---
title: "Solid-State Chemistry: Bands, Defects and Semiconductors"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 22
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/22-solid-state-chemistry-bands-defects-and-semiconductors
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 22 — Solid-State Chemistry: Bands, Defects and Semiconductors

A solar panel on a roof and the light-emitting diode in a lamp are made of the same kind of crystal, silicon or a compound of gallium with nitrogen, arsenic or phosphorus, into which a few impurity atoms per million have been put on purpose. Their working is set by two things a molecule does not have: bands of energy levels, with a gap whose width fixes which light a crystal absorbs or emits, and defects, which carry the charges. This chapter builds the bands from molecular orbitals, counts the carriers of a [semiconductor](#def-b3-solid-state-classes), and treats missing and misplaced atoms as chemical species with their own equilibria, down to the [solid electrolyte](#def-b3-solid-state-ionic-conductor) of the oxygen sensor in a car’s exhaust.

**You already know.**

The Year 1 volume defined crystals, the metallic bond, ionic and covalent crystals, interstitial sites and alloys, and the Nernst equation; the Year 2 volume the Hückel method and the resonance integral $\beta$. [Chapter 9](https://one-course.com/books/chemistry/4/en/chapter/9-crystallography-and-x-ray-diffraction#ch-b3-x-ray-diffraction) gave the [reciprocal lattice](https://one-course.com/books/chemistry/4/en/chapter/9-crystallography-and-x-ray-diffraction#def-b3-x-ray-diffraction-reciprocal) and [Chapter 10](https://one-course.com/books/chemistry/4/en/chapter/10-statistical-thermodynamics-partition-functions#ch-b3-partition-functions) the [Boltzmann distribution](https://one-course.com/books/chemistry/4/en/chapter/10-statistical-thermodynamics-partition-functions#def-b3-partition-functions-boltzmann) and the [statistical entropy](https://one-course.com/books/chemistry/4/en/chapter/10-statistical-thermodynamics-partition-functions#def-b3-partition-functions-statistical-entropy) $S = k\ln W$.

![Solar panels on a tiled roof: each cell is a thin slice of silicon in which light lifts electrons across a band gap of about one electronvolt.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/img-27444a0237f0.jpg)

*Solar panels on a tiled roof: each cell is a thin slice of silicon in which light lifts electrons across a [band gap](#def-b3-solid-state-band) of about one electronvolt.*

## 22.1 From orbitals to bands

Take a chain of $N$ identical atoms, each with one s orbital, in the Hückel approximation: Coulomb integral $\alpha$ on each atom, resonance integral $\beta < 0$ between neighbours, zero otherwise.

**Theorem 22.1 (Levels of a Hückel chain).**

The $N$ molecular orbitals of a linear chain of $N$ atoms have energies

$$
E_k = \alpha + 2\beta\cos\frac{k\pi}{N + 1}, \qquad k = 1, \dots, N,
$$

with coefficients $c_j^{(k)} \propto \sin\bigl(jk\pi/(N + 1)\bigr)$ on atom $j$. As $N \to \infty$ the levels fill the interval from $\alpha + 2\beta$ to $\alpha - 2\beta$ densely: a band of width $4|\beta|$.

**Proof.** The secular equations are $\beta c_{j-1} + (\alpha - E)c_j + \beta c_{j+1} = 0$ for $j = 1, \dots, N$, with $c_0 = c_{N+1} = 0$ (no atom there). Try $c_j = \sin(j\theta)$: since $\sin((j - 1)\theta) + \sin((j + 1)\theta) = 2\cos\theta\sin(j\theta)$, each equation becomes $(\alpha - E + 2\beta\cos\theta)\sin(j\theta) = 0$, so $E = \alpha + 2\beta\cos\theta$. The condition $c_0 = 0$ holds for every $\theta$, and $c_{N+1} = \sin((N + 1)\theta) = 0$ requires $\theta = k\pi/(N + 1)$; $k = 1, \dots, N$ gives $N$ distinct levels (the others repeat them or vanish). The lowest and highest are $\alpha \pm 2\beta\cos(\pi/(N + 1))$, which tend to $\alpha \pm 2\beta$; the spacing between neighbours is at most $2|\beta|\pi/(N + 1)$, which tends to zero. ∎

For $N = 2$ the theorem gives $\alpha \pm \beta$, the orbitals of ethene’s $\pi$ system; for $N = 6$ the open-chain hexatriene. In three dimensions the same happens with every kind of atomic orbital: s orbitals give an s band, p orbitals a p band, and each band holds $2N$ electrons for $N$ atoms.

**Definition 22.2 (Bands).**

An *energy band* of a crystal is a continuous range of allowed one-electron energies, formed from the orbitals of all its atoms. A *band gap* $E_g$ is a range of energies with no levels between two bands. In a [semiconductor](#def-b3-solid-state-classes) or an [insulator](#def-b3-solid-state-classes) at zero temperature, the highest filled band is the *valence band* and the lowest empty one the *conduction band*.

**Definition 22.3 (Density of states and Fermi level).**

The *density of states* $g(E)$ is the number of levels per unit energy (and per atom or per unit volume) near the energy $E$. The *Fermi level* $E_F$ is the energy at which a level has probability $1/2$ of being occupied; at zero temperature all levels below it are filled and all above it empty.

![Hückel levels of chains of 2 to 64 atoms (one short line per level; with < 0 the bonding levels are at the bottom). The levels crowd into a band between + 2 and - 2; on the right, the density of states of the infinite chain, largest at the band edges.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-51635f1c1b27.svg)

*Hückel levels of chains of 2 to 64 atoms (one short line per level; with $\beta < 0$ the bonding levels are at the bottom). The levels crowd into a band between $\alpha + 2\beta$ and $\alpha - 2\beta$; on the right, the [density of states](#def-b3-solid-state-fermi) of the infinite chain, largest at the band edges.*

**Proposition 22.4 (Band filling).**

A crystal whose highest occupied band is partly filled conducts electricity like a metal; a crystal whose bands are either full or empty, with a gap between them, does not conduct at zero temperature.

