Defects — Schottky, Frenkel; non-stoichiometric defects: Exercises

Before any formula appears, let us pin down every symbol we will use, in plain words:

Definition The symbols, defined once

$n$ = the number of defects actually present in the crystal (a plain count, like "37 000 vacancies").
$N$ = the number of lattice sites available (how many "parking spots" exist).
$E_{s}$ = the energy cost to make one Schottky pair (one cation vacancy plus one anion vacancy), measured in joules. It is the pair enthalpy — remember that, it matters in L5.
$k_{B}$ = Boltzmann's constant $= 1.381 \times 1 0^{- 23} J K^{- 1}$ — the exchange rate between temperature and energy. It answers "how much thermal energy does one degree of temperature buy?"
$T$ = absolute temperature in kelvin.
$N_{A}$ = Avogadro's number $= 6.022 \times 1 0^{23} mol^{- 1}$ — how many formula units sit in one mole; it converts the molar mass (grams per mole) into a mass per unit cell (see Density of a Unit Cell).
$e^{- x}$ = the exponential — a number that shrinks fast as $x$ grows. We use it because the count of defects is set by a probability of climbing an energy hill, and that probability is always of the form " $e$ to the minus (hill height over available thermal energy)".

The three point defects at a glance (and one the parent skipped)

The parent note builds Schottky, Frenkel and the non-stoichiometric family. There is a fourth honest member of the "single wrong lattice point" club that exam-setters like to slip in — the antisite defect — so let us picture all of them together before we compute.

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

Level 1 — Recognition

Q1.1

Classify each defect and say whether density goes down, up, or stays the same: (a) NaCl with equal missing Na $^{+}$ and Cl $^{-}$ ; (b) ZnS with a Zn $^{2 +}$ moved into a gap; (c) AgBr; (d) an $A B$ alloy where an $A$ atom sits on a $B$ site.

Recall Solution

(a) Missing cation and anion together ⇒ Schottky. Atoms are gone, volume unchanged ⇒ density decreases. (b) Ion relocated into an interstitial gap, nothing removed ⇒ Frenkel. Density stays the same. (c) AgBr is the famous "both" crystal — it shows Schottky and Frenkel. If forced to name one for density: only its Schottky part lowers density. (d) An atom on the wrong sublattice ⇒ antisite defect. No atoms added or removed ⇒ density stays the same.

Q1.2

$F e_{0.95} O$ — is this stoichiometric or non-stoichiometric, and what kind?

Recall Solution

The subscript $0.95 \neq = 1$ , so the cation:anion ratio has drifted from the ideal $1 : 1$ ⇒ non-stoichiometric. Iron is deficient (there is less than one Fe per O) ⇒ metal-deficiency defect (cation vacancies + some Fe $^{3 +}$ ).

Level 2 — Application

Q2.1

A crystal has $N = 6.02 \times 1 0^{23}$ equal cation and anion sites, Schottky pair energy $E_{s} = 1.6 \times 1 0^{- 19} J$ , at $T = 1000 K$ . Using $n \approx N e^{- E_{s} /2 k_{B} T}$ , estimate $n$ .

Recall Solution

WHAT we do: substitute into the Schottky formula for the equal-sites case.

WHY the $E_{s} /2$ — the split, done inline (no hand-off). A Schottky event is really two independent choices: pull one ion out of the $N$ cation sites and one out of the $N$ anion sites. Statistical mechanics says the equilibrium number of a species costing energy $ε$ each is $\propto e^{- ε / k_{B} T}$ . But here the "unit that costs $E_{s}$ " is the pair, while the count $n$ we want is the number of pairs — and the pair can be assembled from a cation vacancy formed with energy $\frac{E _{s}}{2}$ and an anion vacancy formed with energy $\frac{E _{s}}{2}$ , each drawn from its own $N$ sites. Multiplying two independent Boltzmann factors $e^{- E_{s} /2 k_{B} T} \cdot e^{- E_{s} /2 k_{B} T} = e^{- E_{s} / k_{B} T}$ for the pair, then taking the square root that comes out of minimising $G = H - T S$ with $S = k_{B} ln W$ and $W$ counting both sublattices, leaves each factor as $e^{- E_{s} /2 k_{B} T}$ . In short: the $/2$ is the price of splitting one pair-enthalpy across two independent site-drawings. (Full Stirling algebra is in Entropy and Gibbs Free Energy; the physics is this line.)

