The Solid State — Unit Cells, Packing & Defects

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Solid-state questions are mostly geometry. Count the atoms that belong to one unit cell, relate the edge length to the atomic radius, and the density, packing efficiency and formula all follow. Note: this chapter was dropped from the rationalised NCERT Class 12 book, but it is still asked in TGT/PGT papers and several state boards.

What you will be able to do

• Classify solids as crystalline or amorphous and as molecular, ionic, metallic or covalent
• Count atoms per unit cell and relate edge length to radius for sc, bcc and fcc
• Calculate packing efficiency and density, and find formulas from void occupation
• Describe the NaCl, CsCl, ZnS and CaF₂ structures and use radius ratios
• Explain Schottky, Frenkel and non-stoichiometric defects, semiconductors and magnetic types

The big picture

Count, relate, calculate.

In one line: A crystal is a unit cell repeated in three dimensions; counting the atoms that belong to one cell and relating the edge length to the atomic radius gives density, packing efficiency and formula.

Question type	What you need
Atoms per cell (Z)	Corner ⅛, edge ¼, face ½, body 1
Radius from edge	sc: a = 2r; bcc: √3a = 4r; fcc: √2a = 4r
Density	d = ZM / (a³ Nₐ)
Formula of a compound	Which ions sit in which voids, and what fraction
Defect type	Does density change? Does the ratio of ions change?

Think of it like this: Stack oranges at a fruit stall. However you start, the neatest pile has each orange touching 12 others and fills 74% of the space; the gaps left over are of two sizes — small ones between four oranges and larger ones between six. Every close-packed metal and many ionic crystals are built exactly like that pile.

Crystalline and amorphous solids

Long-range order or only short-range order.

Property	Crystalline	Amorphous
Shape	Definite, regular	Irregular
Order	Long-range	Short-range only
Melting	Sharp m.p.	Softens over a range
Cleavage	Clean, flat faces	Irregular surfaces
Heat of fusion	Definite	Not definite
Physical properties	Anisotropic (differ with direction)	Isotropic
Examples	NaCl, quartz, diamond, ice	Glass, rubber, plastics, quartz glass

• Glass is called a supercooled liquid (pseudo-solid): very old window panes are slightly thicker at the bottom.
• Heating and slow cooling can make an amorphous solid crystalline (old glass turns milky).
• Quartz (crystalline) and quartz glass (amorphous) are both SiO₂ — same composition, different order.

Classification of crystalline solids

What holds the particles together decides the properties.

Type	Particles	Forces	Examples	Properties
Molecular — non-polar	Molecules	Dispersion forces	Ar, CCl₄, H₂, I₂, CO₂	Soft, very low m.p., insulators
Molecular — polar	Molecules	Dipole–dipole	HCl, SO₂	Soft, low m.p., insulators
Molecular — H-bonded	Molecules	Hydrogen bonds	H₂O (ice)	Hard, low m.p., insulators
Ionic	Ions	Electrostatic	NaCl, MgO, ZnS, CaF₂	Hard, brittle, high m.p.; conduct when molten or dissolved
Metallic	Cations in a sea of electrons	Metallic bonding	Fe, Cu, Ag, Mg	Malleable, ductile, conductors
Covalent network	Atoms	Covalent bonds	Diamond, SiO₂, SiC, AlN; graphite	Very hard, very high m.p.; insulators (graphite conducts)

Check yourself

• C1. Why is glass called a supercooled liquid?
• C2. Classify ice, SiC and brass by type of solid.

Answers at the end.

Crystal lattices and unit cells

Seven shapes, fourteen lattices.