**Argued.** In an electric field, electrons gain a little energy and momentum: they must move into empty levels just above the ones they occupy. In a partly filled band such levels lie immediately above the [Fermi level](#def-b3-solid-state-fermi), at no energy cost. In a full band every level is taken and the next empty one is across the gap; moreover a full band carries no net current, since for each electron moving one way another moves the opposite way. ∎

Sodium, with one 3s electron per atom, half-fills its s band: a metal. Magnesium, with two, would fill its s band exactly, but the s and p bands overlap, so it is a metal too. Diamond and silicon have four valence electrons per atom in orbitals that split into a filled bonding band and an empty antibonding band, separated by a gap.

**Definition 22.5 (Semiconductors and insulators).**

A *semiconductor* is a solid with a [band gap](#def-b3-solid-state-band) small enough (up to about $3\,\mathrm{eV}$) that a measurable number of electrons cross it at ordinary temperature or after doping; an *insulator* has a gap so large that it does not conduct.

![Band filling at zero temperature (filled levels shaded). A metal has a partly filled band and its Fermi level inside it; a semiconductor and an insulator have a full valence band and an empty conduction band, with the Fermi level in the gap, which is narrow in the first and wide in the second.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-84ba20562143.svg)

*Band filling at zero temperature (filled levels shaded). A metal has a partly filled band and its [Fermi level](#def-b3-solid-state-fermi) inside it; a [semiconductor](#def-b3-solid-state-classes) and an [insulator](#def-b3-solid-state-classes) have a full [valence band](#def-b3-solid-state-band) and an empty [conduction band](#def-b3-solid-state-band), with the [Fermi level](#def-b3-solid-state-fermi) in the gap, which is narrow in the first and wide in the second.*

## 22.2 Semiconductors

**Definition 22.6 (Carriers).**

An *intrinsic semiconductor* is a pure [semiconductor](#def-b3-solid-state-classes), whose carriers all come from electrons excited across the gap. A *charge carrier* is a mobile particle that carries current: an electron in the [conduction band](#def-b3-solid-state-band), or a *hole*, an empty level in the [valence band](#def-b3-solid-state-band), which moves like a positive charge.

**Theorem 22.7 (Intrinsic carrier concentration).**

With $N_c$ and $N_v$ the effective densities of states of the conduction and [valence bands](#def-b3-solid-state-band) (admitted), the electron and [hole](#def-b3-solid-state-carriers) concentrations of a [semiconductor](#def-b3-solid-state-classes) whose [Fermi level](#def-b3-solid-state-fermi) is more than a few $kT$ from either band edge are

$$
n = N_c\,\eu^{-(E_c - E_F)/kT}, \qquad p = N_v\,\eu^{-(E_F - E_v)/kT},
$$

and in an [intrinsic semiconductor](#def-b3-solid-state-carriers)

$$
n = p = n_i = \sqrt{N_cN_v}\,\eu^{-E_g/2kT}.
$$

**Proof.** The occupancy of a level of energy $E$ is $f(E) = 1/(1 + \eu^{(E - E_F)/kT})$. When $E - E_F \gg kT$, $f \approx \eu^{-(E - E_F)/kT}$, a Boltzmann tail. Summing it over the levels of the [conduction band](#def-b3-solid-state-band), $n = \int g_c(E)\eu^{-(E - E_F)/kT}\,\dd E = \eu^{-(E_c - E_F)/kT}\int g_c(E)\eu^{-(E - E_c)/kT}\,\dd E$, and the last integral is by definition $N_c$. [Holes](#def-b3-solid-state-carriers) are empty levels, with probability $1 - f \approx \eu^{-(E_F - E)/kT}$ in the [valence band](#def-b3-solid-state-band); the same steps give $p$. Then $np = N_cN_v\eu^{-(E_c - E_v)/kT} = N_cN_v\eu^{-E_g/kT}$, and in a pure crystal each excited electron leaves one [hole](#def-b3-solid-state-carriers): $n = p$, so $n = \sqrt{np}$. ∎

**Proposition 22.8 (Mass-action law).**

In any [semiconductor](#def-b3-solid-state-classes) at equilibrium, doped or not, $np = n_i^2$.

**Proof.** The product $np = N_cN_v\eu^{-E_g/kT}$ found in the proof of [Theorem 22.7](#thm-b3-solid-state-intrinsic) does not contain $E_F$: it is the same whatever sets the [Fermi level](#def-b3-solid-state-fermi), and equal to its intrinsic value $n_i^2$. It is the equilibrium constant of $\text{nothing} \rightleftharpoons \mathrm e^- + \mathrm h^+$, like the ionic product of water. ∎

![Left: intrinsic carrier concentrations of silicon and germanium against 1/T (model with the measured gaps and effective densities of states); the slopes are close to -E_g/2k. Right: electrons in silicon doped with 1 × 1015\, cm-3 phosphorus: at low temperature the donors hold their electrons (freeze-out), from about 150 to 500\, K all are ionised (n = N_D), and at high temperature the intrinsic carriers (dashed, n_i) take over.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-f75874d6f50e.svg)

*Left: intrinsic carrier concentrations of silicon and germanium against $1/T$ (model with the measured gaps and effective densities of states); the slopes are close to $-E_g/2k$. Right: electrons in silicon doped with $1 \times 10^{15}\,\mathrm{cm}^{-3}$ phosphorus: at low temperature the donors hold their electrons (freeze-out), from about 150 to $500\,\mathrm{K}$ all are ionised ($n = N_D$), and at high temperature the intrinsic carriers (dashed, $n_i$) take over.*

**Definition 22.9 (Direct and indirect gaps).**

A [semiconductor](#def-b3-solid-state-classes) has a *direct band gap* when the top of its [valence band](#def-b3-solid-state-band) and the bottom of its [conduction band](#def-b3-solid-state-band) occur at the same electron momentum, so that a photon alone can carry an electron across, and an *indirect band gap* when they do not, so that the transition needs a lattice vibration as well.

