Defects — Schottky, Frenkel; non-stoichiometric defects: Worked examples

This page drills the numbers behind the parent topic. We work every kind of question this topic can throw at you: density counting, charge balance, the Boltzmann defect-count formula, and the tricky limiting cases. Each example first asks you to forecast the answer — guess before you compute; that is how the ideas stick.

Before the examples, one symbol everyone will use. $k_{B}$ is the Boltzmann constant — the number that converts a temperature (in kelvin) into an energy (in joules). Think of it as the "exchange rate" between how hot something is and how much jiggling energy each atom carries. Its value:

The scenario matrix

Every question in this topic falls into exactly one of these cells. The examples below are tagged with the cell they cover, so by the end you have seen all of them.

The three "bookkeeping" families and their cells fan out like this — read the flowchart first, then use the table for the fine print:

#	Cell (case class)	What makes it tricky	Example
A	Schottky density drop	atoms actually removed → recount $Z$	E1
B	Frenkel density (control)	nothing removed → density unchanged	E2
C	Charge balance: metal deficiency ( $F e_{0.95} O$ )	1 vacancy needs 2 $F e^{3 +}$	E3
D	Charge balance: metal excess / F-centre	extra electron per anion vacancy	E4
E	Defect count vs temperature	exponential; plug into $n \approx N e^{- E /2 k_{B} T}$	E5
F	Ratio of defects at two temperatures	the $N$ and units cancel	E6
G	Limiting / degenerate: $T \to 0$ and $T \to \infty$	what the formula says at the edges	E7
H	Real-world word problem (colour / semiconductor)	translate words → which defect	E8
I	Exam twist: which $E$ ? ( $E_{s}$ pair vs per-vacancy)	factor-of-2 bookkeeping trap	E9

E1 — Cell A: Schottky lowers the density

Forecast: will the density go up, down, or stay the same? (Look at the mnemonic: Schottky Subtracts.)

Figure — Defects — Schottky, Frenkel; non-stoichiometric defects

Figure s01 — the same-size chalk box with two ions (one cation, one anion) erased: fewer atoms, unchanged volume.

Perfect density. Use the unit-cell density formula from Density of a Unit Cell: $ρ = \frac{Z M}{N _{A} a ^{3}} .$ Why this step? Density is (mass in the box) ÷ (volume of the box). The box holds $Z$ formula units, each of mass $M / N_{A}$ grams; the box volume is $a^{3}$ .
Numbers. $a = 564 pm = 564 \times 1 0^{- 10} cm = 5.64 \times 1 0^{- 8} cm$ , so $a^{3} = 1.794 \times 1 0^{- 22} cm^{3}$ . $ρ_{perfect} = \frac{4 \times 58.5}{6.022 \times 1 0 ^{23} \times 1.794 \times 1 0 ^{- 22}} \approx 2.165 g cm^{- 3} .$ Why this step? Just arithmetic — but note we converted pm→cm so the answer comes out in $g cm^{- 3}$ .
Remove the atoms. A Schottky defect removes one cation AND one anion (the parent's charge-balance rule). Removing $2%$ of the pairs means only $98%$ of the formula units remain: the effective $Z$ drops to $Z^{'} = 4 \times 0.98 = 3.92$ . Why this step? Density fell because fewer atoms sit in the same volume — the edge length $a$ does not change, so $a^{3}$ stays fixed. Look at the figure: the box is the same size, but two chalk circles are erased.
Defective density. $ρ_{def} = \frac{3.92 \times 58.5}{6.022 \times 1 0 ^{23} \times 1.794 \times 1 0 ^{- 22}} \approx 2.122 g cm^{- 3} .$ Why this step? We re-run the same density formula with only $Z$ changed — this isolates the one thing the defect touched (atom count), so any drop we see is purely from missing atoms, not from a changed box.

Verify: $ρ_{def} / ρ_{perfect} = 3.92/4 = 0.98$ — exactly the fraction of atoms remaining. Units: $\frac{g mol ^{- 1}}{mol ^{- 1} \cdot cm ^{3}} = g cm^{- 3}$ . ✓ Density decreased, matching the forecast.

E2 — Cell B: Frenkel does NOT change density (the control)

Forecast: guess before reading. Compare to E1.

