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The Solid State
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Class 12ChemistryChapter 1NCERTCBSE
The Solid State - Class 12 Chemistry Chapter 1
📌 Quick Overview: Chapter 1 deals with the solid state of matter. You will learn why solids have definite shape and volume, the classification of solids (crystalline and amorphous), types of crystal lattices and unit cells, packing efficiency, types of voids, density calculations, and defects in solids. This chapter is highly conceptual and numericals on density and packing efficiency are frequently asked in CBSE board exams.
Topics Covered:
- > General Characteristics of Solids
- > Crystalline vs Amorphous Solids
- > Types of Crystalline Solids
- > Crystal Lattice and Unit Cell
- > Types of Unit Cells
- > Packing in Solids (1D, 2D, 3D)
- > Packing Efficiency
- > Voids (Tetrahedral and Octahedral)
- > Density of Unit Cell
- > Imperfections/Defects in Solids
- > Electrical Properties
- > Magnetic Properties
1. General Characteristics of Solid State
Solids have the following properties:
> Definite shape and volume
> Strong intermolecular forces
> Constituent particles (atoms, molecules, ions) are closely packed
> Particles vibrate about fixed positions but do not move freely
> Incompressible and rigid
> High density compared to liquids and gases
> Definite shape and volume
> Strong intermolecular forces
> Constituent particles (atoms, molecules, ions) are closely packed
> Particles vibrate about fixed positions but do not move freely
> Incompressible and rigid
> High density compared to liquids and gases
2. Crystalline vs Amorphous Solids ⭐
| Property | Crystalline Solids | Amorphous Solids |
|---|---|---|
| Arrangement of particles | Regular, long-range order | Irregular, short-range order |
| Melting point | Sharp, definite melting point | Gradually soften over range |
| Cleavage | Clean, flat cleavage planes | Irregular surfaces |
| Anisotropy | Anisotropic (properties differ with direction) | Isotropic (same in all directions) |
| Heat of fusion | Definite | Not definite |
| Nature | True solids | Pseudo solids / supercooled liquids |
| Examples | NaCl, quartz, diamond, ice | Glass, rubber, plastic, tar |
⭐ Quartz vs Quartz Glass: Quartz (SiO2) is crystalline. When melted and rapidly cooled it forms quartz glass which is amorphous. Same composition, different structure!
3. Types of Crystalline Solids ⭐
| Type | Constituent Particles | Bonding | Properties | Examples |
|---|---|---|---|---|
| Ionic | Cations and anions | Electrostatic (ionic) | Hard, brittle, high mp, conduct electricity in molten/aqueous state | NaCl, MgO, ZnS, CaF2 |
| Covalent (Network) | Atoms | Covalent bonds | Very hard, very high mp, poor conductor (except graphite) | Diamond, SiC, AlN, SiO2, graphite |
| Metallic | Metal cations in electron sea | Metallic bond | Hard to soft, lustrous, good conductor, malleable, ductile | Fe, Cu, Ag, Au, Mg |
| Molecular | Molecules | van der Waals/H-bond/dipole | Soft, low mp, poor conductor | I2, CO2, H2O (ice), naphthalene |
⭐ Exception - Graphite: Graphite is a covalent solid but conducts electricity because electrons in the pi bonds are free to move between layers. It is also soft because layers can slide over each other (weak van der Waals forces between layers).
4. Crystal Lattice and Unit Cell
Crystal Lattice (Space Lattice): A regular 3-dimensional arrangement of points in space showing the positions of constituent particles in a crystal is called a crystal lattice or space lattice.
