Chapter 1 — The Solid State
Welcome to HSLC Guru. This study guide for Class 12 Chemistry Chapter 1 — The Solid State — is prepared strictly according to the ASSEB (Assam State School Education Board) syllabus. Solids are all around us — from the salt we eat to the silicon chips that power our phones. In this chapter you will learn how the orderly arrangement of particles in solids gives rise to their unique properties. We cover crystalline and amorphous solids, classification of solids, unit cells, packing efficiency, density calculations, defects in solids, and electrical and magnetic properties. Each section is supported by clear theory, solved numerical problems, and exam-oriented questions to ensure top performance in your board examination.
Summary
Solids and their classification: Solids are characterised by definite shape, definite volume, rigidity, incompressibility, and short intermolecular distances. Based on the arrangement of constituent particles, solids are classified into two broad types — crystalline solids (long-range order, sharp melting point, anisotropic, true solids; e.g., NaCl, diamond, quartz) and amorphous solids (short-range order, range of melting temperatures, isotropic, pseudo-solids or supercooled liquids; e.g., glass, rubber, plastics). Crystalline solids are further classified on the basis of the nature of intermolecular forces into four categories: ionic solids (NaCl, ZnS) — held by strong electrostatic forces, hard, brittle, high melting point, conduct electricity in molten state; molecular solids — non-polar (Ar, CCl4), polar (HCl, SO2), and hydrogen-bonded (ice); covalent or network solids (diamond, graphite, SiC) — atoms joined by covalent bonds forming a giant network; metallic solids (Cu, Fe, Al) — positive cores in a sea of delocalised electrons giving lustre, malleability, ductility, and high conductivity.
Crystal lattice and unit cells: The regular three-dimensional arrangement of particles in a crystalline solid is called a crystal lattice or space lattice. The smallest repeating portion of a crystal lattice that, when repeated in three dimensions, generates the entire lattice is called the unit cell. A unit cell is characterised by three edge lengths (a, b, c) and three angles (α, β, γ). There are seven crystal systems and fourteen Bravais lattices. Three common cubic unit cells are: simple (primitive) cubic with one atom per unit cell (Z = 1), body-centred cubic (BCC) with two atoms per unit cell (Z = 2), and face-centred cubic (FCC) with four atoms per unit cell (Z = 4). The number of nearest neighbours of a particle is known as its coordination number — 6 for simple cubic, 8 for BCC, and 12 for FCC/HCP.
Close packing, voids and packing efficiency: In two dimensions, particles can be packed in square close packing (coordination 4) or hexagonal close packing (coordination 6). When successive 2D hexagonal layers are stacked, two important 3D arrangements emerge: hexagonal close packing (HCP, ABAB…) as in Mg, Zn, and cubic close packing (CCP/FCC, ABCABC…) as in Cu, Ag, Au. Both have coordination number 12 and packing efficiency 74%. Body-centred cubic packing has 68% efficiency and simple cubic only 52.4%. The empty spaces left between packed spheres are called voids. Tetrahedral voids are surrounded by four spheres and octahedral voids by six spheres. In a CCP/HCP arrangement of N particles, there are N octahedral voids and 2N tetrahedral voids. Many ionic solids can be described in terms of cations occupying voids in a close-packed array of anions.
Density of unit cell, defects and properties: The density of a cubic unit cell is calculated as below. Real crystals exhibit imperfections called defects. Stoichiometric point defects include Schottky defect (equal numbers of cations and anions missing — lowers density; seen in NaCl, KCl) and Frenkel defect (smaller ion displaced to interstitial site — density unchanged; seen in AgBr, ZnS). Non-stoichiometric defects include metal excess defects with anionic vacancies producing F-centres (responsible for the colour of NaCl crystals heated in Na vapour). On the basis of electrical conductivity, solids are conductors (10⁴–10⁷ Ω⁻¹m⁻¹), insulators (10⁻²⁰–10⁻¹⁰ Ω⁻¹m⁻¹) and semiconductors (10⁻⁶–10⁴ Ω⁻¹m⁻¹). Doping silicon with group 15 elements (P, As) gives n-type semiconductors and with group 13 elements (B, Al) gives p-type semiconductors. Magnetic behaviours are: diamagnetic (all electrons paired, weakly repelled), paramagnetic (unpaired electrons, weakly attracted), ferromagnetic (Fe, Co, Ni — strongly attracted, retain magnetism), antiferromagnetic (MnO — net moment zero), and ferrimagnetic (Fe3O4 — unequal opposing moments).
Density formula:
$d = \frac{Z \cdot M}{a^3 \cdot N_A}$
Packing efficiencies:
$\text{P.E.}_{fcc} = 74\%$
$\text{P.E.}_{bcc} = 68\%$
Edge length and radius relations:
$a = 2\sqrt{2}\,r$ (FCC)
$a = \frac{4r}{\sqrt{3}}$ (BCC)
Textbook Questions and Answers
One-Mark Questions
Q1. What is a crystalline solid?
