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State Board · Class 12 · Chemistry

Lesson 1: Solid State

7 concepts · 56 practice questions with answers · Free to read

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Types of solids: crystalline and amorphous

From daily life

A grain of salt shows neat cubic faces, a broken glass piece does not — both solids, different internal order.

Simple definition

Crystalline solids have long-range regular arrangement; amorphous solids have only short-range order and are super-cooled liquids.

TypesCrystalline and amorphous solidsCrystalline (NaCl, diamond)Amorphous (glass, rubber)PropertyCrystallineAmorphousShapedefinite geometricirregularMelting pointsharprangeAnisotropyyesno (isotropic)Naturetrue solidssuper-cooled liquidsCrystals: long-range order; amorphous: short-range order only

In the figure : Crystalline vs amorphous.

Rule 1: Crystalline — Sharp melting point, definite shape, anisotropic, true solids.

  • Easy : NaCl, diamond, ice.
  • Medium : Cleaves along flat planes.
  • Exam-type : Properties vary with direction.

Rule 2: Amorphous — Melting range, irregular, isotropic.

  • Easy : Glass, rubber, plastics.
  • Medium : Breaks with irregular surfaces.
  • Exam-type : Flow very slowly (old window glass thickens).

Rule 3: Anisotropy — In crystals physical properties (refraction, conductivity) depend on direction.

  • Easy : Graphite conducts along layers.
  • Medium : Poor conduction across layers.
  • Exam-type : Glass is isotropic.

Reasoning: why is glass not a true solid?

  • Step 1: Glass has irregular structure (short-range order).
  • Step 2: No sharp melting point; it softens gradually.
  • Step 3: Hence it is treated as a super-cooled liquid.

CET level (MCQ)

  • CET 1) Isotropic solid? 1) glass 2) NaCl 3) diamond 4) ice — Answer: 1).
  • CET 2) Sharp melting point? 1) crystalline 2) amorphous 3) both 4) neither — Answer: 1).

Real-life use: Glassworks, solar cells (amorphous Si), optical fibres.

Remember

Crystalline: long-range order, sharp m.p., anisotropic; amorphous: the reverse.

Common mistake

Treating glass as a true solid.

CET trap

Glass is a super-cooled liquid.

Shortcut

Memory clue: "crystal = disciplined, amorphous = untidy".

Spot the difference

PropertyCrystallineAmorphous
m.p.sharprange
Anisotropyyesno
e.g.NaClglass

Practice questions

Easy
  1. 1What is a crystalline solid?
  2. 2What is an amorphous solid?
  3. 3Explain anisotropy.
Medium
  1. 4Why is glass a super-cooled liquid?
  2. 5Write two differences.
  3. 6Isotropy of glass.
Challenging
  1. 7Give graphite as an example.
  2. 8What is long-range order?
Show answers (try them yourself first)
  1. 1Regular long-range arrangement.
  2. 2Short-range order only.
  3. 3Property varies with direction.
  4. 4No sharp m.p.
  5. 5Shape, m.p., anisotropy.
  6. 6Same in all directions.
  7. 7Conducts along layers.
  8. 8Regularity over millions of particles.

Classification of crystalline solids

From daily life

Salt, sugar, diamond and copper are all crystals, yet differ in conduction because their bonding differs.

Simple definition

Four types: ionic, covalent network, molecular and metallic — based on the constituent particles and the bonding.

ClassificationFour types of crystalline solidsTypeunitsforcee.g.conductsIonicionselectrostaticNaClwhen moltenCovalentatomscovalentdiamondno*Molecularmoleculesvan der WaalsicenoMetalliccations + e⁻metallic bondCuyes*Graphite is an exception: free electrons conductMelting point: molecular < ionic < covalent network

In the figure : Types of crystalline solids.

Rule 1: Ionic and covalent — Ionic: hard, brittle, high m.p., conducts when molten; covalent network: very hard, very high m.p.

  • Easy : NaCl, MgO.
  • Medium : Diamond, SiC, quartz.
  • Exam-type : Diamond does not conduct.

Rule 2: Molecular — Soft, low m.p., insulators (non-polar: dispersion; polar: dipole; H-bonded: ice).

