Part 1 · Chapter 13

The Solid State

Order made visible — how atoms stack into lattices, how we count and weigh a unit cell, and how the tiny flaws inside crystals decide colour, conduction and magnetism

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • The difference between crystalline and amorphous solids, and why glass is a "pseudo-solid".
  • The four classes of crystalline solids — molecular, ionic, covalent and metallic.
  • Lattices, unit cells, the seven crystal systems and the 14 Bravais lattices.
  • How to count atoms in simple, body-centred and face-centred cubic cells.
  • Close packing (hcp/ccp), tetrahedral/octahedral voids, and packing efficiency.
  • The density formula, the radius ratio rule, and point defects: Schottky and Frenkel.
Section 13-1

Crystalline vs Amorphous Solids

A solid has a fixed shape and volume because its particles are locked in place, vibrating about fixed points. But not all solids are built the same. A crystalline solid has long-range order: its particles repeat in a regular, three-dimensional pattern that extends across the whole crystal. An amorphous solid (from Greek a-morphē, "no form") has only short-range order — orderly in small patches but disordered overall, like a frozen snapshot of a liquid.

PropertyCrystallineAmorphous
Orderlong-range, regularshort-range only
Melting pointsharp, definitesoftens over a range
Shape on cleavingclean flat facesirregular fracture
Heat of fusiondefinitenot definite
Anisotropyanisotropicisotropic
Naturetrue solidpseudo-solid / supercooled liquid
Examples\(\ce{NaCl}\), diamond, quartzglass, rubber, plastic
Anisotropy is the giveaway. Because a crystal's arrangement differs along different directions, properties like refractive index or conductivity depend on the direction of measurement — that is anisotropy. Amorphous solids, disordered in every direction, look the same whichever way you probe them (isotropic). Quartz is crystalline \(\ce{SiO2}\); quartz glass is the same compound made amorphous by rapid cooling.
Section 13-2

Classes of Crystalline Solids

Crystalline solids are sorted by the kind of particle at the lattice points and the force holding them together. Four classes cover almost everything, and the binding force predicts every bulk property — hardness, melting point and conductivity.

TypeParticlesForcePropertiesExample
Molecularmoleculesvan der Waals / H-bondsoft, low mp, insulatorice, dry ice \(\ce{CO2}\)
Ionicionselectrostatichard, brittle, high mp\(\ce{NaCl}\), \(\ce{ZnS}\)
Covalentatomscovalent networkvery hard, very high mpdiamond, \(\ce{SiC}\), quartz
Metalliccations + e⁻ seametallic bondmalleable, conducting\(\ce{Cu}\), \(\ce{Fe}\), \(\ce{Ag}\)
Conductivity tells the tale. Ionic solids do not conduct as solids (ions are fixed) but conduct when molten or dissolved (ions are free). Metals conduct in the solid state through mobile electrons. Covalent network solids are insulators — except graphite, whose delocalised electrons make it conduct and whose sliding layers make it slippery.
Section 13-3

Lattice, Unit Cell & the Seven Crystal Systems

A crystal lattice is the regular three-dimensional array of points that represents the repeating pattern of a crystal. The smallest repeating block that, stacked in three dimensions, reproduces the whole lattice is the unit cell. A unit cell is defined by three edge lengths \(a,\,b,\,c\) and three angles \(\alpha,\,\beta,\,\gamma\) between them.

a b c
A unit cell — defined by edges a, b, c and angles α, β, γ

By choosing different combinations of edges and angles, only seven distinct shapes are possible — the seven crystal systems. Adding face-, body- and end-centring gives the 14 Bravais lattices.