Compute the exponent: $\frac{E _{s}}{2 k _{B} T} = \frac{1.6 \times 1 0 ^{- 19}}{2 ( 1.381 \times 1 0 ^{- 23} ) ( 1000 )} = \frac{1.6 \times 1 0 ^{- 19}}{2.762 \times 1 0 ^{- 20}} = 5.793$ Then $e^{- 5.793} = 3.05 \times 1 0^{- 3}$ , so $n \approx 6.02 \times 1 0^{23} \times 3.05 \times 1 0^{- 3} \approx 1.83 \times 1 0^{21} defects.$ So even a tiny energy hill of a fraction of an electron-volt still leaves a huge raw count — because $N$ itself is astronomical.

Q2.2

Same crystal, but now heat it to $T = 2000 K$ . By what factor does $n$ grow?

Recall Solution

WHAT we do: take the ratio $n (2000) / n (1000)$ so $N$ cancels. $\frac{n ( 2000 )}{n ( 1000 )} = \frac{e ^{- E_{s} / (2 k_{B} \cdot 2000)}}{e ^{- E_{s} / (2 k_{B} \cdot 1000)}} = e^{E_{s} /2 k_{B} (1/1000 - 1/2000)} = e^{5.793 (1 - \frac{1}{2})} = e^{2.896}$ $= 18.1.$ Doubling the temperature multiplies defects by ~18, not by 2 — that is the signature of the exponential (linear growth would have given only $\times 2$ ). Look at the steep climb in the figure below.

Level 3 — Analysis

Q3.1

NaCl heated in Na vapour turns yellow. Explain each link in the chain: Na vapour → colour, and estimate the absorbed wavelength by modelling the trapped electron as a particle in a cubic box of side $L = 2.8 \overset{˚}{A}$ (one anion spacing). Name the coloured species.

Recall Solution

Step 1 — extra Na arrives. Na atoms deposit on the crystal surface. Step 2 — an anion migrates out. A $C l^{-}$ leaves its lattice site to pair with surface Na, leaving a Cl $^{-}$ vacancy (an empty, negatively-hungry hole). Step 3 — the electron is stranded. The deposited Na atom ionises to $N a^{+}$ ; its lone electron cannot go with the departed $C l^{-}$ , so it falls into the anion vacancy. Step 4 — WHY colour, made quantitative. A trapped electron is like a marble stuck in a tiny box; a marble in a box can only have certain "allowed" energies (quantised levels). For a particle of mass $m_{e}$ in a cubic box of side $L$ the ground and first excited levels differ by $Δ E = \frac{3 h ^{2}}{8 m _{e} L ^{2}} (the n^{2} jump from 1^{2} + 1^{2} + 1^{2} = 3 to 2^{2} + 1^{2} + 1^{2} = 6 gives a factor 3) .$ Put in $h = 6.626 \times 1 0^{- 34} J s$ , $m_{e} = 9.11 \times 1 0^{- 31} kg$ , $L = 2.8 \times 1 0^{- 10} m$ : $Δ E = \frac{3 ( 6.626 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 2.8 \times 1 0 ^{- 10} ) ^{2}} \approx 2.9 \times 1 0^{- 18} J (\approx 18 eV, order-of-magnitude) .$ The absorbed wavelength is $λ = h c /Δ E$ . This crude "hard box" over-binds the electron, so the number is a few times too large; the point is that shrinking the box to atomic size ( $L \sim$ Å) pushes the level spacing into the range where visible light ( $Δ E \sim 2$ – $3 \times 1 0^{- 19}$ J, $λ \sim 400$ –700 nm) can be absorbed. A slightly larger effective box ( $L \approx 6$ – $7 \overset{˚}{A}$ , because the electron leaks into neighbouring ions) lands $Δ E$ squarely in the visible → the crystal absorbs blue-violet and looks yellow. The coloured species = the F-centre (F for Farbe, colour). Formula drifts to $N a_{1 + x} C l$ — metal excess.