  • Easy : I₂, dry ice.
  • Medium : Ice: H-bonds.
  • Exam-type : Sugar is molecular.

Rule 3: Metallic — Cations in a sea of electrons; conductors, ductile, malleable.

  • Easy : Cu, Fe, Ag.
  • Medium : Lustre and conduction come from electrons.
  • Exam-type : Mercury is a liquid exception.

Reasoning: why does graphite conduct?

  • Step 1: Each C is sp², making three bonds, leaving one electron.
  • Step 2: These electrons are delocalised over the layers (π network).
  • Step 3: Free electrons ⇒ electrical conduction.

CET level (MCQ)

  • CET 1) Particles in diamond? 1) atoms 2) ions 3) molecules 4) cations + e⁻ — Answer: 1).
  • CET 2) Malleable and ductile? 1) metallic 2) ionic 3) molecular 4) network — Answer: 1).

Real-life use: Abrasives (diamond, SiC), wires (Cu), salt industry.

Remember

Ionic: ions; covalent: atoms; molecular: molecules; metallic: cations + e⁻.

Common mistake

Assuming every covalent solid is an insulator.

CET trap

Graphite is an exception — free electrons.

Shortcut

Memory clue: "ion, atom, molecule, metal".

Spot the difference

TypeUnitse.g.
IonicionsNaCl
Covalentatomsdiamond
Molecularmoleculesice
Metalliccations + e⁻Cu

Practice questions

Easy
  1. 1List the four types.
  2. 2Properties of ionic solids.
  3. 3Conduction: diamond vs graphite.
Medium
  1. 4Why is m.p. of molecular solids low?
  2. 5Explain the metallic bond.
  3. 6What type is ice?
Challenging
  1. 7What type is sugar?
  2. 8Examples of network solids.
Show answers (try them yourself first)
  1. 1Ionic, covalent, molecular, metallic.
  2. 2Hard, brittle, high m.p.
  3. 3Free electrons in graphite.
  4. 4Weak intermolecular forces.
  5. 5Cations in an electron sea.
  6. 6Molecular (H-bonded).
  7. 7Molecular.
  8. 8Diamond, SiC, quartz.

Lattice, unit cells and crystal systems

From daily life

Repeat one tile again and again and you cover the floor — the unit cell is that tile.

Simple definition

A crystal lattice is the 3-D arrangement of points; a unit cell is the smallest repeating portion.

Systems7 crystal systems (14 Bravais lattices)Systemedgesanglese.g.nCubica=b=cα=β=γ=90°NaCl, Cu3Tetragonala=b≠cα=β=γ=90°SnO₂, TiO₂2Orthorhombica≠b≠cα=β=γ=90°BaSO₄, S4Monoclinica≠b≠cα=γ=90°, β≠90°Na₂SO₄·10H₂O2Triclinica≠b≠cα≠β≠γ≠90°CuSO₄·5H₂O1Hexagonala=b≠cα=β=90°, γ=120°ZnO, graphite1Rhombohedrala=b=cα=β=γ≠90°calcite, NaNO₃1n = number of Bravais lattices (3+2+4+2+1+1+1 = 14)Unit cell types: primitive (P), body-centred (I), face-centred (F), end-centred (C)

In the figure : Seven crystal systems.

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Lattice, unit cells and crystal systems

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Rule 1: Types of unit cells — P (primitive): corners; I (body-centred): + body centre; F (face-centred): + every face; C (end-centred).

  • Easy : SC = P.
  • Medium : BCC = I.
  • Exam-type : FCC = F.

Rule 2: Number of particles Z — Corner 1/8, edge 1/4, face 1/2, body 1.

  • Easy : SC: 8 × 1/8 = 1.
  • Medium : BCC: 1 + 1 = 2.
  • Exam-type : FCC: 1 + 6 × 1/2 = 4.

Rule 3: Seven systems — Cubic a = b = c, 90°; tetragonal a = b ≠ c; orthorhombic ...; hexagonal γ = 120°; in all 14 Bravais lattices.

  • Easy : Cubic: 3 lattices.
  • Medium : Orthorhombic: 4.
  • Exam-type : Triclinic: 1.

