Solid State and Solutions: JEE Main Guide

Solid State (Class 12, Chapter 1) and Solutions (Class 12, Chapter 2) together contribute 4–8 marks in JEE Main and are among the more calculation-rich early Class 12 chapters. Solid State tests knowledge of crystal structures (unit cells, packing efficiency, density calculation, defects) while Solutions focuses on colligative properties and concentration calculations. Both chapters reward students who understand the underlying principles well enough to handle novel numerical setups — the exact skill JEE Main tests in these chapters.

Solid State: Unit Cells and Packing Efficiency

Crystal systems: 7 crystal systems (cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, rhombohedral/trigonal). For JEE Main, focus on cubic: simple cubic (SC), body-centred cubic (BCC), face-centred cubic (FCC/cubic close-packed CCP). Atoms per unit cell: SC = 1 (8 corners × 1/8), BCC = 2 (8 × 1/8 + 1 body), FCC = 4 (8 × 1/8 + 6 faces × 1/2). Coordination number: SC = 6, BCC = 8, FCC = 12. Packing efficiency: SC = 52.4%, BCC = 68%, FCC/HCP = 74% (most efficient). Radius ratio: for FCC, relation between a (edge length) and r: 4r = a·sqrt(2), so r = a·sqrt(2)/4. For BCC: 4r = a·sqrt(3), so r = a·sqrt(3)/4. Density: rho = (Z × M)/(NA × a³), where Z is atoms per unit cell, M is molar mass in g/mol, NA is Avogadro's number, a is edge length in cm. This density formula is tested numerically in every JEE Main session. For the broader physical chemistry context that includes thermodynamics underpinning solid state transitions, see our Thermodynamics Chemistry Guide.

Tetrahedral and octahedral voids in FCC: FCC packing has 8 tetrahedral voids (each one-eighth of the way from each corner along face diagonal) and 4 octahedral voids (at edge centres and body centre). Total tetrahedral voids = 2Z = 8; octahedral voids = Z = 4 for FCC. Radius ratios for void occupancy: tetrahedral void: r_small/r_large = 0.225; octahedral void: r_small/r_large = 0.414. Rock salt (NaCl) structure: Na+ occupies octahedral voids in Cl- FCC lattice. Zinc blende (ZnS) structure: Zn2+ occupies alternate tetrahedral voids. Fluorite (CaF2) structure: Ca2+ in FCC, F- in all tetrahedral voids. Antifluorite: F positions are cation, Ca positions are anion. These structure types are directly tested in JEE Main.

Crystal Defects and Electrical Properties

Point defects: Schottky defect (equal number of cations and anions missing; density decreases; found in ionic crystals with similar ionic radii, e.g., NaCl, KCl), Frenkel defect (smaller ion displaced to interstitial site; density unchanged; found in crystals with large size difference, e.g., AgCl, ZnS). Impurity defects: substitutional (foreign atom replaces a lattice atom, e.g., nitrogen in steel, 1% Si in Ge makes n-type semiconductor) or interstitial (carbon in steel). Electronic defects: n-type semiconductor (extra electrons from pentavalent dopant, e.g., As/P in Si); p-type semiconductor (holes from trivalent dopant, e.g., B/Al in Si). JEE Main tests: "Schottky defect causes decrease in density because ions are missing from the lattice while the overall lattice dimensions are unchanged." Test your solid state knowledge on our JEE Main chemistry mock tests with numerical and conceptual questions specifically on unit cells and defects.

Bragg's law for X-ray diffraction: n·lambda = 2d·sin(theta), where n is the order of diffraction, lambda is the X-ray wavelength, d is the interplanar spacing, and theta is the glancing angle. This equation is occasionally tested in JEE Main (given lambda and theta, find d or vice versa). Interplanar spacing for cubic crystals: d_hkl = a/sqrt(h² + k² + l²), where (h,k,l) are Miller indices.

Solutions: Colligative Properties

Colligative properties depend on the number of solute particles, not their nature. The four colligative properties: (1) Relative lowering of vapour pressure: (P° − P)/P° = x_solute = n_solute/(n_solute + n_solvent). Raoult's law: P = P°_solvent × x_solvent. (2) Elevation in boiling point: Delta·Tb = Kb × m, where m = molality = moles of solute per kg of solvent. Kb (ebullioscopic constant) = RT²_b·M_solvent/(1000·Delta·H_vap). (3) Depression in freezing point: Delta·Tf = Kf × m. Kf (cryoscopic constant) = RT²_f·M_solvent/(1000·Delta·H_fus). Kf(water) = 1.86 K·kg/mol, Kb(water) = 0.52 K·kg/mol. (4) Osmotic pressure: pi = MRT = (n/V)RT. For iso-osmotic (isotonic) solutions, pi is the same — cells don't shrink or swell. Hypertonic solution: higher pi than cell content — cells lose water (crenation). Hypotonic: lower pi — cells absorb water (lysis). Reverse osmosis: applying pressure greater than osmotic pressure to push solvent through semipermeable membrane from concentrated to dilute side — used for water purification.

Standard JEE Main numerical: "0.5 g of a non-volatile solute is dissolved in 40 g of water. The boiling point of the solution is 100.26°C. Find the molar mass of the solute. (Kb = 0.52 K·kg/mol)" Solution: Delta·Tb = 0.26 = Kb × m = 0.52 × (0.5/M)/(40/1000). Solving: 0.26 = 0.52 × 0.5 × 1000/(M × 40) = 0.52 × 12.5/M. M = 0.52 × 12.5/0.26 = 25 g/mol. This problem type (find molar mass from colligative property) is extremely frequent in JEE Main.

Van't Hoff Factor and Abnormal Colligative Properties

For electrolytes that dissociate, the effective number of particles increases: i = (observed colligative property)/(calculated for no dissociation). For NaCl (dissociates into Na+ and Cl-): i approaches 2 for dilute solutions. For BaCl2 (dissociates into Ba²+ and 2Cl-): i approaches 3. For solutes that associate (e.g., acetic acid in benzene forms dimers): i is less than 1. Degree of dissociation alpha in terms of i: for A → n ions, i = 1 + alpha(n−1), so alpha = (i−1)/(n−1). Degree of association: for n molecules → 1 cluster, i = 1 − beta(1 − 1/n), so beta = (1−i)/(1−1/n) where beta is degree of association. Modified colligative property equations: Delta·Tf = i × Kf × m, pi = i × MRT. JEE Main directly asks: "A 1 molal NaCl solution has a freezing point depression of 3.37°C. Find van't Hoff factor. (Kf = 1.86 K·kg/mol)" Solution: i = Delta·Tf/(Kf × m) = 3.37/(1.86 × 1) = 1.81. This is the standard electrolyte dissociation problem. Register on our platform for targeted Solid State and Solutions practice tests. Our premium plan includes physical chemistry mock tests with full solutions. For electrochemistry — the Class 12 physical chemistry chapter that follows Solutions and builds on ionic concept knowledge — see our Electrochemistry Deep-Dive Guide.

Key study tip: create a single A4 summary sheet for colligative properties with all four property formulas, the van't Hoff factor correction, Kf and Kb values for water (memorise these), and the unit cell density formula for solid state. This 30-minute effort before exam consolidation saves 5–10 minutes of formula recall on exam day.