Class 12 Chemistry · NCERT Chapter 1

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Solid State Class 12 Chemistry Notes

Solid State studies how atoms, ions and molecules pack to build crystals. This chapter covers crystalline and amorphous solids, unit cells, packing efficiency, density calculations, point defects and magnetic behaviour — with board-style numericals and MCQ practice for CBSE Class 12, CUET and NEET.

Class:12Subject:ChemistryCovers:Crystal lattices · Packing · DefectsExams:CBSE · CUET · NEET

6 Key Formulas2 Solved Examples22 Practice MCQs

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

Why is glass called a supercooled liquid?

Glass is an amorphous solid with short-range order whose particles are frozen before they can arrange into a lattice. Like a liquid it is isotropic, flows very slowly and softens over a range of temperature instead of melting sharply.

Classification of Solids

Solids are classified by how their constituent particles are arranged. Crystalline solids have a long-range, orderly repeating arrangement; amorphous solids have only a short-range, irregular arrangement.

Definition

Crystalline solid

A solid whose particles are arranged in a definite, repeating three-dimensional pattern with long-range order, giving sharp melting points and anisotropic properties. Examples: NaCl, quartz, diamond, copper.

Definition

Amorphous solid

A solid whose particles lack long-range order (short-range order only). It is isotropic, softens over a range of temperature and is said to be a pseudo-solid or supercooled liquid. Examples: glass, rubber, plastics.

Crystalline: definite geometry, sharp melting point, anisotropic, cleaves along planes.
Amorphous: irregular geometry, melts over a range, isotropic, breaks into uneven pieces.

Key exception

Glass is described as a supercooled liquid because although it is rigid it is isotropic and flows extremely slowly.

MCQ Questions

Which of the following is an amorphous solid?

Crystalline solids are anisotropic because:

Which pair shows short-range order only?

Crystal Lattices and Unit Cells

The regular three-dimensional arrangement of points representing the positions of constituent particles is a crystal lattice. The smallest repeating portion that reproduces the whole lattice when repeated in three dimensions is the unit cell.

Definition

Unit cell

The smallest building block of a crystal which, repeated along the three axes, generates the entire crystal. It is described by edge lengths a, b, c and axial angles α, β, γ.

Contribution rules for sharing

Number of atoms = \frac{atoms at corners}{8} + \frac{atoms at edges}{4} + \frac{atoms on faces}{2} + atoms in body

Primitive (simple) cell: particles only at the eight corners.
Body-centred (bcc): corners plus one particle at the centre.
Face-centred (fcc): corners plus one particle at each face centre.

MCQ Questions

The number of atoms in a body-centred cubic unit cell is:

In a face-centred cubic cell, the contribution of each face atom is:

The coordination number of a simple cubic lattice is:

Close Packing and Number of Atoms per Unit Cell

Spheres pack most efficiently in three dimensions in a close-packed arrangement. The two common close packings are hexagonal close packing (hcp, ABAB) and cubic close packing (ccp/fcc, ABCABC); both fill 74% of the space.

Simple cubic: z = 1, packing efficiency 52.4%, coordination number 6.
Body-centred cubic: z = 2, packing efficiency 68%, coordination number 8.
Face-centred cubic / ccp / hcp: z = 4, packing efficiency 74%, coordination number 12.

Relation between atomic radius r and edge length a

sc: r = \frac{a}{2}, bcc: r = \frac{3 a}{4}, fcc: r = \frac{2 a}{4}

Remember

The more efficiently spheres pack, the higher the coordination number: sc 6, bcc 8, fcc/hcp 12.

MCQ Questions

The packing efficiency of a body-centred cubic lattice is:

The number of atoms per unit cell in ccp (fcc) is:

For an fcc lattice, the relation between radius r and edge a is:

Density of a Unit Cell

The density of a crystal can be calculated from the mass of the atoms in one unit cell divided by the volume of the cell. Because z, M and a are measurable, this relationship is used to determine Avogadro's number and atomic masses.

Density of a unit cell; a in cm

d = \frac{z M}{a ^{3} N _{A}}

Convert edge length from Å to cm (1 Å = 10⁻⁸ cm) before substitution. Rearranged forms let you find z, M, a or N_A depending on the data given.

Units trap

Forgetting to convert Å to cm is the most common error in density numericals. Always substitute a in cm.