Answer: A solid in which the constituent particles are arranged in a definite, regular, repeating three-dimensional pattern over long range is called a crystalline solid. Examples: NaCl, diamond, sugar.
Q2. Define amorphous solid with one example.
Answer: Solids in which constituent particles have only short-range order and irregular arrangement are amorphous solids. Examples: glass, rubber.
Q3. Why is glass considered a supercooled liquid?
Answer: Glass is amorphous and tends to flow very slowly under gravity, behaving like an extremely viscous liquid. Hence it is called a supercooled liquid or pseudo-solid.
Q4. What is a unit cell?
Answer: The smallest three-dimensional portion of a crystal lattice which, when repeated in space, generates the entire lattice is called a unit cell.
Q5. What is the coordination number of an atom in BCC arrangement?
Answer: 8.
Q6. Name a solid which shows Frenkel defect.
Answer: Silver bromide (AgBr) and zinc sulphide (ZnS).
Q7. What are F-centres?
Answer: Anion vacancies in a crystal that are occupied by unpaired electrons are called F-centres. They are responsible for the colour of non-stoichiometric crystals.
Q8. Give one example of a ferromagnetic substance.
Answer: Iron (Fe), cobalt (Co), nickel (Ni), CrO2.
Q9. What is the number of atoms per unit cell in an FCC lattice?
Answer: 4.
Q10. Name the type of semiconductor produced by doping silicon with arsenic.
Answer: n-type semiconductor.
Two/Three-Mark Questions
Q1. Distinguish between crystalline and amorphous solids (any three points).
Answer: (i) Crystalline solids have long-range order while amorphous solids have only short-range order. (ii) Crystalline solids have sharp, definite melting points whereas amorphous solids melt over a range of temperatures. (iii) Crystalline solids are anisotropic (properties depend on direction) while amorphous solids are isotropic. (iv) Crystalline solids give clean cleavage planes; amorphous solids give irregular cuts.
Q2. Differentiate between Schottky and Frenkel defects.
Answer: Schottky defect arises when equal numbers of cations and anions are missing from their lattice sites, decreasing the density of the crystal. It is shown by ionic solids with similar-sized cations and anions, e.g., NaCl, KCl. Frenkel defect arises when a smaller (usually cationic) ion is displaced from its lattice site to an interstitial position; the density remains unchanged. It is shown by ionic solids with large differences in ionic sizes, e.g., AgBr, ZnS.
Q3. Explain n-type and p-type semiconductors with examples.
Answer: When a Group 14 element (Si or Ge) is doped with a Group 15 element such as P or As, four of the five valence electrons of the dopant form covalent bonds with Si and the fifth electron is delocalised, increasing electrical conductivity. This is an n-type semiconductor (extra negative carrier). When Si is doped with a Group 13 element such as B or Al, only three covalent bonds form and a vacancy or hole is created which behaves as a positive charge carrier. This is a p-type semiconductor.
Q4. What are tetrahedral and octahedral voids? How many of each are present per particle in a close-packed structure?
Answer: A tetrahedral void is a small empty space surrounded by four spheres arranged tetrahedrally; an octahedral void is a larger empty space surrounded by six spheres arranged octahedrally. In a close-packed arrangement of N particles there are 2N tetrahedral voids and N octahedral voids.
Q5. Define ferromagnetism, antiferromagnetism and ferrimagnetism.
Answer: Ferromagnetism: substances with all magnetic moments aligned in the same direction within domains, very strongly attracted by a magnet and retain magnetism (Fe, Co, Ni). Antiferromagnetism: equal numbers of moments aligned in opposite directions giving zero net moment (MnO). Ferrimagnetism: unequal opposing moments giving a small net moment (Fe3O4, ferrites).
Q6. What is meant by coordination number? Give its values for simple cubic, BCC and FCC.
Answer: Coordination number is the number of nearest neighbours of a particle in a crystal lattice. Simple cubic = 6, BCC = 8, FCC/HCP = 12.
Five/Seven-Mark Questions (with Numericals)
Q1. Derive the relation between the edge length (a) and the radius (r) of an atom in a face-centred cubic unit cell.
Answer: In an FCC unit cell, atoms touch each other along the face diagonal. The face diagonal contains four radii (corner-atom radius + diameter of face-centre atom + corner-atom radius). Therefore face diagonal = 4r. By Pythagoras, face diagonal of a cube of side a = a√2. So a√2 = 4r, giving a = 4r/√2 = 2√2 r. Thus $a = 2\sqrt{2}\,r$ for FCC. The packing efficiency works out to 74%.
Q2. Derive the formula for density of a cubic unit cell.