Calculation: A at corners, B at face centres — formula?

  • Step 1: A = 8 × 1/8 = 1.
  • Step 2: B = 6 × 1/2 = 3.
  • Step 3: Formula AB₃.

CET level (MCQ)

  • CET 1) Z in FCC? 1) 4 2) 2 3) 1 4) 8 — Answer: 1).
  • CET 2) Number of Bravais lattices? 1) 14 2) 7 3) 12 4) 10 — Answer: 1).

Real-life use: Crystallography, XRD analysis, gemmology.

Remember

Z = corner/8 + edge/4 + face/2 + body; 7 systems, 14 lattices.

Common mistake

Counting a corner atom as a whole.

CET trap

A corner atom is shared by 8 cells ⇒ 1/8.

Shortcut

Memory clue: "corner eighth, face half, body whole".

Spot the difference

LatticeZ
SC1
BCC2
FCC4

Practice questions

Easy
  1. 1What is a unit cell?
  2. 2Z in SC.
  3. 3Z in BCC.
Medium
  1. 4Z in FCC.
  2. 5Names of the 7 systems.
  3. 6Angles of the cubic system.
Challenging
  1. 7What are Bravais lattices?
  2. 8A at corners, B at faces ⇒ formula.
Show answers (try them yourself first)
  1. 1Smallest repeating unit.
  2. 21.
  3. 32.
  4. 44.
  5. 5Cubic ... triclinic.
  6. 6All 90°.
  7. 7The 14 lattice types.
  8. 8AB₃.

Close packing, voids and packing efficiency

From daily life

Oranges stacked in a crate sit in the hollows of the layer below to fit the most.

Simple definition

Packing efficiency = volume occupied by spheres ÷ unit-cell volume × 100: SC 52.4 %, BCC 68.0 %, FCC 74.0 %.

PackingCubic unit cells and packingSCBCCFCCZ124CN6812PE %52.468.074.0Edge a2r4r/√32√2 rVoid %47.632.026.0hcp (ABAB) and ccp (ABCABC): both CN = 12, PE = 74 %N spheres ⇒ N octahedral and 2N tetrahedral voidsr/R: tetrahedral 0.225–0.414; octahedral 0.414–0.732

In the figure : Cells and packing.

Rule 1: Close packing — hcp: ABAB; ccp (= FCC): ABCABC; both have CN = 12 and PE 74 %.

  • Easy : Mg is hcp.
  • Medium : Cu and Ag are ccp.
  • Exam-type : SC is the least efficient.

Rule 2: Voids — N spheres ⇒ N octahedral + 2N tetrahedral voids; r/R: 0.225–0.414 (tet.), 0.414–0.732 (oct.).

  • Easy : NaCl: r/R = 95/181 = 0.525 ⇒ octahedral.
  • Medium : ZnS is tetrahedral.
  • Exam-type : Tetrahedral 2N.

Rule 3: Finding PE — PE = (Z × 4/3 πr³)/a³ × 100; FCC: a = 2√2 r ⇒ π/(3√2) = 74.0 %.

  • Easy : SC: π/6 = 52.4 %.
  • Medium : BCC: π√3/8 = 68.0 %.
  • Exam-type : Void space in FCC is 26 %.

Calculation: PE of FCC

  • Step 1: Z = 4, a = 2√2 r ⇒ a³ = 16√2 r³.
  • Step 2: sphere volume = 4 × 4/3 πr³ = 16πr³/3.
  • Step 3: PE = (16π/3)/(16√2) = π/(3√2) = 0.7405 = 74.05 %.

CET level (MCQ)

  • CET 1) CN of FCC? 1) 12 2) 8 3) 6 4) 4 — Answer: 1).
  • CET 2) Tetrahedral voids for N spheres? 1) 2N 2) N 3) N/2 4) 4N — Answer: 1).

Real-life use: Ionic structures, alloys, catalysts.

Remember

PE: 52.4, 68.0, 74.0 %; CN: 6, 8, 12; voids 2N and N.

Common mistake

Stating CN of BCC as 12.

CET trap

BCC has CN 8; FCC/hcp 12.