Crystal system	Edges	Angles	Bravais lattices	Example
Cubic	a = b = c	α = β = γ = 90°	sc, bcc, fcc (3)	NaCl, Cu, ZnS
Tetragonal	a = b ≠ c	α = β = γ = 90°	simple, body-centred (2)	SnO₂, TiO₂
Orthorhombic	a ≠ b ≠ c	α = β = γ = 90°	simple, body, face, end (4)	KNO₃, BaSO₄, rhombic S
Hexagonal	a = b ≠ c	α = β = 90°, γ = 120°	simple (1)	Graphite, ZnO, Mg
Rhombohedral	a = b = c	α = β = γ ≠ 90°	simple (1)	Calcite, HgS
Monoclinic	a ≠ b ≠ c	α = γ = 90°, β ≠ 90°	simple, end-centred (2)	Monoclinic S, Na₂SO₄·10H₂O
Triclinic	a ≠ b ≠ c	α ≠ β ≠ γ ≠ 90°	simple (1)	K₂Cr₂O₇, CuSO₄·5H₂O, H₃BO₃

Key idea: 7 crystal systems, 14 Bravais lattices. The triclinic system is the least symmetric (all edges and angles different); cubic is the most symmetric.

Counting atoms in a unit cell

Shared atoms count only for the fraction inside.

Position	Shared by	Contribution
Corner	8 cells	⅛
Edge centre	4 cells	¼
Face centre	2 cells	½
Body centre	1 cell	1

Cubic cell	Atoms per cell (Z)	Relation of a and r	Coordination number	Packing efficiency
Simple (sc)	8 × ⅛ = 1	a = 2r	6	52.4%
Body-centred (bcc)	8 × ⅛ + 1 = 2	√3 a = 4r	8	68%
Face-centred (fcc / ccp)	8 × ⅛ + 6 × ½ = 4	√2 a = 4r	12	74%

Solved example 01 — Formula from positions

In a cubic cell, A atoms are at the corners and B atoms at the face centres. One corner atom is missing. Find the formula.

1. A: 7 corners × ⅛ = 7/8.
2. B: 6 faces × ½ = 3.
3. A : B = 7/8 : 3 = 7 : 24.

Answer: A₇B₂₄.

Solved example 02 — Packing efficiency of bcc

Show that the packing efficiency of a bcc lattice is 68%.

1. Z = 2; body diagonal √3a = 4r → a = 4r/√3.
2. Volume of atoms = 2 × (4/3)πr³ = 8.38r³.
3. Cell volume a³ = (4r/√3)³ = 12.32r³.
4. Efficiency = 8.38 / 12.32.

Answer: ≈ 0.68 → 68%.

Close packing and voids

How spheres stack, and the gaps they leave.

• In two dimensions: square packing (CN 4) or hexagonal packing (CN 6, more efficient).
• In three dimensions: placing a third layer over the holes of the second gives two patterns. ABAB… = hcp (Mg, Zn); ABCABC… = ccp = fcc (Cu, Ag, Au, Al). Both have CN 12 and 74% efficiency.
• Tetrahedral void: surrounded by 4 spheres. Octahedral void: surrounded by 6 spheres.
• For N spheres in close packing: N octahedral voids and 2N tetrahedral voids. In an fcc cell (4 atoms): 4 octahedral voids (body centre + 12 edge centres × ¼) and 8 tetrahedral voids (inside, near each corner).

Radius ratio r⁺/r⁻	Coordination number	Void / shape	Example
0.155 – 0.225	3	Triangular planar	B₂O₃
0.225 – 0.414	4	Tetrahedral	ZnS
0.414 – 0.732	6	Octahedral	NaCl
0.732 – 1.000	8	Cubic	CsCl

Solved example 03 — Formula from voids

Oxide ions form ccp; Al³⁺ ions occupy two-thirds of the octahedral voids. Find the formula.

1. N oxide ions → N octahedral voids.
2. Al³⁺ = ⅔N.
3. Al : O = ⅔ : 1 = 2 : 3.

Answer: Al₂O₃.

Solved example 04 — Two kinds of void

Atoms of Y form ccp; X occupies all tetrahedral voids and Z half the octahedral voids. Formula?

