JEE Main & Advanced

Solid State

Solid State for JEE Main & Advanced

1
Module 1

Crystal Lattice and Unit Cells

Classification of Solids and Crystal SystemsTopic 1

Solids: Substances with definite shape, volume, rigid. Particles closely packed; only vibrational motion.

Classification of Solids:

TypeDescriptionExamples
CrystallineLong-range ordered arrangement; sharp MPNaCl, ice, metals
AmorphousShort-range order only; gradual softeningGlass, rubber, plastics

Properties of crystalline solids:

  • Definite geometric shape
  • Sharp melting point
  • Anisotropic (properties depend on direction)
  • Show cleavage along definite planes

Amorphous solids: Isotropic (same in all directions); also called pseudo-solids or supercooled liquids.

Classification of Crystalline Solids by Bonding:

TypeConstituent particlesBondingExampleProperties
IonicIonsCoulombicNaCl, KClHard, brittle, high MP, conduct when molten/aqueous
MolecularMoleculesVan der Waals/H-bondIce, dry ice, I₂Soft, low MP, poor conductors
Covalent (Network)AtomsCovalent bondsDiamond, SiC, quartzVery hard, very high MP, insulators (except graphite)
MetallicCations + e⁻ seaMetallic bondCu, Fe, AgLustrous, malleable, conduct electricity

Molecular Solids subtype:

  • Non-polar molecular: $H_2, CO_2, I_2$ — only London forces, low MP
  • Polar molecular: $HCl, SO_2$ — dipole-dipole
  • Hydrogen-bonded: $H_2O$ (ice), $NH_3$ — H-bonding, harder

Crystal Lattice: Regular 3D arrangement of points in space. Unit Cell: Smallest repeating unit; when repeated in 3D gives the entire lattice.

Parameters of Unit Cell:

  • Edge lengths: $a, b, c$
  • Angles: $\alpha$ (between b, c), $\beta$ (between a, c), $\gamma$ (between a, b)

Seven Crystal Systems:

SystemEdge lengthsAnglesExample
Cubic$a = b = c$$\alpha = \beta = \gamma = 90°$NaCl, Cu
Tetragonal$a = b \neq c$All $90°$$SnO_2$, white tin
Orthorhombic$a \neq b \neq c$All $90°$$BaSO_4$, rhombic sulfur
Hexagonal$a = b \neq c$$\alpha = \beta = 90°, \gamma = 120°$Graphite, $ZnO$
Trigonal (Rhombohedral)$a = b = c$$\alpha = \beta = \gamma \neq 90°$$CaCO_3$ (calcite)
Monoclinic$a \neq b \neq c$$\alpha = \gamma = 90°, \beta \neq 90°$Monoclinic sulfur
Triclinic$a \neq b \neq c$All angles $\neq 90°$$CuSO_4 \cdot 5H_2O$

Bravais Lattices: $14$ total (combinations of $7$ systems with positions of points: primitive, body-centred, face-centred, end-centred).

Worked Examples
1

Classify based on bonding: Diamond, NaCl, Ice, Cu.

Show solution
  • Diamond: covalent network
  • NaCl: ionic
  • Ice: molecular (H-bonded)
  • Cu: metallic

Final Answer: Covalent, ionic, molecular, metallic respectively.

2

Why is glass considered an amorphous solid even though it appears rigid?

Show solution

Glass has only short-range order — its molecules are arranged randomly over long distances. It doesn't have a sharp MP (softens over a range), and is isotropic. Hence amorphous (supercooled liquid).

Final Answer: Lacks long-range order; isotropic; softens gradually.

✎ Self-Check — 5 questions0 / 5
Q1.

Crystalline solid is characterized by:

Q2.

NaCl is a:

Q3.

Number of Bravais lattices:

Q4.

Cubic system has:

Q5.