Shortcut

Memory clue: "52, 68, 74 — six, eight, twelve".

Spot the difference

LatticeCNPE %
SC652.4
BCC868.0
FCC1274.0

Practice questions

Easy
  1. 1Define PE.
  2. 2hcp vs ccp.
  3. 3Number of voids.
Medium
  1. 4Derive PE of FCC.
  2. 5Derive PE of SC.
  3. 6CN and PE of BCC.
Challenging
  1. 7Which void holds Na⁺ in NaCl?
  2. 8Void % in FCC.
Show answers (try them yourself first)
  1. 1Sphere volume ÷ cell volume.
  2. 2ABAB vs ABCABC.
  3. 3N and 2N.
  4. 474.05 %.
  5. 552.4 %.
  6. 68 and 68 %.
  7. 7Octahedral.
  8. 826 %.

Density and unit cell calculations

From daily life

Density is mass over volume: here, the mass of atoms in one cell over the cell volume.

Simple definition

ρ = Z M / (a³ N_A); a in cm (1 pm = 10⁻¹⁰ cm), M in g mol⁻¹, N_A = 6.022 × 10²³.

DensityDensity ρ = Z M / (a³ NA)Zatoms per cellMg mol⁻¹a³cm³ (1 pm = 10⁻¹⁰ cm)NA = 6.022 × 10²³ mol⁻¹Ex. 1: Cu (FCC), a = 361 pm, M = 63.5ρ = 4 × 63.5 / [(3.61×10⁻⁸)³ × 6.022×10²³] = 8.97 g cm⁻³Ex. 2: NaCl, Z = 4, M = 58.5, a = 564 pmρ = 4 × 58.5 / [(5.64×10⁻⁸)³ × 6.022×10²³] = 2.17 g cm⁻³Reverse: Fe (BCC), ρ = 7.87 ⇒ a = 287 pm, r = 124 pmConvert pm to cm (× 10⁻¹⁰) and cube it

In the figure : Density calculations.

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Density and unit cell calculations

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Rule 1: Finding density — Cu (FCC), a = 361 pm, M = 63.5 ⇒ ρ = 8.97 g cm⁻³.

  • Easy : NaCl: Z = 4, a = 564 pm ⇒ 2.17.
  • Medium : Al (FCC) a = 405 pm ⇒ 2.70.
  • Exam-type : Z comes from the lattice.

Rule 2: Finding a or r — a³ = ZM/(ρ N_A); Fe (BCC), ρ = 7.87 ⇒ a = 287 pm; r = √3 a/4 = 124 pm.

  • Easy : FCC: r = a/(2√2).
  • Medium : SC: r = a/2.
  • Exam-type : BCC: r = √3 a/4.

Rule 3: Finding N_A — N_A = ZM/(ρ a³) from experimental ρ and a.

  • Easy : Cu: ZM = 254.
  • Medium : Unit: mol⁻¹.
  • Exam-type : Used to measure N_A.

Calculation: Cu (FCC), a = 361 pm

  • Step 1: a = 3.61×10⁻⁸ cm ⇒ a³ = 4.70×10⁻²³ cm³.
  • Step 2: Z M = 4 × 63.5 = 254.
  • Step 3: ρ = 254 / (4.70×10⁻²³ × 6.022×10²³) = 8.97 g cm⁻³.

CET level (MCQ)

  • CET 1) Z for NaCl? 1) 4 2) 2 3) 8 4) 1 — Answer: 1).
  • CET 2) r in BCC? 1) √3a/4 2) a/2 3) a/(2√2) 4) √2a/4 — Answer: 1).

Real-life use: Identifying metals, alloys, mineral analysis.

Remember

ρ = ZM/(a³N_A); r = a/2, √3a/4, a/(2√2).

Common mistake

Not converting pm to cm.

CET trap

1 pm = 10⁻¹⁰ cm; cube it for a³.

Shortcut

Memory clue: "Z M over a-cubed N_A".