1. Per N atoms of Y: 2N tetrahedral, N octahedral.
2. X = 2N; Z = N/2.
3. Y : X : Z = 1 : 2 : ½ = 2 : 4 : 1.

Answer: X₄Y₂Z (X : Y : Z = 4 : 2 : 1).

Check yourself

• C3. How many tetrahedral voids are there in one fcc unit cell?
• C4. Radius ratio of a salt is 0.52. What coordination number do you expect?

Answers at the end.

Density of a unit cell

The single formula behind most numericals.

d = Z M / (a³ Nₐ)

Z = atoms (or formula units) per cell; M in g mol⁻¹; a in cm (1 pm = 10⁻¹⁰ cm); Nₐ = 6.022 × 10²³

Solved example 05 — Density of copper

Copper is fcc with a = 361 pm. Find its density. (Cu = 63.5 g mol⁻¹)

1. Z = 4; a = 3.61 × 10⁻⁸ cm → a³ = 4.70 × 10⁻²³ cm³.
2. d = 4 × 63.5 / (4.70 × 10⁻²³ × 6.022 × 10²³) = 254 / 28.3.

Answer: d ≈ 8.97 g cm⁻³ (measured 8.96).

Solved example 06 — Number of atoms in a sample

An element (bcc, a = 288 pm) has density 7.2 g cm⁻³. How many atoms are in 208 g of it?

1. a³ = (2.88 × 10⁻⁸)³ = 2.39 × 10⁻²³ cm³.
2. Mass of one cell = 7.2 × 2.39 × 10⁻²³ = 1.72 × 10⁻²² g, containing 2 atoms.
3. Atoms = (208 / 1.72 × 10⁻²²) × 2.

Answer: ≈ 2.42 × 10²⁴ atoms.

Solved example 07 — Finding the cell type

An element of molar mass 60 g mol⁻¹ has a cubic cell of edge 400 pm and density 6.23 g cm⁻³. Find Z and the cell type.

1. Z = d a³ Nₐ / M = 6.23 × (4.00 × 10⁻⁸)³ × 6.022 × 10²³ / 60.
2. = 6.23 × 6.40 × 10⁻²³ × 6.022 × 10²³ / 60 = 4.0.

Answer: Z = 4 → face-centred cubic.

Solved example 08 — Radius from edge length

Silver is fcc with a = 408.6 pm. Find the atomic radius.

1. √2 a = 4r → r = √2 × 408.6 / 4 = 1.414 × 408.6 / 4.

Answer: r ≈ 144.4 pm.

Ionic structures

Four structures cover most questions.

Structure	Anions	Cations	CN (cation : anion)	Formula units per cell	Examples
Rock salt (NaCl)	fcc (ccp)	All octahedral voids	6 : 6	4	NaCl, KCl, MgO, AgCl
Caesium chloride	Simple cubic	Cube centre	8 : 8	1	CsCl, CsBr, TlCl
Zinc blende (ZnS)	fcc (ccp)	Half the tetrahedral voids	4 : 4	4	ZnS, CuCl, AgI
Fluorite (CaF₂)	Ca²⁺ fcc; F⁻ in all tetrahedral voids	—	8 : 4	4	CaF₂, BaF₂, SrCl₂
Antifluorite (Na₂O)	O²⁻ fcc; Na⁺ in all tetrahedral voids	—	4 : 8	4	Na₂O, K₂O, Li₂O

Key idea: Pressure and temperature change the structure. High pressure raises the coordination number (NaCl, 6:6 → CsCl type, 8:8); high temperature lowers it (CsCl, 8:8 → NaCl type, 6:6 at about 760 K).

Solved example 09 — Edge length of NaCl

Na⁺ = 95 pm and Cl⁻ = 181 pm. Assuming the ions touch along the edge, find the edge length of the NaCl cell.

1. In rock salt, a = 2(r⁺ + r⁻).
2. a = 2 × (95 + 181).

Answer: a = 552 pm.

