Lattice energy oxides

Some properties of selected vanadium compounds are Hsted in Table 1. Detailed solubiUty data are available (3), as are physical constants of other vanadium compounds (4). Included are the lattice energy of several metavanadates and the magnetic susceptibiUty of vanadium bromides, chlorides, fluorides, oxides, and sulfides (5). 

The fluorite stmcture, which has a large crystal lattice energy, is adopted by Ce02 preferentially stahi1i2ing this oxide of the tetravalent cation rather than Ce202. Compounds of cerium(IV) other than the oxide, ceric fluoride [10060-10-3] CeF, and related materials, although less stable can be prepared. For example ceric sulfate [13590-82-4] Ce(S0 2> certain double salts are known. 

More recently considered candidates are large molecular anions with delocalized anionic charge, which offer low lattice energies, relatively small ion-ion interaction, and hence sufficient solubility and relatively large conductivity. Delocalization of the charge is achieved by electron-with drawing substituents such as -F or - CF3. Furthermore, these anions show a good electrochemical stability to oxidation. In contrast to Lewis acid-based salts they are chemically more stable with various solvents and often also show excellent thermal stability. 

With air in contact with the earth s surface, it is not surprising that several metals are found in ores that contain the metal oxide. Metal replacement reactions are possible because of differences in lattice energies and reduction potentials. One interesting reaction of this type is... 

Qualitatively, the dipole-dipole interactions between the macro-molecular chains and the halide salt compensate for the lattice energy of the halide crystal and tend to decrease the interactions existing in the glass between the oxide macroanions. This decrease is probably the reason for the significant drop in the glass transition temperature resulting from the addition of a halide salt (Reggiani et al, 1978). Furthermore this type of reaction is consistent with the fact that dissolution of a halide salt in a vitreous solvent requires the existence of ionic bonds provided by a network modifier. 

The MEG model has been extensively used to determine lattice energies and interionic equilibrium distances in ionic solids (oxides, hydroxides, and fluorides Mackrodt and Stewart, 1979 Tossell, 1981) and defect formation energies (Mack-rodt and Stewart, 1979). Table 1.21 compares the lattice energies and cell edges of various oxides obtained by MEG treatment with experimental values. 

Table 1.21 MEG values of lattice energy and lattice parameter for varions oxides, compared with experimental values. Source of data Mackrod and Stewart (1979). C/ is expressed in kJ/mole (in A) corresponds to the cell edge for cnbic snbstances, whereas it is the lattice parameter in the a plane for AI2O3, Fe203, and Ga203 and it is the lattice parameter parallel to the sixfold axis of the hexagonal unit cell in mtiles CaTi03 and BaTi03.

Compounds of aluminium and magnesium in the lower oxidation states, A1(I) and Mg (I), do not exist under normal conditions. If we make an assumption that the radius of AF or Mg is the same as that of Na (same row of the Periodic Table), then we can also equate the lattice energies, MCI. Use this information in a Born-Haber cycle to calculate... 

The lattice energy of calcium oxide (Born-Lande) is 3554 kJ mol the enthalpy of lattice formation is -3554 - 6 = -3560 kJ mol1. The calculated standard enthalpy of formation of calcium oxide is -677 kJ mol about 7% more exothermic than the observed value, probably due to the inherent error in the calculation. 

In acidic solution the + 2 and + 3 oxidation states are simple hydrated ions, the +4 and + 5 states being oxocations. In alkaline solution the + 2 and +3 states form neutral insoluble oxides (the lattice energies of the oxides giving even more stability than any soluble form in these cases), and the +4 and +5 states exist as oxyanions (dimeric in the +4 case). 

Now if the fluorides are compared with the oxides, the latter have a still greater lattice energy and the V versus n curve should be steeper for the oxides. Thus, if there were no lower fluorides, lower oxides would be still more unstable. From the existence of TiO, it can be concluded that this compound must be stabilized by some kind of non-ionic bonding. The same argument holds for the lower nitride TiN, and for the hydride TiH2. Since the H ion is about the same size as the F ion, the lower hydride could not be stable if the fluoride is not. 

A slightly different problem arises when we consider the lower oxidation states of metals. We know that CaF2 is stable. Why not CaF as well Assuming that CaF would crystallize in the same geometry as KF and that the internuclear distance would be about the same, we can calculate a lattice energy for CaF. U0 = — 795 kJ mol"1. The terms in the Bom-Haher cycle are... 

Thallium has two stable oxidation states, + f and +3. Use the Kapustmskii equation to predict the lattice energies of TIF and TIF. Predict the enthalpies of formation of these compounds. Discuss. 

Explain why the lattice energy of magnesium oxide (3850 kj-mol 1) is greater than that of barium oxide (3114 kj-mol 1), given that they have similar arrangements of ions in the crystal lattice. See Appendix 2D. 

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

---