Silicon ($1.12\,\mathrm{eV}$) and germanium ($0.66\,\mathrm{eV}$) have indirect gaps: they absorb light well enough in a thick layer, which suits a solar cell, but emit it very poorly. Gallium arsenide ($1.42\,\mathrm{eV}$) and gallium nitride (about $3.4\,\mathrm{eV}$) have direct gaps, the basis of infrared and blue light-emitting diodes and lasers; gallium phosphide ($2.26\,\mathrm{eV}$) is indirect but emits green light when doped. The colour of many pigments is also a [band gap](#def-b3-solid-state-band): a crystal absorbs all photons with $h\nu > E_g$, so cadmium sulfide, with a gap in the blue, is yellow.

**Definition 22.10 (Doping).**

A *dopant* is a foreign atom put in a [semiconductor](#def-b3-solid-state-classes) on purpose, in small amounts, to provide carriers. In an *n-type semiconductor* the dopant has one more valence electron than the atom it replaces (phosphorus in silicon) and gives it to the [conduction band](#def-b3-solid-state-band) from a *donor level* just below that band; in a *p-type semiconductor* it has one fewer (boron in silicon) and accepts an electron from the [valence band](#def-b3-solid-state-band) into an *acceptor level* just above it, leaving a [hole](#def-b3-solid-state-carriers).

![Donor and acceptor levels in the gap of silicon (gap not to scale: the dopant levels lie about 0.045 eV from the band edges, the gap is 1.12 eV). A donor gives its electron (dot) to the conduction band; an acceptor takes one from the valence band, leaving a hole (circle).](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-25e7bf3acf3c.svg)

*Donor and [acceptor levels](#def-b3-solid-state-doping) in the gap of silicon (gap not to scale: the [dopant](#def-b3-solid-state-doping) levels lie about 0.045 eV from the band edges, the gap is 1.12 eV). A donor gives its electron (dot) to the [conduction band](#def-b3-solid-state-band); an acceptor takes one from the [valence band](#def-b3-solid-state-band), leaving a [hole](#def-b3-solid-state-carriers) (circle).*

The [donor level](#def-b3-solid-state-doping) of phosphorus lies $0.045\,\mathrm{eV}$ below the [conduction band](#def-b3-solid-state-band) of silicon, less than $2kT$ at room temperature, and boron’s [acceptor level](#def-b3-solid-state-doping) as much above the [valence band](#def-b3-solid-state-band): at $300\,\mathrm{K}$ nearly every [dopant](#def-b3-solid-state-doping) is ionised. That is why one [dopant](#def-b3-solid-state-doping) atom per $10^7$ silicon atoms ($5 \times 10^{15}\,\mathrm{cm}^{-3}$) raises the electron concentration a few hundred thousand times.

**Method 22.11 (Carriers of a doped semiconductor).**

1. Check that the temperature is in the saturation range ( [dopants](#def-b3-solid-state-doping) fully ionised, $N_D \gg n_i$ ): then the majority carriers are $n \approx N_D$ (or $p \approx N_A$ ).
2. Get the minority carriers from the mass-action law: $p = n_i^2/N_D$ .
3. Place the [Fermi level](#def-b3-solid-state-fermi) : $E_c - E_F = kT\ln(N_c/n)$ , or relative to the intrinsic level, $E_F - E_i = kT\ln(n/n_i)$ .
4. If both kinds of [dopant](#def-b3-solid-state-doping) are present, use the difference $N_D - N_A$ .

**Definition 22.12 (p–n junction).**

A *p–n junction* is the boundary, inside one crystal, between a p-type and an n-type region.

At a [p–n junction](#def-b3-solid-state-junction) electrons diffuse from the n side into the p side and [holes](#def-b3-solid-state-carriers) the other way, leaving a thin layer emptied of carriers in which the fixed ionised [dopants](#def-b3-solid-state-doping) set up an electric field; at equilibrium the [Fermi level](#def-b3-solid-state-fermi) is the same on both sides. A voltage applied one way lowers the barrier and current flows; the other way it raises it: a diode. In a light-emitting diode, electrons and [holes](#def-b3-solid-state-carriers) injected across the junction recombine and emit photons of energy close to $E_g$; in a solar cell, photons absorbed near the junction make electron–hole pairs that the field separates, and a current flows in the external circuit.

![A single crystal of silicon pulled from the melt, shown beside a solar cell made from a slice of such a crystal. Photograph: Sebastian Wallroth, public domain.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/img-aae38ca7acf3.jpg)

*A single crystal of silicon pulled from the melt, shown beside a solar cell made from a slice of such a crystal. Photograph: Sebastian Wallroth, public domain.*

**In the lab — Growing a silicon crystal.**

In the Czochralski method, very pure polycrystalline silicon is melted in a silica crucible under argon, at a little above its melting point, with the [dopant](#def-b3-solid-state-doping) added to the melt. A small seed crystal on a rotating rod is dipped into the melt and slowly raised: silicon crystallises on it with the seed’s orientation, and the rate of pulling and rotation sets the diameter. The boule is then sawn into wafers less than a millimetre thick. Silane and the doping gases phosphine and arsine used in later steps are pyrophoric or very toxic, and are handled only in sealed industrial systems.

## 22.3 Point defects

**Definition 22.13 (Point defects).**

A *point defect* is a departure from the perfect crystal localised at one site: a vacancy, an interstitial atom, or a foreign atom. In an ionic crystal, a *Schottky defect* is a pair of vacancies, one cation and one anion, the ions having gone to the surface; a *Frenkel defect* is a pair made of a vacancy and the ion that left it, sitting on an interstitial site.