SystemEdgesAnglesExample
Cubic\(a=b=c\)\(\alpha=\beta=\gamma=90^\circ\)\(\ce{NaCl}\), \(\ce{Cu}\)
Tetragonal\(a=b\ne c\)\(\alpha=\beta=\gamma=90^\circ\)white \(\ce{Sn}\), \(\ce{TiO2}\)
Orthorhombic\(a\ne b\ne c\)\(\alpha=\beta=\gamma=90^\circ\)rhombic \(\ce{S}\), \(\ce{BaSO4}\)
Hexagonal\(a=b\ne c\)\(\alpha=\beta=90^\circ,\ \gamma=120^\circ\)graphite, \(\ce{ZnO}\)
Rhombohedral\(a=b=c\)\(\alpha=\beta=\gamma\ne90^\circ\)calcite, \(\ce{HgS}\)
Monoclinic\(a\ne b\ne c\)\(\alpha=\gamma=90^\circ\ne\beta\)monoclinic \(\ce{S}\)
Triclinic\(a\ne b\ne c\)\(\alpha\ne\beta\ne\gamma\ne90^\circ\)\(\ce{CuSO4.5H2O}\)
Section 13-4

Cubic Unit Cells & Counting Atoms

The three cubic cells dominate the syllabus. To count how many atoms a cell "owns", remember that atoms are shared with neighbouring cells. The share depends on position:

PositionShared betweenContribution
Corner8 cells\(1/8\)
Edge4 cells\(1/4\)
Face centre2 cells\(1/2\)
Body centre1 cell\(1\)
SC · Z=1 BCC · Z=2 FCC · Z=4
Simple cubic, body-centred and face-centred cells — Z = 1, 2, 4
🔢
Atoms per cubic cell (Z)
SC: \(8\times\tfrac18=1\) · BCC: \(8\times\tfrac18+1=2\) · FCC: \(8\times\tfrac18+6\times\tfrac12=4\)

The corners always supply one whole atom; the body centre adds one more for BCC, and the six face centres add three more for FCC.

CellAtoms (Z)Coordination no.\(r\)–\(a\) relation
Simple cubic16\(a=2r\)
Body-centred28\(\sqrt3\,a=4r\)
Face-centred (ccp)412\(\sqrt2\,a=4r\)
Section 13-5

Close Packing & the Voids Between

To pack spheres as tightly as possible, lay down a hexagonal first layer (A), nest the second layer (B) into its hollows, then choose where the third goes. Stacking ABAB… gives hexagonal close packing (hcp); stacking ABCABC… gives cubic close packing (ccp), which is identical to the face-centred cubic lattice. Both reach a coordination number of 12 and the same packing efficiency.

Wherever spheres meet, empty pockets — voids — remain. A pocket touched by four spheres is a tetrahedral void; one touched by six spheres is an octahedral void. For \(N\) close-packed spheres there are exactly \(N\) octahedral voids and \(2N\) tetrahedral voids. Which voids the smaller ions occupy fixes the structure of an ionic compound.

VoidSurrounded byNumber (for N spheres)Limiting radius ratio
Tetrahedral4 spheres\(2N\)\(0.225\)
Octahedral6 spheres\(N\)\(0.414\)
Reading a structure. In rock salt \(\ce{NaCl}\), \(\ce{Cl-}\) ions form a ccp lattice and the smaller \(\ce{Na+}\) ions fill all the octahedral voids — coordination 6:6. In zinc blende \(\ce{ZnS}\), \(\ce{Zn^2+}\) fills half the tetrahedral voids — coordination 4:4. The void a cation chooses follows from its size.
Section 13-6

Packing Efficiency

Packing efficiency is the fraction of a unit cell's volume actually filled by spheres. It rises as the coordination number rises — close packing wastes the least space.

📦
Packing efficiency
\(\text{P.E.}=\dfrac{Z\cdot\frac{4}{3}\pi r^{3}}{a^{3}}\times100\%\)

Substitute the \(r\)–\(a\) relation for each cell. SC gives \(52.4\%\), BCC gives \(68\%\), and ccp/hcp both give \(74\%\) — the densest possible packing of equal spheres.

Structure\(Z\)Packing efficiencyVoid space
Simple cubic152.4 %47.6 %
Body-centred cubic268 %32 %
ccp / fcc & hcp4 (fcc)74 %26 %
Section 13-7

Density of a Unit Cell

Because a unit cell's mass and volume are both known once we know the structure, a crystal's density links directly to atomic mass and edge length. This single formula lets you find Avogadro's number, the type of cell, or an unknown atomic mass from X-ray data.

⚖️
Density formula
\(\rho=\dfrac{Z\,M}{a^{3}\,N_{A}}\)

\(Z\) = atoms per cell, \(M\) = molar mass (g mol⁻¹), \(a\) = edge length (cm), \(N_A=6.022\times10^{23}\). Keep units consistent: if \(a\) is in cm, \(\rho\) comes out in g cm⁻³.

