1. Classification of Solids

Solids are characterized by definite shape, volume, and mass due to strong intermolecular forces and short distances.

Crystalline vs Amorphous

Property	Crystalline	Amorphous
Geometry	Regular, long range order	Irregular, short range
Melting Point	Sharp M.P.	Softens over a range
Nature	Anisotropic (properties vary with direction)	Isotropic (same in all directions)
Examples	NaCl, Quartz, Diamond	Glass, Rubber, Plastic

[Image of Crystalline vs Amorphous structure]

Types of Crystalline Solids

Ionic: Ions held by Coulombic forces. Hard, brittle, high MP. (e.g., NaCl).
Covalent (Network): Atoms held by covalent bonds. Very hard. (e.g., Diamond, SiO₂).
Molecular: Molecules held by Van der Waals/H-bonds. Soft, low MP. (e.g., Ice, Solid CO₂).
Metallic: Positive ions in sea of electrons. Conductive, ductile. (e.g., Fe, Cu).

2. Unit Cells & Packing

Unit Cell: The smallest repeating unit which, when repeated in 3D, generates the entire lattice.

Types of Unit Cells (Cubic)

Simple Cubic (SC): Atoms at corners only. Z = 8 × (1/8) = 1.
Body Centered (BCC): Corners + Body center. Z = 1 + 1 = 2.
Face Centered (FCC/CCP): Corners + Face centers. Z = 1 + 3 = 4.

Packing Efficiency & Formulas

Type	Radius (r) vs Edge (a)	Packing %	Coord. No.
SC	a = 2r	52.4%	6
BCC	4r = √3 a	68%	8
FCC	4r = √2 a	74%	12

Density of Unit Cell (d)
d = (Z × M) / (a³ × N_A)

(Ensure 'a' is in cm for density in g/cm³)

3. Voids in Crystals

Empty spaces inside the packing.

Tetrahedral Void (Td): Surrounded by 4 spheres. Radius ratio = 0.225. (Count = 2N)
Octahedral Void (Oh): Surrounded by 6 spheres. Radius ratio = 0.414. (Count = N)

4. Imperfections in Solids

Stoichiometric Defects

1. Schottky Defect:

Equal number of Cations and Anions missing.

Density: Decreases.
Condition: Similar size of ions (e.g., NaCl, KCl, CsCl).

2. Frenkel Defect (Dislocation):

Smaller ion (cation) dislocated to interstitial site.

Density: Remains same.
Condition: Large difference in ion size (e.g., AgCl, ZnS).

Note: AgBr shows BOTH Schottky and Frenkel defects.

Non-Stoichiometric Defects

Metal Excess (F-Center): Anion vacancy occupied by electron. Causes color (e.g., NaCl yellow, KCl violet).
Metal Deficiency: Cation missing, charge balanced by higher valency cation (e.g., Fe_0.95O).

5. Electrical Properties

Band Theory

Conductors: No gap between Valence Band (VB) and Conduction Band (CB).
Insulators: Large energy gap (Eg > 3 eV). No electron jump.
Semiconductors: Small gap. Electrons jump at high temp.

Doping (Semiconductors)

n-type: Group 14 (Si/Ge) doped with Group 15 (P/As). Electron rich.

p-type: Group 14 (Si/Ge) doped with Group 13 (B/Al). Electron deficient (Holes).

6. Magnetic Properties

1. Paramagnetic:

Weakly attracted by magnetic field. Unpaired electrons present (e.g., O₂, Cu²⁺).

2. Diamagnetic:

Weakly repelled. No unpaired electrons (e.g., H₂O, NaCl).

3. Ferromagnetic:

Strongly attracted. Domains align in same direction. Permanent magnet (e.g., Fe, Ni, Co, CrO₂).

4. Antiferromagnetic:

Domains align oppositely, cancelling magnetic moment (e.g., MnO).

5. Ferrimagnetic:

Domains align unequally in parallel/anti-parallel. Weak attraction (e.g., Fe₃O₄).

Numericals & HOTS

Q1. Density of Unit Cell

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).

Solution:
Formula: d = (Z × M) / (a³ × N_A)
Given:
Lattice = ccp (FCC), so Z = 4.
M = 107.9 g/mol.
a = 408.6 pm = 408.6 × 10^-10 cm.

d = (4 × 107.9) / [(4.086 × 10^-8)³ × 6.022 × 10²³]
d = 431.6 / [68.2 × 10^-24 × 6.022 × 10²³]
d = 431.6 / 41.07
d = 10.5 g/cm³

Q2. Formula from Voids

A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound?

