Solid State
Solid State for JEE Main & Advanced
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:
| Type | Description | Examples |
|---|---|---|
| Crystalline | Long-range ordered arrangement; sharp MP | NaCl, ice, metals |
| Amorphous | Short-range order only; gradual softening | Glass, 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:
| Type | Constituent particles | Bonding | Example | Properties |
|---|---|---|---|---|
| Ionic | Ions | Coulombic | NaCl, KCl | Hard, brittle, high MP, conduct when molten/aqueous |
| Molecular | Molecules | Van der Waals/H-bond | Ice, dry ice, I₂ | Soft, low MP, poor conductors |
| Covalent (Network) | Atoms | Covalent bonds | Diamond, SiC, quartz | Very hard, very high MP, insulators (except graphite) |
| Metallic | Cations + e⁻ sea | Metallic bond | Cu, Fe, Ag | Lustrous, 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:
| System | Edge lengths | Angles | Example |
|---|---|---|---|
| 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).
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.
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.
Crystalline solid is characterized by:
NaCl is a:
Number of Bravais lattices:
Cubic system has:
Diamond is a:
Unit Cells, Packing Efficiency, DensityTopic 2
Types of Cubic Unit Cells:
| Type | Atoms per cell | Coordination # | 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 type | Relation | Derivation |
|---|---|---|
| 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:
$$\text{PE} = \frac{n \times \frac{4}{3}\pi r^3}{a^3} \times 100\%$$
| Cell | Packing 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:
| Crystal | Type | CN |
|---|---|---|
| NaCl (rock salt) | FCC anions + cations in octahedral voids | 6:6 |
| CsCl | BCC-like (Cs at center of Cl cube or vice versa) | 8:8 |
| ZnS (zinc blende) | FCC S + Zn in alternate tetrahedral voids | 4:4 |
| CaF₂ (fluorite) | FCC Ca²⁺ + F⁻ in all tetrahedral voids | 8:4 |
| Na₂O (antifluorite) | Reverse of fluorite | 4:8 |
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.
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).
Number of atoms in FCC unit cell:
Packing efficiency in BCC:
For BCC, $r$ and $a$ relation:
Coordination number in CsCl:
Number of tetrahedral voids for $N$ atoms in CCP:
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).
Differentiate Schottky and Frenkel defects.
Show solution
| Property | Schottky | Frenkel |
|---|---|---|
| Defect | Cation-anion pair missing | Cation displaced to interstitial |
| Effect on density | Decreases | No change |
| Electrical neutrality | Maintained | Maintained |
| Examples | NaCl, KCl | AgCl, ZnS |
| Conditions | Similar-sized ions, high CN | Different-sized ions, low CN |
Final Answer: See table.
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.
Schottky defect decreases:
F-centre is:
Frenkel defect is found in:
Metal deficient compound:
Density unchanged in:
Electrical, Magnetic Properties and Band TheoryTopic 2
Electrical Conductivity: Variation based on solid type.
| Type | Conductivity ($\Omega^{-1}$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):
| Type | Dopant | Behavior |
|---|---|---|
| n-type | Group 15 (P, As, Sb) — extra electron | Conduction by extra electrons |
| p-type | Group 13 (B, Al, Ga) — electron deficient → "hole" | Conduction by hole movement |
Magnetic Properties of Solids:
| Type | Behavior in magnetic field | Cause | Examples |
|---|---|---|---|
| Diamagnetic | Weakly repelled | All electrons paired | NaCl, $H_2O$, Cu, Au, Bi |
| Paramagnetic | Weakly attracted | Unpaired electrons | $O_2$, $Cu^{2+}$, $Fe^{3+}$, NO, $CuSO_4$ |
| Ferromagnetic | Strongly attracted; can be magnetized | Aligned unpaired electrons in domains | Fe, Co, Ni, $CrO_2$ |
| Antiferromagnetic | Net moment $= 0$ | Opposite alignments cancel | MnO, FeO, Mn₂O₃, $V_2O_3$ |
| Ferrimagnetic | Some net moment | Unequal 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: $PbZrO_3$.
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.
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.
Semiconductor band gap:
Si doped with P gives:
Ferromagnetism occurs in:
$Fe_3O_4$ is:
Diamagnetic substance is repelled because:
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