Unit watch. Edge lengths are usually quoted in picometres or ångströms. Convert before substituting: \(1\ \text{pm}=10^{-10}\ \text{cm}\) and \(1\ \text{Å}=10^{-8}\ \text{cm}\). A single missed power of ten throws the density off by orders of magnitude — the most common slip in this chapter.
Section 13-8

Radius Ratio & Ionic Solids

For an ionic crystal, the ratio of the cation radius to the anion radius decides which void the cation fits into, and therefore its coordination number. A cation must be just large enough to touch its surrounding anions without rattling — that lower limit is the radius ratio rule.

\(r_+/r_-\)CoordinationGeometryExample
0.155 – 0.2253trigonal planar\(\ce{B2O3}\)
0.225 – 0.4144tetrahedral\(\ce{ZnS}\)
0.414 – 0.7326octahedral\(\ce{NaCl}\)
0.732 – 1.0008cubic\(\ce{CsCl}\)
Why \(\ce{CsCl}\) breaks the pattern. The large \(\ce{Cs+}\) ion gives a radius ratio above \(0.732\), so it takes 8 anions around it in a body-centred-type arrangement (8:8), unlike the 6:6 octahedral packing of the smaller \(\ce{Na+}\) in rock salt.
Section 13-9

Imperfections & Point Defects

No real crystal is perfect. Above absolute zero, thermodynamics demands some disorder. Point defects are irregularities around a single lattice point. Stoichiometric defects keep the formula intact; non-stoichiometric defects change the ratio of ions.

Schottky Frenkel
Schottky — paired vacancies · Frenkel — an ion shifted to an interstitial site
DefectWhat happensEffect on densityFavoured whenExample
Schottkyequal cation + anion vacanciesdecreasessimilar ion sizes, high coord.\(\ce{NaCl}\), \(\ce{KCl}\), \(\ce{CsCl}\)
Frenkelion shifts to interstitial siteunchangedlarge size difference\(\ce{AgCl}\), \(\ce{ZnS}\), \(\ce{AgBr}\)
Metal-excess colour. Heat \(\ce{NaCl}\) in sodium vapour: extra \(\ce{Na+}\) takes a lattice site and the freed electron sits in an anion vacancy — an F-centre (from German Farbe, colour). These trapped electrons absorb visible light, turning the crystal yellow. The same mechanism gives excess-potassium \(\ce{KCl}\) a violet tint.
Section 13-10

Electrical & Magnetic Properties

The same lattice that fixes a solid's shape also fixes how it carries current and responds to a magnet. Band theory sorts solids by the gap between the filled valence band and the empty conduction band: no gap → conductor, small gap → semiconductor, large gap → insulator.

A pure intrinsic semiconductor like silicon conducts feebly. Doping transforms it: adding a group-15 element (P, As) donates extra electrons to give an n-type semiconductor; adding a group-13 element (B, Al) creates electron "holes" to give a p-type semiconductor. Joining the two builds the p–n junction at the heart of every diode and transistor.

Magnetic classBehaviourCauseExample
Diamagneticweakly repelledall electrons paired\(\ce{NaCl}\), \(\ce{H2O}\)
Paramagneticweakly attractedunpaired electrons\(\ce{O2}\), \(\ce{Cu^2+}\)
Ferromagneticstrongly attracted, permanentaligned domains\(\ce{Fe}\), \(\ce{Co}\), \(\ce{Ni}\)
Antiferromagneticno net momentoppositely aligned spins cancel\(\ce{MnO}\)
Ferrimagneticnet moment, but weakerunequal opposite spins\(\ce{Fe3O4}\)
Worked Examples

Putting It to Work

1 Atoms in an FCC cell

Problem. How many atoms belong to a face-centred cubic unit cell?

Solution. Add the corner and face-centre contributions:

Working
\[ Z = 8\times\tfrac18 + 6\times\tfrac12 = 1 + 3 = 4 \]
2 Edge length from atomic radius (BCC)

Problem. An element crystallises in a BCC lattice with atomic radius \(r=125\ \text{pm}\). Find the edge length \(a\).