Count the atoms after the jump. A Frenkel defect relocates an ion — it stays inside the crystal, just in an interstitial void instead of its lattice site. No atom leaves the box. Why this step? Density only cares about total mass in the box, not where each atom sits. Same atoms, same box.
Effective $Z$ is unchanged: $Z^{'} = 4$ . $ρ_{def} = \frac{4 \times 58.5}{6.022 \times 1 0 ^{23} \times 1.794 \times 1 0 ^{- 22}} = 2.165 g cm^{- 3} .$ Why this step? We deliberately feed the unchanged $Z = 4$ into the identical formula so the comparison with E1 is honest — the only difference between the two examples is whether an atom left the box.

Verify: identical to $ρ_{perfect}$ . This is the contrast the parent stressed: Schottky subtracts mass, Frenkel just rearranges it. ✓

E3 — Cell C: metal deficiency, $F e_{0.95} O$

Forecast: guess — is it around $5%$ , $10%$ , or something bigger?

Figure s02 — charge bookkeeping: one dashed iron vacancy removes +2, patched by two blue Fe³⁺ ions restoring +1 each.

Read the figure first. On the board, one $O^{2 -}$ (pink) sits at left. In the middle is a dashed-yellow iron vacancy — an empty $F e$ site that has removed $+ 2$ of charge from the crystal. At right are two blue $F e^{3 +}$ ions, each of which restores only $+ 1$ . The chalk arrow shows the bookkeeping: the single $+ 2$ hole is patched by $+ 1 + 1 = + 2$ from two oxidised irons. Keep that picture in mind as we do the algebra — it is the whole reason the answer involves a factor of two.

Set up the charge balance. Per $1$ oxygen ( $O^{2 -}$ , charge $- 2$ ) there are $0.95$ iron ions total. Let $x$ = number of $F e^{3 +}$ ; then $F e^{2 +}$ count is $0.95 - x$ . Total positive charge must equal $+ 2$ : $3 x + 2 (0.95 - x) = 2.$ Why this step? The crystal as a whole is neutral. We are doing charge bookkeeping, not atom bookkeeping — the missing iron sites (the dashed vacancy in the figure) left the box short of positive charge, and higher-charge $F e^{3 +}$ ions (the two blue ions) plug that gap without adding atoms.
Solve. $3 x + 1.90 - 2 x = 2 \Rightarrow x = 0.10.$ So $0.10$ of the iron is $F e^{3 +}$ and $0.85$ is $F e^{2 +}$ .
Convert to a percentage of the iron present (total iron $= 0.95$ ): $\frac{0.10}{0.95} \times 100 \approx 10.53%.$ Why this step? The question asks "of the iron," so we divide by the iron present $0.95$ , not by $1$ — dividing by the wrong denominator is the classic slip here, so we make the base explicit.

Verify: count the vacancies (the dashed circle in the figure). Missing iron $= 1 - 0.95 = 0.05$ per oxygen, each removing charge $+ 2$ → deficit $= 0.10$ . Each $F e^{3 +}$ restores $+ 1$ , and we found $0.10$ of them → exactly cancels. One vacancy ⇒ two $F e^{3 +}$ , as the parent's Trap 4 warned ( $0.05$ vacancies ⇒ $0.10 F e^{3 +}$ ). ✓ Total charge: $3 (0.10) + 2 (0.85) = 0.30 + 1.70 = 2.00 =$ oxygen's $∣ - 2 ∣$ . ✓

E4 — Cell D: metal excess, counting F-centres

Forecast: more or fewer than Avogadro's number? Guess the order of magnitude.

What is an F-centre here? Extra $N a$ deposits, a $C l^{-}$ leaves its site to bond with it, and the electron the ionised $N a$ left behind is trapped in the anion vacancy. That trapped electron is the F-centre. Why this step? Trap 3 in the parent: the F-centre is the electron, not a metal atom. One anion vacancy = one trapped electron.
Count vacancies. The formula $N a_{1.0002} C l$ means for every $1 C l$ there is $1.0002 N a$ . But charge neutrality demands equal $+$ and $-$ : each excess $N a^{+}$ ( $0.0002$ of them) needs a compensating electron, and each such electron sits in a $C l^{-}$ vacancy. So there are $0.0002$ F-centres per formula unit. Why this step? Charge bookkeeping again — the excess metal is ionised, so its electrons must live somewhere neutral: the anion holes.
Per mole: $N_{F} = 0.0002 \times N_{A} = 2 \times 1 0^{- 4} \times 6.022 \times 1 0^{23} = 1.2044 \times 1 0^{20} .$ Why this step? "Per formula unit" becomes "per mole" by multiplying by $N_{A}$ — that is the only bridge from a fraction to a countable number of electrons, which is what the question demands.