Solution:
1. Lattice (ccp) is formed by Y. Let number of Y atoms = N.
2. Number of Octahedral voids = N.
3. Atoms X occupy ALL octahedral voids. So, number of X = N.

Ratio X : Y = N : N = 1 : 1.
Formula = XY

Q3. Metal Deficiency (HOTS)

Analysis shows that nickel oxide has the formula Ni_0.98O_1.00. What fraction of nickel exists as Ni²⁺ and Ni³⁺ ions?

Solution:
Assume Total Oxide ions (O^2-) = 100. Charge = -200.
Total Nickel ions = 98. Let Ni²⁺ = x, then Ni³⁺ = 98 - x.

Charge Balance:
2(x) + 3(98 - x) = 200
2x + 294 - 3x = 200
-x = 200 - 294 = -94
x = 94 (Ni²⁺)
Ni³⁺ = 98 - 94 = 4.

Fraction Ni²⁺ = 94/98 × 100 = 96%
Fraction Ni³⁺ = 4/98 × 100 = 4%

Q4. X-Ray Diffraction

Second order Bragg diffraction of X-rays with λ = 1.0 Å from a set of parallel planes in a metal occurs at an angle of 60°. Calculate the distance between the scattering planes.

Solution:
Formula: nλ = 2d sinθ
Given: n = 2, λ = 1.0 Å, θ = 60°.

2 × 1.0 = 2 × d × sin(60°)
2 = 2d × (√3 / 2)
2 = d √3
d = 2 / 1.732
d = 1.15 Å

Q5. BCC Geometry

Sodium crystallizes in a bcc structure with a unit cell edge of 4.29 Å. Calculate the radius of the sodium atom.

Solution:
For BCC Lattice:
Body diagonal = 4r = √3 a

r = (√3 / 4) × a
r = (1.732 / 4) × 4.29
r = 0.433 × 4.29
r = 1.857 Å

Q6. Cation Vacancy (HOTS)

If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies?

Solution:
1 Sr²⁺ replaces 2 Na⁺, creating 1 cation vacancy.
Thus, moles of vacancies = moles of SrCl₂ added.

Concentration = 10^-3 mol % = 10^-3 / 100 = 10^-5 mol fraction.
Vacancies per mole NaCl = 10^-5 × N_A
= 10^-5 × 6.022 × 10²³
= 6.022 × 10¹⁸ vacancies/mol

Q7. Counting Unit Cells

An element crystallizes in a bcc lattice. How many unit cells are present in 208 g of the element? (Atomic mass = 104, Z for bcc = 2).

Solution:
1. Calculate Moles: 208 / 104 = 2 moles.
2. Calculate Atoms: 2 × N_A atoms.

In BCC, 1 unit cell contains 2 atoms.
So, 2 atoms = 1 Unit Cell.
2 × N_A atoms = N_A Unit Cells.

Answer: 6.022 × 10²³ Unit Cells

Q8. Radius Ratio

The radius of the anion in a solid is 100 pm. What is the maximum radius of a cation that can fit into a tetrahedral hole?

Solution:
For Tetrahedral Void, limiting radius ratio (r+/r-) range is 0.225 - 0.414.
To fit exactly (without distorting), we take the lower limit.

r_cation / r_anion = 0.225
r_cation = 100 × 0.225
r_cation = 22.5 pm

Q9. Nearest Neighbor Distance

Xenon crystallizes in the face-centered cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbor distance?

Solution:
In FCC, nearest neighbors are touching along the Face Diagonal.
Face Diagonal = √2 a.
Distance (d) = 2r = Face Diagonal / 2 = (√2 a) / 2 = a / √2.

d = 620 / 1.414
d = 438.5 pm

Q10. Void Occupancy

In a crystalline solid, anions B form ccp arrangement. Cations A occupy 50% of octahedral voids and cations C occupy 50% of tetrahedral voids. What is the formula?

Solution:
Let number of B ions (ccp) = 4 (Standard FCC unit).
Total Octahedral Voids (OV) = 4.
Total Tetrahedral Voids (TV) = 8.

A = 50% of OV = 0.5 × 4 = 2.
C = 50% of TV = 0.5 × 8 = 4.