Diamond is a:

Unit Cells, Packing Efficiency, DensityTopic 2

Types of Cubic Unit Cells:

TypeAtoms per cellCoordination #Atoms at
Simple Cubic (SC)$1$$6$$8$ corners
Body-Centred Cubic (BCC)$2$$8$$8$ corners + $1$ centre
Face-Centred Cubic (FCC) / CCP$4$$12$$8$ corners + $6$ face centres
Hexagonal Closest Packing (HCP)$6$$12$Layered

Contribution to Unit Cell:

  • Corner atom: $1/8$
  • Face-centred: $1/2$
  • Edge-centred: $1/4$
  • Body-centred: $1$ (full)

For SC: $8 \times (1/8) = 1$ atom For BCC: $8 \times (1/8) + 1 = 2$ atoms For FCC: $8 \times (1/8) + 6 \times (1/2) = 4$ atoms

Relation between Edge Length ($a$) and Atomic Radius ($r$):

Cell typeRelationDerivation
SC$a = 2r$Atoms touch along edge
BCC$\sqrt{3}a = 4r \Rightarrow r = a\sqrt{3}/4$Atoms touch along body diagonal
FCC$\sqrt{2}a = 4r \Rightarrow r = a\sqrt{2}/4$Atoms touch along face diagonal

Packing Efficiency:

$$\text{PE} = \frac{n \times \frac{4}{3}\pi r^3}{a^3} \times 100\%$$

CellPacking Efficiency
Simple Cubic$52.4\%$
BCC$68\%$
FCC (CCP)$74\%$
HCP$74\%$

FCC and HCP are closest packings (maximum, $74\%$).

Closest Packing of Spheres:

  • 2D layer: each sphere in contact with 6 others (hexagonal layer)
  • 3D layered: layers stacked in pattern
  • HCP: ABAB... (hexagonal)
  • CCP/FCC: ABCABC... (cubic)

Tetrahedral and Octahedral Voids:

In closest packing:

  • Tetrahedral void: Surrounded by $4$ spheres
  • Octahedral void: Surrounded by $6$ spheres

Number of voids: If $N$ = number of atoms in closest packing:

  • Tetrahedral voids = $2N$
  • Octahedral voids = $N$

Filling of Voids by Cations:

  • Radius ratio rules:
  • $r_+/r_- < 0.225$: linear (CN 2)
  • $0.225 - 0.414$: tetrahedral (CN 4)
  • $0.414 - 0.732$: octahedral (CN 6)
  • $0.732 - 1.0$: BCC (CN 8)

Density of Unit Cell: $$\rho = \frac{Z \times M}{N_A \times a^3}$$

  • $Z$: atoms per unit cell
  • $M$: molar mass
  • $N_A$: Avogadro
  • $a$: edge length

Important Ionic Crystal Structures:

CrystalTypeCN
NaCl (rock salt)FCC anions + cations in octahedral voids6:6
CsClBCC-like (Cs at center of Cl cube or vice versa)8:8
ZnS (zinc blende)FCC S + Zn in alternate tetrahedral voids4:4
CaF₂ (fluorite)FCC Ca²⁺ + F⁻ in all tetrahedral voids8:4
Na₂O (antifluorite)Reverse of fluorite4:8
Worked Examples
1

Cu has FCC; edge length $a = 361$ pm. Find radius of Cu atom.

Show solution

For FCC: $r = a\sqrt{2}/4 = 361 \times 1.414/4 = 510.5/4 = 127.6$ pm.

Final Answer: $r \approx 127.6$ pm.

2

Density of Cu (FCC, $M = 63.5$, $a = 361$ pm, $N_A = 6.022 \times 10^{23}$):

Show solution

$Z = 4$ for FCC. $a^3 = (361 \times 10^{-10})^3$ cm³ $= 4.7 \times 10^{-23}$ cm³. $$\rho = \frac{4 \times 63.5}{6.022 \times 10^{23} \times 4.7 \times 10^{-23}} = \frac{254}{28.3} \approx 8.97 \text{ g/cm}^3$$

Final Answer: $\rho \approx 8.97$ g/cm³ (matches Cu).

