° | 3 (P, I, F) | NaCl, CsCl, ZnS |
| Tetragonal | a = b ≠ c | α = β = γ = 90° | 2 (P, I) | SnO₂, TiO₂ |
| Orthorhombic | a ≠ b ≠ c | α = β = γ = 90° | 4 (P, C, I, F) | KNO₃, K₂SO₄ |
| Hexagonal | a = b ≠ c | α = β = 90°, γ = 120° | 1 (P) | ZnO, Graphite, Mg |
| Rhombohedral | a = b = c | α = β = γ ≠ 90° | 1 (P) | CaCO₃, NaNO₃, Bi |
| Monoclinic | a ≠ b ≠ c | α = γ = 90°, β ≠ 90° | 2 (P, C) | Monoclinic S, Na₂SO₄ |
| Triclinic | a ≠ b ≠ c | α ≠ β ≠ γ ≠ 90° | 1 (P) | K₂Cr₂O₇, H₃BO₃ |
⭐ JEE Key Fact
Total Bravais Lattices = 3 + 2 + 4 + 1 + 1 + 2 + 1 = 14.Primitive (P): 7, Body-centred (I): 3, Face-centred (F): 2, End-centred (C): 2
🧠 Crystal Systems Mnemonic
"Cubic Tigers Often Have Rhombic Monoclinic Teeth"
Cubic → Tetragonal → Orthorhombic → Hexagonal → Rhombohedral → Monoclinic → Triclinic
Cubic → Tetragonal → Orthorhombic → Hexagonal → Rhombohedral → Monoclinic → Triclinic
03
📦 Cubic Unit Cells
Simple cubic, BCC, and FCC — most important for JEE & NEET
Simple Cubic (SC)
Atoms/cell: 1
CN: 6
r = a/2
PE: 52.4%
Body Centred Cubic (BCC)
Atoms/cell: 2
CN: 8
r = a√3/4
PE: 68%
Face Centred Cubic (FCC/CCP)
Atoms/cell: 4
CN: 12
r = a/2√2
PE: 74%
Number of Atoms per Unit Cell — Derivation
| Unit Cell | Corner Atoms (×1/8) | Face Atoms (×1/2) | Body Atom (×1) | Z (total) |
|---|---|---|---|---|
| Simple Cubic | 8 × 1/8 = 1 | — | — | 1 |
| BCC | 8 × 1/8 = 1 | — | 1 × 1 = 1 | 2 |
| FCC | 8 × 1/8 = 1 | 6 × 1/2 = 3 | — | 4 |
| HCP | — | — | — | 6 |
⭐ Important — Contribution of Atoms
Corner: 1/8 · Edge centre: 1/4 · Face centre: 1/2 · Body centre: 1
📏 Density of Unit Cell
Density Formula
ρ = (Z × M) / (Nₐ × a³)
Z = atoms per cell · M = molar mass · Nₐ = 6.022×10²³ · a = edge length
04
🔵 Close Packing in Solids
HCP, CCP arrangements and coordination numbers
One-Dimensional (Row Packing)
Spheres arranged in a row touching each other. Coordination number = 2 (one on each side). This is the simplest close packing along a line.
CN in 1D = 2
Square Close Packing
Rows arranged exactly on top of each other. Less efficient.
- Each sphere touches 4 others
- CN = 4
- Forms Square voids
- Packing fraction = 52.4% (less efficient)
Hexagonal Close Packing
Rows offset by half a sphere. More efficient.
- Each sphere touches 6 others
- CN = 6
- Forms Triangular voids
- Packing fraction = 60.4% (more efficient)
Hexagonal Close Packing (hcp) in 3D — ABAB… Pattern
- Layer A: 1st layer (triangular voids face up and down)
- Layer B: 2nd layer fills alternate voids of A (tetrahedral voids)
- 3rd layer exactly aligns with 1st layer → ABAB pattern
- Coordination number = 12 (6 in same layer + 3 above + 3 below)
- Packing efficiency = 74%
- Examples: Mg, Zn, Ti, Be
In HCP: Atoms per unit cell = 6, CN = 12, c/a = 1.633 (ideal)
Cubic Close Packing (ccp/FCC) in 3D — ABCABC… Pattern
- Layer A: 1st layer
- Layer B: fills alternate tetrahedral voids of A
- Layer C: fills remaining voids → ABCABC pattern
- Coordination number = 12
- Packing efficiency = 74%
- Examples: Cu, Ag, Au, Al, Ni, Pb
Both hcp and ccp have same packing efficiency (74%) and CN = 12. CCP ≡ FCC arrangement.
