Last Updated: April 2026
JEE Main solid state chemistry 2027 aspirants — Solid State (Class 12 NCERT, Chapter 1) is one of the highest-scoring chapters in JEE Main Chemistry. NTA’s Jan 2026 question paper carried 2 direct questions (8 marks) from this chapter — packing efficiency and Schottky vs Frenkel defect. The chapter rewards memorisation: 6 unit cell types, ~12 numerical formulas, 4 defect categories. JEE Main 2026 cut-off (general) was 93.23 percentile — high-yield chapters like this are non-negotiable. This guide covers crystal lattice, packing, defects, and 30 practice MCQs with full solutions.
1. Classification of Solids
- Crystalline: long-range order, sharp melting point, anisotropic (NaCl, diamond, quartz).
- Amorphous: short-range order, gradual softening, isotropic (glass, rubber, plastic).
Crystalline subtypes (by bonding):
| Type | Particles | Forces | Examples |
|---|---|---|---|
| Ionic | Cations + anions | Electrostatic | NaCl, MgO, ZnS |
| Covalent network | Atoms | Covalent bonds | Diamond, SiC, quartz |
| Metallic | Cations + electron sea | Metallic bond | Cu, Fe, Ag |
| Molecular | Molecules | vdW / H-bond | Ice, dry ice, S₈, I₂ |
2. Crystal Lattice and Unit Cell
A unit cell is the smallest 3D repeating unit. There are 7 crystal systems and 14 Bravais lattices. Three cubic types are most important for JEE:
- Simple Cubic (SC): atoms at 8 corners only. Z = 1.
- Body-Centred Cubic (BCC): 8 corners + 1 body centre. Z = 2.
- Face-Centred Cubic (FCC): 8 corners + 6 face centres. Z = 4.
Contributions: corner = 1/8, face = 1/2, edge = 1/4, body = 1.
3. Close Packing in 3D
HCP (ABABAB pattern) and CCP/FCC (ABCABC pattern) — both 74% efficient. BCC is 68% efficient (not close-packed but common in metals like Fe).
Voids in Close Packing
- Tetrahedral voids: 2 per atom in CCP. r_void / r_atom ratio limit = 0.225.
- Octahedral voids: 1 per atom in CCP. Ratio limit = 0.414.
4. Packing Efficiency Master Table
| Type | Z (atoms/cell) | Coord. No. | Packing % | r in terms of a | Examples |
|---|---|---|---|---|---|
| SC | 1 | 6 | 52.4 | r = a/2 | Po (only) |
| BCC | 2 | 8 | 68.0 | r = (√3/4)a | Fe, Na, K, Cr |
| FCC / CCP | 4 | 12 | 74.0 | r = (√2/4)a | Cu, Al, Ag, Au, Pb |
| HCP | 6 | 12 | 74.0 | r = a/2 (a = b) | Mg, Zn, Ti |
Worked Example 1 — Density Formula
ρ = Z·M / (N_A·a³)
Q. Cu has FCC, atomic mass 63.5 g/mol, edge a = 360 pm. Find density.
Sol. Z=4, M=63.5, a=3.6×10⁻⁸ cm, a³ = 4.67×10⁻²³ cm³. ρ = 4×63.5/(6.022×10²³ × 4.67×10⁻²³) = 254/28.12 = 9.03 g/cm³.
Worked Example 2 — Edge Length from Radius
Q. Atomic radius of an FCC metal is 144 pm. Find edge length.
Sol. r = (√2/4)a → a = 4r/√2 = 4×144/1.414 = 407.3 pm.
Worked Example 3 — Packing Efficiency Derivation (BCC)
Body diagonal = √3·a contains 4r → r = (√3/4)a. Volume of 2 atoms = 2·(4/3)πr³. Volume of cell = a³. PE = [2·(4/3)π·(√3/4)³·a³]/a³ × 100 = (√3·π/8)×100 ≈ 68%.
5. Defects in Solids
Point Defects (Stoichiometric)
- Schottky: pair of cation+anion vacancies. Decreases density. Shown by NaCl, KCl, CsCl, AgBr (high coordination, similar ion size).
- Frenkel: cation in interstitial site. No density change. Shown by AgBr, AgCl, ZnS (small cation, large anion).
Non-stoichiometric Defects
- Metal excess (F-centre): anion vacancy fills with electron → colour. NaCl heated in Na vapour turns yellow; KCl turns lilac.
- Metal deficiency: e.g. FeO frequently exists as Fe₀.₉₅O.
Impurity Defects
SrCl₂ in NaCl: each Sr²⁺ replaces 2 Na⁺, leaving 1 cation vacancy.
