CaO lattice

As a first step, CH4 is absorbed through one of its hydrogen atom on 0 sites. The strong electronegative character of O ions in CaO will tend to form OH" ions by an interaction developed between hydrogen atom and O ions over CaO lattice, concomitantly favours the absorption of CH3 species at the neighbouring Ca sites due to the weakening of the Ca " —0 bond, caused by the... 

When ahovalent, ie, different valence, impurities are added to an ionic soHd, the crystal lattice compensates by forming defects that maintain both electrical neutraUty and the anion to cation ratio of the host lattice. For example, addition of x mol of CaO to Zr02 requires the formation of x mol of oxygen vacancies. 

A typical analysis of the iasoluble cerium coaceatrate portioa is by wt % Ce02, - 62 other La oxides, - 10 CaO, 6 other oxides, - 4 and F - 10. The loss on ignition is about 8 wt %. Cerium oxide readily takes F ions iato the lattice. The charge difference is then matched by some Ln " replacing... 

Bancroft et al. (1965), and in the case of CaO and the B1 to B2 transition discovered by Jeanloz and Ahrens (1979), complete reversion of the low-pressure phase occurs upon unloading. These latter transitions involve rearrangement of the lattice which can occur via its deformation rather than complete reconstruction. The volume change in the Si02 transition is much larger than in the case of CaO, as seen in Fig. 4.15. In contrast to the pressure-volume plane when plotted in the Ph-u, plane, the occurrence of these transitions is less striking in this representation (Fig. 4.14). 

To avoid this phase change, zirconia is stabilized in the cubic phase by the addition of a small amount of a divalent or trivalent oxide of cubic symmetry, such as MgO, CaO, or Y2O3. The additive oxide cation enters the crystal lattice and increases the ionic character of the metal-oxygen bonds. The cubic phase is not thermodynamically stable below approximately 1400°C for MgO additions, 1140°C for CaO additions, and below 750°C for Y2O3 additions. However, the diffusion rates for the cations are so low at Xhtstsubsolidus temperatures that the cubic phase can easily be quenched and retained as a metastable phase. Zirconia is commercially applied by thermal spray. It is also readily produced by CVD, mostly on an experimental basis. Its characteristics and properties are summarized in Table 11.8. 

In some ionic crystals (primarily in halides of the alkali metals), there are vacancies in both the cationic and anionic positions (called Schottky defects—see Fig. 2.16). During transport, the ions (mostly of one sort) are shifted from a stable position to a neighbouring hole. The Schottky mechanism characterizes transport in important solid electrolytes such as Nernst mass (Zr02 doped with Y203 or with CaO). Thus, in the presence of 10 mol.% CaO, 5 per cent of the oxygen atoms in the lattice are replaced by vacancies. The presence of impurities also leads to the formation of Schottky defects. Most substances contain Frenkel and Schottky defects simultaneously, both influencing ion transport. 

Alkaline earth oxides (AEO = MgO, CaO, and SrO) doped with 5 mol% Nd203 have been synthesised either by evaporation of nitrate solutions and decomposition, or by sol-gel method. The samples have been characterised by chemical analysis, specific surface area measurement, XRD, CO2-TPD, and FTIR spectroscopy. Their catalytic properties in propane oxidative dehydrogenation have been studied. According to detailed XRD analyses, solid solution formation took place, leading to structural defects which were agglomerated or dispersed, their relative amounts depending on the preparation procedure and on the alkaline-earth ion size match with Nd3+. Relationships between catalyst synthesis conditions, lattice defects, basicity of the solids and catalytic performance are discussed. 

The difference between the CaO and BaO values is because the larger the ion is, the lower the attraction is (greater separation). The lower attraction leads to a lower lattice energy. This size argument will get you 1 point. 