MCQ Questions

Q10

The correct formula for crystal density is:

Q11

If edge length is given in Å, converting to cm requires multiplying by:

Q12

Doubling the edge length of a cubic cell (z, M constant) makes the density:

Imperfections in Solids

Real crystals contain defects where the regular arrangement is disturbed. Point defects — Schottky and Frenkel — involve missing or displaced ions and are more common in ionic solids.

Definition

Schottky defect

A pair of cation and anion vacancies, leaving the lattice electrically neutral. Density decreases. Shown by NaCl, KCl, CsCl, AgBr.

Definition

Frenkel defect

A cation displaced from its lattice site to an interstitial position. Density is unchanged. Shown by ZnS, AgCl, AgBr, AgI.

Schottky: favored by small size difference and high coordination number; density falls.
Frenkel: favored by large size difference and high polarisability; density constant.
F-centres: anionic vacancies filled by trapped electrons, giving colour (e.g. yellow NaCl).

Colour from defects

Excess metal in a crystal creates F-centres whose trapped electrons absorb visible light, so the crystal appears coloured.

MCQ Questions

Q13

Which defect lowers the density of an ionic crystal?

Q14

AgCl is known to show:

Q15

The yellow colour of NaCl is due to:

Electrical and Magnetic Properties

Electrical conduction in solids depends on the availability of free electrons or ions. Metals conduct because of free valence electrons; ionic solids conduct only in the molten state via ion migration.

Conductors: conductivity decreases with rising temperature (metals).
Semiconductors: conductivity increases with temperature (Si, Ge).
Insulators: very large band gap (diamond, quartz).

Diamagnetic: all electrons paired, weakly repelled (TiO₂, benzene).
Paramagnetic: unpaired electrons, weakly attracted (O₂, Cu²⁺ salts).
Ferromagnetic: strong permanent alignment (Fe, Co, Ni).
Antiferromagnetic: equal opposite spins cancel (MnO).
Ferrimagnetic: unequal opposite spins, net magnetic moment (Fe₃O₄).

Trend

Paramagnetism arises only from unpaired electrons; ferromagnetism is a cooperative, permanent alignment of dipoles.

MCQ Questions

Q16

The electrical conductivity of a metal with rising temperature:

Q17

Ferrimagnetism is shown by:

Q18

A substance with all electrons paired is:

NCERT Exercise Questions

From the textbook

NCERT questions for this chapter, in textbook order. Answers are given at a glance below each set.

Exercises (NCERT Q1–Q4)

1
Define the term 'amorphous'. Give a few examples of amorphous solids.
2
What makes glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
3
Analysis shows that nickel oxide has the formula Ni₀.₉₈O₁.₀₀. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
4
Gold (atomic mass = 197 u) crystallises in fcc lattice. If the edge length of the unit cell is 4.07 Å, calculate the density of gold. (N_A = 6.022×10²³ mol⁻¹)

Answers at a glance

1
Amorphous solids have short-range order, are isotropic and melt over a range; e.g. glass, rubber, plastics.
2
Glass is a supercooled liquid (amorphous, short-range order), quartz is crystalline silica (long-range order). Quartz becomes glass on heating strongly and cooling rapidly.
3
Ni²⁺ fraction = 0.94 (94%) and Ni³⁺ fraction = 0.04 (4%).
4
Density of gold ≈ 19.4 g cm⁻³.

CBSE Pattern Questions

How CBSE asks this chapter

Board papers ask roughly 2–3 questions from this chapter: MCQ/VSA on classification and defects, plus one numerical on unit-cell density, packing or radius.

MCQ Questions

Q19

Which of the following solids shows a Schottky defect?

Q20

Which point defect increases the density of a solid?

Q21

Which is NOT a characteristic of ionic solids?

Q22

The number of atoms in an fcc unit cell of a monoatomic substance is:

1 Mark Questions

1
Define amorphous solids. Give two examples.
2
Why is glass considered a supercooled liquid?
3
Give the coordination number of a particle in bcc.

2 Mark Questions

1
Differentiate between crystalline and amorphous solids.
2
State two differences between Schottky and Frenkel defects.
3
Explain why Frenkel defect does not affect density.

3 Mark Questions

1
Calculate the number of unit cells in 8.1 g of aluminium if it crystallises in fcc (M = 27 g mol⁻¹).
2
Analysis shows NiO has the formula Ni₀.₉₈O₁.₀₀. Find the fractions of Ni²⁺ and Ni³⁺.
3
Niobium crystallises in bcc with d = 8.55 g cm⁻³. Calculate its atomic radius (M = 93 u).