Q3.2

$F e O$ that is really $F e_{0.95} O$ : for every 100 O, how many Fe $^{2 +}$ and how many Fe $^{3 +}$ ?

Recall Solution

WHAT we set up: let there be $x$ ions of $F e^{3 +}$ and $y$ of $F e^{2 +}$ per 100 O. Two rules:

Atom count: $x + y = 95$ (total Fe per 100 O, from $F e_{0.95} O$ ).
Charge neutrality (WHY $= 200$ ): the crystal as a whole carries zero net charge, so the total positive charge from all the iron must exactly cancel the total negative charge from oxygen. There are 100 O $^{2 -}$ , contributing $100 \times (- 2) = - 200$ . To cancel that, iron must contribute $+ 200$ . Each Fe $^{3 +}$ gives $+ 3$ and each Fe $^{2 +}$ gives $+ 2$ , so: $3 x + 2 y = 200.$ Solve: subtract $2 \times$ (atom count): $(3 x + 2 y) - (2 x + 2 y) = 200 - 190 \Rightarrow x = 10$ . Then $y = 85$ . Answer: $10$ Fe $^{3 +}$ and $85$ Fe $^{2 +}$ per 100 O. Sanity check with the vacancy logic: there are $100 - 95 = 5$ Fe vacancies per 100 O. Each vacancy needs two Fe $^{3 +}$ to repair its $+ 2$ charge deficit ⇒ $5 \times 2 = 10$ Fe $^{3 +}$ . ✓ (See the charge-balance bars in the figure below.)

Level 4 — Synthesis

Q4.1

A Schottky-defective NaCl has $0.1%$ of its cation sites vacant. NaCl has the rock-salt structure ( $Z = 4$ formula units per cell), molar mass $M = 58.44 g mol^{- 1}$ , edge $a = 5.64 \times 1 0^{- 8} cm$ . Find the percentage drop in density versus the perfect crystal.

Recall Solution

Key idea (link to Density of a Unit Cell): density is $ρ = \frac{Z M}{N _{A} a ^{3}}$ , where $N_{A}$ turns molar mass into mass per cell.

Two assumptions, stated honestly.

Cation and anion vacancies come in equal numbers. This is forced by charge neutrality: a Schottky defect removes one $N a^{+}$ (charge $+ 1$ ) and one $C l^{-}$ (charge $- 1$ ) as a pair, so if $0.1%$ of cation sites are empty, then $0.1%$ of anion sites are empty too, i.e. $0.1%$ of whole formula units are missing. (Without this pairing the crystal would carry net charge — impossible in bulk.)
The cell volume $a^{3}$ is essentially unchanged. In reality the ions bordering a vacancy relax slightly inward, so a real crystal shrinks by a whisker; but the fractional relaxation is far smaller than the $0.1%$ mass effect and averages out over the whole lattice, so treating $a$ as constant is an excellent approximation. (If it changed a lot, you could no longer quote a single edge length $a$ .)

With both assumptions, the mass per cell falls by exactly the vacancy fraction while the volume holds: $\frac{Δ ρ}{ρ} = 0.1% = 0.001.$ Numbers, to make it concrete: Perfect: $ρ_{0} = \frac{4 \times 58.44}{( 6.022 \times 1 0 ^{23} ) ( 5.64 \times 1 0 ^{- 8} ) ^{3}}$ . Volume $a^{3} = (5.64 \times 1 0^{- 8})^{3} = 1.794 \times 1 0^{- 22} cm^{3}$ , so $ρ_{0} = \frac{233.76}{6.022 \times 1 0 ^{23} \times 1.794 \times 1 0 ^{- 22}} = \frac{233.76}{108.0} = 2.164 g cm^{- 3} .$ Defective: $ρ = ρ_{0} \times (1 - 0.001) = 2.164 \times 0.999 = 2.162 g cm^{- 3}$ . Percentage drop $= 0.1%$ .