05
📐 Packing Efficiency
Fraction of space occupied by spheres — critical formula derivation
📌 Formula
Packing Efficiency (PE) = (Volume occupied by spheres / Total volume of unit cell) × 100
Simple Cubic
r = a/2
PE = (1 × 4πr³/3) / a³ = 52.4% | CN = 6 | Least efficient
Body Centred Cubic (BCC)
r = a√3/4
PE = (2 × 4πr³/3) / a³ = 68% | CN = 8 | e.g. Na, K, W, Cr
FCC / CCP / HCP
r = a/2√2
PE = (4 × 4πr³/3) / a³ = 74% | CN = 12 | Most efficient
Summary Table
| Unit Cell | Z | CN | Radius (r) | Packing Efficiency | Examples |
|---|---|---|---|---|---|
| SC | 1 | 6 | a/2 | 52.4% | Polonium (Po) |
| BCC | 2 | 8 | a√3/4 | 68.0% | Na, K, Li, Fe, Cr, W |
| FCC | 4 | 12 | a/2√2 | 74.0% | Cu, Ag, Au, Al, Ni |
| HCP | 6 | 12 | — | 74.0% | Mg, Zn, Ti, Co, Cd |
🧠 Packing Order
SC (52%) < BCC (68%) < FCC=HCP (74%) — Think: "Simple is sad, BCC is better, FCC is fantastic!"
06
🧲 Voids & Radius Ratio
Interstitial voids, their geometry, and ionic structure predictions
Types of Voids
🔺 Tetrahedral Voids
- Formed by 4 spheres in tetrahedral arrangement
- Radius ratio: r/R = 0.225
- In FCC: Number = 2N (N = number of atoms)
- Located at: 1/4 from each corner along body diagonal
- Examples occupied: Zn²⁺ in ZnS, Si in SiO₂
🔷 Octahedral Voids
- Formed by 6 spheres in octahedral arrangement
- Radius ratio: r/R = 0.414
- In FCC: Number = N (body centre + edge centres)
- Located at: body centre + 12 edge centres
- Examples occupied: Na⁺ in NaCl, Ti⁴⁺ in TiO₂
⭐ Void Count in FCC (n atoms)
Tetrahedral voids = 2n · Octahedral voids = n · Ratio T:O = 2:1
Radius Ratio Rules (r⁺/r⁻)
| r⁺/r⁻ Range | Coordination Number | Geometry of Void | Example Ionic Solid |
|---|---|---|---|
| < 0.155 | 2 | Linear | — |
| 0.155 – 0.225 | 3 | Triangular Planar | B₂O₃ |
| 0.225 – 0.414 | 4 | Tetrahedral | ZnS (Zinc Blende) |
| 0.414 – 0.732 | 6 | Octahedral | NaCl (Rock Salt) |
| 0.732 – 1.000 | 8 | Cubic | CsCl |
🧠 Radius Ratio Trick
"LeTe On Cubic" → Linear <0.155 · Tetrahedral 0.225–0.414 · Octahedral 0.414–0.732 · Cubic 0.732–1
Important values: NaCl → r⁺/r⁻ = 0.524 (Octahedral) · CsCl → r⁺/r⁻ = 0.93 (Cubic)
Important values: NaCl → r⁺/r⁻ = 0.524 (Octahedral) · CsCl → r⁺/r⁻ = 0.93 (Cubic)
07
🧊 Ionic Crystal Structures
NaCl, CsCl, ZnS, CaF₂ and more — high-frequency exam topics
🧂 NaCl — Rock Salt Structure
- Cl⁻ ions: FCC arrangement (corner + face)
- Na⁺ ions: All octahedral voids filled
- CN: Na⁺ = 6, Cl⁻ = 6 → (6:6)
- Z (formula units): 4 (NaCl per unit cell)
- Ratio: Na⁺: 12×1/4 + 1 = 4 ; Cl⁻: 8×1/8 + 6×1/2 = 4
- Edge length: a = 2(r⁺ + r⁻)
Examples: NaCl, KCl, KBr, LiCl, AgCl, MgO, FeO, MnS (NiAs type excluded)
r⁺/r⁻ for NaCl = 0.524 (Octahedral range)
r⁺/r⁻ for NaCl = 0.524 (Octahedral range)
🔷 CsCl — Caesium Chloride Structure
- Cl⁻ ions: Simple cubic arrangement (corners)
- Cs⁺ ions: Body centre (cubic void)
- CN: Cs⁺ = 8, Cl⁻ = 8 → (8:8)
- Z (formula units): 1 (CsCl per unit cell)
- Edge length: a = 2(r⁺ + r⁻)/√3