Solution. For BCC the body diagonal carries four radii, \(\sqrt3\,a=4r\):

Working
\[ a=\frac{4r}{\sqrt3}=\frac{4\times125}{1.732}\approx 288.7\ \text{pm} \]
3 Density of an FCC metal

Problem. Copper (\(M=63.5\ \text{g mol}^{-1}\)) is FCC with \(a=361\ \text{pm}\). Find its density.

Solution. Use \(\rho=ZM/(a^3N_A)\) with \(Z=4\) and \(a=3.61\times10^{-8}\ \text{cm}\):

Working
\[ \rho=\frac{4\times63.5}{(3.61\times10^{-8})^{3}\times6.022\times10^{23}}\approx 8.97\ \text{g cm}^{-3} \]
4 Packing efficiency of BCC

Problem. Show that body-centred cubic packing fills \(68\%\) of space.

Solution. With \(Z=2\) and \(r=\sqrt3\,a/4\), substitute into the P.E. formula:

Working
\[ \text{P.E.}=\frac{2\cdot\frac43\pi\left(\frac{\sqrt3\,a}{4}\right)^{3}}{a^{3}}=\frac{\sqrt3\,\pi}{8}\approx 0.68 = 68\% \]
5 Counting voids

Problem. A ccp lattice contains \(0.5\ \text{mol}\) of spheres. How many octahedral and tetrahedral voids are present?

Solution. Octahedral voids \(=N\), tetrahedral voids \(=2N\):

Working
\[ N=0.5\,N_A;\quad \text{oct}=0.5N_A,\ \ \text{tet}=1.0N_A=6.022\times10^{23} \]
6 Identify the defect

Problem. In \(\ce{AgCl}\) a small \(\ce{Ag+}\) ion leaves its site and lodges in an interstitial position while no anion moves. Name the defect and state its effect on density.

Solution. An ion displaced to an interstitial — no mass leaves the crystal:

Working
\[ \Rightarrow\ \textbf{Frenkel defect};\quad \text{density unchanged} \]
Review

Chapter Summary

Two solids

Crystalline = long-range order, sharp mp, anisotropic; amorphous = pseudo-solid, isotropic.

Four classes

Molecular, ionic, covalent, metallic — the binding force sets every bulk property.

Counting Z

Corner \(\tfrac18\), edge \(\tfrac14\), face \(\tfrac12\), body \(1\): SC=1, BCC=2, FCC=4.

Packing

SC 52.4 %, BCC 68 %, ccp/hcp 74 %; voids: \(N\) octahedral, \(2N\) tetrahedral.

Density

\(\rho=ZM/(a^3N_A)\) — links structure to mass and edge length.

Defects

Schottky lowers density; Frenkel leaves it unchanged; F-centres colour crystals.

Practice

Problems

For each item, first decide which idea it tests — structure, packing, density, or defects — then apply the relevant rule. Difficulty rises down the list.

  1. State three differences between crystalline and amorphous solids.
  2. Why are amorphous solids described as isotropic while crystals are anisotropic?
  3. Classify as molecular, ionic, covalent or metallic: dry ice, diamond, \(\ce{MgO}\), brass.
  4. Calculate the number of atoms in a body-centred cubic unit cell, showing each contribution.
  5. An element is FCC with edge length \(400\ \text{pm}\). Find the atomic radius.
  6. Distinguish hexagonal close packing from cubic close packing in terms of stacking and coordination number.
  7. How many octahedral and tetrahedral voids accompany \(1\ \text{mol}\) of close-packed atoms?
  8. Derive the packing efficiency of a simple cubic lattice.
  9. A metal (\(M=56\ \text{g mol}^{-1}\)) is BCC with \(a=287\ \text{pm}\). Calculate its density.
  10. Using the radius ratio rule, predict the coordination number when \(r_+/r_-=0.52\).
  11. Explain why a Schottky defect lowers density but a Frenkel defect does not.
  12. What is an F-centre, and why does it give a crystal colour?
Tip: almost every numerical in this chapter reduces to two relations — the \(r\)–\(a\) link for the cell type and the density formula \(\rho=ZM/(a^3N_A)\). Identify the lattice first (SC, BCC or FCC), write down \(Z\) and the \(r\)–\(a\) relation, and the rest is arithmetic.