Q4.2

A different ionic solid shows only Frenkel defects at the same temperature and same $0.1%$ occupancy of interstitials. What is its density change? Contrast with Q4.1 in one sentence.

Recall Solution

Answer: 0% — no change. In Frenkel the cation merely hops from its site into an interstitial gap; the total atom count and the cell volume are both untouched, so $ρ = \frac{Z M}{N _{A} a ^{3}}$ is unchanged. One-sentence contrast: Schottky removes mass (density falls by the vacancy fraction, Q4.1), while Frenkel only rearranges mass (density flat) — same "0.1%" of defects, opposite density verdicts.

Level 5 — Mastery

Q5.1

Two crystals A and B at the same $T = 1200 K$ have the same number of sites $N$ . Crystal A's Schottky pair energy is $E_{A} = 1.28 \times 1 0^{- 19} J$ ; crystal B's is $E_{B} = 1.92 \times 1 0^{- 19} J$ . Which has more defects, and by what factor $n_{A} / n_{B}$ ?

Recall Solution

WHAT we compute: the ratio, so $N$ cancels: $\frac{n _{A}}{n _{B}} = e^{- E_{A} /2 k_{B} T} / e^{- E_{B} /2 k_{B} T} = e^{(E_{B} - E_{A}) /2 k_{B} T} .$ WHY A wins: lower energy hill ⇒ easier to make defects ⇒ more of them. So $n_{A} > n_{B}$ ; the factor should exceed 1. $\frac{E _{B} - E _{A}}{2 k _{B} T} = \frac{( 1.92 - 1.28 ) \times 1 0 ^{- 19}}{2 ( 1.381 \times 1 0 ^{- 23} ) ( 1200 )} = \frac{0.64 \times 1 0 ^{- 19}}{3.314 \times 1 0 ^{- 20}} = 1.931.$ $\frac{n _{A}}{n _{B}} = e^{1.931} = 6.90.$ Crystal A has about 6.9× more Schottky defects. A modest energy difference ( $50%$ higher hill) causes a big count difference — again the exponential at work.

Q5.2 (subtle bookkeeping)

Textbook X writes $n \approx N e^{- E / k_{B} T}$ with $E = 0.8 \times 1 0^{- 19} J$ "per vacancy". Textbook Y writes $n \approx N e^{- E_{s} /2 k_{B} T}$ with $E_{s}$ "per pair". For the same physical crystal, what value of $E_{s}$ must Y use so both agree? Then confirm the exponents match.

Recall Solution

The trap the question probes: whether you notice that X's " $E$ per vacancy" already is $E_{s} /2$ . WHY matching exponents is the whole job: both formulas have the same prefactor $N$ , so the two predictions $n_{X}$ and $n_{Y}$ are equal if and only if their exponents are equal — $N e^{p} = N e^{q}$ forces $p = q$ . So agreement reduces entirely to equating the two exponents. $\frac{E}{k _{B} T} = \frac{E _{s}}{2 k _{B} T} \Rightarrow E_{s} = 2 E = 2 (0.8 \times 1 0^{- 19}) = 1.6 \times 1 0^{- 19} J .$ So Y must use the pair enthalpy $E_{s} = 1.6 \times 1 0^{- 19} J$ , exactly double X's per-vacancy value. Both then have identical exponents and predict identical $n$ — same physics, different bookkeeping.

Recall One-line self-test (cover the answers)

Schottky density verdict ::: decreases (atoms removed, volume fixed) Frenkel density verdict ::: unchanged (ion only relocated) Antisite density verdict ::: unchanged (atoms only swap sublattices) Fe³⁺ per Fe vacancy in Fe $_{0.95}$ O ::: two (each vacancy removes +2, each Fe³⁺ restores +1) Effect of doubling T on defect count ::: rises by a large exponential factor, not ×2 Meaning of the /2 in $e^{- E_{s} /2 k_{B} T}$ ::: combinatorial split of the pair energy across independent cation+anion vacancies