Examples: CsCl, CsBr, CsI, TlCl, TlBr
r⁺/r⁻ = 0.93 (Cubic range)
r⁺/r⁻ = 0.93 (Cubic range)
💎 ZnS — Two Polymorphic Forms
Zinc Blende (Sphalerite):
- S²⁻: FCC arrangement
- Zn²⁺: 4 alternate tetrahedral voids (half)
- CN: 4:4 (both tetrahedral)
- Z = 4 (ZnS per unit cell)
- Examples: ZnS, CuCl, CuBr, AgI, SiC
Wurtzite:
- S²⁻: HCP arrangement
- Zn²⁺: alternate tetrahedral voids
- CN: 4:4
- Examples: ZnO, ZnS (wurtzite), SiC
🔷 CaF₂ — Fluorite Structure
- Ca²⁺: FCC arrangement
- F⁻: ALL tetrahedral voids filled (2n voids, 2n F⁻)
- CN: Ca²⁺ = 8, F⁻ = 4 → (8:4)
- Z = 4 CaF₂ per unit cell
- Ratio: Cation:Anion = 1:2
Anti-Fluorite Structure (Li₂O, Na₂O):
Positions reversed — anions in FCC, cations fill all tetrahedral voids.
CN: 4:8 (O:Li in Na₂O)
Positions reversed — anions in FCC, cations fill all tetrahedral voids.
CN: 4:8 (O:Li in Na₂O)
🔵 Diamond Cubic
- FCC + half tetrahedral voids filled
- CN = 4 (tetrahedral)
- Z = 8
- Examples: Diamond, Si, Ge, Grey Sn
- PE = 34% (least efficient!)
🔴 Perovskite (CaTiO₃)
- Ca²⁺ at corners (cube)
- Ti⁴⁺ at body centre
- O²⁻ at all 12 edge centres
- Z = 1 formula unit
- Important for superconductors
🟣 Spinel (MgAl₂O₄)
- O²⁻: FCC arrangement
- Mg²⁺: 1/8 tetrahedral voids
- Al³⁺: 1/2 octahedral voids
- Normal spinel vs Inverse spinel
- Fe₃O₄: Inverse spinel
Quick Comparison
| Structure | Anion Lattice | Cation Position | CN (Cat:An) | Z | Examples |
|---|---|---|---|---|---|
| NaCl (Rock Salt) | FCC | All octahedral | 6:6 | 4 | NaCl, KBr, MgO |
| CsCl | SC | Cubic void (body centre) | 8:8 | 1 | CsCl, TlBr |
| ZnS Zinc Blende | FCC | Alt. tetrahedral (1/2) | 4:4 | 4 | ZnS, CuCl |
| ZnS Wurtzite | HCP | Alt. tetrahedral | 4:4 | — | ZnO, BeO |
| CaF₂ (Fluorite) | FCC | All tetrahedral | 8:4 | 4 | CaF₂, ThO₂, ZrO₂ |
| Na₂O (Anti-fluorite) | FCC | All tetrahedral (cation) | 4:8 | 4 | Na₂O, Li₂O, K₂O |
08
⚡ Imperfections (Defects) in Solids
Stoichiometric, non-stoichiometric defects and their effects
📌 Types of Defects
Defects are deviations from the ideal crystal structure. They are of two main types:1. Point Defects — Imperfection at specific lattice points
2. Line Defects — Irregularities along rows of atoms (Dislocations)
A. Stoichiometric Defects
🔷 Schottky Defect
- Equal number of cations AND anions missing
- Electrical neutrality maintained
- Occurs in highly ionic crystals (both ions similar size)
- Density decreases
- Examples: NaCl, KCl, KBr, AgBr (also Frenkel)
- % defect increases with temperature
🔺 Frenkel Defect
- An ion leaves its lattice site → interstitial position
- Usually smaller cation (not anion)
- Density unchanged
- Occurs when ions have very different sizes
- Examples: AgCl, AgBr, AgI, ZnS
- Leads to dielectric constant ↑
🧠 Schottky vs Frenkel
Schottky = Strictly leaves (both ions leave → vacancy) | Frenkel = Finds new place (moves to interstitial)
AgBr shows both Schottky and Frenkel — asked in JEE frequently!