Answer: Let Z = number of atoms per unit cell, M = molar mass (g mol⁻¹), a = edge length (cm), N_A = Avogadro number. Mass of one atom = M/N_A. Mass of unit cell = Z × M/N_A. Volume of unit cell = a³. Density d = mass/volume = ZM/(a³ N_A). Hence $d = \frac{Z \cdot M}{a^3 \cdot N_A}$.
Q3. An element crystallises in a body-centred cubic structure with edge length 288 pm. Its density is 7.2 g cm⁻³. Calculate the atomic mass of the element. (N_A = 6.022 × 10²³)
Answer: Given Z = 2 (BCC), a = 288 pm = 2.88 × 10⁻⁸ cm, d = 7.2 g cm⁻³. Using d = ZM/(a³N_A), M = d × a³ × N_A / Z. a³ = (2.88 × 10⁻⁸)³ = 2.39 × 10⁻²³ cm³. M = (7.2 × 2.39 × 10⁻²³ × 6.022 × 10²³) / 2 = (103.6) / 2 ≈ 51.8 g mol⁻¹. The element is approximately chromium (Cr).
Q4. Silver crystallises in FCC lattice. If edge length of unit cell is 4.077 × 10⁻⁸ cm and density is 10.5 g cm⁻³, calculate the atomic mass of silver.
Answer: Z = 4 for FCC. M = d × a³ × N_A / Z. a³ = (4.077 × 10⁻⁸)³ ≈ 6.776 × 10⁻²³ cm³. M = (10.5 × 6.776 × 10⁻²³ × 6.022 × 10²³) / 4 ≈ (428.4) / 4 ≈ 107.1 g mol⁻¹. This matches the atomic mass of silver (≈ 108).
Q5. Calculate the packing efficiency of a body-centred cubic (BCC) lattice.
Answer: In BCC, atoms touch along the body diagonal. Body diagonal = 4r, also = a√3. So a = 4r/√3, i.e., $a = \frac{4r}{\sqrt{3}}$. Number of atoms per unit cell Z = 2. Volume of 2 atoms = 2 × (4/3)πr³. Volume of unit cell = a³ = (4r/√3)³ = 64r³/3√3. Packing efficiency = (2 × (4/3)πr³) ÷ (64r³/3√3) × 100. On simplification, P.E. = (√3 π)/8 × 100 ≈ 68%. Hence $\text{P.E.}_{bcc} = 68\%$.
Multiple Choice Questions (MCQs)
Q1. Number of atoms per unit cell in FCC is — (a) 1 (b) 2 (c) 4 (d) 8.
Answer: (c) 4.
Q2. Coordination number in BCC is — (a) 4 (b) 6 (c) 8 (d) 12.
Answer: (c) 8.
Q3. Packing efficiency of FCC is — (a) 52.4% (b) 68% (c) 74% (d) 80%.
Answer: (c) 74%.
Q4. Schottky defect lowers — (a) volume (b) density (c) hardness (d) lustre.
Answer: (b) density.
Q5. Frenkel defect is shown by — (a) NaCl (b) KCl (c) AgBr (d) CsCl.
Answer: (c) AgBr.
Q6. Doping Si with B gives — (a) n-type (b) p-type (c) insulator (d) conductor.
Answer: (b) p-type.
Q7. An example of an amorphous solid is — (a) NaCl (b) diamond (c) glass (d) ice.
Answer: (c) glass.
Q8. F-centre is responsible for — (a) hardness (b) magnetism (c) colour (d) density.
Answer: (c) colour.
Q9. Magnetite Fe3O4 is — (a) diamagnetic (b) paramagnetic (c) ferrimagnetic (d) antiferromagnetic.
Answer: (c) ferrimagnetic.
Q10. In CCP/HCP, total voids per atom = (a) 1 (b) 2 (c) 3 (d) 4.
Answer: (c) 3 (1 octahedral + 2 tetrahedral).
Fill in the Blanks
Q1. Glass is an example of __________ solid. Answer: amorphous.
Q2. Number of particles in a BCC unit cell is __________. Answer: 2.
Q3. Packing efficiency of simple cubic lattice is __________ %. Answer: 52.4.
Q4. Doping Si with phosphorus gives __________ semiconductor. Answer: n-type.
Q5. __________ defect is shown by NaCl. Answer: Schottky.
True / False
Q1. Crystalline solids are isotropic. Answer: False.
Q2. Ice is a hydrogen-bonded molecular solid. Answer: True.
Q3. Schottky defect increases the density of a crystal. Answer: False.
Q4. p-type semiconductor is formed by adding Group 13 elements to silicon. Answer: True.
Q5. Antiferromagnetic substances have a non-zero net magnetic moment. Answer: False.
