VKC Notes · Solid state · Civil Services P-I · IFoS P-I · Geo-Scientist P-II

3.1 Crystal Systems, Miller Indices, Bragg’s Law and Radius Ratio

★★★ 2021–2025: 34 parts, 296 marks (Civil Services 7 parts / 90 m · IFoS 10 / 86 m · Geo-Scientist 17 / 120 m). Asked in every paper of every year: one of the surest numericals in all three exams.

Syllabus lines. Civil Services: “Crystal systems; designation of crystal faces, lattice structures and unit cell; Bragg’s law; X-ray diffraction by crystals; close packing, radius ratio rules, calculation of some limiting radius ratio values; structures of NaCl, ZnS, CsCl and CaF₂”. IFoS: the same, plus “law of constancy of interfacial angles, crystal classes, laws of rational indices, CdI₂ and rutile”. Geo-Scientist: “concept of unit cell, different crystal systems, Bravais lattices, law of rational indices, Miller indices, symmetry elements in crystals. X-ray diffraction, Bragg’s law”. Defects, band theory and semiconductors are in 3.2.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

3.1.1 Crystal systems and Bravais lattices ★★★

A lattice is an infinite array of points, each with identical surroundings. The unit cell is the smallest block which, repeated along its three edges a, b, c, builds the whole lattice; it is fixed by the edges and the angles α, β, γ between them CSE 2024 Q1(b)(ii) 5m.SN
Atoms placed on each lattice point (the basis) make the crystal: lattice + basis = crystal structure. ZnS is an F lattice with a two-atom basis.

Fig. 3.1.1 The three cubic Bravais lattices.

Draw it in 60 s: three cubes (back edges dashed); dots at corners; add one centre dot (I) or six face dots (F); write Z and CN under each.

System	Cell edges; angles	Bravais lattices	Essential symmetry	Example
Cubic	a = b = c; all 90°	P, I, F	four 3-fold axes	NaCl, CsCl, Cu
Tetragonal	a = b ≠ c; all 90°	P, I	one 4-fold axis	TiO₂ (rutile), SnO₂
Orthorhombic	a ≠ b ≠ c; all 90°	P, C, I, F	three 2-fold axes	rhombic S, Mg₂SiO₄
Hexagonal	a = b ≠ c; 90°, 90°, 120°	P	one 6-fold axis	graphite, ZnO, Mg
Rhombohedral (trigonal)	a = b = c; all equal ≠ 90°	R	one 3-fold axis	calcite, quartz
Monoclinic	a ≠ b ≠ c; α = γ = 90° ≠ β	P, C	one 2-fold axis	monoclinic S, gypsum
Triclinic	a ≠ b ≠ c; all unequal ≠ 90°	P	none (at most a centre)	CuSO₄·5H₂O, K₂Cr₂O₇

Total 3 + 2 + 4 + 1 + 1 + 2 + 1 = 14 Bravais lattices. Identify the system from the cell constants alone Geo 2021 Q1(d) 5m; tetragonal C reduces to a smaller tetragonal P and tetragonal F to tetragonal I, and a base-centred cube loses its 3-fold axes, so neither is a new lattice Geo 2024 Q1(d) 5m.SN

Orthorhombic symmetry: the minimum (class 222) is three mutually perpendicular 2-fold axes; the maximum (class mmm) is 3 C₂ + 3 mirror planes + a centre = 7 elements. Forsterite Mg₂SiO₄ (olivine) has the maximum Geo 2025 Q1(d) 5m.

3.1.2 Counting atoms in a cell, formula and density ★★★

Position	corner	edge centre	face centre	body centre
Shared by	8 cells	4 cells	2 cells	1 cell
Contribution	⅛	¼	½	1

(3.1.1)

X at corners, Y at face centres: X = 8 × ⅛ = 1, Y = 6 × ½ = 3: XY₃ (Cu₃Au) Geo 2021 Q1(c) 5m. Oxide ccp, ⅛ of tetrahedral holes A²⁺, ½ of octahedral holes B³⁺: A : B : O = ¼ : ½ : 1 = AB₂O₄ (spinel) IFoS 2025 Q1(b) 8m.SN
Solid C₆₀ is fcc: 4 molecules × 60 = 240 C atoms per cell Geo 2023 Q1(c) 5m. FeO, a = 5.0 Å, ρ = 4.0 g cm⁻³: Z = 4.19 ≈ 4 (rock salt) IFoS 2023 Q1(c) 8m.
bcc metal, ρ = 10 300 kg m⁻³, a = 314 pm: M = ρN_Aa³/2 = 0.0960 kg mol⁻¹ (Mo) Geo 2024 Q3(b) 7m. Silicon (diamond, Z = 8), ρ = 2.33 g cm⁻³: a = 5.43 Å Geo 2023 Q11(b) 10m.
Cubic metal, a = 2.88 Å, ρ = 7.20 g cm⁻³: mass of one cell = ρa³ = 1.72 × 10⁻²² g, so 100 g holds 100/(1.72 × 10⁻²²) = 5.81 × 10²³ cells (the lattice type is not needed) Geo 2022 Q11(b) 10m.SN

Solved PYQ IFoS 2023 Q3(b) 10m

LiF (ρ = 2.601 g cm⁻³, Z = 4, cubic): (111) first-order reflection at 8° 44′ with λ = 70.8 pm. Find N_A.

d₁₁₁ = λ/(2 sin θ) = 70.8/(2 × 0.1518) = 233.1 pm.
Cubic: a = d₁₁₁ √3 = 403.8 pm = 4.038 × 10⁻⁸ cm; M = 6.939 + 18.998 = 25.937 g mol⁻¹.
N_A = ZM/(ρa³) = 4 × 25.937/(2.601 × 6.584 × 10⁻²³) = 6.06 × 10²³ mol⁻¹.SN

The same route from NaCl (a = 2 × 281 pm, ρ = 2.165 g cm⁻³) gives N_A = 6.09 × 10²³ mol⁻¹ IFoS 2025 Q3(a)(ii) 10m.