Verify: units — dimensionless fraction × (mol $^{- 1}$ × mol) = pure count. $1.2 \times 1 0^{20}$ is far below $N_{A}$ , sensible for a trace defect. ✓ Each F-centre absorbs visible light → the crystal turns yellow, exactly as the parent describes.

E5 — Cell E: how many Schottky defects at a temperature?

Before we plug numbers in, we must earn the strange $\frac{1}{2}$ that lives in the exponent of the Schottky formula. It is not a typo, and it comes straight from the physics of why defects exist at all — so let us build it on this page rather than sending you back to the parent.

Intuition Two symbols we need:

$H$ and $G$

$H$ = enthalpy — think of it as the energy price to do something. Making a defect costs enthalpy because you must break bonds to pull an ion off its site; call the cost of one full Schottky pair $E_{s}$ .
$G$ = Gibbs free energy — the quantity nature actually minimises, defined as $G = H - T S$ (see Entropy and Gibbs Free Energy). Here $S$ = entropy = a measure of how many ways there are to arrange things, and $T$ = temperature.

Why these two, and not just energy? If crystals only minimised energy ( $H$ ), a perfect crystal (no defects, lowest energy) would always win and defects would never appear. But nature minimises $G = H - T S$ . The $- T S$ term rewards disorder: a few vacancies can be arranged in astronomically many ways, so $S$ jumps, and $- T S$ pulls $G$ down even though $H$ went up. The winning number of defects is the balance point — and finding that balance is what produces the exponential.

Formula Where the

$\frac{1}{2}$ comes from (the origin of $E_{s} /2$ ) A Schottky pair is two independent holes: one on a cation site and one on an anion site. When you minimise $G$ over the number of pairs $n$ , each of the two holes contributes its own combinatorial factor $e^{- E_{s} /2 k_{B} T}$ — and multiplying the cation-hole factor by the anion-hole factor rebuilds the single pair energy $E_{s}$ . In symbols: $cation hole e^{- E_{s} /2 k_{B} T} \times anion hole e^{- E_{s} /2 k_{B} T} = e^{- E_{s} / k_{B} T} (the whole pair) .$ So the per-hole cost is $E_{s} /2$ , and because the defect count $n$ counts pairs (one cation-hole + one anion-hole each), it carries the single factor $n \approx N e^{- E_{s} /2 k_{B} T} .$ The $\frac{1}{2}$ is the price of splitting one pair enthalpy $E_{s}$ across its two independent vacancies. Remember this — E9 is built entirely on it.

Forecast: guess — will it be a huge number close to $N$ , or a tiny number? (The exponent is negative and large...)

Why an exponential (and why $/2$ )? As just derived: minimising $G = H - T S$ produces the Boltzmann factor $e^{- cost / k_{B} T}$ , and because a Schottky pair is two independent holes each carrying $E_{s} /2$ , the pair count is $n \approx N e^{- E_{s} /2 k_{B} T}$ . Why this tool? An exponential is the natural answer whenever a fixed energy price competes against thermal jiggling — it's the same shape as reaction rates and vapour pressures.
Convert energy to joules. $E_{s} = 2.0 eV = 2.0 \times 1.602 \times 1 0^{- 19} = 3.204 \times 1 0^{- 19} J$ . Why this step? $k_{B}$ is in joules per kelvin, so every energy in the exponent must be in joules — never mix eV and J.
Build the exponent. $\frac{E _{s}}{2 k _{B} T} = \frac{3.204 \times 1 0 ^{- 19}}{2 \times 1.38 \times 1 0 ^{- 23} \times 1000} = \frac{3.204 \times 1 0 ^{- 19}}{2.76 \times 1 0 ^{- 20}} \approx 11.61.$ Why this step? We isolate the exponent first because it alone decides the order of magnitude — a large positive number here means a savage suppression, so evaluating it separately lets us sanity-check the size before touching $N$ .
Compute $n$ . $n \approx 6.022 \times 1 0^{23} \times e^{- 11.61} \approx 6.022 \times 1 0^{23} \times 9.05 \times 1 0^{- 6} \approx 5.45 \times 1 0^{18} .$ Why this step? Multiplying the tiny Boltzmann fraction by the huge site count $N$ turns a probability-per-site into an actual number of defect pairs — the two extremes partly cancel, which is why the answer lands at a moderate $1 0^{18}$ rather than at $N$ or at zero.

