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Samacheer Kalvi 12th Chemistry Chapter 6 Exercise Solutions

Tamil Nadu Class 12 Chemistry Chapter 6 Solid State exercise solutions with unit cells, packing efficiency, density formula, crystal defects, Bragg equation, and numerical answer patterns.

Updated 17 June 20267 min read

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Chapter 6 — Solid State Exercise Solutions

Solid State numericals require careful unit conversion. Always convert edge length from pm or Å to cm before substituting in the density formula.

Key Concepts Quick Recap

TermKey Fact
Crystalline solidLong-range order; sharp melting point; anisotropic
Amorphous solidShort-range order; no sharp melting point; isotropic
Unit cellSmallest repeating unit of crystal lattice
SC (Z=1)1 atom per unit cell; packing efficiency 52.4%
BCC (Z=2)2 atoms per unit cell; packing efficiency 68%
FCC (Z=4)4 atoms per unit cell; packing efficiency 74%
Bragg equationnλ = 2d sinθ — used to find inter-planar spacing d
Schottky defectIons missing from lattice → density decreases
Frenkel defectIon displaced to interstitial position → density unchanged
F-centreElectron trapped in anion vacancy → crystal becomes coloured
Packing efficiencyFraction of unit cell volume occupied by atoms

Formula Recap

ConceptFormula
Density of unit cellρ = ZM / (a³Nₐ)
Bragg equationnλ = 2d sinθ
Packing efficiency(Volume of atoms / Volume of unit cell) × 100
Relation (FCC)4r = √2 a → r = a√2/4
Relation (BCC)4r = √3 a → r = a√3/4
Relation (SC)2r = a → r = a/2

I. Choose the Best Answer (MCQs)

1. An element crystallises as FCC. The number of atoms per unit cell is: (a) 1 (b) 2 (c) 4 (d) 6

Answer(c) 4 FCC: 8 corner atoms (×1/8) + 6 face atoms (×1/2) = 1 + 3 = 4 atoms per unit cell.

2. Which crystal defect decreases the density of a crystal? (a) Frenkel defect (b) Interstitial defect (c) Schottky defect (d) Metal excess defect

Answer(c) Schottky defect In Schottky defect, pairs of cation and anion vacancies are created. Mass decreases while volume stays same → density decreases.

3. The Bragg equation is nλ = 2d sinθ. What does 'd' represent? (a) Diameter of atom (b) Inter-planar spacing (c) Edge length (d) Wavelength

Answer(b) Inter-planar spacing 'd' is the distance between successive planes of atoms in the crystal lattice.

4. The packing efficiency of BCC is approximately: (a) 52% (b) 68% (c) 74% (d) 90%

Answer(b) 68% BCC packing efficiency = (2 × 4/3 πr³) / a³ × 100 = 68.04% ≈ 68%.

5. An ionic crystal shows colour when exposed to excess of one ion. This is due to: (a) Schottky defect (b) Frenkel defect (c) F-centre (metal excess defect) (d) Interstitial defect

Answer(c) F-centre NaCl becomes yellow in excess Na vapour — Na deposits and electrons occupy Cl⁻ vacancies (F-centres). These trapped electrons absorb visible light → colour.

6. In a BCC unit cell, the atoms touch along: (a) Face diagonal (b) Edge (c) Body diagonal (d) Corners only

Answer(c) Body diagonal In BCC, the body centre atom touches the 8 corner atoms along the body diagonal → 4r = √3 a.

7. CsCl has a body-centred cubic structure. The coordination number of Cs⁺ is: (a) 4 (b) 6 (c) 8 (d) 12

Answer(c) 8 In CsCl structure, each Cs⁺ is surrounded by 8 Cl⁻ at the corners of a cube → CN = 8. Note: this is NOT true BCC — CsCl has one formula unit per unit cell (Z = 1).

