Class 12 Solid State Guide 2027 — Crystal Lattices, Packing Efficiency, Defects, Band Theory

In short: Solid State (Chapter 1) carries 5 marks in CBSE Class 12 Chemistry 2027 and opens the Physical Chemistry section with the structure of crystalline solids. The key scoring areas are unit cell calculations (atoms per unit cell, coordination number, packing efficiency for SC/BCC/FCC), voids (tetrahedral and octahedral), and point defects (Schottky and Frenkel). Numerical problems on density using the unit cell formula and questions on electrical classification of solids using band theory are board favourites.

Solid State is Chapter 1 in the NCERT textbook and lays the groundwork for understanding material properties through crystal structure. It carries approximately 5 marks on the board paper, often as one 3-mark numerical or conceptual question on unit cells and packing, and one 2-mark question on defects or electrical properties. This chapter has a satisfying structure — the same formula d = ZM/(a³Nₐ) solves most numerical problems, and the logic of packing and voids follows geometric principles that, once understood, are hard to forget.

For the complete marks distribution, see the Class 12 Chemistry chapter-wise weightage. For all formulas and constants, refer to the complete formulas and reactions list.

Types of Solids

Memory trick — "Crystalline = Clean cuts, Clear melting point. Amorphous = All directions same (isotropic), melts messily."

Classification of Crystalline Solids

Memory trick — "Ionic = Ions locked in lattice. Covalent = Continuous network. Metallic = Mobile electrons. Molecular = Molecules held weakly."

Crystal Lattice and Unit Cells

A crystal lattice (space lattice) is the three-dimensional arrangement of points representing the positions of particles in a crystal. A unit cell is the smallest repeating unit of the crystal lattice that, when repeated in all three dimensions, generates the entire crystal.

There are 7 crystal systems and 14 Bravais lattices. For board exams, you need to focus on the cubic system with its three types:

Simple Cubic (SC) / Primitive Cubic

• Atoms at the 8 corners of the cube only
• Each corner atom is shared by 8 unit cells → contribution = 1/8 per atom
• Atoms per unit cell (Z) = 8 × (1/8) = 1
• Coordination number = 6 (each atom touches 6 neighbours)
• Relationship: a = 2r (edge length = 2 × radius)
• Packing efficiency = 52.4%
• Example: Polonium (Po) — the only element with simple cubic structure

Body-Centred Cubic (BCC)

• Atoms at 8 corners + 1 atom at the body centre
• Corner contribution: 8 × (1/8) = 1, Body centre: 1 × 1 = 1
• Atoms per unit cell (Z) = 2
• Coordination number = 8 (body-centre atom touches 8 corner atoms)
• Relationship: √3 a = 4r → a = 4r/√3
• Packing efficiency = 68%
• Examples: Na, K, Fe (α-iron), Cr, W

Face-Centred Cubic (FCC) / Cubic Close-Packed (CCP)

• Atoms at 8 corners + atoms at the centre of each of the 6 faces
• Corner contribution: 8 × (1/8) = 1, Face centre: 6 × (1/2) = 3
• Atoms per unit cell (Z) = 4
• Coordination number = 12 (each atom touches 12 neighbours)
• Relationship: √2 a = 4r → a = 4r/√2 = 2√2 r
• Packing efficiency = 74% (maximum possible for identical spheres)
• Examples: Cu, Ag, Au, Al, Ni

The Master Unit Cell Table

Memory trick — "1, 2, 4 — atoms per unit cell go SC, BCC, FCC. Coordination: 6, 8, 12 — same order."

Packing Efficiency Derivation

Packing efficiency = (Volume occupied by atoms / Volume of unit cell) × 100

For FCC (most commonly asked):

• Z = 4 atoms, each atom is a sphere with volume (4/3)πr³
• Volume occupied = 4 × (4/3)πr³
• Edge length: a = 2√2r → a³ = (2√2r)³ = 16√2 r³
• Packing efficiency = [4 × (4/3)πr³] / [16√2 r³] × 100 = (π)/(3√2) × 100 = 74.05%

For BCC:

• Z = 2, a = 4r/√3 → a³ = 64r³/(3√3)
• Packing efficiency = [2 × (4/3)πr³] / [64r³/(3√3)] × 100 = (√3 π)/8 × 100 = 68.04%

