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ChemistryPhysical ChemistryHigh Yield (1-2 Qs in JEE)

Solid State Chemistry Formulas

Unit cell geometry, density formula, packing efficiency of SC/BCC/FCC, radius ratios, and crystal defects.

7 Formulas•430+ Questions in Bank

Concept Diagram

Cubic Crystal Lattices (SC, BCC, FCC)
Simple Cubica = 2rZ = 1 | 52%BCC√3 a = 4rZ = 2 | 68%FCC / CCP√2 a = 4rZ = 4 | 74%Density d = (Z × M) / (a³ × Nₐ)
Packing geometry and relation between edge length a and atomic radius r for simple cubic, BCC (√3a = 4r), and FCC (√2a = 4r).

Core Formulas & Mathematical Identities

7
01

Unit Cell Density Formula

ρ=Z⋅MNA⋅a3(ρ in g/cm3,  a in cm)\rho = \frac{Z \cdot M}{N_A \cdot a^3} \quad (\rho \text{ in g/cm}^3, \; a \text{ in cm})

Core formula linking atomic mass M, number of atoms per unit cell Z, and edge length a.

Examiner Tip: Remember N_A ≈ 6.022 × 10²³.
02

Lattice Packing Efficiency & Atomic Radii

SC: a=2r  (52.4%),BCC: r=3a4  (68%),FCC: r=a22  (74%)\text{SC: } a = 2r \; (52.4\%), \quad \text{BCC: } r = \frac{\sqrt{3}a}{4} \; (68\%), \quad \text{FCC: } r = \frac{a}{2\sqrt{2}} \; (74\%)

Geometric packing efficiency and nearest neighbor distance.

Examiner Tip: FCC / CCP has the maximum possible packing efficiency of 74%.
03

Limiting Radius Ratios for Voids

r+r−<0.225  (Trigonal),0.225−0.414  (Tetrahedral),0.414−0.732  (Octahedral),0.732−1.0  (Cubic)\frac{r_+}{r_-} < 0.225 \; (\text{Trigonal}), \quad 0.225 - 0.414 \; (\text{Tetrahedral}), \quad 0.414 - 0.732 \; (\text{Octahedral}), \quad 0.732 - 1.0 \; (\text{Cubic})

Predicts coordination number and void geometry in ionic crystals.

Examiner Tip: For N close-packed atoms: Octahedral voids = N, Tetrahedral voids = 2N.
04

Bragg's Law & Interplanar Spacing (Miller Indices)

2dhklsin⁡θ=nλ,dhkl=ah2+k2+l2(Cubic System)2d_{hkl} \sin\theta = n\lambda, \quad d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \quad (\text{Cubic System})

Constructive interference condition for X-rays reflected from crystal planes with Miller indices (h, k, l).

Examiner Tip: For cubic crystals: ratio of spacing d₁₀₀ : d₁₁₀ : d₁₁₁ = 1 : 1/√2 : 1/√3 = 1 : 0.707 : 0.577.
05

Hexagonal Close Packing (HCP) Geometry

Z=6,c=4r23=a83,Vcell=242r3=332a2c,P.E.=74%Z = 6, \quad c = 4r\sqrt{\frac{2}{3}} = a\sqrt{\frac{8}{3}}, \quad V_{\text{cell}} = 24\sqrt{2}r^3 = \frac{3\sqrt{3}}{2}a^2c, \quad \text{P.E.} = 74\%

Parameters of hexagonal closest packed unit cell. Coordination number is 12 (same as FCC/CCP).

Examiner Tip: Number of tetrahedral voids = 12, octahedral voids = 6 per HCP unit cell. c/a ratio = √(8/3) ≈ 1.633.
06

Standard Ionic Crystal Geometries & Radii

NaCl (6:6): r++r−=a2,CsCl (8:8): r++r−=3a2,ZnS (4:4): r++r−=3a4\text{NaCl (6:6): } r_+ + r_- = \frac{a}{2}, \quad \text{CsCl (8:8): } r_+ + r_- = \frac{\sqrt{3}a}{2}, \quad \text{ZnS (4:4): } r_+ + r_- = \frac{\sqrt{3}a}{4}

Distance between nearest cation-anion centers in rock-salt, cesium chloride, and zinc blende crystal lattices.

Examiner Tip: CaF₂ (fluorite) has 8:4 coordination; Na₂O (anti-fluorite) has 4:8 coordination.
07

Point Defects & Density Effects

Schottky: ρactual<ρtheoretical∣Frenkel: ρactual=ρtheoretical\text{Schottky: } \rho_{\text{actual}} < \rho_{\text{theoretical}} \quad | \quad \text{Frenkel: } \rho_{\text{actual}} = \rho_{\text{theoretical}}

Stoichiometric point defects in ionic solids. Schottky forms paired cation-anion vacancies; Frenkel dislocates smaller ion to interstitial site.

Examiner Tip: AgBr shows BOTH Schottky and Frenkel defects (frequent JEE question). Density decreases in Schottky, remains unchanged in Frenkel.

Top Examiner Traps & Pitfalls

Frequently tested misconceptions in Solid State Chemistry

  • •Density calculation units: cell edge length a must be in cm (1 Å = 10^(-8) cm) to match density in g/cm³.
  • •FCC has Z = 4 atoms, BCC has Z = 2 atoms, Simple Cubic has Z = 1 atom.