Verify: the fraction defective is $n / N = e^{- 11.61} \approx 9.05 \times 1 0^{- 6}$ — about $9$ per million sites (so about $18$ vacancies per million sites, since each pair is two holes). Tiny, as forecast, but enough to colour and conduct. ✓

E6 — Cell F: ratio of defects at two temperatures

Forecast: double? ten times? Guess.

Take the ratio — $N$ cancels. $\frac{n _{2}}{n _{1}} = \frac{N e ^{- E_{s} /2 k_{B} T_{2}}}{N e ^{- E_{s} /2 k_{B} T_{1}}} = exp [\frac{E _{s}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})] .$ Why this step? Forming a ratio kills the messy prefactor $N$ (and any factor we're unsure about). This is why exam questions love ratios: fewer constants to plug in.
Plug in. $\frac{E _{s}}{2 k _{B}} = \frac{3.204 \times 1 0 ^{- 19}}{2 \times 1.38 \times 1 0 ^{- 23}} = 11607 K$ . And $\frac{1}{1000} - \frac{1}{1200} = 1.667 \times 1 0^{- 4} K^{- 1}$ . $\frac{n _{2}}{n _{1}} = e^{11607 \times 1.667 \times 1 0^{- 4}} = e^{1.934} \approx 6.92.$ Why this step? Grouping the constants into the single number $11607 K$ (with units of temperature) makes the exponent dimensionless when multiplied by a $1/ T$ difference — a built-in units check before we exponentiate.

Verify: raising $T$ raises defects (positive exponent) — correct direction. Cross-check via E5: $n_{1} = 5.45 \times 1 0^{18}$ , and $n_{2} = N e^{- 11607/1200} = 6.022 \times 1 0^{23} e^{- 9.673} \approx 3.78 \times 1 0^{19}$ ; ratio $3.78 \times 1 0^{19} /5.45 \times 1 0^{18} \approx 6.9$ . ✓ A $200 K$ heat gives a ~ $7 \times$ jump — exponentials are steep.

E7 — Cell G: the limiting cases $T \to 0$ and $T \to \infty$

Forecast: at absolute zero, how many defects? At infinite temperature?

As $T \to 0^{+}$ : the exponent $- \frac{E _{s}}{2 k _{B} T} \to - \infty$ , so $e^{- \infty} = 0$ , giving $n \to 0$ . Why this step? Pushing $T$ to its extreme isolates the dominant term (the exponent explodes and swamps $N$ ) — this is how you read the physics off a formula without any arithmetic. It proves the parent's claim: a perfect crystal exists only at absolute zero, because with no thermal energy the $- T S$ reward vanishes and no defect can pay its enthalpy cost.
As $T \to \infty$ : the exponent $\to 0$ , so $e^{0} = 1$ and $n \to N$ . Why this step? Testing the opposite extreme reveals where the model breaks: the formula assumes $n ≪ N$ (dilute defects, so Stirling's approximation holds); at $n \approx N$ the crystal has melted and the formula no longer applies. The maths hints "lots of defects," but the model dies first.
Degenerate check — $E_{s} = 0$ : then $n = N$ at every $T > 0$ . Why this step? Zeroing the cost input isolates its role: a free defect would fill every site, which is physically impossible but shows that it is precisely the enthalpy price that suppresses defects.

Verify: at $T = 0$ , $e^{- \infty} = 0$ ✓; at $T = 1000 K$ from E5 we got $n / N \approx 9 \times 1 0^{- 6} ≪ 1$ , safely dilute so the approximation is valid there. ✓

E8 — Cell H: real-world word problem

Forecast: does the crystal gain metal or lose metal atoms? Guess before reading.