AgBr shows both Schottky and Frenkel — asked in JEE frequently!
B. Non-Stoichiometric Defects
⬆️ Metal Excess Defect
Type 1 — Anionic vacancies: Anion missing, electron trapped in void (F-centre / colour centre). Solid becomes coloured.
- NaCl in Na vapour → Yellow (Na excess, Cl vacancies)
- KCl in K vapour → Violet/Lilac
- LiCl in Li vapour → Pink
Type 2 — Extra cations in interstitial: Extra metal ion in interstitial + electron to balance. e.g. ZnO (heated → Zn²⁺ excess → yellow, semiconductor).
⬇️ Metal Deficiency Defect
- Less metal than stoichiometric
- Some cations have higher oxidation state to maintain charge balance
- Examples: FeO, FeS, NiO
- FeO exists as Fe₀.₉₃O (Fe²⁺ and Fe³⁺ both present)
- These are often p-type semiconductors
C. Impurity Defects
🔵 Impurity Defects
- NaCl + SrCl₂: Sr²⁺ replaces two Na⁺ → one cation vacancy created → decreases density. Important for solid solution.
- AgCl + CdCl₂: Cd²⁺ replaces two Ag⁺ → vacancy
- Used to introduce defects deliberately (doping)
| Defect | Density Change | Conductivity | Examples |
|---|---|---|---|
| Schottky | Decreases | Ionic conductivity ↑ | NaCl, KCl, KBr |
| Frenkel | Unchanged | Ionic conductivity ↑ | AgCl, ZnS |
| Metal Excess (F-centre) | Decreases slightly | Electronic ↑ (n-type) | NaCl in Na vapour |
| Metal Deficiency | Unchanged | Electronic ↑ (p-type) | FeO, NiO |
| Impurity | Changes | Variable | SrCl₂ in NaCl |
09
💡 Properties of Solids
Electrical, magnetic, and thermal properties — Band theory basics
Electrical Properties — Band Theory
Conductors (Metals)
Conduction Band
(Overlap or empty)
Valence Band
Bands overlap OR conduction band partially filled. Free electron flow. σ decreases with T.
Semiconductors
Conduction Band
Small gap (~1 eV)
Valence Band
Small band gap. σ increases with T. Si, Ge.
Insulators
Conduction Band
Large gap (>3 eV)
Valence Band
Large forbidden gap. No electron flow. Diamond, Glass.
Types of Semiconductors
⬆️ n-type Semiconductor
- Si/Ge doped with Group 15 element (P, As, Sb)
- Extra electron available for conduction
- Majority carriers: Electrons
- Examples: Si doped with P
- Metal excess defects also form n-type (e.g. ZnO)
⬇️ p-type Semiconductor
- Si/Ge doped with Group 13 element (B, Al, Ga)
- Creates "holes" (electron vacancies)
- Majority carriers: Holes
- Examples: Si doped with B
- Metal deficiency defects → p-type (e.g. NiO)
Magnetic Properties
Diamagnetic
↑↓ ↑↓
All electrons paired. Weakly repelled by magnets.
NaCl, H₂O, Cu²⁺
Paramagnetic
↑ ↑ ↑
Unpaired electrons. Weakly attracted. No alignment without field.
O₂, CuO, TiO₃
Ferromagnetic
→ → → →
Domains aligned parallel. Strongly attracted. Retain magnetism.
Fe, Ni, Co, CrO₂
Antiferromagnetic
→ ← → ←
Domains aligned anti-parallel. Net magnetism = 0.
MnO, MnF₂, Cr₂O₃
Ferrimagnetic
→→ ← → →
Unequal anti-parallel domains. Net magnetic moment ≠ 0.
Fe₃O₄, MgFe₂O₄
⭐ NEET Favourite
Ferromagnetism → Antiferromagnetism at Néel Temperature.Ferromagnetism → Paramagnetism at Curie Temperature.
CrO₂ is used in magnetic tapes (ferromagnetic). Fe₃O₄ (magnetite) is ferrimagnetic.