Imperfections in solids

Real crystals are never perfect.

Defect	What happens	Density	Found in
Vacancy	Lattice site empty	Decreases	Non-ionic solids, on heating
Interstitial	Extra particle in a void	Increases	Non-ionic solids
Schottky	Equal numbers of cations and anions missing	Decreases	NaCl, KCl, CsCl, AgBr (similar-size ions)
Frenkel	A smaller ion (usually the cation) moves to an interstitial site	Unchanged	ZnS, AgCl, AgBr, AgI (large size difference)

• AgBr shows both Schottky and Frenkel defects.
• Schottky defects in NaCl: about 10⁶ pairs per mole at room temperature.
• Impurity defects: adding SrCl₂ to NaCl — each Sr²⁺ replaces two Na⁺, leaving one cation vacancy. CdCl₂ in AgCl behaves the same way.
• Metal excess by anion vacancies (F-centres): heating NaCl in Na vapour creates Cl⁻ vacancies filled by electrons, which absorb light — NaCl turns yellow, LiCl pink, KCl violet.
• Metal excess by interstitial cations: ZnO turns yellow on heating as it loses oxygen: ZnO → Zn²⁺ + ½O₂ + 2e⁻; Zn²⁺ and electrons occupy interstitial sites.
• Metal deficiency: FeO is really about Fe₀.₉₅O — some Fe²⁺ are missing and others become Fe³⁺ to keep the crystal neutral.

Solved example 10 — Impurity vacancies

NaCl is doped with 10⁻³ mol % SrCl₂. Find the concentration of cation vacancies per mole of NaCl.

1. 10⁻³ mol % = 10⁻⁵ mol of SrCl₂ per mol of NaCl.
2. Each Sr²⁺ creates one cation vacancy.
3. Vacancies = 10⁻⁵ × 6.022 × 10²³.

Answer: 6.02 × 10¹⁸ vacancies per mole.

Check yourself

• C5. Which defect lowers the density of a crystal?
• C6. Why does ZnO turn yellow on heating?

Answers at the end.

Electrical properties

Conductors, insulators and semiconductors.

• Band theory: in metals the valence band is partly filled or overlaps the conduction band — conductors. In insulators a wide forbidden gap separates them. In semiconductors the gap is small, so conductivity increases with temperature.
• n-type: Si or Ge doped with a group 15 element (P, As). The fifth electron is free → negative carriers.
• p-type: Si or Ge doped with a group 13 element (B, Al, Ga). An electron-deficient “hole” → positive carriers.
• A p–n junction is a diode (rectifier); n–p–n and p–n–p arrangements make transistors.
• Compound semiconductors: 13–15 (InSb, AlP, GaAs) and 12–16 (ZnS, CdS, CdSe, HgTe).

Magnetic properties

Unpaired electrons and how they line up.

Type	Alignment of magnetic moments	Examples
Diamagnetic	No unpaired electrons; weakly repelled	H₂O, NaCl, C₆H₆
Paramagnetic	Unpaired, randomly oriented; weakly attracted	O₂, Cu²⁺, Fe³⁺, Cr³⁺
Ferromagnetic	Domains aligned in one direction; strongly attracted, permanent magnets	Fe, Co, Ni, Gd, CrO₂
Antiferromagnetic	Equal moments aligned opposite; net zero	MnO
Ferrimagnetic	Unequal moments aligned opposite; net moment	Fe₃O₄, ferrites MgFe₂O₄, ZnFe₂O₄

• Heating a ferromagnetic or ferrimagnetic solid above its Curie temperature makes it paramagnetic (Fe₃O₄ above 850 K).
• CrO₂ is used in audio and video cassette tapes.

Advanced problems

Each needs two ideas from the chapter.

Solved example 11 — Fe³⁺ in a non-stoichiometric oxide

A sample of ferrous oxide has the formula Fe₀.₉₃O. What percentage of the iron is present as Fe³⁺?