Unit Cell: The smallest repeating structural unit of a crystal lattice that, when repeated in all three dimensions, generates the entire crystal. A unit cell is characterised by:
> Edge lengths: a, b, c
> Angles between edges: alpha, beta, gamma
Unit Cell: The smallest repeating structural unit of a crystal lattice that, when repeated in all three dimensions, generates the entire crystal. A unit cell is characterised by:
> Edge lengths: a, b, c
> Angles between edges: alpha, beta, gamma
Number of Atoms per Unit Cell (Z)
Contribution of atoms at different positions:
> Corner atom: shared by 8 unit cells → contributes 1/8
> Face-centred atom: shared by 2 unit cells → contributes 1/2
> Edge-centred atom: shared by 4 unit cells → contributes 1/4
> Body-centred atom: completely inside → contributes 1
> Corner atom: shared by 8 unit cells → contributes 1/8
> Face-centred atom: shared by 2 unit cells → contributes 1/2
> Edge-centred atom: shared by 4 unit cells → contributes 1/4
> Body-centred atom: completely inside → contributes 1
5. Types of Cubic Unit Cells ⭐
| Unit Cell | Atoms per Cell (Z) | Coordination Number | Packing Efficiency | Relation a and r |
|---|---|---|---|---|
| Simple Cubic (SC) | 1 (8 corners x 1/8) | 6 | 52.4% | a = 2r |
| Body Centred Cubic (BCC) | 2 (8x1/8 + 1) | 8 | 68% | 4r = a sqrt(3) |
| Face Centred Cubic (FCC) | 4 (8x1/8 + 6x1/2) | 12 | 74% | 4r = a sqrt(2) |
⭐ Memory Trick for Z values:
SC = 1, BCC = 2, FCC = 4
HCP (hexagonal close packing) also has Z = 4 and packing efficiency = 74%
SC = 1, BCC = 2, FCC = 4
HCP (hexagonal close packing) also has Z = 4 and packing efficiency = 74%
6. Packing Efficiency ⭐
Packing Efficiency = (Volume occupied by atoms in unit cell / Total volume of unit cell) x 100%
6.1 Simple Cubic
r = a/2, Volume of atom = (4/3)pi r^3 = (4/3)pi (a/2)^3
Z = 1
PE = [1 x (4/3)pi(a/2)^3] / a^3 x 100 = (pi/6) x 100 = 52.4%
Z = 1
PE = [1 x (4/3)pi(a/2)^3] / a^3 x 100 = (pi/6) x 100 = 52.4%
6.2 Body Centred Cubic (BCC)
4r = a sqrt(3) → r = a sqrt(3)/4
Z = 2
PE = [2 x (4/3)pi(a sqrt(3)/4)^3] / a^3 x 100 = (pi sqrt(3)/8) x 100 = 68%
Z = 2
PE = [2 x (4/3)pi(a sqrt(3)/4)^3] / a^3 x 100 = (pi sqrt(3)/8) x 100 = 68%
6.3 Face Centred Cubic (FCC)
4r = a sqrt(2) → r = a sqrt(2)/4 = a/(2 sqrt(2))
Z = 4
PE = [4 x (4/3)pi(a/(2 sqrt(2)))^3] / a^3 x 100 = (pi/(3 sqrt(2))) x 100 = 74%
Z = 4
PE = [4 x (4/3)pi(a/(2 sqrt(2)))^3] / a^3 x 100 = (pi/(3 sqrt(2))) x 100 = 74%
7. Voids (Interstitial Sites) ⭐
The empty spaces between the closely packed spheres in a crystal are called voids or holes.
| Property | Tetrahedral Void | Octahedral Void |
|---|---|---|
| Surrounded by | 4 spheres (tetrahedron) | 6 spheres (octahedron) |
| Number per sphere | 2 per sphere | 1 per sphere |
| Radius ratio (r/R) | 0.225 | 0.414 |
| In FCC with N atoms | 2N tetrahedral voids | N octahedral voids |
| Location in FCC | Inside the unit cell (8 positions) | Edge centres + body centre (total 4) |
⭐ Key Relation: If there are N atoms (spheres) in a close-packed structure:
> Number of octahedral voids = N
> Number of tetrahedral voids = 2N
Radius Ratios for different coordination numbers:
CN = 3: r/R = 0.155 (triangular void)
CN = 4: r/R = 0.225 (tetrahedral void)
CN = 6: r/R = 0.414 (octahedral void)
CN = 8: r/R = 0.732 (cubic void)
> Number of octahedral voids = N
> Number of tetrahedral voids = 2N
Radius Ratios for different coordination numbers:
CN = 3: r/R = 0.155 (triangular void)
CN = 4: r/R = 0.225 (tetrahedral void)
CN = 6: r/R = 0.414 (octahedral void)
CN = 8: r/R = 0.732 (cubic void)
8. Density of Unit Cell ⭐ (Numerical Formula)
DENSITY FORMULA:
d = (Z x M) / (NA x a^3)
Where:
d = density of unit cell (g/cm^3)
Z = number of atoms per unit cell
M = molar mass (g/mol)
NA = Avogadro's number (6.022 x 10^23)
a = edge length of unit cell (cm)
a^3 = volume of unit cell
Example 1: An element has BCC structure with edge length 288 pm. Atomic mass = 52 g/mol. Find density.