✎ Self-Check — 5 questions0 / 5
Q1.

Number of atoms in FCC unit cell:

Q2.

Packing efficiency in BCC:

Q3.

For BCC, $r$ and $a$ relation:

Q4.

Coordination number in CsCl:

Q5.

Number of tetrahedral voids for $N$ atoms in CCP:

2
Module 2

Defects and Properties

Point Defects, Stoichiometric and Non-StoichiometricTopic 1

Defects are deviations from perfect crystal arrangement. Real crystals always have defects.

Types of Defects:

  • Point defects: Around a single point
  • Line defects: Along a line/edge
  • Plane defects: Across a plane

Point Defects:

A. Stoichiometric Defects (do not change the ratio of cations to anions):

1. Schottky Defect:

  • Equal pairs of cations and anions missing from lattice
  • Maintains electrical neutrality
  • Decreases density
  • Found in: NaCl, KCl, CsCl, AgBr — ionic compounds with similar-sized ions and high CN

2. Frenkel Defect:

  • Cation displaced from regular site to interstitial site
  • Density remains unchanged
  • Found in: AgCl, AgBr, AgI, ZnS — ionic compounds with low CN and high size difference between cation and anion (small cation, large anion)

B. Non-Stoichiometric Defects (ratio of ions differs from formula):

1. Metal Excess Defect:

  • (a) Due to anion vacancy: missing anion; electron occupies the vacant site → F-centre (Farben = color)
  • NaCl heated with Na vapor → yellow color due to F-centres (e⁻ traps absorb light)
  • Similarly, KCl → violet, LiCl → pink
  • (b) Due to extra cation in interstitial position: e.g., $ZnO$ heated → loses O, gives $Zn^{2+}$ in interstitial, with electron(s); becomes yellow when hot, white when cool. ZnO is a non-stoichiometric solid.

2. Metal Deficiency Defect:

  • Less cations than required by stoichiometry
  • Some cations exhibit higher oxidation state to compensate
  • Example: $Fe_{0.95}O$, $Cu_{1.97}S$. The $Fe^{2+}$ deficiency is compensated by $Fe^{3+}$ ions.

Effect of Defects on Properties:

  • Schottky: lowers density, affects ionic conductivity
  • Frenkel: increases ionic conductivity
  • F-centres: cause color (excited electrons absorb visible light)
  • Metal deficiency: causes semiconducting behavior

Impurity Defects: Adding aliovalent ions creates vacancies (e.g., $SrCl_2$ in NaCl creates one cation vacancy per $Sr^{2+}$ added).

Worked Examples
1

Differentiate Schottky and Frenkel defects.

Show solution
PropertySchottkyFrenkel
DefectCation-anion pair missingCation displaced to interstitial
Effect on densityDecreasesNo change
Electrical neutralityMaintainedMaintained
ExamplesNaCl, KClAgCl, ZnS
ConditionsSimilar-sized ions, high CNDifferent-sized ions, low CN

Final Answer: See table.

2

Why does heating NaCl in Na vapor give a yellow color?

Show solution

Excess Na atoms diffuse into NaCl lattice. Cl⁻ vacancies form; electrons from Na occupy these vacancies (anion sites with electrons) — these are F-centres. F-centres absorb light in visible region, giving yellow color to NaCl crystal. This is metal excess defect due to anion vacancy.

Final Answer: F-centres absorb visible light → color.

✎ Self-Check — 5 questions0 / 5
Q1.

Schottky defect decreases:

Q2.

F-centre is:

Q3.

Frenkel defect is found in:

Q4.

Metal deficient compound:

Q5.

Density unchanged in:

Electrical, Magnetic Properties and Band TheoryTopic 2

Electrical Conductivity: Variation based on solid type.