1. Let x = Fe³⁺ per formula unit; Fe²⁺ = 0.93 − x.
2. Charge balance with one O²⁻: 3x + 2(0.93 − x) = 2 → x = 0.14.
3. Fraction Fe³⁺ = 0.14 / 0.93.

Answer: About 15% of the iron is Fe³⁺.

Solved example 12 — Nearest-neighbour distance

Iron is bcc with a = 286.6 pm. Find the distance between nearest neighbours and the atomic radius.

1. In bcc the nearest atoms touch along the body diagonal: distance = √3a / 2.
2. = 1.732 × 286.6 / 2 = 248.2 pm.
3. r = half of that.

Answer: Nearest-neighbour distance ≈ 248.2 pm; r ≈ 124.1 pm.

Solved example 13 — Voids and formula with two cations

In a ccp lattice of oxide ions, A²⁺ occupies one-eighth of the tetrahedral voids and B³⁺ half of the octahedral voids. Find the formula.

1. N oxide ions → 2N tetrahedral and N octahedral voids.
2. A²⁺ = 2N/8 = N/4; B³⁺ = N/2.
3. A : B : O = ¼ : ½ : 1 = 1 : 2 : 4. Check charge: +2 + 2(+3) − 8 = 0.

Answer: AB₂O₄ — the spinel structure (e.g. MgAl₂O₄).

Solved example 14 — Edge length from density

KCl has the rock-salt structure and density 1.98 g cm⁻³. Find the edge length. (KCl = 74.5 g mol⁻¹)

1. Z = 4 formula units.
2. a³ = ZM / (d Nₐ) = 4 × 74.5 / (1.98 × 6.022 × 10²³) = 2.50 × 10⁻²² cm³.
3. a = (2.50 × 10⁻²²)^1/3 cm = 6.30 × 10⁻⁸ cm.

Answer: a ≈ 630 pm.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	Simple cubic has 8 atoms per cell.	1 atom (8 × ⅛).	Each corner atom is shared by 8 cells.
2	For fcc, a = 2r.	fcc: √2a = 4r (a = 2√2 r); a = 2r is simple cubic.	In fcc the atoms touch along the face diagonal.
3	hcp is more efficient than ccp.	Both are 74% with CN 12.	Only the stacking order differs.
4	There are N tetrahedral voids for N atoms.	2N tetrahedral, N octahedral.	A tetrahedral void is formed by 4 spheres in two layers.
5	Frenkel defects lower the density.	Frenkel: unchanged. Schottky: lower.	In Frenkel no ion leaves the crystal.
6	Schottky defects occur in AgCl.	AgCl shows Frenkel; AgBr shows both.	Large size difference favours Frenkel.
7	Forgetting to convert pm to cm in d = ZM/a³Nₐ.	1 pm = 10⁻¹⁰ cm; cube after converting.	Otherwise the answer is off by 10³⁰.
8	Silicon doped with boron is n-type.	p-type (holes).	Boron has only three valence electrons.
9	CsCl is body-centred cubic.	It is simple cubic with a two-ion basis (Cl⁻ at corners, Cs⁺ at centre).	A true bcc needs identical particles at corners and centre.
10	In CaF₂, F⁻ forms the fcc lattice.	Ca²⁺ forms the fcc lattice; F⁻ fill all tetrahedral voids.	Antifluorite (Na₂O) is the reverse.
11	Conductivity of semiconductors falls with temperature.	It rises; that of metals falls.	More electrons cross the small band gap.
12	Fe₃O₄ is ferromagnetic.	Fe₃O₄ is ferrimagnetic.	Unequal opposing moments.