Spot the difference

MetalLatticeρ (g cm⁻³)
CuFCC8.97
AlFCC2.70
NaClZ = 42.17

Practice questions

Easy
  1. 1Formula of ρ.
  2. 2Density of Cu.
  3. 3Density of Al (a = 405 pm).
Medium
  1. 4a and r of Fe (BCC).
  2. 5How is N_A measured?
  3. 6pm ⇒ cm.
Challenging
  1. 7r and a in FCC.
  2. 8Z in NaCl.
Show answers (try them yourself first)
  1. 1ZM/(a³N_A).
  2. 28.97.
  3. 32.70.
  4. 4287 pm, 124 pm.
  5. 5ZM/(ρa³).
  6. 6× 10⁻¹⁰.
  7. 7a/(2√2).
  8. 84.

Defects in crystals

From daily life

A sack of rice is never perfect; crystals too have missing or misplaced particles.

Simple definition

Defects are deviations from the ideal arrangement. Schottky: equal cation and anion vacancies; Frenkel: cation in an interstitial site; F-centre: colour.

DefectsCrystal defects: Schottky and Frenkel+–+––□–++□+––+–++–+–Schottky+–+––+–++□+––+–++–+–Frenkelboth ions missing ⇒ density ↓cation moves to interstitial ⇒ density sameSchottky: NaCl, KCl, CsCl; Frenkel: AgCl, ZnS (large size difference)F-centre: anion vacancy + trapped e⁻ ⇒ colour; impurity defect: SrCl₂ in NaCl

In the figure : Schottky and Frenkel.

Rule 1: Schottky defect — Equal numbers of cations and anions missing ⇒ density ↓; NaCl, KCl, CsCl.

  • Easy : Ions of similar size.
  • Medium : Large coordination number.
  • Exam-type : Electrical neutrality is kept.

Rule 2: Frenkel defect — A small cation moves to an interstitial site ⇒ density unchanged; AgCl, ZnS.

  • Easy : Large size difference.
  • Medium : Low coordination number.
  • Exam-type : AgBr shows both.

Rule 3: Other defects — F-centre (anion vacancy + e⁻ gives colour; yellow NaCl); impurity defects (SrCl₂ in NaCl); metal excess/deficiency.

  • Easy : KCl turns violet.
  • Medium : Sr²⁺ replaces two Na⁺, leaving a vacancy.
  • Exam-type : ZnO turns yellow on heating.

Reasoning: why density falls in Schottky but not in Frenkel

  • Step 1: Schottky: ions leave the crystal ⇒ mass ↓, volume same.
  • Step 2: Frenkel: the ion stays inside ⇒ mass same.
  • Step 3: ρ = m/V ⇒ ρ falls in Schottky, stays in Frenkel.

CET level (MCQ)

  • CET 1) Defect lowering density? 1) Schottky 2) Frenkel 3) neither 4) not F-centre — Answer: 1).
  • CET 2) Shows Frenkel defect? 1) AgCl 2) NaCl 3) KCl 4) CsCl — Answer: 1).

Real-life use: Coloured gems, semiconductor doping, ionic conduction.

Remember

Schottky: ρ ↓, large CN; Frenkel: ρ same, size difference; F-centre: colour.

Common mistake

Saying Frenkel lowers density.

CET trap

In Frenkel no ion leaves the crystal.

Shortcut

Memory clue: "Schottky = gone, Frenkel = moved".

Spot the difference

DefectDensitye.g.
SchottkydecreasesNaCl
FrenkelsameAgCl
F-centre—yellow NaCl

Practice questions

Easy
  1. 1Explain Schottky defect.
  2. 2Explain Frenkel defect.
  3. 3Differences between them.
Medium
  1. 4What is an F-centre?
  2. 5Which defects occur in AgBr?
  3. 6Impurity defect with SrCl₂.
Challenging
  1. 7Condition for Schottky.
  2. 8Condition for Frenkel.
Show answers (try them yourself first)
  1. 1Ion vacancies, ρ ↓.
  2. 2Interstitial ion.
  3. 3ρ falls vs same.
  4. 4Electron in an anion vacancy.
  5. 5Both.
  6. 6Creates Na⁺ vacancies.
  7. 7Similar sizes, large CN.
  8. 8Large size difference, small CN.