TypeConductivity ($\Omega^{-1}$m⁻¹)
Insulators$10^{-20}$ to $10^{-10}$
Semiconductors$10^{-6}$ to $10^{4}$
Conductors (metals)$10^{6}$ to $10^{8}$

Band Theory of Solids:

  • Atomic orbitals combine to form molecular orbitals, then bands of energy levels
  • Valence band: Lower energy filled band
  • Conduction band: Higher energy band, may be empty or partially filled
  • Energy gap: Between valence and conduction bands

Classification:

  • Conductors: Bands overlap; electrons free to move; very high conductivity
  • Insulators: Large band gap (> $5$ eV); electrons cannot reach conduction band
  • Semiconductors: Small band gap (< $3$ eV); electrons thermally excited to conduction band

Intrinsic Semiconductors: Pure Si, Ge — low conductivity at low T, increases with T.

Extrinsic Semiconductors (Doped):

TypeDopantBehavior
n-typeGroup 15 (P, As, Sb) — extra electronConduction by extra electrons
p-typeGroup 13 (B, Al, Ga) — electron deficient → "hole"Conduction by hole movement

Magnetic Properties of Solids:

TypeBehavior in magnetic fieldCauseExamples
DiamagneticWeakly repelledAll electrons pairedNaCl, $H_2O$, Cu, Au, Bi
ParamagneticWeakly attractedUnpaired electrons$O_2$, $Cu^{2+}$, $Fe^{3+}$, NO, $CuSO_4$
FerromagneticStrongly attracted; can be magnetizedAligned unpaired electrons in domainsFe, Co, Ni, $CrO_2$
AntiferromagneticNet moment $= 0$Opposite alignments cancelMnO, FeO, Mn₂O₃, $V_2O_3$
FerrimagneticSome net momentUnequal opposing alignments$Fe_3O_4$ (magnetite), ferrites

Curie temperature: Above this $T$, ferromagnetic materials become paramagnetic (thermal energy disrupts domain alignment).

  • Fe: $1043$ K
  • Co: $1394$ K
  • Ni: $631$ K

Dielectric Properties: Insulators that align dipoles in electric field. Used in capacitors.

Piezoelectricity: Some crystals (e.g., quartz, Rochelle salt) generate electric current when pressed; conversely vibrate when current applied.

Pyroelectricity: Generate small electric current on heating.

Ferroelectricity: Have permanent dipoles even without external field; align in same direction. Example: $BaTiO_3$, Rochelle salt. (Above critical T, become paraelectric.)

Antiferroelectricity: Dipoles align in alternating opposite directions; net dipole zero. Example: $PbZrO_3$.

Worked Examples
1

Classify: $H_2O, O_2, Cu, Fe_3O_4, MnO$ based on magnetic behavior.

Show solution
  • $H_2O$: diamagnetic (all paired)
  • $O_2$: paramagnetic (2 unpaired in MOT)
  • Cu: diamagnetic ($3d^{10}4s^1$ — but bulk Cu metal is diamagnetic)
  • $Fe_3O_4$: ferrimagnetic (mixed valence; net moment)
  • MnO: antiferromagnetic

Final Answer: As above.

2

Why does Si doped with B make a p-type semiconductor?

Show solution

Si is Group 14 (4 valence e⁻). B is Group 13 (3 valence e⁻). When B replaces Si in lattice, it has one fewer electron — creates a "hole" (positive vacancy). Conduction occurs via hole movement: adjacent electrons fill holes, propagating the hole. Hence p-type.

Final Answer: Boron's electron deficiency creates positive holes — p-type.

✎ Self-Check — 5 questions0 / 5
Q1.

Semiconductor band gap:

Q2.

Si doped with P gives:

Q3.

Ferromagnetism occurs in:

Q4.

$Fe_3O_4$ is:

Q5.

Diamagnetic substance is repelled because:

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