For SC:

• Z = 1, a = 2r → a³ = 8r³
• Packing efficiency = [(4/3)πr³] / [8r³] × 100 = π/6 × 100 = 52.36%

Voids in Close-Packed Structures

When spheres are packed in close-packed layers (ABAB for HCP or ABCABC for CCP/FCC), two types of voids are formed:

Tetrahedral Voids

• Formed when a sphere of the second layer sits in the depression of three spheres of the first layer
• Surrounded by 4 atoms (tetrahedral arrangement)
• Number of tetrahedral voids = 2N (where N = number of atoms in packing)
• Smaller void — radius ratio r/R ≈ 0.225

Octahedral Voids

• Formed between two sets of three atoms in adjacent layers
• Surrounded by 6 atoms (octahedral arrangement)
• Number of octahedral voids = N (where N = number of atoms in packing)
• Larger void — radius ratio r/R ≈ 0.414

Memory trick — "Tetrahedral = 2N (TWO for T). Octahedral = N (ONE for O). T-voids are smaller, O-voids are bigger."

For FCC with 4 atoms per unit cell:

• Tetrahedral voids = 2 × 4 = 8
• Octahedral voids = 4

Density Formula (Most Important Numerical Formula)

$$d = \frac{Z \times M}{a^3 \times N_A}$$

Where:

• d = density (g/cm³)
• Z = number of atoms per unit cell (1, 2, or 4)
• M = molar mass (g/mol)
• a = edge length (cm) — often given in pm, convert: 1 pm = 10⁻¹⁰ cm
• Nₐ = Avogadro's number (6.022 × 10²³)

This single formula solves 90% of Solid State numericals. You may be asked to find d, a, Z, or r using this equation along with the edge-radius relationships.

Point Defects in Crystals

Real crystals are never perfect. Point defects are irregularities at individual lattice points.

Stoichiometric Defects (ratio of cations to anions stays the same)

Schottky Defect:

• Equal number of cations AND anions are missing from the lattice
• Found in ionic compounds where cation and anion are of similar size
• Decreases density (atoms are missing)
• Examples: NaCl, KCl, CsCl, AgBr

Frenkel Defect:

• A smaller ion (usually cation) leaves its normal position and occupies an interstitial site
• No atoms leave the crystal — just displacement
• Density stays the same
• Found when there is a large difference in size between cation and anion
• Examples: ZnS, AgCl, AgBr, AgI

Memory trick — "Schottky = Something's missing (vacancy, lower density). Frenkel = Finds a new position (displacement, same density)."

AgBr shows BOTH Schottky and Frenkel defects — this is a favourite board exam fact.

Non-Stoichiometric Defects

Metal excess defect:

• (a) Anion vacancy: An anion is missing, and an electron occupies the void (F-centre). The trapped electrons absorb light and give colour. Example: NaCl heated in Na vapour turns yellow (excess Na⁺, electrons in anion vacancies absorb blue light).
• (b) Extra cation: An extra cation occupies an interstitial site with an electron for charge balance.

Metal deficiency defect:

• A cation is missing, and a neighbouring cation has a higher oxidation state to maintain electrical neutrality.
• Example: FeO (often exists as Fe₀.₉₅O — some Fe²⁺ replaced by Fe³⁺)

Electrical Properties and Band Theory

Classification of Solids by Conductivity

Band Theory Basics

In a crystal with N atoms, the atomic orbitals combine to form bands containing N closely spaced energy levels.

• Valence band: The highest energy band that is fully occupied by electrons at 0 K
• Conduction band: The next higher energy band (empty at 0 K)
• Band gap (Eg): The energy difference between the top of the valence band and the bottom of the conduction band

Conductors: Valence band and conduction band overlap — electrons move freely. Insulators: Large band gap — electrons cannot jump to the conduction band. Semiconductors: Small band gap — at room temperature, some electrons gain enough thermal energy to jump to the conduction band. Conductivity increases with temperature (opposite of metals).