![Point defects in a layer of a rock-salt crystal (small dots: cations; large circles: anions; dashed: vacancies). Left: a Schottky pair, one cation and one anion missing. Right: a Frenkel pair, a silver ion moved from its site to an interstitial position.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-8e890c81000b.svg)

*[Point defects](#def-b3-solid-state-defects) in a layer of a rock-salt crystal (small dots: cations; large circles: anions; dashed: vacancies). Left: a Schottky pair, one cation and one anion missing. Right: a Frenkel pair, a silver ion moved from its site to an interstitial position.*

**Theorem 22.14 (Equilibrium concentration of Schottky defects).**

In a crystal MX of $N$ formula units, the number $n$ of Schottky pairs at equilibrium, with formation enthalpy $\Delta H_S$ per pair and $n \ll N$, is

$$
\frac{n}{N} = \eu^{-\Delta H_S/2kT}.
$$

**Proof.** Creating $n$ pairs costs $n\,\Delta H_S$ and gives a configurational entropy, from the $\binom{N}{n}$ ways of choosing the cation vacancies and as many for the anions: $S = 2k\ln\frac{N!}{n!(N - n)!}$. With Stirling’s formula $\ln x! \approx x\ln x - x$,

$$
G(n) = n\,\Delta H_S - 2kT\bigl[N\ln N - n\ln n - (N - n)\ln(N - n)\bigr].
$$

At equilibrium $\dd G/\dd n = 0$: $\Delta H_S - 2kT\ln\frac{N - n}{n} = 0$, so $n/(N - n) = \eu^{-\Delta H_S/2kT}$, and $n/N$ for $n \ll N$. The vibrational entropy of the defects, neglected here, multiplies the result by a constant factor. ∎

**Proposition 22.15 (Frenkel defects).**

With $N$ ions of the moving kind, $N_i$ interstitial sites available to them and $\Delta H_F$ the formation enthalpy of a pair, the number of Frenkel pairs is $n =
\sqrt{NN_i}\,\eu^{-\Delta H_F/2kT}$ (for $n \ll N, N_i$).

**Proof.** The entropy is now $S = k\ln\binom{N}{n} + k\ln\binom{N_i}{n}$, and the same minimisation gives

$$
\Delta H_F = kT\ln\frac{(N - n)(N_i - n)}{n^2} \approx kT\ln\frac{NN_i}{n^2},
$$

hence the result. ∎

The square root shows that defects form in pairs: like $n_i$ in a [semiconductor](#def-b3-solid-state-classes), their concentration has half the formation energy in its exponent. Alkali halides form mainly [Schottky defects](#def-b3-solid-state-defects); silver halides, whose small, polarisable $\ce{Ag+}$ fits between the anions, mainly [Frenkel defects](#def-b3-solid-state-defects).

**Definition 22.16 (Kröger–Vink notation).**

*Kröger–Vink notation* writes a [point defect](#def-b3-solid-state-defects) as $\mathrm{A_S^c}$: A is the species on the site (V for a vacancy, $i$ as the site for an interstitial), S the site it occupies in the perfect crystal, and c its charge relative to that site: $\bullet$ for each positive unit, $'$ for each negative unit, $\times$ for none.

Thus $\mathrm{V_{Na}'}$ is a sodium vacancy (the missing $+1$ leaves a relative charge $-1$), $\mathrm{V_{Cl}^{\bullet}}$ a chloride vacancy, $\mathrm{Ag_i^{\bullet}}$ an interstitial silver ion, $\mathrm{Ca_{Na}^{\bullet}}$ a calcium ion on a sodium site, $\mathrm{O_O^{\times}}$ an oxide ion in place. The Schottky equilibrium of NaCl is $\text{nil} \rightleftharpoons \mathrm{V_{Na}' + V_{Cl}^{\bullet}}$, the Frenkel equilibrium of AgCl $\mathrm{Ag_{Ag}^{\times}} \rightleftharpoons \mathrm{Ag_i^{\bullet} + V_{Ag}'}$.

**Method 22.17 (Writing a defect equation).**

1. Write what is added to the host and the defects it makes.
2. Site balance: the ratio of cation to anion sites of the host is kept (vacancies count as sites).
3. Mass balance: the same atoms on both sides (vacancies have no mass; electrons $e'$ and [holes](#def-b3-solid-state-carriers) $h^\bullet$ none either).
4. Charge balance: the sums of the effective charges are equal.

**Example 22.18 (Yttria in zirconia).**

Dissolving $\ce{Y2O3}$ in $\ce{ZrO2}$ puts two $\ce{Y^3+}$ on $\ce{Zr^4+}$ sites, each of relative charge $-1$, and three oxide ions on oxygen sites; the host’s ratio of one cation to two oxygen sites requires four oxygen sites for the two cations, so one stays empty:

$$
\ce{Y2O3} \xrightarrow{\ce{ZrO2}} \mathrm{2\,Y_{Zr}' + 3\,O_O^{\times} + V_O^{\bullet\bullet}}.
$$

Sites: 2 cation, 4 oxygen; mass: 2 Y and 3 O; charge: $2(-1) + 2 = 0$. One oxide vacancy for each two yttrium ions: this is what makes yttria-stabilised zirconia an oxide-ion conductor.

**Definition 22.19 (Colour centre).**

A *colour centre* is a [point defect](#def-b3-solid-state-defects) that absorbs visible light, such as an electron trapped in an anion vacancy of an alkali halide (an F centre), which colours a crystal that is otherwise transparent.

Heating sodium chloride in sodium vapour or irradiating it makes F centres: the trapped electron behaves like a [particle in a box](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-box) the size of the vacancy, and absorbs in the visible, turning the crystal yellow-brown.

## 22.4 Non-stoichiometry

**Definition 22.20 (Non-stoichiometric compounds).**

A *non-stoichiometric compound* is a solid whose composition varies over a range around a simple ratio of whole numbers, the difference being taken up by [point defects](#def-b3-solid-state-defects).

Iron(II) oxide is never exactly $\ce{FeO}$: it is always short of iron, $\mathrm{Fe_{1-x}O}$, with $x$ set by the temperature and the oxygen pressure at which it was made. Each missing $\ce{Fe^2+}$ is compensated by two $\ce{Fe^3+}$, which in defect language are [holes](#def-b3-solid-state-carriers) on iron sites.