6. Magnetic Properties
| Type | Spin alignment | Behaviour | Example |
|---|---|---|---|
| Diamagnetic | All paired | Weakly repelled | NaCl, H₂O, Cu |
| Paramagnetic | Some unpaired | Weakly attracted | O₂, Cu²⁺, Fe³⁺ |
| Ferromagnetic | Aligned domains | Strongly attracted; permanent | Fe, Co, Ni, CrO₂ |
| Antiferromagnetic | Anti-parallel cancel | No net moment | MnO |
| Ferrimagnetic | Anti-parallel unequal | Net moment, weaker than ferro | Fe₃O₄, ferrites |
7. Conduction in Solids
Band theory: gap between valence and conduction band determines class.
- Metal: bands overlap → free electrons → high conductivity.
- Insulator: gap > 3 eV → no conduction (diamond gap ~6 eV).
- Semiconductor: gap ~ 1 eV (Si 1.12, Ge 0.66). n-type (As-doped Si), p-type (B-doped Si).
8. 30 JEE Practice MCQs
- Q1. In FCC, number of atoms = 4 (8×1/8 + 6×1/2).
- Q2. Coordination number of BCC = 8.
- Q3. NaCl has — FCC of Cl⁻ with Na⁺ in all octahedral voids.
- Q4. ZnS has — FCC of S²⁻ with Zn²⁺ in alternate tetrahedral voids (Z=4).
- Q5. CaF₂ has — FCC of Ca²⁺ with F⁻ in all tetrahedral voids (Z=4 Ca, 8 F).
- Q6. Antifluorite (Na₂O) — FCC of O²⁻, all tetrahedral voids by Na⁺.
- Q7. Schottky defect — equal cation+anion vacancies; density ↓.
- Q8. Frenkel defect — small cation in interstice; density unchanged.
- Q9. Schottky shown by — NaCl, KCl, CsCl, AgBr (high coord, similar size).
- Q10. Frenkel shown by — AgBr, AgCl, ZnS (small cation).
- Q11. Yellow colour of non-stoichiometric NaCl — F-centre (anion vacancy + trapped electron).
- Q12. Doping Si with P gives — n-type semiconductor.
- Q13. Doping Si with B gives — p-type semiconductor.
- Q14. Cu has FCC. Density given M, a, N_A, Z=4 → use ρ = ZM/(N_A·a³).
- Q15. Number of tetrahedral voids in 0.5 mol of FCC = 0.5 × 2 × N_A = N_A.
- Q16. In CCP r/R = 0.414 limit → ions in octahedral voids.
- Q17. Atomic mass of metal: ρ = ZM/(N_A·a³). Solve for M when ρ, Z, a known.
- Q18. Distance between nearest neighbours in BCC = (√3/2)a.
- Q19. Distance between nearest neighbours in FCC = a/√2 = (√2/2)a.
- Q20. Edge of NaCl unit cell with r(Na+) = 95 pm, r(Cl−) = 181 pm: a = 2(r₊+r₋) = 552 pm.
- Q21. Number of atoms per unit cell of HCP = 6.
- Q22. CsCl structure — simple cubic of Cl⁻, body centre Cs⁺ (coord no. 8).
- Q23. Number of octahedral voids in 1 mol FCC = N_A.
- Q24. Number of tetrahedral voids in 1 mol FCC = 2N_A.
- Q25. Curie temperature — temp above which ferromagnetic → paramagnetic.
- Q26. Hexagonal axial ratio c/a for HCP ≈ 1.633.
- Q27. Total volume occupied in BCC = 2·(4/3)πr³ where r = (√3/4)a.
- Q28. O₂ is paramagnetic due to 2 unpaired electrons in π* MOs.
- Q29. Density of Fe (BCC, M=55.85, a=287 pm) ≈ 7.86 g/cm³.
- Q30. Diamond structure: each C tetrahedrally bonded; effectively 8 C per cubic cell.
9. JEE Main Strategy for Solid State
This chapter is formula-driven — master the density equation, packing-efficiency derivation (BCC/FCC), and defects table. Devote 2-3 hours; expect 1-2 sure-shot questions per shift. Use JEE Gurukul Free Resources, follow our JEE 2027 plan, and join our chemistry intensive for live numerical drills.
FAQ
Q1. Why is FCC packing more efficient than BCC?
FCC arranges spheres in ABCABC layers with no wasted gap; coordination number 12. BCC has only 8 nearest neighbours and corner-to-body-centre packing leaves more void → 68% vs 74%.
Q2. Schottky vs Frenkel — quickest distinguishing rule?
Schottky decreases density; Frenkel does not. Schottky: vacancies of both ions. Frenkel: ion shifted to interstitial site (no net loss).
Q3. Why does NaCl turn yellow on heating in sodium vapour?
Excess Na atoms diffuse in, lose electrons that occupy anion vacancies. These F-centres absorb visible light (~580 nm) → yellow colour.
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