Solid State

The solid state is characterized by fixed positions of constituent particles. Crystalline solids have long-range ordered arrangement, sharp melting points, are anisotropic (direction-dependent properties), and show clean cleavage. Amorphous solids have only short-range order, melt over a range, are isotropic, and show irregular fracture. Examples: quartz (crystalline) vs glass (amorphous).

Crystal lattice and unit cells: A crystal lattice is a regular 3D arrangement of points, and a unit cell is the smallest repeating unit. There are 7 crystal systems and 14 Bravais lattices. For NEET, the three cubic unit cells are most important:

• Simple Cubic (SC): Atoms at corners only. Z = 8 × ($\frac{1}{8}$) = 1 atom/cell. Coordination number = 6. Packing efficiency = π/6 = 52.4%. Edge-radius: 2r = a, so r = a/2.
• Body-Centered Cubic (BCC): Corners + 1 body center. Z = 8 × ($\frac{1}{8}$) + 1 = 2 atoms/cell. CN = 8. Packing efficiency = π√$\frac{3}{8}$ = 68%. Atoms touch along body diagonal: 4r = a√3, so r = a√$\frac{3}{4}$.
• Face-Centered Cubic (FCC): Corners + face centers. Z = 8 × ($\frac{1}{8}$) + 6 × ($\frac{1}{2}$) = 4 atoms/cell. CN = 12. Packing efficiency = π/(3√2) = 74%. Atoms touch along face diagonal: 4r = a√2, so r = a√$\frac{2}{4}$ = a/(2√2).

Close-packed structures: HCP (ABAB stacking) and CCP (ABCABC stacking = FCC). Both have 74% packing efficiency.

Voids in close packing: For n atoms: n octahedral voids (radius ratio r/R = 0.414) and 2n tetrahedral voids (radius ratio r/R = 0.225). Why? Each atom is surrounded by 2 tetrahedral voids (one above, one below in adjacent layers) and 1 octahedral void per atom (between layers).

Density formula derivation: Mass of unit cell = Z × M/Nₐ (Z atoms, each of molar mass M/Nₐ grams). Volume = a³. Therefore: ρ = ZM/(a³Nₐ). Dimensional analysis: (atoms × g/mol) / (cm³ × atoms/mol) = g/cm³ ✓

Solved Example 1: Silver (FCC, a = 4.077 Å = 4.077 × 10⁻⁸ cm). Calculate density. ρ = ZM/(a³Nₐ) = (4 × 108) / ((4.077 × 10⁻⁸)³ × 6.022 × 10²³) = 432 / (6.776 × 10⁻²³ × 6.022 × 10²³) = 432 / 40.80 = 10.59 g/cm³

Solved Example 2: Edge-radius relationships. FCC: Atoms touch along face diagonal → face diagonal = a√2 = 4r → r = a/(2√2) BCC: Atoms touch along body diagonal → body diagonal = a√3 = 4r → r = a√$\frac{3}{4}$

Ionic crystal structures:

• NaCl type: FCC of Cl⁻; Na⁺ occupies ALL octahedral voids. CN = 6:6. r⁺/r⁻ = 0.414–0.732. Examples: NaCl, KCl, MgO, CaO.
• CsCl type: BCC-like arrangement. CN = 8:8. r⁺/r⁻ > 0.732. Examples: CsCl, CsBr, CsI.
• ZnS (zinc blende) type: FCC of S²⁻; Zn²⁺ occupies only HALF of the tetrahedral voids. CN = 4:4. r⁺/r⁻ = 0.225–0.414. Examples: ZnS, CuCl, SiC.

Solved Example 3: X forms FCC, Y fills ALL octahedral voids. Find the formula. FCC has 4 atoms of X. Number of octahedral voids = 4 (= n). Y fills all 4 voids. Formula = X₄Y₄ = XY (e.g., NaCl structure).

Crystal defects (point defects):

• Schottky defect: Equal number of cation and anion vacancies. Density decreases (missing ion pairs). Found in highly ionic compounds with similar ion sizes: NaCl, KCl, CsCl, AgBr.
• Frenkel defect: Cation displaced to an interstitial site. Density unchanged (no atoms leave the crystal). Found in compounds with large size difference between cation and anion: ZnS, AgCl, AgBr, AgI.
• AgBr shows BOTH Schottky and Frenkel defects — a classic NEET exception.
• Non-stoichiometric defects: Metal excess type — F-centres (electrons trapped in anion vacancies, give colour: NaCl → yellow, KCl → violet). Metal deficiency type — cation vacancies with higher charge on adjacent cations (e.g., FeO, FeS).

Electrical properties (band theory): Conductors (no band gap), semiconductors (small gap ~1 eV), insulators (large gap >3 eV). n-type semiconductors are doped with Group 15 elements (P, As, Sb) — extra electron, majority carriers are electrons. p-type semiconductors are doped with Group 13 elements (B, Ga, In) — electron hole, majority carriers are holes.

Magnetic properties: Diamagnetic (all paired, weakly repelled — NaCl, benzene), paramagnetic (unpaired electrons, weakly attracted — O₂, Cu²⁺), ferromagnetic (domains aligned same direction, strongly attracted — Fe, Co, Ni), antiferromagnetic (domains antiparallel, cancel — MnO), ferrimagnetic (domains antiparallel but unequal, net moment — Fe₃O₄, ferrites).

The key testable concept is unit cell density calculations using ρ = ZM/(a³Nₐ) and distinguishing Schottky (density decreases) from Frenkel (density unchanged) defects.