Electrical and magnetic properties of solids

From daily life

Copper conducts, glass does not, silicon conducts a little — the difference is the band gap.

Simple definition

Band theory: the gap between valence and conduction bands is zero in conductors, small in semiconductors (Si ≈ 1.1 eV) and large in insulators.

BandsBand theory and magnetismconductionvalenceConductorconductionvalenceSemiconductorconductionvalenceInsulatorgap: 0 = conductor; Si ≈ 1.1 eV; insulator > 3 eVn-type: Si + P (free e⁻); p-type: Si + B (holes)Diamagneticall e⁻ pairedNaCl, H₂OParamagneticunpaired e⁻O₂, Cu²⁺Ferromagneticaligned, permanentFe, Co, NiAntiferro.opposed, net 0MnOFerrimagneticunequal opposedFe₃O₄

In the figure : Bands and magnetism.

Rule 1: Semiconductors — Intrinsic: pure Si; n-type: doped with P (free electrons); p-type: doped with B (holes).

  • Easy : T↑ ⇒ conductivity ↑.
  • Medium : p–n junction: diode.
  • Exam-type : Ge, Si, GaAs.

Rule 2: Magnetic properties — Diamagnetic (all paired), paramagnetic (unpaired), ferromagnetic (aligned).

  • Easy : O₂ and Cu²⁺ are paramagnetic.
  • Medium : Fe, Co, Ni are ferromagnetic.
  • Exam-type : NaCl is diamagnetic.

Rule 3: Antiferro and ferri — Antiferromagnetic: equal opposed ⇒ net 0 (MnO); ferrimagnetic: unequal opposed ⇒ some magnetism (Fe₃O₄).

  • Easy : Heating turns ferri into para.
  • Medium : Magnetic tape: Fe₃O₄/CrO₂.
  • Exam-type : Ferrites.

Reasoning: why P makes Si n-type

  • Step 1: Si has 4 valence electrons; P has 5.
  • Step 2: Four of P's electrons bond, one is left free.
  • Step 3: Free electrons carry negative charge ⇒ n-type.

CET level (MCQ)

  • CET 1) Dopant for p-type Si? 1) B 2) P 3) As 4) Sb — Answer: 1).
  • CET 2) Ferromagnetic? 1) Fe 2) NaCl 3) H₂O 4) Zn — Answer: 1).

Real-life use: Transistors, solar cells, magnetic storage, MRI.

Remember

Gap: conductor 0, semiconductor ≈ 1 eV, insulator > 3 eV; n = electrons, p = holes.

Common mistake

Thinking positive ions carry current in p-type.

CET trap

A hole is a missing electron, not an ion.

Shortcut

Memory clue: "P has five ⇒ n; B makes holes ⇒ p".

Spot the difference

TypeMagnetism
Diamagneticpaired
Paramagneticunpaired
Ferromagneticaligned

Practice questions

Easy
  1. 1State band theory.
  2. 2Conductor, semiconductor, insulator.
  3. 3n-type and p-type.
Medium
  1. 4What is paramagnetic?
  2. 5Ferromagnetic examples.
  3. 6Antiferromagnetic MnO.
Challenging
  1. 7Ferrimagnetic example.
  2. 8Effect of temperature.
Show answers (try them yourself first)
  1. 1Valence–conduction gap.
  2. 2Gap 0 / small / large.
  3. 3P ⇒ n; B ⇒ p.
  4. 4Unpaired electrons.
  5. 5Fe, Co, Ni.
  6. 6Net zero.
  7. 7Fe₃O₄.
  8. 8Semiconductor ↑, metal ↓.

Quick revision of the whole lesson

ConceptKey formula / fact
Particles Zcorner/8 + face/2 + body
PE of FCCπ/(3√2) = 74 %
PE of BCCπ√3/8 = 68 %
PE of SCπ/6 = 52.4 %
a and r (FCC)a = 2√2 r
a and r (BCC)a = 4r/√3
Densityρ = ZM/(a³N_A)
VoidsN oct + 2N tet
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Crystalline, amorphous.
  • Four types.
  • Unit cells.
  • Packing, PE.
  • Density problems.
  • Defects.
  • Bands, magnetism.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.