II. Short Answer Questions

Q1. Distinguish between crystalline and amorphous solids with two examples each.
Answer
PropertyCrystallineAmorphous
OrderLong-range periodic orderShort-range order only
Melting pointSharp, definiteRange of temperatures
AnisotropyAnisotropicIsotropic
CleavageClean, regularIrregular fracture
ExamplesNaCl, diamond, quartzGlass, rubber, plastics
Q2. Calculate the number of atoms per unit cell in SC, BCC, and FCC.
Simple Cubic (SC): Corner atoms: 8 × (1/8) = 1 Total Z = 1 Body-Centred Cubic (BCC): Corner: 8 × (1/8) = 1 Body centre: 1 × 1 = 1 Total Z = 2 Face-Centred Cubic (FCC): Corner: 8 × (1/8) = 1 Face centres: 6 × (1/2) = 3 Total Z = 4
Q3. What is the difference between Schottky and Frenkel defects?
Answer
FeatureSchottky DefectFrenkel Defect
NatureBoth cation and anion vacanciesIon displaced to interstitial site
Effect on densityDensity decreasesDensity unchanged
Crystal typeIonic with similar sized ions (NaCl, KCl)Ionic with large size difference (AgBr, ZnS)
Electrical conductivitySlightly increasedSlightly increased
Q4. What is packing efficiency? Compare SC, BCC, and FCC.
Answer

Packing efficiency = (volume occupied by atoms / total volume of unit cell) × 100

CrystalZAtom volumePacking efficiency
SC11 × 4/3πr³52.4%
BCC22 × 4/3πr³68%
FCC44 × 4/3πr³74%

FCC (and HCP) give the closest packing — used by many metals (Cu, Al, Ag, Au).

Worked Examples

Example 1 — Density of a unit cell

An element (molar mass 60 g/mol) crystallises in FCC structure with edge length 400 pm. Find its density.

Given: Z = 4 (FCC), M = 60 g/mol, a = 400 pm = 4×10⁻⁸ cm Nₐ = 6.022×10²³ mol⁻¹ ρ = ZM / (a³ × Nₐ) = (4 × 60) / ((4×10⁻⁸)³ × 6.022×10²³) = 240 / (6.4×10⁻²³ × 6.022×10²³) = 240 / 38.54 ≈ 6.23 g cm⁻³

Example 2 — Edge length from density

An FCC metal (M = 108 g/mol, ρ = 10.5 g/cm³). Find edge length.

ρ = ZM / (a³ × Nₐ) → a³ = ZM / (ρ × Nₐ) a³ = (4 × 108) / (10.5 × 6.022×10²³) = 432 / (6.323×10²⁴) = 6.83×10⁻²³ cm³ a = (6.83×10⁻²³)^(1/3) = 4.08×10⁻⁸ cm = 408 pm (This matches silver: Z=47, M=108 g/mol)

Example 3 — Packing efficiency of FCC

In FCC: atoms touch along face diagonal 4r = √2 a → r = a√2/4 Volume of 4 atoms = 4 × (4/3)πr³ = 4 × (4/3)π(a√2/4)³ = 4 × (4/3)π × (a³ × 2√2/64) = π a³ √2 / 6 Packing efficiency = (π√2/6) / 1 × 100 = 74.05% ≈ 74%

Example 4 — Bragg equation

X-rays of wavelength 154 pm hit a crystal. Reflection at θ = 45°. Find d (n=1).

nλ = 2d sinθ 1 × 154 = 2 × d × sin 45° 154 = 2 × d × 0.7071 d = 154 / 1.4142 = 108.9 pm ≈ 109 pm

Example 5 — Ionic crystal radius ratio

Radius ratio (r⁺/r⁻) determines coordination number and structure: r⁺/r⁻ CN Structure < 0.155 2 Linear 0.155–0.225 3 Triangular planar 0.225–0.414 4 Tetrahedral (ZnS type) 0.414–0.732 6 Octahedral (NaCl type) 0.732–1.000 8 Cubic (CsCl type)

Common Mistakes

  • Using wrong Z value (SC=1, BCC=2, FCC=4).
  • Not converting edge length from pm to cm (1 pm = 10⁻¹⁰ m = 10⁻⁸ cm).
  • Forgetting to cube the edge length in the density formula.
  • Confusing Schottky (density decreases) with Frenkel (density unchanged).
  • Using radius in density formula instead of edge length.

Exam Checklist

  • Crystal type and Z are identified first.
  • Edge length is converted to cm before use in formula.
  • All steps shown: formula → substitution → calculation → unit.
  • Defect type is matched to its effect on density.
  • Final answer has correct unit (g cm⁻³ for density, pm for length).

Frequently Asked Questions

Density formula ρ = ZM / (a³Nₐ) is most important. Also know packing efficiency: SC 52%, BCC 68%, FCC/HCP 74%.