**Proposition 22.21 (Conductivity and oxygen pressure).**

For a metal-deficient oxide $\mathrm{M_{1-x}O}$ whose defects are doubly charged metal vacancies compensated by [holes](#def-b3-solid-state-carriers), the [hole](#def-b3-solid-state-carriers) concentration, and the conductivity, vary as $p(\ce{O2})^{1/6}$. For an oxygen-deficient oxide compensated by electrons, they vary as $p(\ce{O2})^{-1/6}$.

**Proof.** Taking up oxygen creates metal vacancies:

$$
\tfrac12\ce{O2} \rightleftharpoons \mathrm{O_O^{\times} + V_M'' + 2\,h^{\bullet}}, \qquad K = \frac{[\mathrm{V_M''}][\mathrm h^\bullet]^2}{p(\ce{O2})^{1/2}}
$$

(the activity of $\mathrm{O_O^{\times}}$ is 1). The charge balance is $[\mathrm h^\bullet] = 2[\mathrm{V_M''}]$, so $[\mathrm h^\bullet]^3 = 2K\,p(\ce{O2})^{1/2}$ and $[\mathrm h^\bullet] \propto p(\ce{O2})^{1/6}$. For the oxygen-deficient case, $\mathrm{O_O^{\times}} \rightleftharpoons \tfrac12\ce{O2} + \mathrm{V_O^{\bullet\bullet} + 2\,e'}$, $[\mathrm e'] = 2[\mathrm{V_O^{\bullet\bullet}}]$, and the same steps give $[\mathrm e'] \propto p(\ce{O2})^{-1/6}$. The conductivity is proportional to the carrier concentration. ∎

A plot of $\log\sigma$ against $\log p(\ce{O2})$ thus tells which defects dominate: its slope, $\pm 1/4$ or $\pm 1/6$, depends on their charges.

## 22.5 Ionic conductors

**Definition 22.22 (Ionic conductors).**

An *ionic conductor* is a solid in which the current is carried by ions moving through the lattice. A *solid electrolyte* is an ionic conductor whose electronic conductivity is negligible, so that it can separate the two electrodes of an electrochemical cell.

**Proposition 22.23 (Arrhenius law of ionic conduction).**

For ions that move by hopping between neighbouring sites over a barrier $E_a$, $\sigma T = A\,\eu^{-E_a/kT}$.

**Partial proof.** An ion attempts jumps at a frequency $\nu_0$ and succeeds with probability $\eu^{-E_a/kT}$, so its diffusion coefficient is $D = \tfrac16\nu_0 z a^2\eu^{-E_a/kT}$ for $z$ neighbouring sites at distance $a$ (random walk). The Nernst–Einstein relation $\sigma = nq^2D/kT$ (admitted), for $n$ mobile ions of charge $q$ per unit volume, gives $\sigma T = (nq^2\nu_0za^2/6k)\,\eu^{-E_a/kT}$. The concentration $n$ is fixed by doping in an extrinsic conductor; when it is itself thermally created, $E_a$ also contains half the formation enthalpy. ∎

Three [solid electrolytes](#def-b3-solid-state-ionic-conductor) show the range. In yttria-stabilised zirconia the oxide vacancies made by the [dopant](#def-b3-solid-state-doping) let $\ce{O^2-}$ ions hop; it conducts usefully only when hot, at several hundred degrees Celsius. In $\beta$-alumina, sodium ions move in loosely packed planes between spinel blocks, the basis of sodium–sulfur batteries. Silver iodide becomes a superionic conductor above $146\,{}^{\circ}\mathrm{C}$, where its silver ions are spread over many more sites than there are ions, almost like a liquid within a rigid iodide lattice. Lithium-ion conductors (lithium lanthanum zirconate garnets, sulfides) are the electrolytes of the all-solid batteries now under development.

![The zirconia oxygen sensor in cross-section: porous platinum electrodes on the two faces of the solid electrolyte, one in the exhaust and one in air. Oxide ions carry the current through the electrolyte; the voltage measures the ratio of the two oxygen pressures.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-solid-state/fig-a8270a9d2bbd.svg)

*The zirconia oxygen sensor in cross-section: porous platinum electrodes on the two faces of the [solid electrolyte](#def-b3-solid-state-ionic-conductor), one in the exhaust and one in air. Oxide ions carry the current through the electrolyte; the voltage measures the ratio of the two oxygen pressures.*

**History — The transistor and the pulled crystal.**

In 1947 John Bardeen and Walter Brattain, in William Shockley’s group at Bell Laboratories, made the first transistor from a crystal of germanium; the three shared the 1956 Nobel Prize in Physics. The crystals came from a method Jan Czochralski had found in 1916 while measuring how fast metals crystallise, pulling a thread of tin out of the melt: adapted to germanium and then silicon, it still makes the wafers of nearly all integrated circuits.

## 22.6 Exercises

**Exercise 22.1 ★.**

Give the six Hückel levels of a chain of six atoms in units of $|\beta|$, and the width of the range they span.

**Solution of Exercise 22.1.**

$(E - \alpha)/|\beta| = -2\cos(k\pi/7)$: $-1.802$, $-1.247$, $-0.445$, $+0.445$, $+1.247$, $+1.802$. They span $3.604|\beta|$, already 90 % of the band width $4|\beta|$ of the infinite chain.

**Exercise 22.2 ★.**

How many electrons does a band built from the s orbitals of $N$ atoms hold? Why are sodium and magnesium both metals?

**Solution of Exercise 22.2.**

$2N$ ($N$ orbitals, two electrons each). Sodium, with one 3s electron per atom, half fills its s band: a metal. Magnesium has two and would fill the s band exactly, but the 3s and 3p bands overlap, so the highest occupied levels belong to a partly filled combined band: a metal too.