Memory hook: “8 corners × ⅛ = 1; add 1 for bcc; add 3 for fcc.” · “sc 52, bcc 68, fcc 74.” · “N octahedral, 2N tetrahedral.” · “Schottky shrinks density; Frenkel keeps it.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Atoms per unit cell in fcc: (Easy)

• (a) 1
• (b) 2
• (c) 4
• (d) 6

Q02. Coordination number in ccp: (Easy)

• (a) 6
• (b) 8
• (c) 10
• (d) 12

Q03. Packing efficiency of simple cubic: (Easy)

• (a) 52.4%
• (b) 68%
• (c) 74%
• (d) 80%

Q04. An example of an amorphous solid: (Easy)

• (a) quartz
• (b) NaCl
• (c) glass
• (d) diamond

Q05. Number of Bravais lattices: (Easy)

• (a) 7
• (b) 12
• (c) 14
• (d) 32

Q06. Frenkel defect is shown by: (Easy)

• (a) NaCl
• (b) KCl
• (c) AgCl
• (d) CsCl

Q07. Schottky defect causes density to: (Easy)

• (a) increase
• (b) decrease
• (c) remain same
• (d) double

Q08. In bcc, the relation between a and r is: (Moderate)

• (a) a = 2r
• (b) √2a = 4r
• (c) √3a = 4r
• (d) a = 4r

Q09. Number of tetrahedral voids in a ccp arrangement of N atoms: (Moderate)

• (a) N
• (b) 2N
• (c) N/2
• (d) 4N

Q10. Coordination number of Na⁺ in NaCl: (Moderate)

• (a) 4
• (b) 6
• (c) 8
• (d) 12

Q11. Coordination numbers in CaF₂ (Ca²⁺ : F⁻): (Moderate)

• (a) 4 : 8
• (b) 8 : 4
• (c) 6 : 6
• (d) 8 : 8

Q12. Si doped with As gives: (Moderate)

• (a) p-type
• (b) n-type
• (c) intrinsic
• (d) insulator

Q13. MnO is: (Moderate)

• (a) ferromagnetic
• (b) paramagnetic
• (c) antiferromagnetic
• (d) ferrimagnetic

Q14. Crystal system with a ≠ b ≠ c and α = β = γ = 90°: (Moderate)

• (a) tetragonal
• (b) orthorhombic
• (c) monoclinic
• (d) triclinic

Q15. NaCl heated in Na vapour turns yellow because of: (Moderate)

• (a) Frenkel defects
• (b) F-centres
• (c) interstitial Na⁺
• (d) Schottky defects only

Q16. A radius ratio of 0.30 suggests a coordination number of: (Moderate)

• (a) 3
• (b) 4
• (c) 6
• (d) 8

Q17. Oxide ions in ccp with M in ⅔ of octahedral voids gives: (Moderate)

• (a) MO
• (b) M₂O₃
• (c) M₃O₂
• (d) MO₂

Q18. Fe₀.₉₅O contains: (Moderate)

• (a) only Fe²⁺
• (b) Fe²⁺ and Fe³⁺
• (c) Fe²⁺ and Fe
• (d) only Fe³⁺

Q19. Copper (fcc, a = 361 pm) has a density of about: (Tough)

• (a) 4.5 g cm⁻³
• (b) 8.97 g cm⁻³
• (c) 17.9 g cm⁻³
• (d) 2.24 g cm⁻³

Q20. Under high pressure NaCl changes to a structure with CN: (Tough)

• (a) 4
• (b) 6
• (c) 8
• (d) 12

Q21. A⁺ ions at the corners and B⁻ at the body centre of a cube. Formula and CN of A⁺: (Tough)

• (a) AB, 8
• (b) AB₂, 6
• (c) A₂B, 4
• (d) AB, 6

Q22. An element (M = 60, a = 400 pm, d = 6.23 g cm⁻³) has Z: (Tough)

• (a) 1
• (b) 2
• (c) 4
• (d) 6

Q23. Packing efficiency of bcc: (Tough)