Z = 2 (BCC), a = 288 pm = 288 x 10^(-10) cm
d = (2 x 52) / (6.022 x 10^23 x (288 x 10^(-10))^3)
= 104 / (6.022 x 10^23 x 2.39 x 10^(-23))
= 104 / 14.39
= 7.22 g/cm^3
Z = 2 (BCC), a = 288 pm = 288 x 10^(-10) cm
d = (2 x 52) / (6.022 x 10^23 x (288 x 10^(-10))^3)
= 104 / (6.022 x 10^23 x 2.39 x 10^(-23))
= 104 / 14.39
= 7.22 g/cm^3
Example 2: Silver has FCC structure, density = 10.5 g/cm^3, atomic mass = 108 g/mol. Find edge length.
Z = 4 (FCC)
a^3 = ZM / (NA x d) = (4 x 108) / (6.022 x 10^23 x 10.5)
= 432 / (6.32 x 10^24) = 6.84 x 10^(-23) cm^3
a = (6.84 x 10^(-23))^(1/3) = 4.09 x 10^(-8) cm = 409 pm
Z = 4 (FCC)
a^3 = ZM / (NA x d) = (4 x 108) / (6.022 x 10^23 x 10.5)
= 432 / (6.32 x 10^24) = 6.84 x 10^(-23) cm^3
a = (6.84 x 10^(-23))^(1/3) = 4.09 x 10^(-8) cm = 409 pm
9. Imperfections (Defects) in Solids ⭐
9.1 Point Defects
| Defect | Description | Effect on Density | Example |
|---|---|---|---|
| Vacancy Defect | Atom missing from its lattice site | Decreases | Any solid on heating |
| Interstitial Defect | Extra atom in interstitial site | Increases | Any solid |
| Schottky Defect | Equal cations and anions missing (ionic crystals) | Decreases | NaCl, KCl, KBr |
| Frenkel Defect | Cation leaves normal site, occupies interstitial site | No change | ZnS, AgCl, AgBr, AgI |
| Metal Excess Defect | Extra cations with electrons in holes (F-centres) | Slightly changes | NaCl (yellow colour), KCl (violet) |
| Metal Deficiency Defect | Missing cation compensated by higher charge cation | Slight change | FeO, FeS |
⭐ Key Points on Defects:
> Schottky defect: large difference in ionic sizes (both ions missing) — NaCl, KCl
> Frenkel defect: large difference in ionic sizes (smaller ion displaced) — ZnS, AgCl
> AgBr shows BOTH Schottky AND Frenkel defects.
> F-centres (Farbe centres) are responsible for colour in metal excess defects — trapped electrons absorb light.
> Schottky defect: large difference in ionic sizes (both ions missing) — NaCl, KCl
> Frenkel defect: large difference in ionic sizes (smaller ion displaced) — ZnS, AgCl
> AgBr shows BOTH Schottky AND Frenkel defects.
> F-centres (Farbe centres) are responsible for colour in metal excess defects — trapped electrons absorb light.
10. Electrical Properties of Solids
| Type | Conductivity (S/m) | Examples | Reason |
|---|---|---|---|
| Conductors | 10^4 to 10^7 | Metals (Cu, Ag) | Overlapping valence and conduction bands |
| Insulators | 10^(-20) to 10^(-10) | Wood, rubber, diamond | Large energy gap between bands |
| Semiconductors | 10^(-6) to 10^4 | Si, Ge, GaAs | Small energy gap; conductivity increases with temperature |
10.1 Types of Semiconductors
Intrinsic Semiconductor: Pure semiconductor (Si, Ge). Conducts due to thermal excitation.