3.1.3 Close packing, packing fraction and voids ★★★

Cell	Atoms touch along	a in terms of r	Z	CN	Packing fraction
Simple cubic	edge	a = 2r	1	6	π/6 = 0.524
bcc	body diagonal	√3 a = 4r	2	8	√3π/8 = 0.680
fcc (ccp)	face diagonal	√2 a = 4r	4	12	π/(3√2) = 0.740

(3.1.2)

Compactness fcc = hcp (74 %) > bcc (68 %) > sc (52.4 %): more nearest neighbours fill space better Geo 2022 Q3 10m.SN
In N close-packed spheres there are N octahedral and 2N tetrahedral holes. Largest sphere that fits: tetrahedral 0.225R, octahedral 0.414R, cubic (centre of a simple cubic cell) 0.732R.SN

Solved PYQ CSE 2021 Q2(c) 20m

bcc iron, ρ = 7.86 g cm⁻³, M = 55.845. Atomic radius? Does iron expand or contract on going to fcc above 910 °C (same radius)?

a³ = 2 × 55.845/(7.86 × 6.022 × 10²³) = 2.360 × 10⁻²³ cm³; a = 286.8 pm.
bcc contact: √3a = 4r, so r = 124.2 pm.
fcc with the same r: a′ = 4r/√2 = 351.3 pm. Volume per atom: bcc a³/2 = 11.80 Å³; fcc a′³/4 = 10.84 Å³.
Ratio 0.919: iron contracts by 8.1 %, because fcc packs better (0.740 against 0.680).SN

Primitive cubic, R = 178.1 pm: 1 lattice point, CN 6, void 1 − π/6 = 47.6 %; the body-centre sphere has r = (√3 − 1)R = 130.4 pm and CN 8 CSE 2023 Q2(c) 10m.
Simple cubic, (100) at θ = 14.88° with 1.541 Å: a = 2.998 Å, R = 1.499 Å, cubic hole 0.732R = 1.097 Å > 1.08 Å: the atom fits without distortion CSE 2022 Q2(c) 20m.
fcc of A with B in the body-centre octahedral hole (r_B = 0.414r_A): void fraction = 1 − (π/12√2)[4 + (√2 − 1)³] = 0.246 IFoS 2021 Q2(c) 15m.

3.1.4 Radius ratio rules and ionic structures ★★★

Fig. 3.1.2 Limiting radius ratios for CN 3 and CN 6.

Draw it in 60 s: three touching circles with a tiny dot in the gap (CN 3); four circles on a square with a dot in the centre and the diagonal dashed (CN 6).

At the limit the cation touches every anion and the anions just touch each other; a smaller cation would rattle, so a lower CN is chosen.SN
CN 3 (triangle): the centre-to-corner distance 2r⁻/√3 = r⁺ + r⁻, so r⁺/r⁻ = 2/√3 − 1 = 0.155 CSE 2024 Q2(c)(i) 10m IFoS 2022 Q2(c)(i) 10m.
CN 6 (square cross-section of the octahedron): diagonal 2(r⁺ + r⁻) = 2√2 r⁻, so r⁺/r⁻ = √2 − 1 = 0.414.
CN 4 (tetrahedron): √(3/2) − 1 = 0.225; CN 8 (cube): √3 − 1 = 0.732.

r⁺/r⁻	0.155–0.225	0.225–0.414	0.414–0.732	0.732–1
CN, shape	3, triangle	4, tetrahedral	6, octahedral	8, cubic
Example	BO₃ units in B₂O₃	ZnS	NaCl	CsCl

ZnS: 0.74/1.84 = 0.402 → tetrahedral, 4 : 4. S²⁻ ccp, Zn²⁺ in half the tetrahedral holes (zinc blende, 4 ZnS per cell); √3a/4 = r⁺ + r⁻ gives a = 5.96 Å CSE 2025 Q2(c) 15m. It is an F lattice because the translations are fcc; the tetrahedral bonding is the local environment of the basis IFoS 2022 Q2(c)(ii) 5m.SN
NaCl 6 : 6 (two interpenetrating fcc, 4 NaCl per cell, a = 2 × interionic distance); CsCl 8 : 8 (simple cubic lattice, 1 per cell); CaF₂ 8 : 4 (Ca²⁺ fcc, F⁻ in all 8 tetrahedral holes); rutile TiO₂ 6 : 3; CdI₂ layers (Cd²⁺ in alternate layers of octahedral holes of hcp I⁻).
NaCl, ρ = 2.17 g cm⁻³: a = 563.5 pm, Na–Cl = 281.7 pm. RbBr: 1.47/1.95 = 0.754 → CN 8 predicted; RbI: 0.681 → CN 6. (Real RbBr is rock salt: the rule ignores polarisation and covalency, so treat it as a guide) IFoS 2023 Q3(a) 10m.SN
Kinds of crystal: SiC covalent network; S₈ molecular (van der Waals); KBr and LiCl ionic; Mg metallic Geo 2021 Q11(b) 5m.

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