• (a) 52.4%
• (b) 68%
• (c) 74%
• (d) 90%

Q24. Doping NaCl with 10⁻³ mol % SrCl₂ gives cation vacancies per mole: (Tough)

• (a) 6.02 × 10¹⁸
• (b) 6.02 × 10²⁰
• (c) 6.02 × 10¹⁶
• (d) 6.02 × 10²³

Q25. Fe₃O₄ above 850 K becomes: (Tough)

• (a) ferromagnetic
• (b) paramagnetic
• (c) diamagnetic
• (d) antiferromagnetic

Answer key with explanations

1. (c) 4 — 8 × ⅛ + 6 × ½.
2. (d) 12 — Close packing.
3. (a) 52.4% — Least efficient.
4. (c) glass — Short-range order only.
5. (c) 14 — Fourteen.
6. (c) AgCl — Large size difference.
7. (b) decrease — Ions leave the crystal.
8. (c) √3a = 4r — Atoms touch along the body diagonal.
9. (b) 2N — Two per atom.
10. (b) 6 — Octahedral voids.
11. (b) 8 : 4 — Fluorite.
12. (b) n-type — Extra electron from group 15.
13. (c) antiferromagnetic — Opposing equal moments.
14. (b) orthorhombic — Three unequal edges, right angles.
15. (b) F-centres — Electrons in anion vacancies.
16. (b) 4 — 0.225 – 0.414.
17. (b) M₂O₃ — ⅔ : 1.
18. (b) Fe²⁺ and Fe³⁺ — Charge balance for missing Fe²⁺.
19. (b) 8.97 g cm⁻³ — d = ZM/a³Nₐ with Z = 4.
20. (c) 8 — CsCl type.
21. (a) AB, 8 — CsCl type.
22. (c) 4 — Z = da³Nₐ/M.
23. (b) 68% — Z = 2, √3a = 4r.
24. (a) 6.02 × 10¹⁸ — 10⁻⁵ mol × Nₐ.
25. (b) paramagnetic — Above its Curie temperature.

Check-yourself answers

• C1. It has short-range order like a liquid and flows extremely slowly; it does not melt sharply.
• C2. Ice: molecular (H-bonded); SiC: covalent network; brass: metallic.
• C3. Eight.
• C4. 6 (0.414 – 0.732, octahedral).
• C5. Schottky (and vacancy) defects.
• C6. It loses oxygen: Zn²⁺ ions and electrons occupy interstitial sites; the electrons absorb blue light.

One-page revision

Read this page the night before the exam.

Types of solid

• Crystalline: sharp m.p., anisotropic
• Amorphous: glass, rubber; isotropic
• Molecular, ionic, metallic, covalent
• 7 systems, 14 Bravais lattices

Unit cells

• sc 1, bcc 2, fcc 4 atoms
• sc a = 2r; bcc √3a = 4r; fcc √2a = 4r
• CN 6, 8, 12
• Efficiency 52.4, 68, 74%

Packing and voids

• hcp ABAB; ccp ABCABC; both 74%
• N octahedral, 2N tetrahedral
• Radius ratio 0.225 / 0.414 / 0.732
• fcc cell: 4 octa, 8 tetra voids

Density and structures

• d = ZM / (a³Nₐ); pm → cm
• NaCl 6:6, 4 f.u.; CsCl 8:8, 1 f.u.
• ZnS 4:4; CaF₂ 8:4; Na₂O 4:8
• Pressure ↑ CN; temperature ↓ CN

Defects

• Schottky: density ↓ (NaCl, KCl)
• Frenkel: density same (AgCl, ZnS)
• AgBr both; SrCl₂ in NaCl → vacancies
• F-centres; ZnO yellow; Fe₀.₉₅O

Electrical and magnetic

• n-type: group 15 dopant; p-type: group 13
• Semiconductor σ rises with T
• Ferro Fe Co Ni CrO₂; antiferro MnO
• Ferri Fe₃O₄; Curie T → paramagnetic