Glossary
| Term | Meaning |
|---|---|
| Crystalline solid | Solid with long-range orderly arrangement of particles |
| Amorphous solid | Solid with short-range order (e.g., glass, rubber) |
| Crystal lattice | 3D regular arrangement of points representing particles |
| Unit cell | Smallest repeating part of a crystal lattice |
| Coordination number | Number of nearest neighbours of a particle |
| Tetrahedral void | Void surrounded by four spheres |
| Octahedral void | Void surrounded by six spheres |
| Schottky defect | Equal cation and anion vacancies — lowers density |
| Frenkel defect | Ion displaced to interstitial site — density unchanged |
| F-centre | Anion vacancy occupied by an electron causing colour |
| n-type semiconductor | Si/Ge doped with Group 15 element (extra electrons) |
| p-type semiconductor | Si/Ge doped with Group 13 element (electron holes) |
| Ferromagnetic | Strongly attracted, retains magnetism (Fe, Co, Ni) |
| Antiferromagnetic | Equal opposite moments — net moment zero (MnO) |
| Ferrimagnetic | Unequal opposite moments — small net moment (Fe3O4) |
Formula Table
| Formula | Use |
|---|---|
| d = ZM / (a³ N_A) | Density of a cubic unit cell |
| a = 2√2 r | Edge length of FCC in terms of radius |
| a = 4r / √3 | Edge length of BCC in terms of radius |
| a = 2r | Edge length of simple cubic in terms of radius |
| P.E. (FCC/HCP) = 74% | Maximum packing efficiency |
| P.E. (BCC) = 68% | BCC packing efficiency |
| P.E. (simple cubic) = 52.4% | Simple cubic packing efficiency |
| Z (SC) = 1, Z (BCC) = 2, Z (FCC) = 4 | Atoms per cubic unit cell |
| Tetrahedral voids = 2N, Octahedral voids = N | Voids in close packing of N atoms |
Quick Recap of Crystal Systems and Bravais Lattices
| Crystal System | Edge lengths | Angles | Bravais Lattices | Example |
|---|---|---|---|---|
| Cubic | a = b = c | α = β = γ = 90° | Primitive, BCC, FCC | NaCl, Cu |
| Tetragonal | a = b ≠ c | α = β = γ = 90° | Primitive, BCC | SnO2 |
| Orthorhombic | a ≠ b ≠ c | α = β = γ = 90° | P, BCC, FCC, End-centred | Rhombic sulphur |
| Hexagonal | a = b ≠ c | α = β = 90°, γ = 120° | Primitive | Graphite, ZnO |
| Trigonal/Rhombohedral | a = b = c | α = β = γ ≠ 90° | Primitive | Calcite |
| Monoclinic | a ≠ b ≠ c | α = γ = 90°, β ≠ 90° | Primitive, End-centred | Monoclinic sulphur |
| Triclinic | a ≠ b ≠ c | α ≠ β ≠ γ ≠ 90° | Primitive | K2Cr2O7 |
Important Points to Remember
1. Number of atoms contributed by particles at different positions of a cubic unit cell — corner: 1/8, face: 1/2, edge: 1/4, body centre: 1.
2. For simple cubic: Z = 8 × (1/8) = 1; for BCC: Z = 8 × (1/8) + 1 = 2; for FCC: Z = 8 × (1/8) + 6 × (1/2) = 4.
3. The volume occupied per atom inside a unit cell = (4/3)πr³. Packing efficiency = (Z × volume of one atom × 100) / (volume of unit cell).
4. Density formula reminder: $d = \frac{Z \cdot M}{a^3 \cdot N_A}$. Always express edge length in cm and molar mass in g mol⁻¹ to obtain density in g cm⁻³.
5. Schottky defect is more common in highly ionic compounds with high coordination numbers (NaCl, KCl, CsCl, AgBr — note that AgBr shows BOTH Schottky and Frenkel).
6. The colour of NaCl heated in sodium vapour is yellow, of KCl heated in K vapour is violet (lilac) and of LiCl heated in Li vapour is pink — all due to F-centres.
7. The ratio of conductivity to that of conductors places semiconductors in between insulators and metals; their conductivity increases with rise in temperature, unlike metals.
8. Examples to remember: 12-17 compounds — ZnS, CdS, CdSe, HgTe; 13-15 compounds — InSb, AlP, GaAs. These are useful semiconductors for solar cells and LEDs.
Additional Practice Problem
Q. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of a copper atom?
Answer: For an FCC lattice, $a = 2\sqrt{2}\,r$, so r = a / (2√2) = 361 / (2 × 1.414) pm = 361 / 2.828 pm ≈ 127.6 pm. Hence the radius of a copper atom is about 128 pm.
Continue your preparation with the next ASSEB Class 12 Chemistry chapter on Solutions for a smooth conceptual flow. Stay with HSLC Guru for chapter-wise notes, solved numericals and previous-year board-style practice for every ASSEB topic.