n-type Semiconductor: Doped with group 15 element (P, As). Extra electrons carry current. (n = negative)
p-type Semiconductor: Doped with group 13 element (B, Al). Creates positive holes that carry current. (p = positive)
n-type Semiconductor: Doped with group 15 element (P, As). Extra electrons carry current. (n = negative)
p-type Semiconductor: Doped with group 13 element (B, Al). Creates positive holes that carry current. (p = positive)
11. Magnetic Properties of Solids ⭐
| Type | Behaviour in Magnetic Field | Unpaired Electrons | Examples |
|---|---|---|---|
| Diamagnetic | Weakly repelled | None (all paired) | NaCl, TiO2, H2O, Cu^+ |
| Paramagnetic | Weakly attracted | Present | O2, Cu^2+, Fe^3+, CuO |
| Ferromagnetic | Strongly attracted, can be permanently magnetised | Many, aligned parallel | Fe, Co, Ni, Gd, CrO2 |
| Antiferromagnetic | Net magnetism = 0 (moments cancel) | Equal, aligned antiparallel | MnO, MnO2, Cr2O3 |
| Ferrimagnetic | Net magnetism (unequal moments) | Unequal, antiparallel | Fe3O4, ferrites (MgFe2O4) |
12. Important Board Exam Questions
Q1. What is the difference between crystalline and amorphous solids? Give two examples of each.
Crystalline solids have long-range regular arrangement of particles, sharp melting point, and are anisotropic. Examples: NaCl, quartz.
Amorphous solids have short-range irregular arrangement, no sharp melting point, and are isotropic. Examples: glass, rubber.
Amorphous solids have short-range irregular arrangement, no sharp melting point, and are isotropic. Examples: glass, rubber.
Q2. How many atoms are present per unit cell in (a) Simple Cubic (b) BCC (c) FCC?
(a) SC: 8 corners x 1/8 = 1 atom
(b) BCC: 8 x 1/8 + 1 (body centre) = 1 + 1 = 2 atoms
(c) FCC: 8 x 1/8 + 6 x 1/2 = 1 + 3 = 4 atoms
(b) BCC: 8 x 1/8 + 1 (body centre) = 1 + 1 = 2 atoms
(c) FCC: 8 x 1/8 + 6 x 1/2 = 1 + 3 = 4 atoms
Q3. An element with molar mass 27 g/mol forms an FCC crystal with edge length 405 pm. Find its density.
Z = 4, M = 27 g/mol, a = 405 pm = 405 x 10^(-10) cm
d = ZM / (NA x a^3)
= (4 x 27) / (6.022 x 10^23 x (405 x 10^(-10))^3)
= 108 / (6.022 x 10^23 x 6.65 x 10^(-23))
= 108 / 40.05
= 2.70 g/cm^3 (This is Aluminium!)
d = ZM / (NA x a^3)
= (4 x 27) / (6.022 x 10^23 x (405 x 10^(-10))^3)
= 108 / (6.022 x 10^23 x 6.65 x 10^(-23))
= 108 / 40.05
= 2.70 g/cm^3 (This is Aluminium!)
Q4. What are Schottky and Frenkel defects? How do they affect the density of solids?
Schottky Defect: Equal numbers of cations and anions are missing from their lattice sites. This decreases the density of the solid. Common in NaCl, KBr.
Frenkel Defect: A cation is displaced from its normal lattice site to an interstitial position. Since no particles are lost, density remains unchanged. Common in ZnS, AgCl.
Frenkel Defect: A cation is displaced from its normal lattice site to an interstitial position. Since no particles are lost, density remains unchanged. Common in ZnS, AgCl.
Q5. What is meant by n-type and p-type semiconductors? How are they formed?
n-type: When Si or Ge is doped with a group 15 element (like P or As), which has 5 valence electrons, 4 form covalent bonds and the 5th is free to conduct electricity. Conductivity is due to extra electrons.
p-type: When Si or Ge is doped with a group 13 element (like B or Al), which has 3 valence electrons, only 3 bonds form and a positive hole is created. Conductivity is due to movement of these positive holes.
p-type: When Si or Ge is doped with a group 13 element (like B or Al), which has 3 valence electrons, only 3 bonds form and a positive hole is created. Conductivity is due to movement of these positive holes.