Translate colour + n-type into a defect. Colour from a heated ionic solid = trapped electrons (F-centres) absorbing visible light. n-type = extra mobile electrons are the charge carriers (link: Semiconductors and Band Theory). Why this step? Both clues point at excess electrons trapped in anion vacancies — a metal-excess defect, not a cation-vacancy (that would be p-type and needs multiple oxidation states, which $Z n^{2 +}$ lacks).
Mechanism. Hot $Z n O$ loses oxygen: $O^{2 -}$ leaves as $O_{2}$ gas, dumping its two electrons into the vacancy it leaves. Those trapped electrons are excited by heat → yellow, and are loosely bound → n-type conduction. $Z n O Δ Z n_{1 + x} O + \frac{x}{2} O_{2} ↑ .$ Why this step? Writing the balanced loss of oxygen makes the electron bookkeeping visible — the two electrons of each departing $O^{2 -}$ have to go somewhere, and the vacancy is that somewhere.
Formula and sign. Since oxygen is lost (anion-deficient = metal-excess), the drifted formula is $Z n_{1 + x} O$ with $x > 0$ . Why this step? Naming the sign of $x$ turns a qualitative story into a testable statement: metal-excess must give $x > 0$ , so an examiner's " $Z n_{1 - x} O$ " would be wrong by inspection.

Verify: charge check per lost $O^{2 -}$ : removing $O^{2 -}$ removes charge $- 2$ ; the two trapped electrons supply $- 2$ back → neutral. ✓ Consistent with the parent: anion vacancy is the dominant defect in oxygen-deficient ZnO.

E9 — Cell I: the exam twist — which $E$ are they giving you?

Forecast: is Book B's $E$ equal to $2.0 eV$ , $1.0 eV$ , or $4.0 eV$ ?

Match the exponents. For identical $n$ we need the exponents equal: $\frac{E _{s}}{2 k _{B} T} = \frac{E}{k _{B} T} \Rightarrow E = \frac{E _{s}}{2} = 1.0 eV .$ Why this step? This is Trap 5 in the parent, and it is exactly the $\frac{1}{2}$ we derived in E5: Book B has already absorbed the "split across two holes" factor into its $E$ , so its $E$ is the per-vacancy energy $E_{s} /2$ . The physics is one number; the "2" is just where you write the split.
Check both give the same $n$ at $1000 K$ . Book A exponent (from E5) $= 11.61$ . Book B: $\frac{E}{k _{B} T} = \frac{1.602 \times 1 0 ^{- 19}}{1.38 \times 1 0 ^{- 23} \times 1000} = 11.61$ . Same. $n = 6.022 \times 1 0^{23} e^{- 11.61} \approx 5.45 \times 1 0^{18} (both books) .$ Why this step? Re-deriving $n$ from Book B's convention and landing on E5's number is the acid test that the two notations are the same physics — if the counts differed, one book would be wrong.

Verify: identical exponent → identical $n$ . ✓ The moral: always read whether " $E$ " is the pair enthalpy $E_{s}$ or the per-vacancy energy $E_{s} /2$ before dividing by 2.

Recall Quick self-test

Schottky removes 3% of pairs — density becomes what fraction of perfect? ::: $0.97 \times$ (Cell A logic). Frenkel removes... how much density? ::: None — density unchanged (Cell B). In $F e_{0.90} O$ , what percent of iron is $F e^{3 +}$ ? ::: deficit $= 2 (0.10) = 0.20$ ; $x /0.90 = 0.20/0.90 \approx 22.2%$ . As $T \to 0$ , $n \to$ ? ::: $0$ — perfect crystal only at absolute zero (Cell G). Where does the $\frac{1}{2}$ in $e^{- E_{s} /2 k_{B} T}$ come from? ::: a Schottky pair is two independent holes, each carrying $E_{s} /2$ of the pair enthalpy. One Schottky defect = how many vacancies? ::: two — one cation-vacancy plus one anion-vacancy. Book gives " $E$ " not " $E_{s}$ " — do you still divide by 2? ::: No — per-vacancy $E$ already includes the halving (Cell I).