Ionic compounds lattice enthalpies

To date there is no evidence that sodium forms any chloride other than NaCl indeed the electronic theory of valency predicts that Na" and CU, with their noble gas configurations, are likely to be the most stable ionic species. However, since some noble gas atoms can lose electrons to form cations (p. 354) we cannot rely fully on this theory. We therefore need to examine the evidence provided by energetic data. Let us consider the formation of a number of possible ionic compounds and first, the formation of sodium dichloride , NaCl2. The energy diagram for the formation of this hypothetical compound follows the pattern of that for NaCl but an additional endothermic step is added for the second ionisation energy of sodium. The lattice energy is calculated on the assumption that the compound is ionic and that Na is comparable in size with Mg ". The data are summarised below (standard enthalpies in kJ) ... 

The enthalpy of solution is quite small for many simple ionic compounds and can be either positive or negative. It is the difference between two large quantities, the sum of the hydration enthalpies and the lattice energy. 

Prediction of solubility for simple ionic compounds is difficult since we need to know not only values of hydration and lattice enthalpies but also entropy changes on solution before any informed prediction can be given. Even then kinetic factors must be considered. 

Because the fluoride ion is so small, the lattice enthalpies of its ionic compounds tend to be high (see Table 6.6). As a result, fluorides are less soluble than other halides. This difference in solubility is one of the reasons why the oceans are salty with chlorides rather than fluorides, even though fluorine is more abundant than chlorine in the Earth s crust. Chlorides are more readily dissolved and washed out to sea. There are some exceptions to this trend in solubilities, including AgF, which is soluble the other silver halides are insoluble. The exception arises because the covalent character of the silver halides increases from AgCl to Agl as the anion becomes larger and more polarizable. Silver fluoride, which contains the small and almost unpolarizable fluoride ion, is freely soluble in water because it is predominantly ionic. 

Although there is no space to develop a detailed discussion of the solubilities of compounds of the transition elements, the general insolubility of their + 2 and + 3 hydroxides is important. The rationale underlying their insolubility can be summarized (i) the hydroxide ion is relatively small (152 pm ionic radius) and the ions of the +2 and +3 transition metals assume a similar size if their radii are increased by 60-80 pm, and (ii) the enthalpy of hydration of the hydroxide ion (—519 kJ mol ) is sufficiently negative to represent a reasonable degree of competition with the metal ions for the available water molecules, thus preventing the metal ions from becoming fully hydrated. Such effects combine to allow the lattice enthalpies of the hydroxides to become dominant. 

The enthalpy of formation of an ionic compound can be calculated with an accuracy of a few percent by means of the Born-Land6 equation (Eq. 4.13) and the Bcrn-Haber cycle. Consider NaCI. for example. We have seen that by using the predicted internuclear distance of 283 pm (or the experimental value of 281.4 pm), the Madelung constant of 1.748, the Rorn exponent, n, and various constants, a value of — 755 kJ mol-1 could be calculated for the lattice energy. The heat capacity correction is 2.1 kJ mol-1, which yields U 9i = —757 kJ mol-1. The Bom-Haber summation is then... 

In order for an ionic compound to dissolve, the Madelung energy or electrostatic attraction between the ions in the lattice must be overcome. In a solution in which the ions are separated by molecules of a solvent with a high dielectric constant ( H 0 81.7 ) the attractive force will be considerably less. The process of solution of an ionic compound in water may be considered by a Bom-Haber type of cycle. The overall enthalpy of the process is the sum of two terms, the enthalpy of dissociating the ions from the lattice (the lattice energy) and the enthalpy of introducing the dissociated ions into the solvent (the solvation energy) ... 

Table 4.12 presents the calculated energies of formation for the solid neutral species and salts based on the CBS-4M method (see Ch. 4.2.1). Furthermore we see from table 4.12 that for the nitronium ([N02]+) species the covalently bound form is favored over the ionic salt by 26.9 kcal mol 1 while for the nitrosonium species ([NO]+) the salt is favored over the covalent isomer by 10.5 kcal mol-1. This change from the preferred covalent form of —N02 compound (actually a nitrato ester) to the ionic nitronium salt can be attributed almost exclusively to the increased lattice enthalpy of the (smaller ) N0+ species (AffL(N0+ - N02 salt) = 31.4 kcal mol-1) (N.B. The difference in the ionization potentials of NO (215 kcal mol-1) and N02 (221 kcal mol-1) is only marginal). 

The atomic and ionic properties of an element, particularly IE, ionic radius and electronegativity, underly its chemical behaviour and determine the types of compound it can form. The simplest type of compound an element can form is a binary compound, one in which it is combined with only one other element. The transition elements form binary compounds with a wide variety of non-metals, and the stoichiometries of these compounds will depend upon the thermodynamics of the compound-forming process. Binary oxides, fluorides and chlorides of the transition elements reveal the oxidation states available to them and, to some extent, reflect trends in IE values. However, the lEs of the transition elements are by no means the only contributors to the thermodynamics of compound formation. Other factors such as lattice enthalpy and the extent of covalency in bonding are important. In this chapter some examples of binary transition element compounds will be used to reveal the factors which determine the stoichiometry of compounds. 

Lattice enthalpies are important thermodynamic parameters and depend upon the solid state structures of the compound and hence the ionic radius of the metal ion. 

The values of A//l are proportional to the product of the charges on ions divided by the sum of their ionic radii. So the 3+/3- system will get a x9 multiplier, whereas the 2+/2- will have a x4 on the 1+Zl- system. Within compounds that have the same charges, tbe bigger ions lead to larger sums on the radii and smaller lattice energies. Considering both of these factors, one can substantiate the trend shown above for increasing lattice enthalpies. 

How does each of the following affect the solubility of an ionic compound (a) lattice energy, (b) solvent (polar versus nonpolar), (c) enthalpies of hydration of cation and anion... 

Note Using the tabulated ionic radius of Ca (i.e. that of Ca2+) would be less valid than using the atomic radius of a neighboring monovalent ion, for the problem asks about a hypothetical compound of monovalent calcium. Predictions with the smaller Ca2+ radius (100 pm) differ substantially from those listed above the expected structure changes to rock-salt, the lattice enthalpy to 758 kJmol-1, Af// (CaCl) to —446kJ mol-1 and the final reaction enthalpy to +96 kJ mol-1. 

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