Doping of Semiconductors

n-type semiconductor: Silicon doped with Group 15 element (P, As). The extra electron from the Group 15 atom becomes a free electron (negative charge carrier).

p-type semiconductor: Silicon doped with Group 13 element (B, Al, Ga). One bond is incomplete, creating a "hole" (positive charge carrier).

Memory trick — "n-type = Negative extra electron = Group 15 dopant. p-type = Positive hole = Group 13 dopant."

Board Exam Question Types

Type 1: Unit Cell Calculation (3 marks)

"Silver crystallises in FCC. If the edge length is 408.6 pm, calculate the density. (Atomic mass of Ag = 108, Nₐ = 6.022 × 10²³)"

Approach: Use d = ZM/(a³Nₐ). For FCC, Z = 4. Convert a from pm to cm (408.6 pm = 408.6 × 10⁻¹⁰ cm). Substitute and calculate. Show units clearly.

Type 2: Defects (2-3 marks)

"Explain Schottky and Frenkel defects. Which ionic compounds show each?"

Approach: Define each defect with a diagram if possible. State the effect on density (Schottky decreases density, Frenkel does not). Give examples. Mention that AgBr shows both.

Type 3: Packing and Voids (2-3 marks)

"In a close-packed structure, the number of tetrahedral voids is twice the number of atoms. If the number of atoms is N, how many octahedral voids are present?"

Approach: State the formulas: tetrahedral voids = 2N, octahedral voids = N. For FCC with Z = 4: 8 tetrahedral and 4 octahedral voids per unit cell.

Type 4: Electrical Properties (2 marks)

"Distinguish between n-type and p-type semiconductors."

Approach: State the dopant group (15 for n-type, 13 for p-type), the majority charge carrier (electrons for n-type, holes for p-type), and give examples.

Common Mistakes That Cost Marks

1. Using wrong Z value for the unit cell type. SC = 1, BCC = 2, FCC = 4. Students often confuse these, especially under pressure. Remember: count corner atoms (÷8), face atoms (÷2), body-centre atoms (÷1), edge atoms (÷4).

2. Forgetting to convert pm to cm in the density formula. The density formula requires a in cm. 1 pm = 10⁻¹⁰ cm. If a = 400 pm, then a = 400 × 10⁻¹⁰ cm = 4 × 10⁻⁸ cm. Getting this conversion wrong gives a density off by many orders of magnitude.

3. Confusing Schottky and Frenkel defects. Schottky = vacancy (both ions missing, density decreases). Frenkel = displacement (ion moves to interstitial site, density unchanged). The key distinction is whether atoms leave the crystal or just move within it.

4. Mixing up tetrahedral and octahedral void counts. Tetrahedral voids = 2N, octahedral = N. Students often reverse these. Remember: the SMALLER void (tetrahedral) is MORE numerous.

5. Confusing n-type and p-type doping. n-type = Group 15 element (extra electron). p-type = Group 13 element (electron deficit = hole). Think: "n for negative (electron), p for positive (hole)."

6. Writing that semiconductor conductivity decreases with temperature. The opposite is true — semiconductor conductivity INCREASES with temperature because more electrons gain enough energy to cross the band gap. Metal conductivity decreases with temperature.

4-Week Practice Plan

Week 1: Types of Solids and Crystal Structures

• Read NCERT Chapter 1, Sections 1.1-1.5
• Learn the classification of crystalline solids with examples
• Study SC, BCC, FCC unit cells — draw each one and label atoms
• Memorise Z, coordination number, and edge-radius relationships for all three
• Reference: NCERT Exercise Q1-Q5

Week 2: Packing Efficiency and Density Problems

• Derive packing efficiency for FCC, BCC, SC (do all three from scratch)
• Solve 10 density calculation problems using d = ZM/(a³Nₐ)
• Study close-packing (HCP and CCP), ABAB vs ABCABC stacking
• Learn tetrahedral and octahedral voids with numbers
• Reference: NCERT Exercise Q6-Q12

Week 3: Defects, Electrical Properties, Band Theory

• Study all point defects (Schottky, Frenkel, non-stoichiometric)
• Learn F-centres and colour due to metal excess defect
• Study band theory (conductors, semiconductors, insulators)
• Understand n-type and p-type doping with diagrams
• Reference: NCERT Exercise Q13-Q20