Ratio A : B : C = 2 : 4 : 4
Simplify: 1 : 2 : 2
Formula = AB₂C₂

Important Formulae

1. Density of Unit Cell (d)

d = (Z × M) / (a³ × N_A)

Z = No. of atoms per unit cell
M = Molar Mass (g/mol)
a = Edge length (cm)
N_A = 6.022 × 10²³

*Remember: If 'a' is in pm, convert to cm (1 pm = 10^-10 cm).

2. Radius (r) vs Edge (a)

Simple Cubic	a = 2r
Face Centered (FCC)	a = 2√2 r
Body Centered (BCC)	a = 4r / √3

3. Packing Efficiency (P.E.)

Percentage of total space filled:

FCC / CCP / HCP: 74% (Most efficient)
BCC: 68%
Simple Cubic: 52.4%

Void Space = 100 - P.E.

4. Calculation of Voids

If 'N' is the number of atoms in packing:

Octahedral Voids = N

Tetrahedral Voids = 2N

Ionic Crystals: Anions form the lattice (N), Cations occupy voids.

5. Bragg's Equation

nλ = 2d sinθ

n = Order, d = Interplanar distance

20 Golden Facts (NEET)

1. Supercooled Liquids: Amorphous solids (like Glass) have a tendency to flow very slowly, causing old window panes to become thicker at the bottom.
2. AgBr Exception: Silver Bromide (AgBr) is unique as it shows BOTH Schottky (vacancy) and Frenkel (dislocation) defects.
3. F-Centers: The anionic sites occupied by unpaired electrons are called F-centers (Farbenzenter). They impart color (e.g., NaCl turns yellow, KCl turns violet).
4. ZnO Heating: Zinc Oxide (white) turns Yellow on heating due to metal excess defect (loss of oxygen leaving Zn²⁺ and electrons in voids).
5. Coordination Number (CN):
NaCl structure (Rock Salt): Na⁺(6) : Cl^-(6)
CsCl structure: Cs⁺(8) : Cl^-(8)
ZnS structure: Zn²⁺(4) : S^2-(4)
6. Pressure Effect: High pressure increases Coordination Number. NaCl type (6:6) → CsCl type (8:8).
7. Temperature Effect: High temperature decreases Coordination Number. CsCl type (8:8) → NaCl type (6:6).
8. Ferrimagnetism: Magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers. Example: Fe₃O₄ (Magnetite). Becomes paramagnetic on heating.
9. Quartz vs Glass: Quartz is crystalline SiO₂ (ordered). Glass is amorphous SiO₂ (random). Quartz can be converted to glass by melting and rapid cooling.
10. 12-16 Compounds: Semiconductors formed by Group 12 and 16 elements (e.g., ZnS, CdS, HgTe). They have average valency of 4.
11. Diamond: Covalent solid. Each Carbon is sp³ hybridized. Hardest substance. Insulator (no free electrons).
12. Graphite: Covalent solid but layered. Each Carbon is sp² hybridized. Good conductor (free 4th electron between layers). Soft lubricant.
13. Distance of Nearest Neighbor (d):
SC: d = a
BCC: d = √3 a / 2
FCC: d = a / √2
14. Radius Ratio Rule:
0.155 - 0.225 → Triangular (CN 3)
0.225 - 0.414 → Tetrahedral (CN 4)
0.414 - 0.732 → Octahedral (CN 6)
0.732 - 1.000 → Cubic (CN 8)
15. CrO₂ Application: Chromium Dioxide (CrO₂) is ferromagnetic and is used to make magnetic tapes for audio recording.
16. Rectifier: p-n junction allows current to flow only in one direction, acting as a rectifier.
17. HCP vs CCP: Both have 74% efficiency.
HCP (Mg, Zn): ABABAB pattern.
CCP/FCC (Cu, Ag, Au): ABCABC pattern.
18. Cation Vacancy: Fe_xO (where x ≈ 0.95) exists because some Fe²⁺ are missing and charge is balanced by Fe³⁺.
19. Pyroelectricity: Some polar crystals produce a small electric current when heated (e.g., Tourmaline).
20. Molecular Solid Types:
Non-polar: Ar, CCl₄ (Dispersion forces).
Polar: HCl, SO₂ (Dipole-dipole).
H-bonded: H₂O (Hydrogen bonds).

📱 Practice MCQs for this topic inside our App