**Exercise 22.3 ★.**

Estimate the wavelength of the light emitted by diodes of gallium phosphide ($2.26\,\mathrm{eV}$) and gallium nitride ($3.4\,\mathrm{eV}$), using $\lambda \approx hc/E_g$.

**Solution of Exercise 22.3.**

$\lambda \approx 1239.8\,\mathrm{eV}\,\mathrm{nm}/E_g$: gallium phosphide $549\,\mathrm{nm}$ (green); gallium nitride $365\,\mathrm{nm}$ (near ultraviolet). Blue diodes use gallium nitride alloyed with indium, whose gap is smaller.

**Exercise 22.4 ★.**

Write the Kröger–Vink equations for dissolving $\ce{CaCl2}$ in $\ce{NaCl}$ and $\ce{Y2O3}$ in $\ce{ZrO2}$, and check each balance.

**Solution of Exercise 22.4.**

$\ce{CaCl2} \xrightarrow{\ce{NaCl}} \mathrm{Ca_{Na}^{\bullet} + V_{Na}' + 2\,Cl_{Cl}^{\times}}$: two cation and two anion sites, as in NaCl; one Ca and two Cl; charge $+1 - 1 = 0$. $\ce{Y2O3} \xrightarrow{\ce{ZrO2}} \mathrm{2\,Y_{Zr}' + 3\,O_O^{\times} + V_O^{\bullet\bullet}}$: two cation and four oxygen sites; two Y and three O; charge $-2 + 2 = 0$.

**Exercise 22.5 ★★.**

With $N_c = 6.2 \times 10^{15}\,T^{3/2}$, $N_v = 3.5 \times 10^{15}\,T^{3/2}$ (in $\mathrm{cm}^{-3}$) and $E_g = 1.125\,\mathrm{eV}$ for silicon at $300\,\mathrm{K}$, and $N_c = 1.98 \times 10^{15}\,T^{3/2}$, $N_v = 9.6 \times 10^{14}\,T^{3/2}$, $E_g = 0.661\,\mathrm{eV}$ for germanium, compute $n_i$ for each, and their ratio.

**Solution of Exercise 22.5.**

$T^{3/2} = 5196$ at $300\,\mathrm{K}$, $kT = 0.025\,85\,\mathrm{eV}$. Silicon: $\sqrt{N_cN_v} = 2.42 \times 10^{19}\,\mathrm{cm}^{-3}$, $\eu^{-1.125/0.05170} = 3.55 \times 10^{-10}$, $n_i = 8.6 \times 10^{9}\,\mathrm{cm}^{-3}$. Germanium: $\sqrt{N_cN_v} = 7.16 \times 10^{18}\,\mathrm{cm}^{-3}$, $\eu^{-0.661/0.05170} = 2.80 \times 10^{-6}$, $n_i = 2.0 \times 10^{13}\,\mathrm{cm}^{-3}$. Ratio $4.3 \times 10^{-4}$: the smaller gap of germanium gives it over two thousand times more carriers.

**Exercise 22.6 ★★.**

Silicon is doped with $1.0 \times 10^{16}\,\mathrm{cm}^{-3}$ phosphorus. With $n_i = 1.0 \times 10^{10}\,\mathrm{cm}^{-3}$ at $300\,\mathrm{K}$, give the electron and [hole](#def-b3-solid-state-carriers) concentrations.

**Solution of Exercise 22.6.**

$n = N_D = 1.0 \times 10^{16}\,\mathrm{cm}^{-3}$; $p = n_i^2/n = 1.0 \times 10^{20}/1.0 \times 10^{16} = 1.0 \times 10^{4}\,\mathrm{cm}^{-3}$.

**Exercise 22.7 ★★.**

For the crystal of [Exercise 22.6](#exo-b3-solid-state-6), how far above the intrinsic level is the [Fermi level](#def-b3-solid-state-fermi)?

**Solution of Exercise 22.7.**

$E_F - E_i = kT\ln(n/n_i) = 0.025\,85\,\mathrm{eV} \times \ln(10^6) = 0.36\,\mathrm{eV}$.

**Exercise 22.8 ★★.**

The formation enthalpy of a Schottky pair in NaCl is about $2.3\,\mathrm{eV}$ (data of the exercise). Compute the fraction of vacant sites at $800\,\mathrm{K}$, and the number of pairs in a mole.

**Solution of Exercise 22.8.**

$kT = 0.068\,94\,\mathrm{eV}$ at $800\,\mathrm{K}$; $n/N = \eu^{-2.3/0.1379} = 5.7 \times 10^{-8}$; in a mole $6.02 \times 10^{23} \times 5.7 \times 10^{-8} = 3.4 \times 10^{16}$ pairs.

**Exercise 22.9 ★★.**

A sample of iron(II) oxide (rock-salt structure, four oxygen sites per cell) has $a = 430.0\,\mathrm{pm}$ and a density of $5.70\,\mathrm{g}\,\mathrm{cm}^{-3}$ (data of the exercise). Assuming the oxygen sites full, find $x$ in $\mathrm{Fe_{1-x}O}$.

**Solution of Exercise 22.9.**

The mass of a cell is $\rho a^3 = 5.70\,\mathrm{g}\,\mathrm{cm}^{-3} \times (4.300 \times 10^{-8}\,\mathrm{cm})^3 = 4.532 \times 10^{-22}\,\mathrm{g}$, that is $4.532 \times 10^{-22}\,\mathrm{g} \times N_A = 272.9\,\mathrm{g}$ per mole of cells, or $68.23\,\mathrm{g}$ per oxygen. Then $(1 - x) \times 55.8 + 16.0 = 68.23$ gives $1 - x = 0.936$: $x = 0.064$, $\mathrm{Fe_{0.936}O}$.