Week 4: Board Paper Practice

• Solve Solid State questions from the last 5 years of CBSE board papers
• Time yourself: numerical problems in 5-6 minutes
• Focus on conceptual + numerical hybrid questions
• Final drill: write the master unit cell table from memory, then solve one numerical for each type

Quick Revision Card for Exam Day

• SC: Z=1, CN=6, a=2r, 52.4% packed | BCC: Z=2, CN=8, a=4r/√3, 68%
• FCC: Z=4, CN=12, a=2√2r, 74% packed (maximum for identical spheres)
• Density: d = ZM/(a³Nₐ) — convert a from pm to cm (×10⁻¹⁰)
• Voids: Tetrahedral = 2N, Octahedral = N (T is smaller but more numerous)
• Schottky: Both ions missing → density decreases (NaCl, KCl)
• Frenkel: Ion displaced to interstitial → density same (ZnS, AgCl)
• AgBr shows BOTH Schottky and Frenkel defects
• F-centre: Electron in anion vacancy → colour (NaCl + Na → yellow)
• n-type: Group 15 dopant, electrons | p-type: Group 13 dopant, holes
• Semiconductor conductivity increases with temperature (opposite of metals)

For guided practice on Solid State numericals and crystal structures, try Padhaao's AI tutor.

Frequently Asked Questions

How many marks does Solid State carry in CBSE Class 12 Chemistry 2027?

Solid State carries approximately 5 marks in the CBSE Class 12 Chemistry 2027 board paper. Expect one 3-mark question involving a numerical problem (density calculation using the unit cell formula) or a conceptual question on packing efficiency and voids. A 2-mark question on point defects (Schottky vs Frenkel) or electrical properties (n-type vs p-type semiconductors) is also common.

What is the most important formula in the Solid State chapter?

The density formula d = ZM/(a³Nₐ) is the single most important formula. It connects the unit cell type (Z), molar mass (M), edge length (a), and density (d). Almost every numerical problem in this chapter can be solved using this formula along with the appropriate edge-radius relationship (a = 2r for SC, a = 4r/√3 for BCC, a = 2√2r for FCC).

What is the difference between Schottky and Frenkel defects?

In a Schottky defect, equal numbers of cations and anions are missing from the lattice, creating vacancies. This decreases the density. It occurs in compounds where cation and anion are of similar size (NaCl, KCl). In a Frenkel defect, a smaller ion (usually cation) moves from its normal position to an interstitial site. No atoms leave the crystal, so density remains unchanged. It occurs when there is a large size difference between cation and anion (ZnS, AgCl).

How do you calculate the number of tetrahedral and octahedral voids?

In a close-packed structure with N atoms, the number of tetrahedral voids is 2N and the number of octahedral voids is N. For an FCC unit cell with 4 atoms, there are 8 tetrahedral voids and 4 octahedral voids. Tetrahedral voids are smaller (radius ratio 0.225) and octahedral voids are larger (radius ratio 0.414).

Why does semiconductor conductivity increase with temperature while metal conductivity decreases?

In semiconductors, the band gap between the valence band and conduction band is small. As temperature increases, more electrons gain enough thermal energy to jump across the band gap to the conduction band, increasing the number of charge carriers and hence conductivity. In metals, the conduction band is already full of free electrons. Increasing temperature causes more lattice vibrations that scatter these electrons, increasing resistance and decreasing conductivity.

What are F-centres and why do they cause colour?

F-centres (from the German word Farbe, meaning colour) are electrons trapped in anion vacancies of a crystal. When NaCl is heated in sodium vapour, sodium atoms deposit on the surface, and Cl⁻ ions migrate to the surface to form NaCl with the deposited Na⁺. The anion vacancies left behind trap electrons. These trapped electrons absorb visible light (blue light in the case of NaCl) and the crystal appears the complementary colour (yellow). The F-centre electrons undergo electronic transitions when they absorb light.

Solid State is the chapter that connects chemistry to materials science. The beauty of this chapter is its internal logic — once you understand how atoms pack and what determines the unit cell, the properties and defects follow naturally. Master the density formula, understand the three unit cells, and learn the defects — the 5 marks are built on these pillars.