5 Mark Questions

1
Explain with examples: (i) Schottky defect (ii) Frenkel defect (iii) F-centres (iv) effect of defects on density.
2
Gold (M = 197 u) crystallises in fcc with a = 4.07 Å. Calculate the density of gold (N_A = 6.022×10²³ mol⁻¹).

PDF: Assignments & Tests

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Class 12 Chemistry — Solid State assignments and CBSE-pattern tests (PDF).

Assignment 1 — Solids, Unit Cells & Packing

Topic-wise practice on classification, unit cells and packing efficiency with answers.

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Assignment 2 — Density, Defects & Magnetism

Numericals on density and practice on point defects and magnetic properties.

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CBSE Pattern Test 1

Full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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CBSE Pattern Test 2

Second full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Density of a unit cell

d = \frac{z M}{a ^{3} N _{A}}

z = atoms per cell, M = molar mass, a = edge length, N_A = 6.022×10²³.

Packing efficiency

η = \frac{z \frac{4}{3} π r ^{3}}{a ^{3}} \times 100%

sc 52.4%, bcc 68%, fcc/hcp 74%.

Radius-edge relations

sc: a = 2 r, bcc: 3 a = 4 r, fcc: 2 a = 4 r

Corner atoms touch along these directions.

Coordination number

sc = 6, bcc = 8, fcc = 12

Number of nearest neighbours of any particle.

Atoms per cell (z)

sc = 1, bcc = 2, fcc = 4

Corner 1/8, edge 1/4, face 1/2, body 1.

Frenkel vs Schottky

Schottky \Rightarrow d ↓, Frenkel \Rightarrow d unchanged

Frenkel is favoured by large size difference and high polarisability.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Numericals on unit-cell density, edge length, radius and Avogadro's number appear every year in CBSE 3/5-mark sections.
Draw and label simple cubic, bcc and fcc unit cells; state z and coordination number for each.
Schottky vs Frenkel defects: definition, effect on density and example — a standard 2 or 3-mark question (NaCl, AgCl, ZnS).
Packing efficiency: know the values (52.4%, 68%, 74%) and which lattice gives 74% (fcc/ccp and hcp).
Classification of solids (ionic, covalent, metallic, molecular, amorphous) with properties is a frequent MCQ and 1-mark topic.
F-centres explain the colour of crystals such as NaCl and are asked as a short-answer point.
Magnetic classification (diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, ferrimagnetic) with one example each.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Density of a metal with an fcc cell

Problem StatementSilver crystallises in fcc with edge length 4.07 Å. Find its density (M = 108 g mol⁻¹, N_A = 6.022×10²³ mol⁻¹).

Step-by-Step Solution

1For fcc, z = 4. Convert a to cm: a = 4.07 × 10⁻⁸ cm.
2a³ = (4.07×10⁻⁸)³ = 6.74 × 10⁻²³ cm³.
3d = zM / (a³ N_A) = (4 × 108) / (6.74×10⁻²³ × 6.022×10²³) = 432 / 40.6.

Final Answer

d ≈ 10.6 g cm⁻³

Radius from edge length (bcc)

Problem StatementA metal has bcc packing with edge length 4.29 Å. Find the atomic radius.

Step-by-Step Solution

1For bcc, √3 a = 4r, so r = (√3 a) / 4.
2r = (1.732 × 4.29) / 4 = 1.86 Å.

Final Answer

r ≈ 1.86 Å

FAQ

Frequently asked questions

Is Solid State still in the CBSE Class 12 Chemistry syllabus?

Solid State was rationalised out of the NCERT Class 12 textbook, but it remains foundational for entrance exams and is retained in many reference editions. These notes cover it in full.

Which solids show both Schottky and Frenkel defects?

AgBr shows both. Both defects are favoured by ionic compounds with high polarisation and comparable cation–anion sizes.

Why does the density stay the same in Frenkel defect but fall in Schottky defect?

Frenkel simply displaces a cation to an interstitial site (no ion lost), so mass and volume are unchanged. Schottky removes a cation–anion pair, so mass decreases while the cell volume stays roughly constant.

What is the difference between ccp and hcp?

Both give 74% packing. ccp (fcc) stacks layers ABCABC and has face-centred cubic symmetry; hcp stacks ABAB and has hexagonal symmetry.

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