10
🧮 Master Formula Sheet
All critical formulas for rapid revision — print-ready cheat sheet
Density of Crystal
ρ = ZM / (Nₐ × a³)
Z=atoms/cell · M=molar mass · a=edge length · Nₐ=Avogadro number
Radius — Simple Cubic
r = a/2
Contact along edge. PE = (π/6) × 100 = 52.4%
Radius — BCC
r = a√3/4
Contact along body diagonal. 4r = a√3. PE = 68%
Radius — FCC
r = a/(2√2)
Contact along face diagonal. 4r = a√2. PE = 74%
Packing Efficiency
PE = (Z × (4πr³/3)) / a³
SC=52.4% · BCC=68% · FCC/HCP=74% · Diamond=34%
Tetrahedral Void Radius
r_void/R = 0.225
r = radius of void atom · R = radius of sphere in lattice
Octahedral Void Radius
r_void/R = 0.414
More commonly occupied than tetrahedral in ionic crystals
Void Count in FCC
T = 2n, O = n
T=Tetrahedral voids · O=Octahedral voids · n=no. of atoms
Relation in FCC
d = a/√2
d = face diagonal = 4r for FCC. Body diagonal = a√3.
NaCl Edge Length
a = 2(r⁺ + r⁻)
r⁺=Na⁺ radius · r⁻=Cl⁻ radius. Contact along edge.
CsCl Body Diagonal
a√3 = 2(r⁺ + r⁻)
Contact along body diagonal in CsCl type structure.
HCP c/a Ratio
c/a = √(8/3) ≈ 1.633
Ideal HCP ratio. For Mg: 1.633, Zn: 1.856 (non-ideal)
📊 Constants You Must Remember
| Quantity | Value | Use Case |
|---|---|---|
| Avogadro Number (Nₐ) | 6.022 × 10²³ mol⁻¹ | Density calculations |
| Cube root of 2 | 1.414 | FCC face diagonal |
| Cube root of 3 | 1.732 | BCC body diagonal |
| π/6 | 0.5236 | SC packing fraction |
| π√2/6 | 0.7405 | FCC packing fraction |
| π√3/8 | 0.6802 | BCC packing fraction |
11
🎯 Practice Quiz — Solid State
15 IIT-JEE & NEET pattern questions with explanations
Quiz Complete! 🎉
12
📜 Previous Year Questions
Handpicked PYQs from JEE Mains, JEE Advanced & NEET
13
🔢 Solved Numerical Problems
Step-by-step solutions to high-frequency calculation types in JEE & NEET
Problem 1 — Density of BCC Crystal
Easy · JEE/NEET
Iron has BCC crystal structure with edge length a = 286 pm. Molar mass of Fe = 56 g/mol. Calculate the density of iron. (Nₐ = 6.022 × 10²³)
1
Identify Z (atoms per unit cell)
BCC structure → Z = 2 atoms per unit cell
2
Convert edge length to cm
a = 286 pm = 286 × 10⁻¹⁰ cm = 2.86 × 10⁻⁸ cm
3
Calculate a³
a³ = (2.86 × 10⁻⁸)³ = 23.39 × 10⁻²⁴ cm³
4
Apply density formula
ρ = Z × M / (Nₐ × a³)
ρ = (2 × 56) / (6.022 × 10²³ × 23.39 × 10⁻²⁴)
ρ = 112 / (14.09) = 7.95 g/cm³
✅ Density of Iron ≈ 7.95 g/cm³ (Standard value: 7.874 g/cm³)
Problem 2 — Number of Atoms & Voids in FCC
Easy · NEET
In a FCC crystal structure, find (a) number of atoms per unit cell, (b) number of tetrahedral voids, and (c) number of octahedral voids per unit cell.
1
Count atoms (Z) in FCC
Corners: 8 × 1/8 = 1 | Face centres: 6 × 1/2 = 3
Z = 1 + 3 = 4 atoms per unit cell
2
Tetrahedral Voids
In close-packed arrangement, tetrahedral voids = 2 × Z
Tetrahedral voids = 2 × 4 = 8
These 8 voids are located at 1/4 of each body diagonal from each corner — i.e., at 8 positions inside the cube.