**Exercise 22.10 ★★★.**

A lithium-ion conductor has $\sigma = 1.0 \times 10^{-4}\,\mathrm{S}\,\mathrm{cm}^{-1}$ at $300\,\mathrm{K}$, $7.8 \times 10^{-4}\,\mathrm{S}\,\mathrm{cm}^{-1}$ at $350\,\mathrm{K}$ and $3.6 \times 10^{-3}\,\mathrm{S}\,\mathrm{cm}^{-1}$ at $400\,\mathrm{K}$ (data of the exercise). Find its activation energy.

**Solution of Exercise 22.10.**

$\ln(\sigma T)$: $-3.51$, $-1.30$, $0.36$ at $1/T = 3.333$, $2.857$, $2.500 \times 10^{-3}\,\mathrm{K}^{-1}$; slope between the end points $(0.36 + 3.51)/(-8.33 \times 10^{-4}\,\mathrm{K}^{-1}) = -4.65 \times 10^{3}\,\mathrm{K}$ (the middle point lies on the line), so $E_a = 4.65 \times 10^{3}\,\mathrm{K} \times 8.617 \times 10^{-5}\,\mathrm{eV}\,\mathrm{K}^{-1} = 0.40\,\mathrm{eV}$.

**Exercise 22.11 ★★★.**

Show that in an oxide whose dominant defects are singly charged metal vacancies $\mathrm{V_M'}$ compensated by [holes](#def-b3-solid-state-carriers), the conductivity varies as $p(\ce{O2})^{1/4}$.

**Solution of Exercise 22.11.**

$\tfrac12\ce{O2} \rightleftharpoons \mathrm{O_O^{\times} + V_M' + h^{\bullet}}$, $K = [\mathrm{V_M'}][\mathrm h^\bullet]/p(\ce{O2})^{1/2}$. Charge balance $[\mathrm h^\bullet] = [\mathrm{V_M'}]$, so $[\mathrm h^\bullet]^2 = K\,p(\ce{O2})^{1/2}$ and $[\mathrm h^\bullet] \propto p(\ce{O2})^{1/4}$; the conductivity follows the [holes](#def-b3-solid-state-carriers).

**Exercise 22.12 ★★★.**

In AgCl the Frenkel pair has a formation enthalpy of about $1.4\,\mathrm{eV}$, and there are two interstitial sites per silver ion (data of the exercise). Compute the fraction of silver ions on interstitial sites at $600\,\mathrm{K}$.

**Solution of Exercise 22.12.**

$n/N = \sqrt{N_i/N}\,\eu^{-\Delta H_F/2kT} = \sqrt 2\,\eu^{-1.4/(2 \times 0.05170)} = 1.414 \times 1.32 \times 10^{-6} = 1.9 \times 10^{-6}$.

## 22.7 Problem: The Oxygen Sensor in an Exhaust Pipe

**Problem 22.1.**

Weekend problem — the oxygen sensor in an exhaust pipe: the defects of yttria-stabilised zirconia, its ionic conductivity and operating temperature, the Nernst voltage of the cell, and the switch between lean and rich exhaust

The sensor’s electrolyte is zirconia with 8.0 mol % $\ce{Y2O3}$, of fluorite structure with four cation and eight oxygen sites per cubic cell, $a = 514\,\mathrm{pm}$; a disc $1.0\,\mathrm{mm}$ thick and $0.20\,\mathrm{cm}^{2}$ in area separates the exhaust from air. Its conductivity follows $\sigma T = A\eu^{-E_a/kT}$ with $E_a = 1.0\,\mathrm{eV}$ and $\sigma = 0.030\,\mathrm{S}\,\mathrm{cm}^{-1}$ at $800\,{}^{\circ}\mathrm{C}$. Lean exhaust has $p(\ce{O2}) = 0.010\,\mathrm{bar}$, rich exhaust $p(\ce{O2}) = 1.0 \times 10^{-20}\,\mathrm{bar}$ (all data of the problem).

**Part I — The electrolyte.**

1. Write the Kröger–Vink equation for dissolving $\ce{Y2O3}$ in $\ce{ZrO2}$ .
2. Check its site, mass and charge balance.
3. How many oxide vacancies are created per yttrium ion?
4. Write the composition as $\mathrm{Zr_{1-x}Y_xO_{2-x/2}}$ and compute $x$ .
5. Compute the fraction of oxygen sites that are vacant.
6. Compute the number of vacancies per cell.
7. Compute the vacancy concentration in $\mathrm{cm}^{-3}$ .

**Part II — Conductivity and temperature.**

8. Why does the conductivity rise steeply with temperature?
9. Compute $A$ .
10. Compute $\sigma$ at $700\,{}^{\circ}\mathrm{C}$ .
11. Compute $\sigma$ at $300\,{}^{\circ}\mathrm{C}$ .
12. Compute the resistance of the disc at these two temperatures.
13. The electronics need a resistance below $1\,\mathrm{k}\Omega$ : find the lowest working temperature.
14. Why are such sensors fitted with a heater?

**Part III — The Nernst voltage.**

15. Write the electrode reaction on the air side, in ordinary and in [Kröger–Vink notation](#def-b3-solid-state-kroger-vink) .
16. Show that the cell voltage is $E = (RT/4F)\ln\bigl(p_{\mathrm{air}}/p_{\mathrm{exh}}\bigr)$ .
17. Compute $RT/4F$ at $700\,{}^{\circ}\mathrm{C}$ .
18. What is the voltage if the exhaust contained air?
19. Compute the change of voltage per factor of ten in $p(\ce{O2})$ .

**Part IV — Lean, rich and the switch.**

20. Compute the voltage in lean exhaust at $700\,{}^{\circ}\mathrm{C}$ .
21. Compute the voltage in rich exhaust at $700\,{}^{\circ}\mathrm{C}$ .
22. Why is $p(\ce{O2})$ so small in rich exhaust?
23. The engine control switches at $0.45\,\mathrm{V}$ : what oxygen pressure does that correspond to?
24. Why does the voltage jump at the stoichiometric air-to-fuel ratio, and how does the engine control use it?
25. State the result: the sensor voltage in rich exhaust at $700\,{}^{\circ}\mathrm{C}$ .

