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Concept 06 / 07Solid State

Imperfections in Solids: Schottky, Frenkel, Non-Stoichiometric and Impurity Defects

Point and line defects, stoichiometric defects (vacancy, interstitial, Schottky and Frenkel) with their effect on density, impurity defects created by aliovalent doping, and non-stoichiometric defects: metal excess by anion vacancies (F-centres) and by interstitial cations, and metal deficiency.

Figure 1
Real crystals are never perfect. Defects are irregularities in the arrangement of particles. Point defects involve a single site; line defects involve rows of sites. Point defects form more readily at higher temperatures and on rapid crystallisation, and they strongly affect properties. Stoichiometric defects (composition unchanged). Vacancy defect (non-ionic solids): some lattice sites are empty; density decreases. Produced on heating. Interstitial defect (non-ionic): extra particles occupy interstitial sites; density increases. Schottky defect (ionic): equal numbers of cations and anions are missing, keeping the crystal neutral (see figure). Density decreases. Found in highly ionic compounds with cations and anions of similar size: NaCl, KCl, CsCl, AgBr. At room temperature, NaCl has about 10610^{6} Schottky pairs per cm3\mathrm{cm^{3}}. Frenkel (dislocation) defect (ionic): the smaller ion (usually the cation) leaves its site and occupies an interstitial site, creating a vacancy there. Density is unchanged. Found where the ions differ greatly in size: ZnS, AgCl, AgBr, AgI. AgBr shows both Schottky and Frenkel defects. Impurity defects. Adding an ion of different charge creates vacancies to maintain neutrality. If NaCl is crystallised with a little SrCl2\mathrm{SrCl_{2}}, each Sr2+\mathrm{Sr^{2+}} replaces two Na+\mathrm{Na^{+}}, occupying one site and leaving the other vacant. So the number of cation vacancies equals the number of Sr2+\mathrm{Sr^{2+}} ions. Similarly CdCl2\mathrm{CdCl_{2}} in AgCl. Non-stoichiometric defects (composition changes). Metal excess due to anionic vacancies (F-centres): when NaCl crystals are heated in sodium vapour, Na atoms deposit on the surface, Cl−\mathrm{Cl^{-}} ions diffuse out to combine with them, and the Na atoms ionise, releasing electrons that diffuse in and occupy the anion vacancies. These trapped electrons are F-centres (Farbenzentren, colour centres); they absorb visible light, making NaCl yellow, KCl violet and LiCl pink. Such crystals are also paramagnetic. Metal excess due to extra cations at interstitial sites: ZnO is white when cold and yellow when hot, as it loses oxygen, ZnO→Zn2++12O2+2e−\mathrm{ZnO \rightarrow Zn^{2+} + \frac{1}{2}O_{2} + 2e^{-}}; the excess Zn2+\mathrm{Zn^{2+}} ions sit in interstitial sites with electrons nearby, giving colour and conductivity. Metal deficiency: some cations are missing, and neighbouring cations take a higher charge to compensate. FeO is really Fe0.95O\mathrm{Fe_{0.95}O} (range Fe0.93O\mathrm{Fe_{0.93}O} to Fe0.96O\mathrm{Fe_{0.96}O}), with some Fe3+\mathrm{Fe^{3+}} ions; similarly FeS and NiO. Key Takeaways: - Schottky: cation and anion vacancies in pairs, density falls (similar-size ions). - Frenkel: cation to interstitial site, density unchanged (very different sizes). - Sr2+\mathrm{Sr^{2+}} in NaCl creates one cation vacancy per Sr2+\mathrm{Sr^{2+}}. - F-centres: electrons in anion vacancies; colour and paramagnetism. - Metal deficiency: Fe0.95O\mathrm{Fe_{0.95}O} with some Fe3+\mathrm{Fe^{3+}}. - Pitfall: Frenkel defects do not change density because no ion leaves the crystal. Brain-Teaser 1: NaCl is doped with 10−310^{-3} mol percent of SrCl2\mathrm{SrCl_{2}}. How many cation vacancies are there per mole of NaCl? Solution: 10−310^{-3} mol %\% means 10−5 mol10^{-5}\,\mathrm{mol} SrCl2\mathrm{SrCl_{2}} per mole of NaCl. Each Sr2+\mathrm{Sr^{2+}} creates one vacancy: 10−5×6.022×1023=6.02×101810^{-5} \times 6.022 \times 10^{23} = 6.02 \times 10^{18} cation vacancies. Brain-Teaser 2: Analysis shows a nickel oxide sample to be Ni0.98O\mathrm{Ni_{0.98}O}. What fraction of the nickel is present as Ni3+\mathrm{Ni^{3+}}? Solution: Let xx be the number of Ni3+\mathrm{Ni^{3+}} per formula unit. Charge balance: 3x+2(0.98−x)=23x + 2(0.98 - x) = 2, so x+1.96=2x + 1.96 = 2 and x=0.04x = 0.04. Fraction =0.040.98=4.1%= \frac{0.04}{0.98} = 4.1\%. Brain-Teaser 3: Why does ZnO turn yellow on heating and white again on cooling? Solution: On heating, ZnO loses a little oxygen: ZnO→Zn2++12O2+2e−\mathrm{ZnO \rightarrow Zn^{2+} + \frac{1}{2}O_{2} + 2e^{-}}. The extra Zn2+\mathrm{Zn^{2+}} ions move to interstitial sites and the electrons occupy neighbouring interstitial sites (metal excess defect); these electrons absorb blue light, so the solid looks yellow. On cooling, oxygen is taken up again and the defect disappears.