Q6. The density of chromium is 7.2 g/cm^3. If the unit cell is BCC, find the edge length. (Atomic mass of Cr = 52 g/mol)
Z = 2, M = 52, d = 7.2, NA = 6.022 x 10^23
a^3 = ZM/(NA x d) = (2 x 52)/(6.022 x 10^23 x 7.2)
= 104/(4.336 x 10^24) = 2.399 x 10^(-23) cm^3
a = (2.399 x 10^(-23))^(1/3) = 2.88 x 10^(-8) cm = 288 pm
a^3 = ZM/(NA x d) = (2 x 52)/(6.022 x 10^23 x 7.2)
= 104/(4.336 x 10^24) = 2.399 x 10^(-23) cm^3
a = (2.399 x 10^(-23))^(1/3) = 2.88 x 10^(-8) cm = 288 pm
13. Key Formulas and Facts at a Glance
UNIT CELL ATOMS (Z):
SC = 1, BCC = 2, FCC = 4
COORDINATION NUMBERS:
SC = 6, BCC = 8, FCC = 12
PACKING EFFICIENCY:
SC = 52.4%, BCC = 68%, FCC = HCP = 74%
RADIUS RELATIONS:
SC: a = 2r
BCC: 4r = a.sqrt(3) → r = a.sqrt(3)/4
FCC: 4r = a.sqrt(2) → r = a.sqrt(2)/4
VOIDS (for N atoms):
Tetrahedral voids = 2N
Octahedral voids = N
DENSITY:
d = (Z x M) / (NA x a^3)
RADIUS RATIOS:
CN=4 (tetrahedral): r/R = 0.225
CN=6 (octahedral): r/R = 0.414
CN=8 (cubic): r/R = 0.732
DEFECTS:
Schottky → density decreases (NaCl, KCl)
Frenkel → density unchanged (ZnS, AgCl)
AgBr → shows BOTH defects
14. MCQ Practice (1 Mark)
1. The number of atoms per unit cell in a BCC lattice is:
(a) 1 (b) 2 (c) 4 (d) 6
Answer: (b) 2
2. Which defect causes a decrease in the density of a crystal?
(a) Frenkel defect (b) Metal excess defect (c) Schottky defect (d) Interstitial defect
Answer: (c) Schottky defect
3. In an FCC unit cell, the number of octahedral voids is:
(a) 4 (b) 8 (c) 2 (d) 6
Answer: (a) 4 — octahedral voids = Z = 4 for FCC
4. The packing efficiency of FCC structure is:
(a) 52.4% (b) 68% (c) 74% (d) 100%
Answer: (c) 74%
5. Which of the following is an amorphous solid?
(a) Diamond (b) NaCl (c) Quartz (d) Glass
Answer: (d) Glass
6. In which of the following defects do both cations and anions go missing?
(a) Frenkel defect (b) Schottky defect (c) Metal excess (d) Vacancy defect
Answer: (b) Schottky defect
15. Exam Tips
- For density numericals: always convert edge length from pm to cm (1 pm = 10^(-10) cm) before calculating a^3.
- Z values SC=1, BCC=2, FCC=4 must be memorised — these appear in every numerical.
- Don't confuse Schottky (both ions missing, density decreases) with Frenkel (cation displaced, density unchanged).
- Graphite is covalent solid but conductor — this is a common tricky MCQ.
- AgBr shows BOTH Schottky and Frenkel defects — very commonly asked.
- For coordination number questions: SC=6, BCC=8, FCC=12 — link these to packing efficiency (higher CN = more efficient packing).
- F-centres give colour to crystals — NaCl becomes yellow, KCl becomes violet when heated in sodium/potassium vapour.
- Ferromagnetic materials (Fe, Co, Ni) can be permanently magnetised — distinguish from paramagnetic (only temporarily attracted).
✅ Summary: The Solid State is a highly conceptual chapter with important numericals. Focus on: (1) crystalline vs amorphous differences, (2) Z values and packing efficiency for all three cubic cells, (3) density formula and numericals, (4) Schottky vs Frenkel defects, and (5) electrical and magnetic properties. These topics together cover nearly all the marks from this chapter in CBSE board exams.
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