**Solution of Problem 22.1.**

**1.** $\ce{Y2O3} \xrightarrow{\ce{ZrO2}} \mathrm{2\,Y_{Zr}' + 3\,O_O^{\times} + V_O^{\bullet\bullet}}$.

**2.** Sites: two cation sites and four oxygen sites (three filled, one empty), the 1 : 2 ratio of $\ce{ZrO2}$. Mass: 2 Y and 3 O on each side. Charge: $2 \times (-1) + 2 = 0$.

**3.** One vacancy for two yttrium ions: one half per yttrium.

**4.** Per $(\ce{ZrO2})_{0.92}(\ce{Y2O3})_{0.08}$ there are 0.92 Zr and 0.16 Y, 1.08 cations: $x = 0.16/1.08 = 0.148$, $\mathrm{Zr_{0.852}Y_{0.148}O_{1.926}}$.

**5.** $(x/2)/2 = 0.0370$: 3.7 % of the oxygen sites are empty.

**6.** $8 \times 0.0370 = 0.296$ vacancies per cell.

**7.** $a^3 = (5.14 \times 10^{-8}\,\mathrm{cm})^3 = 1.358 \times 10^{-22}\,\mathrm{cm}^{3}$: $0.296/1.358 \times 10^{-22}\,\mathrm{cm}^{3} = 2.2 \times 10^{21}\,\mathrm{cm}^{-3}$.

**8.** Each oxide ion must hop over a barrier of $1.0\,\mathrm{eV}$ into a neighbouring vacancy; the fraction of attempts that succeed, $\eu^{-E_a/kT}$, grows very fast with $T$.

**9.** At $1073.15\,\mathrm{K}$, $kT = 0.092\,48\,\mathrm{eV}$: $A = \sigma T\eu^{E_a/kT} = 0.030 \times 1073.15 \times \eu^{10.81} = 1.6 \times 10^{6}\,\mathrm{S}\,\mathrm{K}\,\mathrm{cm}^{-1}$.

**10.** At $973.15\,\mathrm{K}$: $\sigma = (A/T)\eu^{-11.92} = 0.011\,\mathrm{S}\,\mathrm{cm}^{-1}$.

**11.** At $573.15\,\mathrm{K}$: $\sigma = (A/T)\eu^{-20.25} = 4.5 \times 10^{-6}\,\mathrm{S}\,\mathrm{cm}^{-1}$.

**12.** $R = L/(\sigma S) = 0.10\,\mathrm{cm}/(\sigma \times 0.20\,\mathrm{cm}^{2})$: $46\,\Omega$ at $700\,{}^{\circ}\mathrm{C}$, $1.1 \times 10^{5}\,\Omega$ at $300\,{}^{\circ}\mathrm{C}$.

**13.** $R < 1\,\mathrm{k}\Omega$ needs $\sigma > 5.0 \times 10^{-4}\,\mathrm{S}\,\mathrm{cm}^{-1}$; solving $(A/T)\eu^{-E_a/kT} = 5.0 \times 10^{-4}$ (by trial) gives $T \approx 760\,\mathrm{K}$, about $490\,{}^{\circ}\mathrm{C}$.

**14.** Exhaust is cold after a start and at low load; the heater brings the electrolyte to its working temperature within seconds, so that the control works from the start, when most pollutants are emitted.

**15.** $\ce{O2 + 4e- -> 2O^2-}$; in [Kröger–Vink notation](#def-b3-solid-state-kroger-vink) $\ce{O2} + \mathrm{2\,V_O^{\bullet\bullet} + 4\,e'} \longrightarrow \mathrm{2\,O_O^{\times}}$.

**16.** Oxygen is reduced on the air side and the oxide ions are oxidised back to oxygen on the exhaust side: the cell transfers $\ce{O2}$ from air to exhaust, with $\Delta_rG = RT\ln(p_{\mathrm{exh}}/p_{\mathrm{air}})$ per mole of $\ce{O2}$ and four electrons: $E = -\Delta_rG/4F = (RT/4F)\ln(p_{\mathrm{air}}/p_{\mathrm{exh}})$.

**17.** $RT/4F = 8.3145 \times 973.15/(4 \times 96485) = 0.020\,96\,\mathrm{V}$.

**18.** Zero: the two pressures are equal.

**19.** $(RT/4F)\ln 10 = 48.3\,\mathrm{mV}$ per decade.

**20.** $0.020\,96\,\mathrm{V} \times \ln(0.2095/0.010) = 0.020\,96\,\mathrm{V} \times 3.04 = 0.064\,\mathrm{V}$.

**21.** $0.020\,96\,\mathrm{V} \times \ln(0.2095/1.0 \times 10^{-20}) = 0.020\,96\,\mathrm{V} \times 44.49 = 0.93\,\mathrm{V}$.

**22.** Rich exhaust contains carbon monoxide, hydrogen and unburnt hydrocarbons; on the platinum electrode the little oxygen left reacts with them, and $p(\ce{O2})$ is set by the equilibria $\ce{2CO + O2 <=> 2CO2}$ and $\ce{2H2 + O2 <=> 2H2O}$, which leave a tiny value at $700\,{}^{\circ}\mathrm{C}$.

**23.** $p = 0.2095\,\eu^{-0.45/0.02096} = 1.0 \times 10^{-10}\,\mathrm{bar}$.

**24.** Around the stoichiometric ratio, the exhaust passes from a slight excess of oxygen to a slight excess of CO and hydrogen: $p(\ce{O2})$ falls by some eighteen powers of ten for a small change in fuel, and the voltage jumps from below $0.1\,\mathrm{V}$ to about $0.9\,\mathrm{V}$. The engine control adds fuel when the voltage is low and removes it when high, holding the mixture at the stoichiometric point where the three-way catalyst converts CO, hydrocarbons and nitrogen oxides at once.

**25.** In rich exhaust at $700\,{}^{\circ}\mathrm{C}$ the sensor gives about $0.93\,\mathrm{V}$.