3
Octahedral Voids
Octahedral voids = Z = 4 (body centre: 1, edge centres: 12 × 1/4 = 3)
Total octahedral voids = 1 + 3 = 4
✅ Z = 4 | Tetrahedral voids = 8 | Octahedral voids = 4 Ratio T:O = 2:1
Problem 3 — Packing Efficiency of BCC
Medium · JEE Mains
Derive the packing efficiency of a Body-Centred Cubic (BCC) unit cell, given that atoms touch along the body diagonal.
1
Atoms touch along body diagonal
Body diagonal = a√3 = 4r → r = a√3/4
2
Volume of atoms in unit cell (Z=2)
V_atoms = 2 × (4/3)πr³ = (8/3)π × (a√3/4)³
= (8/3)π × (3√3/64)a³ = (π√3/8)a³
3
Packing Efficiency
PE = V_atoms / V_cell × 100 = (π√3/8) × 100
= (3.14159 × 1.732 / 8) × 100 ≈ 68.02%
✅ Packing Efficiency of BCC = π√3/8 × 100 ≈ 68%
Problem 4 — Determining Crystal Structure from Density
Medium · JEE Advanced
An element with molar mass 63.5 g/mol and density 8960 kg/m³ forms a cubic unit cell with edge length 361.6 pm. Identify the unit cell type.
1
Convert units consistently
ρ = 8960 kg/m³ = 8.96 g/cm³ | a = 361.6 pm = 3.616 × 10⁻⁸ cm
2
Use density formula, solve for Z
Z = ρ × Nₐ × a³ / M
Z = 8.96 × 6.022×10²³ × (3.616×10⁻⁸)³ / 63.5
Z = 8.96 × 6.022×10²³ × 47.28×10⁻²⁴ / 63.5
3
Calculate numerator
= 8.96 × 6.022 × 47.28 × 10⁻¹ / 63.5 ≈ 254.1 / 63.5 ≈ 4.00
4
Identify from Z value
Z = 4 → Face Centred Cubic (FCC) structure
✅ The element is Copper (Cu) with FCC structure. Z = 4 confirms face-centred cubic.
Problem 5 — Radius Ratio & Structure Prediction
Hard · JEE Advanced
The radii of Na⁺ and Cl⁻ ions are 102 pm and 181 pm respectively. (a) Calculate the radius ratio. (b) Predict the coordination number. (c) Identify the crystal structure. (d) Calculate the edge length of the unit cell.
1
Calculate radius ratio (r⁺/r⁻)
r⁺/r⁻ = r(Na⁺)/r(Cl⁻) = 102/181 = 0.5635
2
Identify void type from ratio
0.5635 lies in range 0.414 – 0.732 → Octahedral voids
Coordination Number = 6 (Octahedral)
3
Crystal structure
CN = 6:6 with octahedral arrangement → NaCl (Rock Salt) structure
4
Edge length of unit cell
In NaCl structure, ions touch along the edge:
a = 2(r⁺ + r⁻) = 2(102 + 181) = 2 × 283 = 566 pm
✅ r⁺/r⁻ = 0.56 → Octahedral → NaCl structure → a = 566 pm
Problem 6 — Formula from Unit Cell Occupancy
Medium · NEET/JEE Mains
In a cubic unit cell, atoms X are at corners, Y atoms are at face centres and Z atoms are at the body centre. What is the formula of the compound?
1
Count X (corner atoms)
X = 8 corners × (1/8) = 1
2
Count Y (face centre atoms)
Y = 6 faces × (1/2) = 3
3
Count Z (body centre atom)
Z = 1 × 1 = 1
4
Write the formula
X:Y:Z = 1:3:1 → Formula = XY₃Z
Real analogy: This is similar to Perovskite structure (ABO₃) like CaTiO₃ where Ca at corners, O at face centres, Ti at body centre.
✅ Formula = XY₃Z | Example: Perovskite (CaTiO₃)
14
🃏 Interactive Flashcards
Click any card to reveal the answer — perfect for last-minute revision
Click cards to flip → 0 / 24 revealed
15
🗺 Chapter Mind Map
Visual overview of all topics and their interconnections
🗺 How to Use the Mind Map
The central node connects to 8 major sub-topics. Follow any branch to study that topic. Colour coding matches the sidebar navigation. Use this for last-day rapid revision — it gives you the chapter's complete skeleton at a glance!