We can introduce vacancies into a crystal by doping it with a selected impurity. For instance, if we add CaCl2 to a NaCl crystal, each Ca ion replaces fwoNa ions in order to preserve electrical neutrality, and so one cation vacancy is created. Such created vacancies are known as extrinsic. An important example that you will meet later in the chapter is that of zirconia, ZrOz. This structure can be stabilised by doping with CaO, where the Ca ions replace the Zr(IV) atoms in the lattice. The charge compensation here is achieved by the production of anion vacancies on the oxide sublattice. 

XPS has been used by Inoue and Yasumori (425a) to study the surface of MgO, CaO, and BaO after exposure to water vapor followed by in situ dehydration at various temperatures. Signals attributed to O2- lattice ions were observed after all treatments with peaks at 531.6 (MgO), 529.6 (CaO), and 528.5 (BaO) eV. Immediately after hydration, a second series of peaks at 533.7 (MgO), 532.2 (CaO), and 532.1 (BaO) eV were observed, which progressively disappeared as the dehydration temperature increased these were assigned to OH- ions. In addition, peaks at 532.1 (CaO) and 530.6 (BaO) eV which appeared during dehydration were assigned to O- ions formed from the following reaction ... 

If the oxygen in CaO is replaced by the group CO , then a lattice essentially similar to that of CaO is obtained. It is not exactly the same, as the shape of the CO - is different from that of the 02 ion, but each CO ion is still surrounded by six Ca2+ ions. Consequently, there is no basis whatever for employing the formula... 

Schrevelius asserts that in terms of the law of Vegard (1917) determining the lattice parameter a, we can derive the formula for a solid solution which is an abrasive material, e.g., (Ti0.oo6Al0.994.)203. Exactly the same can be done by an accurate measurement of hardness. Since all bauxite synthesized alumina contain up to some 3% Ti02, it is important to determine how much titanium is contained in the solid solution of the abrasive and how much is bonded into a soft aluminate CaO 6 [(Al, Ti)203] or is exuded as free rutile Ti02. 

Which metal oxide, BaO, SrO, MgO, or CaO, will have the lowest lattice energy, given that they all crystallize into the same structural type ... 

Order the following compounds according to their expected lattice energies AlBr3, MgBr2, LiBr, CaO. 

Rb02, and Cs02. Under the same conditions, however, the lighter group 1A and 2A 453-455. metals form normal oxides, such as Li20, MgO, CaO, and SrO. All these compounds are ionic solids, and the nature of the product obtained depends on the amount of 02 present, the temperature, the sizes of the ions and how they pack together, and the resultant lattice energies of the various crystalline solids (Section 6.6). 

Such vibrational structure in the spectra of the Bi3+ ion has also been observed in a few other host lattices, viz. CaO and SrO (rock salt) [8,9], NaLn02 (Ln = Sc, Y, Gd, Lu, ordered rock salt) [10], CaS [11], YA13B4012 [12], Ca3(P04)2 [13], and CaS04 [14]. A structural requirement for the occurrence of this vibrational structure seems to be that the Bi3 + ion occupies a relatively small six-coordinated site, with CaS04 as an exception (eight coordination). Due to a low site symmetry and/or the simultaneous occurrence of... 

Many oxide minerals can be visualized as a face-centered oxide ion lattice with cations distributed within the tetrahedral and octahedral holes. Calculate the lattice constant, a, for a face-centered O2- lattice. If cations occupy all the octahedral holes in MgO and CaO, calculate a for these minerals. Use data in Table 10-1. 

These ideas are illustrated in Figure 2 for CaO. The intrinsic defects are Ca and 0 vacancies. When a Eu3+ ion enters the lattice, it substitutes for a Ca2+ and a charge compensation is required. 

Using the same arguments as we used in CaO, one would expect that an increased concentration of lanthanide dopant should introduce a like number of fluoride interstitials into the lattice. If the pairing of the interstitials with the lanthanide is not complete, the number of free interstitials would increase. The larger number of interstitials should lead to an increase in the number of lanthanides that have the fluoride interstitial charge compensation in a nearby position relative to the ones that have the interstitial distant according to the law of mass action. Laser spectroscopy shows the opposite effect though. 

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