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🏆 Exam Strategy & Tips
Marks analysis, common mistakes, and how to score maximum in Solid State
📊 Topic Weightage Analysis
2–3
Questions in JEE Mains
1–2
Questions in NEET
8–12
Marks in JEE Mains
4–8
Marks in NEET
74%
FCC packing — most asked value
📈 Sub-topic Weightage in JEE
⚠️ Common Mistakes — Don't Do These!
❌ Mistake #1
Confusing BCC and FCC touch direction: BCC — atoms touch along body diagonal. FCC — atoms touch along face diagonal. Mixing these gives wrong radius formulas.
❌ Mistake #2
Schottky vs Frenkel density: Schottky → density decreases. Frenkel → density unchanged. Many students get this reversed in exams!
❌ Mistake #3
Wrong formula for CaF₂: Students write 4:4 CN for CaF₂. Correct CN is Ca²⁺=8, F⁻=4 (8:4). Always check anion:cation stoichiometry.
❌ Mistake #4
Forgetting unit conversions: In density formulas, always convert pm to cm (1 pm = 10⁻¹⁰ cm) and check g/cm³ vs kg/m³.
❌ Mistake #5
Void count confusion: T:O voids = 2:1 in FCC, NOT equal. For Z=4 (FCC): 8 tetrahedral and 4 octahedral voids.
❌ Mistake #6
HCP vs CCP packing: Both have same PE (74%) and CN (12) but different stacking (ABAB vs ABCABC). Don't think HCP is less efficient!
✅ High-Scoring Tips
✅ Tip #1 — Master Density Formula
The single formula ρ = ZM/(Nₐa³) can solve or verify most unit-cell numericals. Practise rearranging it to find Z, M, or a.
✅ Tip #2 — Memorise All Packing Efficiencies
SC=52.4%, BCC=68%, FCC/HCP=74%, Diamond=34%. These appear as direct MCQs and are needed in derivations.
✅ Tip #3 — Ionic Structures: CN is the Key
Know CN for every major structure: NaCl(6:6), CsCl(8:8), ZnS(4:4), CaF₂(8:4), Diamond(4). CN unlocks the packing type.
✅ Tip #4 — Defects: Focus on Real-World Effects
Examiners love linking defects to real effects — F-centre colour, density change, conductivity. Know why not just what.
💡 Tip #5 — JEE Advanced Pattern
JEE Advanced often asks about derivation of PE, proofs for radius ratios, or multi-step structure problems. Practice the complete derivations, not just final answers.
💡 Tip #6 — NEET Focus Areas
NEET emphasises: types of solids, crystalline vs amorphous, packing efficiency values, defect names, magnetic property classification, and conductivity basics.
⭐ High-Priority Topics
Must-master: Z values for all cubic cells · NaCl structure · density formula · Schottky vs Frenkel · packing efficiencies · radius ratio table. These alone cover 70% of exam questions!
⭐ Last-Day Revision
Review: formula sheet (Section 10) → mind map (Section 15) → flashcards (Section 14) → review 3 PYQs you got wrong. 2 hours of targeted revision is better than 6 hours of re-reading.
🔁 Quick Facts — One-Liners to Remember
| # | One-Line Fact | Exam Tag |
|---|---|---|
| 1 | Polonium is the only element with Simple Cubic structure at room temperature. | ★ BOTH |
| 2 | AgBr shows both Schottky and Frenkel defects simultaneously. | JEE |
| 3 | ZnO is yellow when hot due to metal excess defect (Zn²⁺ in interstitial, extra electrons). | ★ BOTH |
| 4 | Diamond has the lowest packing efficiency (34%) — less than even Simple Cubic! | JEE |
| 5 | Graphite is a good electrical conductor — exceptional covalent solid (delocalised π electrons). | NEET |
| 6 | Ferromagnetic → Paramagnetic at Curie temperature. Above it, thermal agitation disrupts domain alignment. | ★ BOTH |
| 7 | SrCl₂ doped into NaCl creates cationic vacancies (each Sr²⁺ replaces 2 Na⁺ → one vacancy). | JEE |
| 8 | CrO₂ is ferromagnetic — used in making magnetic recording tapes. | NEET |
| 9 | In CaF₂ (fluorite), all tetrahedral voids are occupied by F⁻. (Anti-fluorite: all T-voids by cations) | ★ BOTH |
| 10 | Nearest neighbours in BCC = 8; Next nearest = 6 (along face diagonals). | JEE |